3.6.100 \(\int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [600]

Optimal. Leaf size=444 \[ -\frac {(a-b) \left (a^2+4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]

[Out]

-1/4*a^(5/2)*(15*a^4+46*a^2*b^2+63*b^4)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(7/2)/(a^2+b^2)^3/d+1/2*(a-
b)*(a^2+4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/2*(a-b)*(a^2+4*a*b+b^2)*arctan(
1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/4*(a+b)*(a^2-4*a*b+b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(
d*x+c))/(a^2+b^2)^3/d*2^(1/2)-1/4*(a+b)*(a^2-4*a*b+b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^3/
d*2^(1/2)+1/4*(15*a^4+31*a^2*b^2+8*b^4)*tan(d*x+c)^(1/2)/b^3/(a^2+b^2)^2/d-1/2*a^2*tan(d*x+c)^(5/2)/b/(a^2+b^2
)/d/(a+b*tan(d*x+c))^2-1/4*a^2*(5*a^2+13*b^2)*tan(d*x+c)^(3/2)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.74, antiderivative size = 444, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3646, 3726, 3728, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} -\frac {(a-b) \left (a^2+4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 d \left (a^2+b^2\right )^2}-\frac {a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{7/2} d \left (a^2+b^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(9/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((a - b)*(a^2 + 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d)) + ((a - b)*(
a^2 + 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - (a^(5/2)*(15*a^4 + 46*a
^2*b^2 + 63*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(7/2)*(a^2 + b^2)^3*d) + ((a + b)*(a^2 - 4
*a*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((a + b)*(a^2 -
4*a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) + ((15*a^4 + 31*a
^2*b^2 + 8*b^4)*Sqrt[Tan[c + d*x]])/(4*b^3*(a^2 + b^2)^2*d) - (a^2*Tan[c + d*x]^(5/2))/(2*b*(a^2 + b^2)*d*(a +
b*Tan[c + d*x])^2) - (a^2*(5*a^2 + 13*b^2)*Tan[c + d*x]^(3/2))/(4*b^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3726

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Ta
n[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3734

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (\frac {5 a^2}{2}-2 a b \tan (c+d x)+\frac {1}{2} \left (5 a^2+4 b^2\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {\sqrt {\tan (c+d x)} \left (\frac {3}{4} a^2 \left (5 a^2+13 b^2\right )-4 a b^3 \tan (c+d x)+\frac {1}{4} \left (15 a^4+31 a^2 b^2+8 b^4\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {-\frac {1}{8} a \left (15 a^4+31 a^2 b^2+8 b^4\right )+b^3 \left (a^2-b^2\right ) \tan (c+d x)-\frac {1}{8} a \left (15 a^4+31 a^2 b^2+24 b^4\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{b^3 \left (a^2+b^2\right )^2}\\ &=\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {b^4 \left (3 a^2-b^2\right )+a b^3 \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{b^3 \left (a^2+b^2\right )^3}-\frac {\left (a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right )\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 b^3 \left (a^2+b^2\right )^3}\\ &=\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {2 \text {Subst}\left (\int \frac {b^4 \left (3 a^2-b^2\right )+a b^3 \left (a^2-3 b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^3 \left (a^2+b^2\right )^3 d}-\frac {\left (a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 b^3 \left (a^2+b^2\right )^3 d}\\ &=\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}-\frac {\left (a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 b^3 \left (a^2+b^2\right )^3 d}\\ &=-\frac {a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 d}+\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=-\frac {a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}\\ &=-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 5.35, size = 403, normalized size = 0.91 \begin {gather*} \frac {2 b^2 \tan ^{\frac {11}{2}}(c+d x)+\frac {2 a^2 \left (15 a^2+16 b^2\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}{b^3}+\frac {2 a \left (5 a^2+4 b^2\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{b^2}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{b}+2 a \tan ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))-2 b \tan ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))-\frac {a (a+b \tan (c+d x)) \left (a \sqrt {b} \left (a^2+b^2\right ) \left (15 a^4+31 a^2 b^2+24 b^4\right ) \sqrt {\tan (c+d x)}+\left (-4 (-1)^{3/4} (a+i b)^3 b^{7/2} \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )-4 \sqrt [4]{-1} b^{7/2} (i a+b)^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right ) (a+b \tan (c+d x))\right )}{b^{7/2} \left (a^2+b^2\right )^2}}{4 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(9/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

(2*b^2*Tan[c + d*x]^(11/2) + (2*a^2*(15*a^2 + 16*b^2)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x]))/b^3 + (2*a*(5*a
^2 + 4*b^2)*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))/b^2 - (2*a^2*Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x]))/b +
2*a*Tan[c + d*x]^(7/2)*(a + b*Tan[c + d*x]) - 2*b*Tan[c + d*x]^(9/2)*(a + b*Tan[c + d*x]) - (a*(a + b*Tan[c +
d*x])*(a*Sqrt[b]*(a^2 + b^2)*(15*a^4 + 31*a^2*b^2 + 24*b^4)*Sqrt[Tan[c + d*x]] + (-4*(-1)^(3/4)*(a + I*b)^3*b
^(7/2)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + a^(5/2)*(15*a^4 + 46*a^2*b^2 + 63*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan
[c + d*x]])/Sqrt[a]] - 4*(-1)^(1/4)*b^(7/2)*(I*a + b)^3*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])*(a + b*Tan[c +
d*x])))/(b^(7/2)*(a^2 + b^2)^2))/(4*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2)

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Maple [A]
time = 0.15, size = 356, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b^{3}}-\frac {2 a^{3} \left (\frac {\left (-\frac {9}{8} a^{4} b -\frac {13}{4} a^{2} b^{3}-\frac {17}{8} b^{5}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {a \left (7 a^{4}+22 a^{2} b^{2}+15 b^{4}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 a^{4}+46 a^{2} b^{2}+63 b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3} \left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(356\)
default	\(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b^{3}}-\frac {2 a^{3} \left (\frac {\left (-\frac {9}{8} a^{4} b -\frac {13}{4} a^{2} b^{3}-\frac {17}{8} b^{5}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {a \left (7 a^{4}+22 a^{2} b^{2}+15 b^{4}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 a^{4}+46 a^{2} b^{2}+63 b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3} \left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(356\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/b^3*tan(d*x+c)^(1/2)-2*a^3/b^3/(a^2+b^2)^3*(((-9/8*a^4*b-13/4*a^2*b^3-17/8*b^5)*tan(d*x+c)^(3/2)-1/8*a*
(7*a^4+22*a^2*b^2+15*b^4)*tan(d*x+c)^(1/2))/(a+b*tan(d*x+c))^2+1/8*(15*a^4+46*a^2*b^2+63*b^4)/(a*b)^(1/2)*arct
an(b*tan(d*x+c)^(1/2)/(a*b)^(1/2)))+2/(a^2+b^2)^3*(1/8*(3*a^2*b-b^3)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+t
an(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*t
an(d*x+c)^(1/2)))+1/8*(a^3-3*a*b^2)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^
(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))))

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Maxima [A]
time = 0.50, size = 434, normalized size = 0.98 \begin {gather*} -\frac {\frac {{\left (15 \, a^{7} + 46 \, a^{5} b^{2} + 63 \, a^{3} b^{4}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (9 \, a^{5} b + 17 \, a^{3} b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (7 \, a^{6} + 15 \, a^{4} b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{6} b^{3} + 2 \, a^{4} b^{5} + a^{2} b^{7} + {\left (a^{4} b^{5} + 2 \, a^{2} b^{7} + b^{9}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{4} + 2 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )} - \frac {8 \, \sqrt {\tan \left (d x + c\right )}}{b^{3}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/4*((15*a^7 + 46*a^5*b^2 + 63*a^3*b^4)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^6*b^3 + 3*a^4*b^5 + 3*a^2*
b^7 + b^9)*sqrt(a*b)) - (2*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*
x + c)))) + 2*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) -
sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(a^3 - 3*
a^2*b - 3*a*b^2 + b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6
) - ((9*a^5*b + 17*a^3*b^3)*tan(d*x + c)^(3/2) + (7*a^6 + 15*a^4*b^2)*sqrt(tan(d*x + c)))/(a^6*b^3 + 2*a^4*b^5
+ a^2*b^7 + (a^4*b^5 + 2*a^2*b^7 + b^9)*tan(d*x + c)^2 + 2*(a^5*b^4 + 2*a^3*b^6 + a*b^8)*tan(d*x + c)) - 8*sq
rt(tan(d*x + c))/b^3)/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 11339 vs. \(2 (392) = 784\).
time = 18.47, size = 22791, normalized size = 51.33 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/16*(16*sqrt(2)*((a^22*b^3 + 3*a^20*b^5 - 17*a^18*b^7 - 123*a^16*b^9 - 342*a^14*b^11 - 546*a^12*b^13 - 546*a
^10*b^15 - 342*a^8*b^17 - 123*a^6*b^19 - 17*a^4*b^21 + 3*a^2*b^23 + b^25)*d^5*cos(d*x + c)^4 + 2*(3*a^20*b^5 +
26*a^18*b^7 + 99*a^16*b^9 + 216*a^14*b^11 + 294*a^12*b^13 + 252*a^10*b^15 + 126*a^8*b^17 + 24*a^6*b^19 - 9*a^
4*b^21 - 6*a^2*b^23 - b^25)*d^5*cos(d*x + c)^2 + (a^18*b^7 + 9*a^16*b^9 + 36*a^14*b^11 + 84*a^12*b^13 + 126*a^
10*b^15 + 126*a^8*b^17 + 84*a^6*b^19 + 36*a^4*b^21 + 9*a^2*b^23 + b^25)*d^5 + 4*((a^21*b^4 + 8*a^19*b^6 + 27*a
^17*b^8 + 48*a^15*b^10 + 42*a^13*b^12 - 42*a^9*b^16 - 48*a^7*b^18 - 27*a^5*b^20 - 8*a^3*b^22 - a*b^24)*d^5*cos
(d*x + c)^3 + (a^19*b^6 + 9*a^17*b^8 + 36*a^15*b^10 + 84*a^13*b^12 + 126*a^11*b^14 + 126*a^9*b^16 + 84*a^7*b^1
8 + 36*a^5*b^20 + 9*a^3*b^22 + a*b^24)*d^5*cos(d*x + c))*sin(d*x + c))*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 +
20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 + 2*(3*a^17*b + 8*a^15*b^3 - 12*a^13*b^5 - 72*a^11*b^7 - 110*a^9*b
^9 - 72*a^7*b^11 - 12*a^5*b^13 + 8*a^3*b^15 + 3*a*b^17)*d^2*sqrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b
^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30
*a^2*b^10 + b^12))*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/((
a^24 + 12*a^22*b^2 + 66*a^20*b^4 + 220*a^18*b^6 + 495*a^16*b^8 + 792*a^14*b^10 + 924*a^12*b^12 + 792*a^10*b^14
+ 495*a^8*b^16 + 220*a^6*b^18 + 66*a^4*b^20 + 12*a^2*b^22 + b^24)*d^4))*(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 +
20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))^(3/4)*arctan(((a^24 - 6*a^22*b^2 - 84*a^20*b^4 - 322*a^18*
b^6 - 603*a^16*b^8 - 540*a^14*b^10 + 540*a^10*b^14 + 603*a^8*b^16 + 322*a^6*b^18 + 84*a^4*b^20 + 6*a^2*b^22 -
b^24)*d^4*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/((a^24 + 12
*a^22*b^2 + 66*a^20*b^4 + 220*a^18*b^6 + 495*a^16*b^8 + 792*a^14*b^10 + 924*a^12*b^12 + 792*a^10*b^14 + 495*a^
8*b^16 + 220*a^6*b^18 + 66*a^4*b^20 + 12*a^2*b^22 + b^24)*d^4))*sqrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a
^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) - sqrt(2)*((3*a^26*b + 35*a^24*b^3 + 186*a^22*b^5 + 594*a^20*b^
7 + 1265*a^18*b^9 + 1881*a^16*b^11 + 1980*a^14*b^13 + 1452*a^12*b^15 + 693*a^10*b^17 + 165*a^8*b^19 - 22*a^6*b
^21 - 30*a^4*b^23 - 9*a^2*b^25 - b^27)*d^7*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8
- 30*a^2*b^10 + b^12)/((a^24 + 12*a^22*b^2 + 66*a^20*b^4 + 220*a^18*b^6 + 495*a^16*b^8 + 792*a^14*b^10 + 924*a
^12*b^12 + 792*a^10*b^14 + 495*a^8*b^16 + 220*a^6*b^18 + 66*a^4*b^20 + 12*a^2*b^22 + b^24)*d^4))*sqrt(1/((a^12
+ 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) - (a^21 + 6*a^19*b^2 + 9*a^17*
b^4 - 24*a^15*b^6 - 126*a^13*b^8 - 252*a^11*b^10 - 294*a^9*b^12 - 216*a^7*b^14 - 99*a^5*b^16 - 26*a^3*b^18 - 3
*a*b^20)*d^5*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/((a^24 +
12*a^22*b^2 + 66*a^20*b^4 + 220*a^18*b^6 + 495*a^16*b^8 + 792*a^14*b^10 + 924*a^12*b^12 + 792*a^10*b^14 + 495
*a^8*b^16 + 220*a^6*b^18 + 66*a^4*b^20 + 12*a^2*b^22 + b^24)*d^4)))*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*
a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 + 2*(3*a^17*b + 8*a^15*b^3 - 12*a^13*b^5 - 72*a^11*b^7 - 110*a^9*b^9
- 72*a^7*b^11 - 12*a^5*b^13 + 8*a^3*b^15 + 3*a*b^17)*d^2*sqrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6
+ 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^
2*b^10 + b^12))*sqrt(((a^18 - 27*a^16*b^2 + 168*a^14*b^4 + 224*a^12*b^6 - 366*a^10*b^8 - 366*a^8*b^10 + 224*a^
6*b^12 + 168*a^4*b^14 - 27*a^2*b^16 + b^18)*d^2*sqrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*
b^8 + 6*a^2*b^10 + b^12)*d^4))*cos(d*x + c) + sqrt(2)*((a^21 - 30*a^19*b^2 + 249*a^17*b^4 - 280*a^15*b^6 - 103
8*a^13*b^8 + 732*a^11*b^10 + 1322*a^9*b^12 - 504*a^7*b^14 - 531*a^5*b^16 + 82*a^3*b^18 - 3*a*b^20)*d^3*sqrt(1/
((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))*cos(d*x + c) - (3*a^14*b
- 91*a^12*b^3 + 795*a^10*b^5 - 1611*a^8*b^7 + 1217*a^6*b^9 - 345*a^4*b^11 + 33*a^2*b^13 - b^15)*d*cos(d*x + c
))*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 + 2*(3*a^17*b + 8*a^15*b
^3 - 12*a^13*b^5 - 72*a^11*b^7 - 110*a^9*b^9 - 72*a^7*b^11 - 12*a^5*b^13 + 8*a^3*b^15 + 3*a*b^17)*d^2*sqrt(1/(
(a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)))/(a^12 - 30*a^10*b^2 + 25
5*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12))*sqrt(sin(d*x + c)/cos(d*x + c))*(1/((a^12 + 6*a^1
0*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))^(1/4) + (a^12 - 30*a^10*b^2 + 255*a^8*
b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)*sin(d*x + c))/cos(d*x + c))*(1/((a^12 + 6*a^10*b^2 + 15*
a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))^(3/4) - sqrt(2)*((3*a^32*b - 10*a^30*b^3 - 294*a^
28*b^5 - 1674*a^26*b^7 - 4890*a^24*b^9 - 8370*a...

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(9/2)/(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

Mupad [B]
time = 20.80, size = 2500, normalized size = 5.63 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(9/2)/(a + b*tan(c + d*x))^3,x)

[Out]

(log(- (((((((((64*a*b*(15*a^4 + 2*b^4 + 41*a^2*b^2))/d - 256*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2
*(1i/(d^2*(a*1i - b)^6))^(1/2))*(1i/(d^2*(a*1i - b)^6))^(1/2))/2 - (8*a*tan(c + d*x)^(1/2)*(225*a^14 - 184*b^1
4 + 608*a^2*b^12 - 272*a^4*b^10 + 3937*a^6*b^8 + 5804*a^8*b^6 + 4006*a^10*b^4 + 1380*a^12*b^2))/(b^4*d^2*(a^2
+ b^2)^4))*(1i/(d^2*(a*1i - b)^6))^(1/2))/2 + (2*a^2*(1125*a^14 + 16*b^14 + 6112*a^2*b^12 - 17727*a^4*b^10 - 2
3239*a^6*b^8 - 11174*a^8*b^6 + 2930*a^10*b^4 + 3525*a^12*b^2))/(b^4*d^3*(a^2 + b^2)^6))*(1i/(d^2*(a*1i - b)^6)
)^(1/2))/2 + (tan(c + d*x)^(1/2)*(32*b^18 - 225*a^18 + 128*a^2*b^16 + 192*a^4*b^14 - 3841*a^6*b^12 + 18050*a^8
*b^10 + 26801*a^10*b^8 + 16860*a^12*b^6 + 4049*a^14*b^4 - 30*a^16*b^2))/(b^5*d^4*(a^2 + b^2)^8))*(1i/(d^2*(a*1
i - b)^6))^(1/2))/2 - (a^3*(225*a^12 + 504*b^12 + 872*a^2*b^10 + 4457*a^4*b^8 + 5916*a^6*b^6 + 4006*a^8*b^4 +
1380*a^10*b^2))/(2*b^5*d^5*(a^2 + b^2)^8))*(-1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*
d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2))/2 - log(- (((((((((64*a*b*(15*a^4 + 2*b^4 + 41*a^2*b^2))/d
+ 256*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*(1i/(d^2*(a*1i - b)^6))^(1/2))*(1i/(d^2*(a*1i - b)^6))
^(1/2))/2 + (8*a*tan(c + d*x)^(1/2)*(225*a^14 - 184*b^14 + 608*a^2*b^12 - 272*a^4*b^10 + 3937*a^6*b^8 + 5804*a
^8*b^6 + 4006*a^10*b^4 + 1380*a^12*b^2))/(b^4*d^2*(a^2 + b^2)^4))*(1i/(d^2*(a*1i - b)^6))^(1/2))/2 + (2*a^2*(1
125*a^14 + 16*b^14 + 6112*a^2*b^12 - 17727*a^4*b^10 - 23239*a^6*b^8 - 11174*a^8*b^6 + 2930*a^10*b^4 + 3525*a^1
2*b^2))/(b^4*d^3*(a^2 + b^2)^6))*(1i/(d^2*(a*1i - b)^6))^(1/2))/2 - (tan(c + d*x)^(1/2)*(32*b^18 - 225*a^18 +
128*a^2*b^16 + 192*a^4*b^14 - 3841*a^6*b^12 + 18050*a^8*b^10 + 26801*a^10*b^8 + 16860*a^12*b^6 + 4049*a^14*b^4
- 30*a^16*b^2))/(b^5*d^4*(a^2 + b^2)^8))*(1i/(d^2*(a*1i - b)^6))^(1/2))/2 - (a^3*(225*a^12 + 504*b^12 + 872*a
^2*b^10 + 4457*a^4*b^8 + 5916*a^6*b^6 + 4006*a^8*b^4 + 1380*a^10*b^2))/(2*b^5*d^5*(a^2 + b^2)^8))*(-1/(4*(b^6*
d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i)))^(1/2)
- atan(((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b
^2*d^2)))^(1/2)*((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i +
15*a^4*b^2*d^2)))^(1/2)*((2250*a^20*b*d^2 + 32*a^2*b^19*d^2 + 12288*a^4*b^17*d^2 - 10974*a^6*b^15*d^2 - 10516
2*a^8*b^13*d^2 - 150758*a^10*b^11*d^2 - 85314*a^12*b^9*d^2 - 3578*a^14*b^7*d^2 + 22210*a^16*b^5*d^2 + 11550*a^
18*b^3*d^2)/(b^21*d^5 + 8*a^2*b^19*d^5 + 28*a^4*b^17*d^5 + 56*a^6*b^15*d^5 + 70*a^8*b^13*d^5 + 56*a^10*b^11*d^
5 + 28*a^12*b^9*d^5 + 8*a^14*b^7*d^5 + a^16*b^5*d^5) + (-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6
i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^
5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*((128*a*b^26*d^4 + 3648*a^3*b^24*d^4 +
25536*a^5*b^22*d^4 + 88320*a^7*b^20*d^4 + 182784*a^9*b^18*d^4 + 244608*a^11*b^16*d^4 + 217728*a^13*b^14*d^4 +
128256*a^15*b^12*d^4 + 48000*a^17*b^10*d^4 + 10304*a^19*b^8*d^4 + 960*a^21*b^6*d^4)/(b^21*d^5 + 8*a^2*b^19*d^
5 + 28*a^4*b^17*d^5 + 56*a^6*b^15*d^5 + 70*a^8*b^13*d^5 + 56*a^10*b^11*d^5 + 28*a^12*b^9*d^5 + 8*a^14*b^7*d^5
+ a^16*b^5*d^5) + (tan(c + d*x)^(1/2)*(-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^
2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*(512*b^30*d^4 + 4608*a^2*b^28*d^4 + 17920*a^4*b^26*d^4 + 38400*a
^6*b^24*d^4 + 46080*a^8*b^22*d^4 + 21504*a^10*b^20*d^4 - 21504*a^12*b^18*d^4 - 46080*a^14*b^16*d^4 - 38400*a^1
6*b^14*d^4 - 17920*a^18*b^12*d^4 - 4608*a^20*b^10*d^4 - 512*a^22*b^8*d^4))/(b^21*d^4 + 8*a^2*b^19*d^4 + 28*a^4
*b^17*d^4 + 56*a^6*b^15*d^4 + 70*a^8*b^13*d^4 + 56*a^10*b^11*d^4 + 28*a^12*b^9*d^4 + 8*a^14*b^7*d^4 + a^16*b^5
*d^4)) - (tan(c + d*x)^(1/2)*(1800*a^23*b*d^2 - 1472*a*b^23*d^2 - 1024*a^3*b^21*d^2 + 8448*a^5*b^19*d^2 + 4608
8*a^7*b^17*d^2 + 177344*a^9*b^15*d^2 + 402912*a^11*b^13*d^2 + 541632*a^13*b^11*d^2 + 455472*a^15*b^9*d^2 + 248
064*a^17*b^7*d^2 + 87008*a^19*b^5*d^2 + 18240*a^21*b^3*d^2))/(b^21*d^4 + 8*a^2*b^19*d^4 + 28*a^4*b^17*d^4 + 56
*a^6*b^15*d^4 + 70*a^8*b^13*d^4 + 56*a^10*b^11*d^4 + 28*a^12*b^9*d^4 + 8*a^14*b^7*d^4 + a^16*b^5*d^4))) + (tan
(c + d*x)^(1/2)*(32*b^18 - 225*a^18 + 128*a^2*b^16 + 192*a^4*b^14 - 3841*a^6*b^12 + 18050*a^8*b^10 + 26801*a^1
0*b^8 + 16860*a^12*b^6 + 4049*a^14*b^4 - 30*a^16*b^2))/(b^21*d^4 + 8*a^2*b^19*d^4 + 28*a^4*b^17*d^4 + 56*a^6*b
^15*d^4 + 70*a^8*b^13*d^4 + 56*a^10*b^11*d^4 + 28*a^12*b^9*d^4 + 8*a^14*b^7*d^4 + a^16*b^5*d^4))*1i - (-1i/(4*
(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*
((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)
))^(1/2)*((2250*a^20*b*d^2 + 32*a^2*b^19*d^2 + 12288*a^4*b^17*d^2 - 10974*a^6*b^15*d^2 - 105162*a^8*b^13*d^2 -
150758*a^10*b^11*d^2 - 85314*a^12*b^9*d^2 - 35...

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