3.6.91 \(\int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [591]

Optimal. Leaf size=399 \[ -\frac {\left (a^2+2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {a^{7/2} \left (5 a^2+9 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2-2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 \left (a^2+b^2\right ) d}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \]

[Out]

a^(7/2)*(5*a^2+9*b^2)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(7/2)/(a^2+b^2)^2/d+1/2*(a^2+2*a*b-b^2)*arcta
n(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)+1/2*(a^2+2*a*b-b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a
^2+b^2)^2/d*2^(1/2)+1/4*(a^2-2*a*b-b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^2/d*2^(1/2)-1/4*(a
^2-2*a*b-b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^2/d*2^(1/2)-a*(5*a^2+4*b^2)*tan(d*x+c)^(1/2)
/b^3/(a^2+b^2)/d+1/3*(5*a^2+2*b^2)*tan(d*x+c)^(3/2)/b^2/(a^2+b^2)/d-a^2*tan(d*x+c)^(5/2)/b/(a^2+b^2)/d/(a+b*ta
n(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.68, antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3646, 3728, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} -\frac {\left (a^2+2 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 )^2}+\frac {\left (a^2+2 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 )^2}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 d \left (a^2+b^2\right )}+\frac {\left (a^2-2 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 )^2}-\frac {\left (a^2-2 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 )^2}-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 d \left (a^2+b^2\right )}+\frac {a^{7/2} \left (5 a^2+9 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{7/2} d \left (a^2+b^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 + 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*d)) + ((a^2 + 2*a*b - b^
2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*d) + (a^(7/2)*(5*a^2 + 9*b^2)*ArcTan[(Sqrt[b
]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(b^(7/2)*(a^2 + b^2)^2*d) + ((a^2 - 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c +
d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) - ((a^2 - 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] +
Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) - (a*(5*a^2 + 4*b^2)*Sqrt[Tan[c + d*x]])/(b^3*(a^2 + b^2)*d) + ((5
*a^2 + 2*b^2)*Tan[c + d*x]^(3/2))/(3*b^2*(a^2 + b^2)*d) - (a^2*Tan[c + d*x]^(5/2))/(b*(a^2 + b^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 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))^2} \, dx &=-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (\frac {5 a^2}{2}-a b \tan (c+d x)+\frac {1}{2} \left (5 a^2+2 b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {2 \int \frac {\sqrt {\tan (c+d x)} \left (-\frac {3}{4} a \left (5 a^2+2 b^2\right )-\frac {3}{2} b^3 \tan (c+d x)-\frac {3}{4} a \left (5 a^2+4 b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{3 b^2 \left (a^2+b^2\right )}\\ &=-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 \left (a^2+b^2\right ) d}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {4 \int \frac {\frac {3}{8} a^2 \left (5 a^2+4 b^2\right )+\frac {3}{4} a b^3 \tan (c+d x)+\frac {3}{8} \left (5 a^4+4 a^2 b^2-2 b^4\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{3 b^3 \left (a^2+b^2\right )}\\ &=-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 \left (a^2+b^2\right ) d}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {4 \int \frac {\frac {3 a b^4}{2}+\frac {3}{4} b^3 \left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{3 b^3 \left (a^2+b^2\right )^2}+\frac {\left (a^4 \left (5 a^2+9 b^2\right )\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 b^3 \left (a^2+b^2\right )^2}\\ &=-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 \left (a^2+b^2\right ) d}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {8 \text {Subst}\left (\int \frac {\frac {3 a b^4}{2}+\frac {3}{4} b^3 \left (a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{3 b^3 \left (a^2+b^2\right )^2 d}+\frac {\left (a^4 \left (5 a^2+9 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{2 b^3 \left (a^2+b^2\right )^2 d}\\ &=-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 \left (a^2+b^2\right ) d}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (a^2-2 a b-b^2\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 )^2 d}+\frac {\left (a^2+2 a b-b^2\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 )^2 d}+\frac {\left (a^4 \left (5 a^2+9 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^3 \left (a^2+b^2\right )^2 d}\\ &=\frac {a^{7/2} \left (5 a^2+9 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 d}-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 \left (a^2+b^2\right ) d}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (a^2-2 a b-b^2\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 )^2 d}+\frac {\left (a^2-2 a b-b^2\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 )^2 d}+\frac {\left (a^2+2 a b-b^2\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 )^2 d}+\frac {\left (a^2+2 a b-b^2\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 )^2 d}\\ &=\frac {a^{7/2} \left (5 a^2+9 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 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 )^2 d}-\frac {\left (a^2-2 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 )^2 d}-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 \left (a^2+b^2\right ) d}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (a^2+2 a b-b^2\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 )^2 d}-\frac {\left (a^2+2 a b-b^2\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 )^2 d}\\ &=-\frac {\left (a^2+2 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 )^2 d}+\frac {\left (a^2+2 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 )^2 d}+\frac {a^{7/2} \left (5 a^2+9 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 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 )^2 d}-\frac {\left (a^2-2 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 )^2 d}-\frac {a \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{b^3 \left (a^2+b^2\right ) d}+\frac {\left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.91, size = 442, normalized size = 1.11 \begin {gather*} \frac {b^2 \tan ^{\frac {11}{2}}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {-\frac {b \tan ^{\frac {9}{2}}(c+d x)}{d}+\frac {2 \left (\frac {9 a b \tan ^{\frac {7}{2}}(c+d x)}{2 d}+\frac {2 \left (-\frac {63 a^2 b \tan ^{\frac {5}{2}}(c+d x)}{4 d}+\frac {2 \left (\frac {2 \left (\frac {2 \left (\frac {2 \left (-\frac {945}{32} a^3 b^6+\frac {945}{64} a^3 b^4 \left (5 a^2+4 b^2\right )+\frac {945}{64} a^3 b^2 \left (5 a^4+4 a^2 b^2-2 b^4\right )\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \left (a^2+b^2\right ) d}+\frac {-\frac {\sqrt [4]{-1} \left (\frac {945 a^2 b^6}{16}-\frac {945}{32} i a (a-b) b^5 (a+b)\right ) \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {\sqrt [4]{-1} \left (\frac {945 a^2 b^6}{16}+\frac {945}{32} i a (a-b) b^5 (a+b)\right ) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}}{a^2+b^2}\right )}{b}-\frac {945 a^2 b \left (5 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{16 d}\right )}{3 b}+\frac {105 a b \left (5 a^2+2 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{8 d}\right )}{5 b}\right )}{7 b}\right )}{9 b}}{a \left (a^2+b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*Tan[c + d*x]^(11/2))/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])) + (-((b*Tan[c + d*x]^(9/2))/d) + (2*((9*a*b*T
an[c + d*x]^(7/2))/(2*d) + (2*((-63*a^2*b*Tan[c + d*x]^(5/2))/(4*d) + (2*((2*((2*((2*((-945*a^3*b^6)/32 + (945
*a^3*b^4*(5*a^2 + 4*b^2))/64 + (945*a^3*b^2*(5*a^4 + 4*a^2*b^2 - 2*b^4))/64)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]
])/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*(a^2 + b^2)*d) + (-(((-1)^(1/4)*((945*a^2*b^6)/16 - ((945*I)/32)*a*(a - b)*b^5*(
a + b))*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d) - ((-1)^(1/4)*((945*a^2*b^6)/16 + ((945*I)/32)*a*(a - b)*b^5
*(a + b))*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d)/(a^2 + b^2)))/b - (945*a^2*b*(5*a^2 + 4*b^2)*Sqrt[Tan[c +
d*x]])/(16*d)))/(3*b) + (105*a*b*(5*a^2 + 2*b^2)*Tan[c + d*x]^(3/2))/(8*d)))/(5*b)))/(7*b)))/(9*b))/(a*(a^2 +
b^2))

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Maple [A]
time = 0.14, size = 311, normalized size = 0.78

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 a \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{b^{3}}+\frac {2 a^{4} \left (\frac {\left (-\frac {a^{2}}{2}-\frac {b^{2}}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{a +b \tan \left (d x +c \right )}+\frac {\left (5 a^{2}+9 b^{2}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3} \left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}+\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(311\)
default	\(\frac {-\frac {2 \left (-\frac {b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 a \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{b^{3}}+\frac {2 a^{4} \left (\frac {\left (-\frac {a^{2}}{2}-\frac {b^{2}}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{a +b \tan \left (d x +c \right )}+\frac {\left (5 a^{2}+9 b^{2}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3} \left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}+\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(311\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2/b^3*(-1/3*b*tan(d*x+c)^(3/2)+2*a*tan(d*x+c)^(1/2))+2*a^4/b^3/(a^2+b^2)^2*((-1/2*a^2-1/2*b^2)*tan(d*x+c
)^(1/2)/(a+b*tan(d*x+c))+1/2*(5*a^2+9*b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2)))+2/(a^2+b^2)^2*(
1/4*a*b*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)))+1/8*(a^2-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 = 0.51, size = 311, normalized size = 0.78 \begin {gather*} -\frac {\frac {12 \, a^{4} \sqrt {\tan \left (d x + c\right )}}{a^{3} b^{3} + a b^{5} + {\left (a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )} - \frac {12 \, {\left (5 \, a^{6} + 9 \, a^{4} b^{2}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt {a b}} - \frac {3 \, {\left (2 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\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^{2} + 2 \, a b - b^{2}\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^{2} - 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {8 \, {\left (b \tan \left (d x + c\right )^{\frac {3}{2}} - 6 \, a \sqrt {\tan \left (d x + c\right )}\right )}}{b^{3}}}{12 \, 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))^2,x, algorithm="maxima")

[Out]

-1/12*(12*a^4*sqrt(tan(d*x + c))/(a^3*b^3 + a*b^5 + (a^2*b^4 + b^6)*tan(d*x + c)) - 12*(5*a^6 + 9*a^4*b^2)*arc
tan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^4*b^3 + 2*a^2*b^5 + b^7)*sqrt(a*b)) - 3*(2*sqrt(2)*(a^2 + 2*a*b - b^2)
*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a^2 + 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqr
t(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(a^2 - 2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)
+ sqrt(2)*(a^2 - 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^4 + 2*a^2*b^2 + b^4) -
8*(b*tan(d*x + c)^(3/2) - 6*a*sqrt(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. 7246 vs. \(2 (353) = 706\).
time = 14.90, size = 14604, normalized size = 36.60 \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))^2,x, algorithm="fricas")

[Out]

[1/12*(12*sqrt(2)*((a^14*b^3 + 5*a^12*b^5 + 9*a^10*b^7 + 5*a^8*b^9 - 5*a^6*b^11 - 9*a^4*b^13 - 5*a^2*b^15 - b^
17)*d^5*cos(d*x + c)^3 + 2*(a^13*b^4 + 6*a^11*b^6 + 15*a^9*b^8 + 20*a^7*b^10 + 15*a^5*b^12 + 6*a^3*b^14 + a*b^
16)*d^5*cos(d*x + c)^2*sin(d*x + c) + (a^12*b^5 + 6*a^10*b^7 + 15*a^8*b^9 + 20*a^6*b^11 + 15*a^4*b^13 + 6*a^2*
b^15 + b^17)*d^5*cos(d*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 + 4*(a^11*b + 3*a^9*b^3 + 2
*a^7*b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/
(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((
a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^
4))*(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))^(3/4)*arctan(((a^16 - 20*a^12*b^4 - 64*a^10*b^6
- 90*a^8*b^8 - 64*a^6*b^10 - 20*a^4*b^12 + b^16)*d^4*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/(
(a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^14 + b^16)*d
^4))*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + sqrt(2)*(2*(a^17*b + 8*a^15*b^3 + 28*a^13
*b^5 + 56*a^11*b^7 + 70*a^9*b^9 + 56*a^7*b^11 + 28*a^5*b^13 + 8*a^3*b^15 + a*b^17)*d^7*sqrt((a^8 - 12*a^6*b^2
+ 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 +
28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^4))*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (a^14 +
5*a^12*b^2 + 9*a^10*b^4 + 5*a^8*b^6 - 5*a^6*b^8 - 9*a^4*b^10 - 5*a^2*b^12 - b^14)*d^5*sqrt((a^8 - 12*a^6*b^2
+ 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 +
28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^4)))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 + 4*(a^11*b + 3*a^
9*b^3 + 2*a^7*b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8
)*d^4)))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b
^6 + 15*a^4*b^8 - 10*a^2*b^10 + b^12)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))*cos(d*
x + c) + sqrt(2)*((a^14 - 11*a^12*b^2 + 25*a^10*b^4 + 37*a^8*b^6 - 37*a^6*b^8 - 25*a^4*b^10 + 11*a^2*b^12 - b^
14)*d^3*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))*cos(d*x + c) - 2*(a^9*b - 12*a^7*b^3 + 3
8*a^5*b^5 - 12*a^3*b^7 + a*b^9)*d*cos(d*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 + 4*(a^11*
b + 3*a^9*b^3 + 2*a^7*b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b
^6 + b^8)*d^4)))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*(1/((a^8
+ 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))^(1/4) + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)*si
n(d*x + c))/cos(d*x + c))*(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))^(3/4) - sqrt(2)*(2*(a^21*b
+ 2*a^19*b^3 - 19*a^17*b^5 - 104*a^15*b^7 - 238*a^13*b^9 - 308*a^11*b^11 - 238*a^9*b^13 - 104*a^7*b^15 - 19*a
^5*b^17 + 2*a^3*b^19 + a*b^21)*d^7*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2
+ 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^4))*sqrt(1/((a^8
+ 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (a^18 - a^16*b^2 - 20*a^14*b^4 - 44*a^12*b^6 - 26*a^10*b^8
+ 26*a^8*b^10 + 44*a^6*b^12 + 20*a^4*b^14 + a^2*b^16 - b^18)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*
b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^
14 + b^16)*d^4)))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 + 4*(a^11*b + 3*a^9*b^3 + 2*a^7*b^5 - 2*
a^5*b^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/(a^8 - 12*a^6
*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^
2*b^6 + b^8)*d^4))^(3/4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)) + 12*sqrt(2)*((a^14*b^3 + 5*a^12
*b^5 + 9*a^10*b^7 + 5*a^8*b^9 - 5*a^6*b^11 - 9*a^4*b^13 - 5*a^2*b^15 - b^17)*d^5*cos(d*x + c)^3 + 2*(a^13*b^4
+ 6*a^11*b^6 + 15*a^9*b^8 + 20*a^7*b^10 + 15*a^5*b^12 + 6*a^3*b^14 + a*b^16)*d^5*cos(d*x + c)^2*sin(d*x + c) +
(a^12*b^5 + 6*a^10*b^7 + 15*a^8*b^9 + 20*a^6*b^11 + 15*a^4*b^13 + 6*a^2*b^15 + b^17)*d^5*cos(d*x + c))*sqrt((
a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 + 4*(a^11*b + 3*a^9*b^3 + 2*a^7*b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*
b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a
^2*b^6 + b^8))*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*
a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^4))*(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4
+ 4*a^2*b^6 + b^8)*d^4))^(3/4)*arctan(-((a^16 - 20*a^12*b^4 - 64*a^10*b^6 - 90*a^8*b^8 - 64*a^6*b^10 - 20*a^4*
b^12 + b^16)*d^4*sqrt((a^8 - 12*a^6*b^2 + 38*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))**2,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))^2,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 9.32, size = 2500, normalized size = 6.27 \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))^2,x)

[Out]

(2*tan(c + d*x)^(3/2))/(3*b^2*d) - atan((((((1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*
d^2*6i))^(1/2)*(((1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*((((8*(64*a^
2*b^18*d^4 + 480*a^4*b^16*d^4 + 1440*a^6*b^14*d^4 + 2240*a^8*b^12*d^4 + 1920*a^10*b^10*d^4 + 864*a^12*b^8*d^4
+ 160*a^14*b^6*d^4))/(b^13*d^5 + 4*a^2*b^11*d^5 + 6*a^4*b^9*d^5 + 4*a^6*b^7*d^5 + a^8*b^5*d^5) - (8*tan(c + d*
x)^(1/2)*(1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*(32*b^22*d^4 + 160*a
^2*b^20*d^4 + 288*a^4*b^18*d^4 + 160*a^6*b^16*d^4 - 160*a^8*b^14*d^4 - 288*a^10*b^12*d^4 - 160*a^12*b^10*d^4 -
32*a^14*b^8*d^4))/(b^13*d^4 + 4*a^2*b^11*d^4 + 6*a^4*b^9*d^4 + 4*a^6*b^7*d^4 + a^8*b^5*d^4))*(1/(a^4*d^2*1i +
b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2))/2 + (16*tan(c + d*x)^(1/2)*(60*a*b^17*d^2 +
200*a^17*b*d^2 + 52*a^3*b^15*d^2 - 72*a^5*b^13*d^2 - 56*a^7*b^11*d^2 + 660*a^9*b^9*d^2 + 2020*a^11*b^7*d^2 + 2
288*a^13*b^5*d^2 + 1120*a^15*b^3*d^2))/(b^13*d^4 + 4*a^2*b^11*d^4 + 6*a^4*b^9*d^4 + 4*a^6*b^7*d^4 + a^8*b^5*d^
4)))/2 - (8*(4*a*b^15*d^2 + 400*a^15*b*d^2 + 16*a^3*b^13*d^2 + 600*a^5*b^11*d^2 - 240*a^7*b^9*d^2 - 1612*a^9*b
^7*d^2 - 144*a^11*b^5*d^2 + 1040*a^13*b^3*d^2))/(b^13*d^5 + 4*a^2*b^11*d^5 + 6*a^4*b^9*d^5 + 4*a^6*b^7*d^5 + a
^8*b^5*d^5)))/2 - (16*tan(c + d*x)^(1/2)*(2*b^14 - 25*a^14 + 4*a^2*b^12 + 2*a^4*b^10 + 81*a^8*b^6 + 9*a^10*b^4
- 65*a^12*b^2))/(b^13*d^4 + 4*a^2*b^11*d^4 + 6*a^4*b^9*d^4 + 4*a^6*b^7*d^4 + a^8*b^5*d^4))*(1/(a^4*d^2*1i + b
^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*1i)/2 - ((((1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^
3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*(((1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b
^2*d^2*6i))^(1/2)*((((8*(64*a^2*b^18*d^4 + 480*a^4*b^16*d^4 + 1440*a^6*b^14*d^4 + 2240*a^8*b^12*d^4 + 1920*a^1
0*b^10*d^4 + 864*a^12*b^8*d^4 + 160*a^14*b^6*d^4))/(b^13*d^5 + 4*a^2*b^11*d^5 + 6*a^4*b^9*d^5 + 4*a^6*b^7*d^5
+ a^8*b^5*d^5) + (8*tan(c + d*x)^(1/2)*(1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6
i))^(1/2)*(32*b^22*d^4 + 160*a^2*b^20*d^4 + 288*a^4*b^18*d^4 + 160*a^6*b^16*d^4 - 160*a^8*b^14*d^4 - 288*a^10*
b^12*d^4 - 160*a^12*b^10*d^4 - 32*a^14*b^8*d^4))/(b^13*d^4 + 4*a^2*b^11*d^4 + 6*a^4*b^9*d^4 + 4*a^6*b^7*d^4 +
a^8*b^5*d^4))*(1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2))/2 - (16*tan(c
+ d*x)^(1/2)*(60*a*b^17*d^2 + 200*a^17*b*d^2 + 52*a^3*b^15*d^2 - 72*a^5*b^13*d^2 - 56*a^7*b^11*d^2 + 660*a^9*b
^9*d^2 + 2020*a^11*b^7*d^2 + 2288*a^13*b^5*d^2 + 1120*a^15*b^3*d^2))/(b^13*d^4 + 4*a^2*b^11*d^4 + 6*a^4*b^9*d^
4 + 4*a^6*b^7*d^4 + a^8*b^5*d^4)))/2 - (8*(4*a*b^15*d^2 + 400*a^15*b*d^2 + 16*a^3*b^13*d^2 + 600*a^5*b^11*d^2
- 240*a^7*b^9*d^2 - 1612*a^9*b^7*d^2 - 144*a^11*b^5*d^2 + 1040*a^13*b^3*d^2))/(b^13*d^5 + 4*a^2*b^11*d^5 + 6*a
^4*b^9*d^5 + 4*a^6*b^7*d^5 + a^8*b^5*d^5)))/2 + (16*tan(c + d*x)^(1/2)*(2*b^14 - 25*a^14 + 4*a^2*b^12 + 2*a^4*
b^10 + 81*a^8*b^6 + 9*a^10*b^4 - 65*a^12*b^2))/(b^13*d^4 + 4*a^2*b^11*d^4 + 6*a^4*b^9*d^4 + 4*a^6*b^7*d^4 + a^
8*b^5*d^4))*(1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*1i)/2)/(((((1/(a^
4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*(((1/(a^4*d^2*1i + b^4*d^2*1i + 4*a
*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*((((8*(64*a^2*b^18*d^4 + 480*a^4*b^16*d^4 + 1440*a^6*b^14*d^4
+ 2240*a^8*b^12*d^4 + 1920*a^10*b^10*d^4 + 864*a^12*b^8*d^4 + 160*a^14*b^6*d^4))/(b^13*d^5 + 4*a^2*b^11*d^5 +
6*a^4*b^9*d^5 + 4*a^6*b^7*d^5 + a^8*b^5*d^5) - (8*tan(c + d*x)^(1/2)*(1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2
- 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*(32*b^22*d^4 + 160*a^2*b^20*d^4 + 288*a^4*b^18*d^4 + 160*a^6*b^16*d^4
- 160*a^8*b^14*d^4 - 288*a^10*b^12*d^4 - 160*a^12*b^10*d^4 - 32*a^14*b^8*d^4))/(b^13*d^4 + 4*a^2*b^11*d^4 + 6*
a^4*b^9*d^4 + 4*a^6*b^7*d^4 + a^8*b^5*d^4))*(1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*
d^2*6i))^(1/2))/2 + (16*tan(c + d*x)^(1/2)*(60*a*b^17*d^2 + 200*a^17*b*d^2 + 52*a^3*b^15*d^2 - 72*a^5*b^13*d^2
- 56*a^7*b^11*d^2 + 660*a^9*b^9*d^2 + 2020*a^11*b^7*d^2 + 2288*a^13*b^5*d^2 + 1120*a^15*b^3*d^2))/(b^13*d^4 +
4*a^2*b^11*d^4 + 6*a^4*b^9*d^4 + 4*a^6*b^7*d^4 + a^8*b^5*d^4)))/2 - (8*(4*a*b^15*d^2 + 400*a^15*b*d^2 + 16*a^
3*b^13*d^2 + 600*a^5*b^11*d^2 - 240*a^7*b^9*d^2 - 1612*a^9*b^7*d^2 - 144*a^11*b^5*d^2 + 1040*a^13*b^3*d^2))/(b
^13*d^5 + 4*a^2*b^11*d^5 + 6*a^4*b^9*d^5 + 4*a^6*b^7*d^5 + a^8*b^5*d^5)))/2 - (16*tan(c + d*x)^(1/2)*(2*b^14 -
25*a^14 + 4*a^2*b^12 + 2*a^4*b^10 + 81*a^8*b^6 + 9*a^10*b^4 - 65*a^12*b^2))/(b^13*d^4 + 4*a^2*b^11*d^4 + 6*a^
4*b^9*d^4 + 4*a^6*b^7*d^4 + a^8*b^5*d^4))*(1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^
2*6i))^(1/2))/2 - (16*(25*a^12 - 18*a^4*b^8 - 10*a^6*b^6 + 81*a^8*b^4 + 90*a^10*b^2))/(b^13*d^5 + 4*a^2*b^11*d
^5 + 6*a^4*b^9*d^5 + 4*a^6*b^7*d^5 + a^8*b^5*d^5) + ((((1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2
- a^2*b^2*d^2*6i))^(1/2)*(((1/(a^4*d^2*1i + b^...

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