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

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

[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]))

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Rubi [A]  time = 1.08647, antiderivative size = 444, normalized size of antiderivative = 1., 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 = {3565, 3645, 3647, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{a^{5/2} \left (46 a^2 b^2+15 a^4+63 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{7/2} d \left (a^2+b^2\right )^3}-\frac{a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{4 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (31 a^2 b^2+15 a^4+8 b^4\right ) \sqrt{\tan (c+d x)}}{4 b^3 d \left (a^2+b^2\right )^2}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

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 3565

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 3645

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*T
an[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 3647

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 3653

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]

Rule 3534

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 1168

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 1162

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 617

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 204

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

Rule 1165

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 628

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 3634

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 63

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 205

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

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 \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 5.12525, size = 403, normalized size = 0.91 \[ \frac{\frac{2 a \left (5 a^2+4 b^2\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{b^2}+\frac{2 a^2 \left (15 a^2+16 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))}{b^3}-\frac{a (a+b \tan (c+d x)) \left (a \sqrt{b} \left (a^2+b^2\right ) \left (31 a^2 b^2+15 a^4+24 b^4\right ) \sqrt{\tan (c+d x)}+(a+b \tan (c+d x)) \left (a^{5/2} \left (46 a^2 b^2+15 a^4+63 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )-4 (-1)^{3/4} b^{7/2} (a+i b)^3 \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )-4 \sqrt [4]{-1} b^{7/2} (b+i a)^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )\right )\right )}{b^{7/2} \left (a^2+b^2\right )^2}-\frac{2 a^2 \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))}{b}-2 b \tan ^{\frac{9}{2}}(c+d x) (a+b \tan (c+d x))+2 a \tan ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))+2 b^2 \tan ^{\frac{11}{2}}(c+d x)}{4 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]

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 [B]  time = 0.046, size = 920, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

2/d/b^3*tan(d*x+c)^(1/2)+9/4/d*a^7/b^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)+13/2/d*a^5/(a^2+b^2)^3/
(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)+17/4/d*a^3*b^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)+7/4/d*a^8/b
^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)+11/2/d*a^6/b/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2
)+15/4/d*a^4*b/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)-15/4/d*a^7/b^3/(a^2+b^2)^3/(a*b)^(1/2)*arctan(t
an(d*x+c)^(1/2)*b/(a*b)^(1/2))-23/2/d*a^5/b/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))-63/
4/d*a^3*b/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))+3/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^
(1/2)*tan(d*x+c)^(1/2))*a^2*b-1/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+3/2/d/(a^2+b^2)
^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^2*b-1/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(
1/2)*b^3+3/4/d/(a^2+b^2)^3*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)))*a^2*b-1/4/d/(a^2+b^2)^3*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)))*b^3+1/4/d/(a^2+b^2)^3*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)+ta
n(d*x+c)))*2^(1/2)*a^3-3/4/d/(a^2+b^2)^3*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^(1/2)*a*b^2+1/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^3-3/2/d/(a^2+b^2)^
3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2+1/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/
2)*a^3-3/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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