3.156 \(\int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=179 \[ -\frac{8 (-1)^{3/4} a^3 d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{8 i a^3 d^2 \sqrt{d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}-\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{8 i a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac{8 a^3 d (d \tan (e+f x))^{3/2}}{3 f} \]

[Out]

(-8*(-1)^(3/4)*a^3*d^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - ((8*I)*a^3*d^2*Sqrt[d*Tan[e
+ f*x]])/f + (8*a^3*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (((8*I)/5)*a^3*(d*Tan[e + f*x])^(5/2))/f - (40*a^3*(d*Ta
n[e + f*x])^(7/2))/(63*d*f) - (2*(d*Tan[e + f*x])^(7/2)*(a^3 + I*a^3*Tan[e + f*x]))/(9*d*f)

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Rubi [A]  time = 0.318803, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3556, 3592, 3528, 3533, 205} \[ -\frac{8 (-1)^{3/4} a^3 d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{8 i a^3 d^2 \sqrt{d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}-\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{8 i a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac{8 a^3 d (d \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x])^3,x]

[Out]

(-8*(-1)^(3/4)*a^3*d^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - ((8*I)*a^3*d^2*Sqrt[d*Tan[e
+ f*x]])/f + (8*a^3*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (((8*I)/5)*a^3*(d*Tan[e + f*x])^(5/2))/f - (40*a^3*(d*Ta
n[e + f*x])^(7/2))/(63*d*f) - (2*(d*Tan[e + f*x])^(7/2)*(a^3 + I*a^3*Tan[e + f*x]))/(9*d*f)

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3592

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

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 (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx &=-\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+i a^3 \tan (e+f x)\right )}{9 d f}+\frac{(2 a) \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) (8 a d+10 i a d \tan (e+f x)) \, dx}{9 d}\\ &=-\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}-\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+i a^3 \tan (e+f x)\right )}{9 d f}+\frac{(2 a) \int (d \tan (e+f x))^{5/2} \left (18 a^2 d+18 i a^2 d \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac{8 i a^3 (d \tan (e+f x))^{5/2}}{5 f}-\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}-\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+i a^3 \tan (e+f x)\right )}{9 d f}+\frac{(2 a) \int (d \tan (e+f x))^{3/2} \left (-18 i a^2 d^2+18 a^2 d^2 \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac{8 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (d \tan (e+f x))^{5/2}}{5 f}-\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}-\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+i a^3 \tan (e+f x)\right )}{9 d f}+\frac{(2 a) \int \sqrt{d \tan (e+f x)} \left (-18 a^2 d^3-18 i a^2 d^3 \tan (e+f x)\right ) \, dx}{9 d}\\ &=-\frac{8 i a^3 d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{8 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (d \tan (e+f x))^{5/2}}{5 f}-\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}-\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+i a^3 \tan (e+f x)\right )}{9 d f}+\frac{(2 a) \int \frac{18 i a^2 d^4-18 a^2 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{9 d}\\ &=-\frac{8 i a^3 d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{8 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (d \tan (e+f x))^{5/2}}{5 f}-\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}-\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+i a^3 \tan (e+f x)\right )}{9 d f}-\frac{\left (144 a^5 d^7\right ) \operatorname{Subst}\left (\int \frac{1}{18 i a^2 d^5+18 a^2 d^4 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=-\frac{8 (-1)^{3/4} a^3 d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{8 i a^3 d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{8 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (d \tan (e+f x))^{5/2}}{5 f}-\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}-\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+i a^3 \tan (e+f x)\right )}{9 d f}\\ \end{align*}

Mathematica [A]  time = 3.54353, size = 150, normalized size = 0.84 \[ \frac{a^3 d^2 \sqrt{d \tan (e+f x)} \left (10080 i \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )-i \sqrt{i \tan (e+f x)} \sec ^4(e+f x) (570 i \sin (2 (e+f x))+555 i \sin (4 (e+f x))+4900 \cos (2 (e+f x))+1547 \cos (4 (e+f x))+3633)\right )}{1260 f \sqrt{i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x])^3,x]

[Out]

(a^3*d^2*((10080*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] - I*Sec[e + f*x]^4*(36
33 + 4900*Cos[2*(e + f*x)] + 1547*Cos[4*(e + f*x)] + (570*I)*Sin[2*(e + f*x)] + (555*I)*Sin[4*(e + f*x)])*Sqrt
[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(1260*f*Sqrt[I*Tan[e + f*x]])

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Maple [B]  time = 0.021, size = 454, normalized size = 2.5 \begin{align*}{\frac{-{\frac{2\,i}{9}}{a}^{3}}{f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}}-{\frac{6\,{a}^{3}}{7\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{{\frac{8\,i}{5}}{a}^{3}}{f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{8\,{a}^{3}d}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{8\,i{a}^{3}{d}^{2}}{f}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{i{a}^{3}{d}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }+{\frac{2\,i{a}^{3}{d}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{2\,i{a}^{3}{d}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{{a}^{3}{d}^{3}\sqrt{2}}{f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-2\,{\frac{{a}^{3}{d}^{3}\sqrt{2}}{f\sqrt [4]{{d}^{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }+2\,{\frac{{a}^{3}{d}^{3}\sqrt{2}}{f\sqrt [4]{{d}^{2}}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x)

[Out]

-2/9*I/f*a^3/d^2*(d*tan(f*x+e))^(9/2)-6/7*a^3*(d*tan(f*x+e))^(7/2)/d/f+8/5*I*a^3*(d*tan(f*x+e))^(5/2)/f+8/3*a^
3*d*(d*tan(f*x+e))^(3/2)/f-8*I*a^3*d^2*(d*tan(f*x+e))^(1/2)/f+I/f*a^3*d^2*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)
+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+
(d^2)^(1/2)))+2*I/f*a^3*d^2*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*I/f*a^3*d
^2*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/f*a^3*d^3/(d^2)^(1/4)*2^(1/2)*ln(
(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(
1/2)*2^(1/2)+(d^2)^(1/2)))-2/f*a^3*d^3/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+
2/f*a^3*d^3/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)

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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((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.35588, size = 1373, normalized size = 7.67 \begin{align*} -\frac{315 \, \sqrt{\frac{64 i \, a^{6} d^{5}}{f^{2}}}{\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-8 i \, a^{3} d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{64 i \, a^{6} d^{5}}{f^{2}}}{\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} d^{2}}\right ) - 315 \, \sqrt{\frac{64 i \, a^{6} d^{5}}{f^{2}}}{\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-8 i \, a^{3} d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{64 i \, a^{6} d^{5}}{f^{2}}}{\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} d^{2}}\right ) -{\left (-16816 i \, a^{3} d^{2} e^{\left (8 i \, f x + 8 i \, e\right )} - 43760 i \, a^{3} d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 58128 i \, a^{3} d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 34640 i \, a^{3} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 7936 i \, a^{3} d^{2}\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{1260 \,{\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/1260*(315*sqrt(64*I*a^6*d^5/f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e)
+ 4*f*e^(2*I*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d^3*e^(2*I*f*x + 2*I*e) + sqrt(64*I*a^6*d^5/f^2)*(I*f*e^(2*I
*f*x + 2*I*e) + I*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a
^3*d^2)) - 315*sqrt(64*I*a^6*d^5/f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*
e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d^3*e^(2*I*f*x + 2*I*e) + sqrt(64*I*a^6*d^5/f^2)*(-I*f*e^(
2*I*f*x + 2*I*e) - I*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)
/(a^3*d^2)) - (-16816*I*a^3*d^2*e^(8*I*f*x + 8*I*e) - 43760*I*a^3*d^2*e^(6*I*f*x + 6*I*e) - 58128*I*a^3*d^2*e^
(4*I*f*x + 4*I*e) - 34640*I*a^3*d^2*e^(2*I*f*x + 2*I*e) - 7936*I*a^3*d^2)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d
)/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f
*e^(2*I*f*x + 2*I*e) + f)

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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((d*tan(f*x+e))**(5/2)*(a+I*a*tan(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.34322, size = 301, normalized size = 1.68 \begin{align*} -\frac{8 \, \sqrt{2} a^{3} d^{\frac{5}{2}} \arctan \left (\frac{16 \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{70 i \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{4} + 270 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{3} - 504 i \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{2} - 840 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right ) + 2520 i \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8}}{315 \, d^{18} f^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-8*sqrt(2)*a^3*d^(5/2)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqr
t(d)))/(f*(I*d/sqrt(d^2) + 1)) - 1/315*(70*I*sqrt(d*tan(f*x + e))*a^3*d^20*f^8*tan(f*x + e)^4 + 270*sqrt(d*tan
(f*x + e))*a^3*d^20*f^8*tan(f*x + e)^3 - 504*I*sqrt(d*tan(f*x + e))*a^3*d^20*f^8*tan(f*x + e)^2 - 840*sqrt(d*t
an(f*x + e))*a^3*d^20*f^8*tan(f*x + e) + 2520*I*sqrt(d*tan(f*x + e))*a^3*d^20*f^8)/(d^18*f^9)