3.163 \(\int \frac{(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=159 \[ -\frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{9/2} f}+\frac{8 i a^3}{d^4 f \sqrt{d \tan (e+f x)}}+\frac{8 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}-\frac{32 i a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}} \]

[Out]

(-8*(-1)^(1/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(9/2)*f) - (((32*I)/35)*a^3)/(d^2*f*(
d*Tan[e + f*x])^(5/2)) + (8*a^3)/(3*d^3*f*(d*Tan[e + f*x])^(3/2)) + ((8*I)*a^3)/(d^4*f*Sqrt[d*Tan[e + f*x]]) -
(2*(a^3 + I*a^3*Tan[e + f*x]))/(7*d*f*(d*Tan[e + f*x])^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.314688, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3553, 3591, 3529, 3533, 205} \[ -\frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{9/2} f}+\frac{8 i a^3}{d^4 f \sqrt{d \tan (e+f x)}}+\frac{8 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}-\frac{32 i a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(9/2),x]

[Out]

(-8*(-1)^(1/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(9/2)*f) - (((32*I)/35)*a^3)/(d^2*f*(
d*Tan[e + f*x])^(5/2)) + (8*a^3)/(3*d^3*f*(d*Tan[e + f*x])^(3/2)) + ((8*I)*a^3)/(d^4*f*Sqrt[d*Tan[e + f*x]]) -
(2*(a^3 + I*a^3*Tan[e + f*x]))/(7*d*f*(d*Tan[e + f*x])^(7/2))

Rule 3553

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

Rule 3591

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

Rule 3529

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

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 \frac{(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx &=-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}-\frac{2 \int \frac{(a+i a \tan (e+f x)) \left (-8 i a^2 d+6 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{7/2}} \, dx}{7 d^2}\\ &=-\frac{32 i a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}-\frac{2 \int \frac{14 a^3 d^2+14 i a^3 d^2 \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{7 d^4}\\ &=-\frac{32 i a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}+\frac{8 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}-\frac{2 \int \frac{14 i a^3 d^3-14 a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{7 d^6}\\ &=-\frac{32 i a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}+\frac{8 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac{8 i a^3}{d^4 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}-\frac{2 \int \frac{-14 a^3 d^4-14 i a^3 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{7 d^8}\\ &=-\frac{32 i a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}+\frac{8 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac{8 i a^3}{d^4 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}-\frac{\left (112 a^6\right ) \operatorname{Subst}\left (\int \frac{1}{-14 a^3 d^5+14 i a^3 d^4 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=-\frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{9/2} f}-\frac{32 i a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}+\frac{8 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac{8 i a^3}{d^4 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\\ \end{align*}

Mathematica [B]  time = 9.45519, size = 416, normalized size = 2.62 \[ \frac{8 e^{-3 i e} \sqrt{-\frac{i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}} \cos ^3(e+f x) \tan ^{\frac{9}{2}}(e+f x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) (a+i a \tan (e+f x))^3}{f \sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} (\cos (f x)+i \sin (f x))^3 (d \tan (e+f x))^{9/2}}+\frac{\sin ^3(e+f x) \tan ^2(e+f x) (a+i a \tan (e+f x))^3 \left (\left (-\frac{2}{7} \cos (3 e)+\frac{2}{7} i \sin (3 e)\right ) \csc ^4(e+f x)+i \csc (e) \left (\frac{6}{5} \cos (3 e)-\frac{6}{5} i \sin (3 e)\right ) \sin (f x) \csc ^3(e+f x)+\csc (e) (63 \cos (e)+170 i \sin (e)) \left (-\frac{2}{105} \sin (3 e)-\frac{2}{105} i \cos (3 e)\right ) \csc ^2(e+f x)-i \csc (e) \left (\frac{46}{5} \cos (3 e)-\frac{46}{5} i \sin (3 e)\right ) \sin (f x) \csc (e+f x)+i \csc (e) (483 \cos (e)+155 i \sin (e)) \left (\frac{2}{105} \cos (3 e)-\frac{2}{105} i \sin (3 e)\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (d \tan (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(9/2),x]

[Out]

((Csc[e]*Csc[e + f*x]^2*(63*Cos[e] + (170*I)*Sin[e])*(((-2*I)/105)*Cos[3*e] - (2*Sin[3*e])/105) + I*Csc[e]*(48
3*Cos[e] + (155*I)*Sin[e])*((2*Cos[3*e])/105 - ((2*I)/105)*Sin[3*e]) + Csc[e + f*x]^4*((-2*Cos[3*e])/7 + ((2*I
)/7)*Sin[3*e]) + I*Csc[e]*Csc[e + f*x]^3*((6*Cos[3*e])/5 - ((6*I)/5)*Sin[3*e])*Sin[f*x] - I*Csc[e]*Csc[e + f*x
]*((46*Cos[3*e])/5 - ((46*I)/5)*Sin[3*e])*Sin[f*x])*Sin[e + f*x]^3*Tan[e + f*x]^2*(a + I*a*Tan[e + f*x])^3)/(f
*(Cos[f*x] + I*Sin[f*x])^3*(d*Tan[e + f*x])^(9/2)) + (8*Sqrt[((-I)*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(
e + f*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]]*Cos[e + f*x]^3*Tan[e + f*x]^(9
/2)*(a + I*a*Tan[e + f*x])^3)/(E^((3*I)*e)*Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]*f*(Cos[f
*x] + I*Sin[f*x])^3*(d*Tan[e + f*x])^(9/2))

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Maple [B]  time = 0.028, size = 436, normalized size = 2.7 \begin{align*}{\frac{-{\frac{6\,i}{5}}{a}^{3}}{f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}-{\frac{2\,{a}^{3}}{7\,fd} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{7}{2}}}}+{\frac{8\,i{a}^{3}}{{d}^{4}f}{\frac{1}{\sqrt{d\tan \left ( fx+e \right ) }}}}+{\frac{8\,{a}^{3}}{3\,{d}^{3}f} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{{a}^{3}\sqrt{2}}{f{d}^{5}}\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 ) }+2\,{\frac{{a}^{3}\sqrt [4]{{d}^{2}}\sqrt{2}}{f{d}^{5}}\arctan \left ({\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }-2\,{\frac{{a}^{3}\sqrt [4]{{d}^{2}}\sqrt{2}}{f{d}^{5}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }+{\frac{i{a}^{3}\sqrt{2}}{{d}^{4}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}}}}}+{\frac{2\,i{a}^{3}\sqrt{2}}{{d}^{4}f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{2\,i{a}^{3}\sqrt{2}}{{d}^{4}f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.19595, size = 1404, normalized size = 8.83 \begin{align*} \frac{105 \,{\left (d^{5} f e^{\left (8 i \, f x + 8 i \, e\right )} - 4 \, d^{5} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{5} f e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} f\right )} \sqrt{-\frac{64 i \, a^{6}}{d^{9} f^{2}}} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} 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}} \sqrt{-\frac{64 i \, a^{6}}{d^{9} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 105 \,{\left (d^{5} f e^{\left (8 i \, f x + 8 i \, e\right )} - 4 \, d^{5} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{5} f e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} f\right )} \sqrt{-\frac{64 i \, a^{6}}{d^{9} f^{2}}} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} -{\left (d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} 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}} \sqrt{-\frac{64 i \, a^{6}}{d^{9} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \,{\left (319 \, a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} - 327 \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - 95 \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 387 \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 164 \, a^{3}\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}}}{420 \,{\left (d^{5} f e^{\left (8 i \, f x + 8 i \, e\right )} - 4 \, d^{5} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{5} f e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

1/420*(105*(d^5*f*e^(8*I*f*x + 8*I*e) - 4*d^5*f*e^(6*I*f*x + 6*I*e) + 6*d^5*f*e^(4*I*f*x + 4*I*e) - 4*d^5*f*e^
(2*I*f*x + 2*I*e) + d^5*f)*sqrt(-64*I*a^6/(d^9*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (d^5*f*e^(2*I*f
*x + 2*I*e) + d^5*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-64*I*a^6/(d^9*f^2)
))*e^(-2*I*f*x - 2*I*e)/a^3) - 105*(d^5*f*e^(8*I*f*x + 8*I*e) - 4*d^5*f*e^(6*I*f*x + 6*I*e) + 6*d^5*f*e^(4*I*f
*x + 4*I*e) - 4*d^5*f*e^(2*I*f*x + 2*I*e) + d^5*f)*sqrt(-64*I*a^6/(d^9*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x +
2*I*e) - (d^5*f*e^(2*I*f*x + 2*I*e) + d^5*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*
sqrt(-64*I*a^6/(d^9*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(319*a^3*e^(8*I*f*x + 8*I*e) - 327*a^3*e^(6*I*f*x +
6*I*e) - 95*a^3*e^(4*I*f*x + 4*I*e) + 387*a^3*e^(2*I*f*x + 2*I*e) - 164*a^3)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) +
I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^5*f*e^(8*I*f*x + 8*I*e) - 4*d^5*f*e^(6*I*f*x + 6*I*e) + 6*d^5*f*e^(4*I*f*x
+ 4*I*e) - 4*d^5*f*e^(2*I*f*x + 2*I*e) + d^5*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((a+I*a*tan(f*x+e))**3/(d*tan(f*x+e))**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.29683, size = 211, normalized size = 1.33 \begin{align*} \frac{8 i \, \sqrt{2} a^{3} \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 )}{d^{\frac{9}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{-840 i \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 280 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} + 126 i \, a^{3} d^{3} \tan \left (f x + e\right ) + 30 \, a^{3} d^{3}}{105 \, \sqrt{d \tan \left (f x + e\right )} d^{7} f \tan \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x, algorithm="giac")

[Out]

8*I*sqrt(2)*a^3*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))/
(d^(9/2)*f*(I*d/sqrt(d^2) + 1)) - 1/105*(-840*I*a^3*d^3*tan(f*x + e)^3 - 280*a^3*d^3*tan(f*x + e)^2 + 126*I*a^
3*d^3*tan(f*x + e) + 30*a^3*d^3)/(sqrt(d*tan(f*x + e))*d^7*f*tan(f*x + e)^3)