3.2.63 \(\int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx\) [163]

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

[Out]

-8*(-1)^(1/4)*a^3*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(9/2)/f+8*I*a^3/d^4/f/(d*tan(f*x+e))^(1/2)
-32/35*I*a^3/d^2/f/(d*tan(f*x+e))^(5/2)+8/3*a^3/d^3/f/(d*tan(f*x+e))^(3/2)-2/7*(a^3+I*a^3*tan(f*x+e))/d/f/(d*t
an(f*x+e))^(7/2)

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Rubi [A]
time = 0.21, antiderivative size = 159, normalized size of antiderivative = 1.00, 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 = {3634, 3672, 3610, 3614, 211} \begin {gather*} -\frac {8 \sqrt [4]{-1} a^3 \text {ArcTan}\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}} \end {gather*}

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 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 3610

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 3614

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 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] /;
FreeQ[{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] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3672

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]

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 ) \text {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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(416\) vs. \(2(159)=318\).
time = 7.70, size = 416, normalized size = 2.62 \begin {gather*} \frac {\left (\csc (e) \csc ^2(e+f x) (63 \cos (e)+170 i \sin (e)) \left (-\frac {2}{105} i \cos (3 e)-\frac {2}{105} \sin (3 e)\right )+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 )+\csc ^4(e+f x) \left (-\frac {2}{7} \cos (3 e)+\frac {2}{7} i \sin (3 e)\right )+i \csc (e) \csc ^3(e+f x) \left (\frac {6}{5} \cos (3 e)-\frac {6}{5} i \sin (3 e)\right ) \sin (f x)-i \csc (e) \csc (e+f x) \left (\frac {46}{5} \cos (3 e)-\frac {46}{5} i \sin (3 e)\right ) \sin (f x)\right ) \sin ^3(e+f x) \tan ^2(e+f x) (a+i a \tan (e+f x))^3}{f (\cos (f x)+i \sin (f x))^3 (d \tan (e+f x))^{9/2}}+\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)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) \cos ^3(e+f x) \tan ^{\frac {9}{2}}(e+f x) (a+i a \tan (e+f x))^3}{\sqrt {\frac {-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}} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (132 ) = 264\).
time = 0.12, size = 341, normalized size = 2.14

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}-\frac {3 i}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {d}{7 \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {4 i}{d^{2} \sqrt {d \tan \left (f x +e \right )}}+\frac {4}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,d^{2}}\)	\(341\)
default	\(\frac {2 a^{3} \left (\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}-\frac {3 i}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {d}{7 \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {4 i}{d^{2} \sqrt {d \tan \left (f x +e \right )}}+\frac {4}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,d^{2}}\)	\(341\)

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,method=_RETURNVERBOSE)

[Out]

2/f*a^3/d^2*(1/d^2*(1/2/d*(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*arctan(2^(1/2)/(d^2)^(1/4)*(d*t
an(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))+1/2*I/(d^2)^(1/4)*2^(1/2)*(ln((d*ta
n(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*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*ta
n(f*x+e))^(1/2)+1)))-3/5*I/(d*tan(f*x+e))^(5/2)-1/7*d/(d*tan(f*x+e))^(7/2)+4*I/d^2/(d*tan(f*x+e))^(1/2)+4/3/d/
(d*tan(f*x+e))^(3/2))

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Maxima [A]
time = 0.54, size = 245, normalized size = 1.54 \begin {gather*} \frac {\frac {105 \, a^{3} {\left (\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d^{3}} - \frac {2 \, {\left (-420 i \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 140 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} + 63 i \, a^{3} d^{3} \tan \left (f x + e\right ) + 15 \, a^{3} d^{3}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} d^{3}}}{105 \, d f} \end {gather*}

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]

1/105*(105*a^3*((2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(
d) + (2*I + 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (I -
1)*sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + (I - 1)*sqrt(2)*log(d*tan(
f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/d^3 - 2*(-420*I*a^3*d^3*tan(f*x + e)^3 - 140*a^3
*d^3*tan(f*x + e)^2 + 63*I*a^3*d^3*tan(f*x + e) + 15*a^3*d^3)/((d*tan(f*x + e))^(7/2)*d^3))/(d*f)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 549 vs. \(2 (137) = 274\).
time = 0.41, size = 549, normalized size = 3.45 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a^{3} \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\right )\, dx\right ) \end {gather*}

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]

-I*a**3*(Integral(I/(d*tan(e + f*x))**(9/2), x) + Integral(-3*tan(e + f*x)/(d*tan(e + f*x))**(9/2), x) + Integ
ral(tan(e + f*x)**3/(d*tan(e + f*x))**(9/2), x) + Integral(-3*I*tan(e + f*x)**2/(d*tan(e + f*x))**(9/2), x))

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Giac [A]
time = 0.91, size = 156, normalized size = 0.98 \begin {gather*} \frac {8 i \, \sqrt {2} a^{3} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{d^{\frac {9}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (-420 i \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 140 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} + 63 i \, a^{3} d^{3} \tan \left (f x + e\right ) + 15 \, a^{3} d^{3}\right )}}{105 \, \sqrt {d \tan \left (f x + e\right )} d^{7} f \tan \left (f x + e\right )^{3}} \end {gather*}

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(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(
d^(9/2)*f*(I*d/sqrt(d^2) + 1)) - 2/105*(-420*I*a^3*d^3*tan(f*x + e)^3 - 140*a^3*d^3*tan(f*x + e)^2 + 63*I*a^3*
d^3*tan(f*x + e) + 15*a^3*d^3)/(sqrt(d*tan(f*x + e))*d^7*f*tan(f*x + e)^3)

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Mupad [B]
time = 5.01, size = 119, normalized size = 0.75 \begin {gather*} -\frac {\frac {2\,a^3}{7\,d\,f}+\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,6{}\mathrm {i}}{5\,d\,f}-\frac {8\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{3\,d\,f}-\frac {a^3\,{\mathrm {tan}\left (e+f\,x\right )}^3\,8{}\mathrm {i}}{d\,f}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{{\left (-d\right )}^{9/2}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^3/(d*tan(e + f*x))^(9/2),x)

[Out]

(16i^(1/2)*a^3*atan((16i^(1/2)*(d*tan(e + f*x))^(1/2))/(4*(-d)^(1/2)))*2i)/((-d)^(9/2)*f) - ((2*a^3)/(7*d*f) +
(a^3*tan(e + f*x)*6i)/(5*d*f) - (8*a^3*tan(e + f*x)^2)/(3*d*f) - (a^3*tan(e + f*x)^3*8i)/(d*f))/(d*tan(e + f*
x))^(7/2)

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