3.2.62 \(\int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx\) [162]

Optimal. Leaf size=132 \[ \frac {8 (-1)^{3/4} a^3 \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}-\frac {8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}+\frac {8 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, 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 (-1)^{3/4} a^3 \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}+\frac {8 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]

[Out]

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

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))^{7/2}} \, dx &=-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}-\frac {2 \int \frac {(a+i a \tan (e+f x)) \left (-6 i a^2 d+4 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{5/2}} \, dx}{5 d^2}\\ &=-\frac {8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}-\frac {2 \int \frac {10 a^3 d^2+10 i a^3 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{5 d^4}\\ &=-\frac {8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}+\frac {8 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}-\frac {2 \int \frac {10 i a^3 d^3-10 a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{5 d^6}\\ &=-\frac {8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}+\frac {8 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}+\frac {\left (80 a^6\right ) \text {Subst}\left (\int \frac {1}{10 i a^3 d^4+10 a^3 d^3 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {8 (-1)^{3/4} a^3 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}-\frac {8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}+\frac {8 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\\ \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. \(377\) vs. \(2(132)=264\).
time = 7.16, size = 377, normalized size = 2.86 \begin {gather*} \frac {\left (\csc (e) (21 \cos (e)+5 i \sin (e)) \left (\frac {2}{5} \cos (3 e)-\frac {2}{5} i \sin (3 e)\right )+\csc (e) \csc ^2(e+f x) (\cos (e)+5 i \sin (e)) \left (-\frac {2}{5} \cos (3 e)+\frac {2}{5} i \sin (3 e)\right )+\csc (e) \csc ^3(e+f x) \left (\frac {2}{5} \cos (3 e)-\frac {2}{5} i \sin (3 e)\right ) \sin (f x)+\csc (e) \csc (e+f x) \left (-\frac {42}{5} \cos (3 e)+\frac {42}{5} i \sin (3 e)\right ) \sin (f x)\right ) \sin ^3(e+f x) \tan (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))^{7/2}}-\frac {8 i 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 {7}{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))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]

[Out]

((Csc[e]*(21*Cos[e] + (5*I)*Sin[e])*((2*Cos[3*e])/5 - ((2*I)/5)*Sin[3*e]) + Csc[e]*Csc[e + f*x]^2*(Cos[e] + (5
*I)*Sin[e])*((-2*Cos[3*e])/5 + ((2*I)/5)*Sin[3*e]) + Csc[e]*Csc[e + f*x]^3*((2*Cos[3*e])/5 - ((2*I)/5)*Sin[3*e
])*Sin[f*x] + Csc[e]*Csc[e + f*x]*((-42*Cos[3*e])/5 + ((42*I)/5)*Sin[3*e])*Sin[f*x])*Sin[e + f*x]^3*Tan[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])^(7/2)) - ((8*I)*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]^(7/2)*(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])^(7/2))

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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. 324 vs. \(2 (110 ) = 220\).
time = 0.12, size = 325, normalized size = 2.46

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}}{d}-\frac {i}{\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {d}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {4}{d \sqrt {d \tan \left (f x +e \right )}}\right )}{f \,d^{2}}\)	\(325\)
default	\(\frac {2 a^{3} \left (\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}}{d}-\frac {i}{\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {d}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {4}{d \sqrt {d \tan \left (f x +e \right )}}\right )}{f \,d^{2}}\)	\(325\)

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))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.57, size = 228, normalized size = 1.73 \begin {gather*} \frac {\frac {5 \, 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^{2}} + \frac {2 \, {\left (20 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} - 5 i \, a^{3} d^{2} \tan \left (f x + e\right ) - a^{3} d^{2}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}}{5 \, 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))^(7/2),x, algorithm="maxima")

[Out]

1/5*(5*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^2 + 2*(20*a^3*d^2*tan(f*x + e)^2 - 5*I*a^3*d^2*ta
n(f*x + e) - a^3*d^2)/((d*tan(f*x + e))^(5/2)*d^2))/(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. 492 vs. \(2 (115) = 230\).
time = 0.38, size = 492, normalized size = 3.73 \begin {gather*} \frac {5 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{7} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d^{4} 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^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 5 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{7} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{4} 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^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (-13 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 6 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 11 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, 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}}}{20 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} 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))^(7/2),x, algorithm="fricas")

[Out]

1/20*(5*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(6
4*I*a^6/(d^7*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (I*d^4*f*e^(2*I*f*x + 2*I*e) + I*d^4*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^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) -
5*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(64*I*a^
6/(d^7*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (-I*d^4*f*e^(2*I*f*x + 2*I*e) - I*d^4*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^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(-
13*I*a^3*e^(6*I*f*x + 6*I*e) + 6*I*a^3*e^(4*I*f*x + 4*I*e) + 11*I*a^3*e^(2*I*f*x + 2*I*e) - 8*I*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^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I
*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*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 {7}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{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))**(7/2),x)

[Out]

-I*a**3*(Integral(I/(d*tan(e + f*x))**(7/2), x) + Integral(-3*tan(e + f*x)/(d*tan(e + f*x))**(7/2), x) + Integ
ral(tan(e + f*x)**3/(d*tan(e + f*x))**(7/2), x) + Integral(-3*I*tan(e + f*x)**2/(d*tan(e + f*x))**(7/2), x))

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Giac [A]
time = 0.85, size = 139, normalized size = 1.05 \begin {gather*} \frac {8 i \, \sqrt {2} a^{3} \arctan \left (\frac {8 i \, \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 {7}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 \, {\left (20 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} - 5 i \, a^{3} d^{2} \tan \left (f x + e\right ) - a^{3} d^{2}\right )}}{5 \, \sqrt {d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}} \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))^(7/2),x, algorithm="giac")

[Out]

8*I*sqrt(2)*a^3*arctan(8*I*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^(7/2)*f*(-I*d/sqrt(d^2) + 1)) + 2/5*(20*a^3*d^2*tan(f*x + e)^2 - 5*I*a^3*d^2*tan(f*x + e) - a^3*d^2)/(sqr
t(d*tan(f*x + e))*d^5*f*tan(f*x + e)^2)

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Mupad [B]
time = 4.66, size = 95, normalized size = 0.72 \begin {gather*} -\frac {\frac {2\,a^3}{5\,d\,f}-\frac {8\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{d\,f}+\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}}{d\,f}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}-\frac {2\,\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atanh}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {d}}\right )}{d^{7/2}\,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))^(7/2),x)

[Out]

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

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