3.2.55 \(\int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{7/2}} \, dx\) [155]

Optimal. Leaf size=118 \[ \frac {4 (-1)^{3/4} a^2 \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}-\frac {2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac {4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {4 a^2}{d^3 f \sqrt {d \tan (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3623, 3610, 3614, 211} \begin {gather*} \frac {4 (-1)^{3/4} a^2 \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}+\frac {4 a^2}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac {2 a^2}{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])^2/(d*Tan[e + f*x])^(7/2),x]

[Out]

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

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 3623

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(b*c - a*d)^2*((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*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, 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))^2}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac {2 a^2}{5 d f (d \tan (e+f x))^{5/2}}+\frac {\int \frac {2 i a^2 d-2 a^2 d \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{d^2}\\ &=-\frac {2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac {4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {\int \frac {-2 a^2 d^2-2 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{d^4}\\ &=-\frac {2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac {4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {4 a^2}{d^3 f \sqrt {d \tan (e+f x)}}+\frac {\int \frac {-2 i a^2 d^3+2 a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{d^6}\\ &=-\frac {2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac {4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {4 a^2}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {\left (8 a^4\right ) \text {Subst}\left (\int \frac {1}{-2 i a^2 d^4-2 a^2 d^3 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {4 (-1)^{3/4} a^2 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}-\frac {2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac {4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {4 a^2}{d^3 f \sqrt {d \tan (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(381\) vs. \(2(118)=236\).
time = 7.02, size = 381, normalized size = 3.23 \begin {gather*} \frac {\left (\csc (e) (33 \cos (e)+10 i \sin (e)) \left (\frac {2}{15} \cos (2 e)-\frac {2}{15} i \sin (2 e)\right )+\csc (e) \csc ^2(e+f x) (3 \cos (e)+10 i \sin (e)) \left (-\frac {2}{15} \cos (2 e)+\frac {2}{15} i \sin (2 e)\right )+\csc (e) \csc ^3(e+f x) \left (\frac {2}{5} \cos (2 e)-\frac {2}{5} i \sin (2 e)\right ) \sin (f x)+\csc (e) \csc (e+f x) \left (-\frac {22}{5} \cos (2 e)+\frac {22}{5} i \sin (2 e)\right ) \sin (f x)\right ) \sin ^2(e+f x) \tan ^2(e+f x) (a+i a \tan (e+f x))^2}{f (\cos (f x)+i \sin (f x))^2 (d \tan (e+f x))^{7/2}}-\frac {4 i e^{-2 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 ^2(e+f x) \tan ^{\frac {7}{2}}(e+f x) (a+i a \tan (e+f x))^2}{\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} f (\cos (f x)+i \sin (f x))^2 (d \tan (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Csc[e]*(33*Cos[e] + (10*I)*Sin[e])*((2*Cos[2*e])/15 - ((2*I)/15)*Sin[2*e]) + Csc[e]*Csc[e + f*x]^2*(3*Cos[e]
+ (10*I)*Sin[e])*((-2*Cos[2*e])/15 + ((2*I)/15)*Sin[2*e]) + Csc[e]*Csc[e + f*x]^3*((2*Cos[2*e])/5 - ((2*I)/5)
*Sin[2*e])*Sin[f*x] + Csc[e]*Csc[e + f*x]*((-22*Cos[2*e])/5 + ((22*I)/5)*Sin[2*e])*Sin[f*x])*Sin[e + f*x]^2*Ta
n[e + f*x]^2*(a + I*a*Tan[e + f*x])^2)/(f*(Cos[f*x] + I*Sin[f*x])^2*(d*Tan[e + f*x])^(7/2)) - ((4*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]^2*Tan[e + f*x]^(7/2)*(a + I*a*Tan[e + f*x])^2)/(E^((2*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])^2*(d*Tan[e + f*x])^(7/2))

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (97 ) = 194\).
time = 0.10, size = 327, normalized size = 2.77

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}-\frac {1}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {2 i}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}\right )}{f d}\)	\(327\)
default	\(\frac {2 a^{2} \left (\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}-\frac {1}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {2 i}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}\right )}{f d}\)	\(327\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^2/(d*tan(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (100) = 200\).
time = 0.52, size = 228, normalized size = 1.93 \begin {gather*} \frac {\frac {15 \, a^{2} {\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 {4 \, {\left (30 \, a^{2} d^{2} \tan \left (f x + e\right )^{2} - 10 i \, a^{2} d^{2} \tan \left (f x + e\right ) - 3 \, a^{2} d^{2}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}}{30 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2/(d*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

1/30*(15*a^2*(-(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 + 4*(30*a^2*d^2*tan(f*x + e)^2 - 10*I*a^2*d^2
*tan(f*x + e) - 3*a^2*d^2)/((d*tan(f*x + e))^(5/2)*d^2))/(d*f)

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 492 vs. \(2 (100) = 200\).
time = 0.39, size = 492, normalized size = 4.17 \begin {gather*} \frac {15 \, {\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 {16 i \, a^{4}}{d^{7} f^{2}}} \log \left (\frac {{\left (-4 i \, a^{2} 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 {16 i \, a^{4}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 15 \, {\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 {16 i \, a^{4}}{d^{7} f^{2}}} \log \left (\frac {{\left (-4 i \, a^{2} 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 {16 i \, a^{4}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 8 \, {\left (-43 i \, a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 11 i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 31 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 23 i \, a^{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}}}{60 \, {\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))^2/(d*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/60*(15*(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(
16*I*a^4/(d^7*f^2))*log(1/2*(-4*I*a^2*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(16*I*a^4/(d^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) -
15*(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(16*I*
a^4/(d^7*f^2))*log(1/2*(-4*I*a^2*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(16*I*a^4/(d^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) - 8*(
-43*I*a^2*e^(6*I*f*x + 6*I*e) + 11*I*a^2*e^(4*I*f*x + 4*I*e) + 31*I*a^2*e^(2*I*f*x + 2*I*e) - 23*I*a^2)*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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \left (- \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\right )\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {2 i \tan {\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))**2/(d*tan(f*x+e))**(7/2),x)

[Out]

-a**2*(Integral(-1/(d*tan(e + f*x))**(7/2), x) + Integral(tan(e + f*x)**2/(d*tan(e + f*x))**(7/2), x) + Integr
al(-2*I*tan(e + f*x)/(d*tan(e + f*x))**(7/2), x))

________________________________________________________________________________________

Giac [A]
time = 0.79, size = 139, normalized size = 1.18 \begin {gather*} -\frac {4 i \, \sqrt {2} a^{2} \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 (30 \, a^{2} d^{2} \tan \left (f x + e\right )^{2} - 10 i \, a^{2} d^{2} \tan \left (f x + e\right ) - 3 \, a^{2} d^{2}\right )}}{15 \, \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))^2/(d*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-4*I*sqrt(2)*a^2*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)) + 2/15*(30*a^2*d^2*tan(f*x + e)^2 - 10*I*a^2*d^2*tan(f*x + e) - 3*a^2*d^2
)/(sqrt(d*tan(f*x + e))*d^5*f*tan(f*x + e)^2)

________________________________________________________________________________________

Mupad [B]
time = 4.66, size = 95, normalized size = 0.81 \begin {gather*} -\frac {\frac {2\,a^2}{5\,d\,f}-\frac {4\,a^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{d\,f}+\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}}{3\,d\,f}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}-\frac {2\,\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atanh}\left (\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\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)^2/(d*tan(e + f*x))^(7/2),x)

[Out]

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

________________________________________________________________________________________