3.1025 \(\int \frac{(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=204 \[ -\frac{10 i a^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac{5 i a^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^2 f}+\frac{10 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

((-10*I)*a^(7/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(c^(3/2)*f
) - (((2*I)/3)*a*(a + I*a*Tan[e + f*x])^(5/2))/(f*(c - I*c*Tan[e + f*x])^(3/2)) + (((10*I)/3)*a^2*(a + I*a*Tan
[e + f*x])^(3/2))/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) + ((5*I)*a^3*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e
+ f*x]])/(c^2*f)

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Rubi [A]  time = 0.190154, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ -\frac{10 i a^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac{5 i a^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^2 f}+\frac{10 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-10*I)*a^(7/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(c^(3/2)*f
) - (((2*I)/3)*a*(a + I*a*Tan[e + f*x])^(5/2))/(f*(c - I*c*Tan[e + f*x])^(3/2)) + (((10*I)/3)*a^2*(a + I*a*Tan
[e + f*x])^(3/2))/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) + ((5*I)*a^3*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e
+ f*x]])/(c^2*f)

Rule 3523

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 11.7989, size = 348, normalized size = 1.71 \[ \frac{\cos ^3(e+f x) (a+i a \tan (e+f x))^{7/2} \left (\cos (4 f x) \left (\frac{2 \sin (e)}{3 c^2}-\frac{2 i \cos (e)}{3 c^2}\right )+\cos (2 f x) \left (\frac{10 \sin (e)}{3 c^2}+\frac{10 i \cos (e)}{3 c^2}\right )+\sin (2 f x) \left (-\frac{10 \cos (e)}{3 c^2}+\frac{10 i \sin (e)}{3 c^2}\right )+\sin (4 f x) \left (\frac{2 \cos (e)}{3 c^2}+\frac{2 i \sin (e)}{3 c^2}\right )+\frac{5 \sin (3 e)}{c^2}+\frac{5 i \cos (3 e)}{c^2}\right ) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^3}-\frac{10 i \sqrt{e^{i f x}} e^{-i (4 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2}}{c f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{7}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-10*I)*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x))]*(a + I*a*Tan[
e + f*x])^(7/2))/(c*E^(I*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e + f*x]^(7/2)*(Cos[f*x] + I*Sin
[f*x])^(7/2)) + (Cos[e + f*x]^3*(((5*I)*Cos[3*e])/c^2 + Cos[4*f*x]*((((-2*I)/3)*Cos[e])/c^2 + (2*Sin[e])/(3*c^
2)) + Cos[2*f*x]*((((10*I)/3)*Cos[e])/c^2 + (10*Sin[e])/(3*c^2)) + (5*Sin[3*e])/c^2 + ((-10*Cos[e])/(3*c^2) +
(((10*I)/3)*Sin[e])/c^2)*Sin[2*f*x] + ((2*Cos[e])/(3*c^2) + (((2*I)/3)*Sin[e])/c^2)*Sin[4*f*x])*Sqrt[Sec[e + f
*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^(7/2))/(f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [B]  time = 0.037, size = 379, normalized size = 1.9 \begin{align*}{\frac{{a}^{3}}{3\,f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 45\,i\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{2}ac+3\,i\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( \tan \left ( fx+e \right ) \right ) ^{3}+15\,\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{3}ac-15\,i\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) ac-57\,i\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) -45\,\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) \tan \left ( fx+e \right ) ac-37\,\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( \tan \left ( fx+e \right ) \right ) ^{2}+23\,\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

1/3/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^2*(45*I*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x
+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)^2*a*c+3*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+
e)^3+15*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)^3*a*c-15*I*ln((a*
c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c-57*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2)
)^(1/2)*tan(f*x+e)-45*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)*a*c
-37*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2+23*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1
+tan(f*x+e)^2))^(1/2)/(tan(f*x+e)+I)^3/(a*c)^(1/2)

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Maxima [B]  time = 1.97373, size = 900, normalized size = 4.41 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-((90*a^3*cos(2*f*x + 2*e) + 90*I*a^3*sin(2*f*x + 2*e) + 90*a^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (90*a^3*cos(2*f*x + 2*e) + 90*I*a
^3*sin(2*f*x + 2*e) + 90*a^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (24*a^3*cos(2*f*x + 2*e) + 24*I*a^3*sin(2*f*x + 2*e) + 24*a^3)*cos(
3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (144*a^3*cos(2*f*x + 2*e) + 144*I*a^3*sin(2*f*x + 2*e) + 18
0*a^3)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-45*I*a^3*cos(2*f*x + 2*e) + 45*a^3*sin(2*f*x +
2*e) - 45*I*a^3)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (45*I*a^3*cos(2*f*x +
2*e) - 45*a^3*sin(2*f*x + 2*e) + 45*I*a^3)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/
2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1)
- (-24*I*a^3*cos(2*f*x + 2*e) + 24*a^3*sin(2*f*x + 2*e) - 24*I*a^3)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))) - (144*I*a^3*cos(2*f*x + 2*e) - 144*a^3*sin(2*f*x + 2*e) + 180*I*a^3)*sin(1/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-18*I*c^2*cos(2*f*x + 2*e) + 18*c^2*sin(2*f*x + 2*e) - 18*I*c^2)*f
)

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Fricas [B]  time = 1.59054, size = 926, normalized size = 4.54 \begin{align*} -\frac{15 \, \sqrt{\frac{a^{7}}{c^{3} f^{2}}} c^{2} f \log \left (\frac{2 \,{\left (4 \,{\left (a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} +{\left (2 i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, c^{2} f\right )} \sqrt{\frac{a^{7}}{c^{3} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, \sqrt{\frac{a^{7}}{c^{3} f^{2}}} c^{2} f \log \left (\frac{2 \,{\left (4 \,{\left (a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} +{\left (-2 i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c^{2} f\right )} \sqrt{\frac{a^{7}}{c^{3} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) -{\left (-8 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 40 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 60 i \, a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}}{6 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/6*(15*sqrt(a^7/(c^3*f^2))*c^2*f*log(2*(4*(a^3*e^(2*I*f*x + 2*I*e) + a^3)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*
sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + (2*I*c^2*f*e^(2*I*f*x + 2*I*e) - 2*I*c^2*f)*sqrt(a^7/(c^3*
f^2)))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - 15*sqrt(a^7/(c^3*f^2))*c^2*f*log(2*(4*(a^3*e^(2*I*f*x + 2*I*e) + a^3
)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + (-2*I*c^2*f*e^(2*I*f*x
+ 2*I*e) + 2*I*c^2*f)*sqrt(a^7/(c^3*f^2)))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - (-8*I*a^3*e^(4*I*f*x + 4*I*e) +
40*I*a^3*e^(2*I*f*x + 2*I*e) + 60*I*a^3)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*
e^(I*f*x + I*e))/(c^2*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))**(7/2)/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(3/2), x)