3.810 \(\int \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=226 \[ -\frac{2 a^{5/2} (4 B+i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac{a^2 (4 B+i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^2 f}+\frac{2 a (4 B+i A) (a+i a \tan (e+f x))^{3/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

(-2*a^(5/2)*(I*A + 4*B)*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) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a*(I*A + 4*B)*(a +
I*a*Tan[e + f*x])^(3/2))/(3*c*f*Sqrt[c - I*c*Tan[e + f*x]]) + (a^2*(I*A + 4*B)*Sqrt[a + I*a*Tan[e + f*x]]*Sqr
t[c - I*c*Tan[e + f*x]])/(c^2*f)

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Rubi [A]  time = 0.31489, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3588, 78, 47, 50, 63, 217, 203} \[ -\frac{2 a^{5/2} (4 B+i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac{a^2 (4 B+i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^2 f}+\frac{2 a (4 B+i A) (a+i a \tan (e+f x))^{3/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(-2*a^(5/2)*(I*A + 4*B)*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) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a*(I*A + 4*B)*(a +
I*a*Tan[e + f*x])^(3/2))/(3*c*f*Sqrt[c - I*c*Tan[e + f*x]]) + (a^2*(I*A + 4*B)*Sqrt[a + I*a*Tan[e + f*x]]*Sqr
t[c - I*c*Tan[e + f*x]])/(c^2*f)

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((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)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

Mathematica [A]  time = 13.928, size = 227, normalized size = 1. \[ \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \left (\sqrt{\sec (e+f x)} \sqrt{c-i c \tan (e+f x)} ((4 A-13 i B) \sin (2 (e+f x))+(11 B+2 i A) \cos (2 (e+f x))+2 i A+8 B)-\frac{6 i c (A-4 i B) e^{-3 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt{\frac{c}{1+e^{2 i (e+f x)}}}}\right )}{3 c^2 f \sec ^{\frac{7}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x])*(((-6*I)*(A - (4*I)*B)*c*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)
*(e + f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^((3*I)*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) + Sqrt[Sec[e +
f*x]]*((2*I)*A + 8*B + ((2*I)*A + 11*B)*Cos[2*(e + f*x)] + (4*A - (13*I)*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan
[e + f*x]]))/(3*c^2*f*Sec[e + f*x]^(7/2)*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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Maple [B]  time = 0.178, size = 667, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(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^2/c^2*(-12*I*B*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+9*I*A*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*A*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+36*I*B*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+29*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2+36*B*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*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*
tan(f*x+e)^3-3*I*A*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-12*I*A*(a*c*(
1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-9*A*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-8*A*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-19*I*B*(a*c*(1+tan(f*x+
e)^2))^(1/2)*(a*c)^(1/2)-12*B*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-45
*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)+4*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+t
an(f*x+e)^2))^(1/2)/(tan(f*x+e)+I)^3/(a*c)^(1/2)

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Maxima [B]  time = 2.15514, size = 1114, normalized size = 4.93 \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))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-(((18*A - 72*I*B)*a^2*cos(2*f*x + 2*e) - 18*(-I*A - 4*B)*a^2*sin(2*f*x + 2*e) + (18*A - 72*I*B)*a^2)*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
) + ((18*A - 72*I*B)*a^2*cos(2*f*x + 2*e) - 18*(-I*A - 4*B)*a^2*sin(2*f*x + 2*e) + (18*A - 72*I*B)*a^2)*arctan
2(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) + ((12*A - 12*I*B)*a^2*cos(2*f*x + 2*e) - 12*(-I*A - B)*a^2*sin(2*f*x + 2*e) + (12*A - 12*I*B)*a^2)*cos(3
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - ((36*A - 108*I*B)*a^2*cos(2*f*x + 2*e) + 36*(I*A + 3*B)*a^2*
sin(2*f*x + 2*e) + (36*A - 144*I*B)*a^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (9*(-I*A - 4*B
)*a^2*cos(2*f*x + 2*e) + (9*A - 36*I*B)*a^2*sin(2*f*x + 2*e) + 9*(-I*A - 4*B)*a^2)*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(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (9*(I*A + 4*B)*a^2*cos(2*f*x + 2*e) - (9*A - 36*I*B)*a^2*sin(2*f*x
+ 2*e) + 9*(I*A + 4*B)*a^2)*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) - (12*(-I*A -
B)*a^2*cos(2*f*x + 2*e) + (12*A - 12*I*B)*a^2*sin(2*f*x + 2*e) + 12*(-I*A - B)*a^2)*sin(3/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) - (36*(I*A + 3*B)*a^2*cos(2*f*x + 2*e) - (36*A - 108*I*B)*a^2*sin(2*f*x + 2*e) + 36
*(I*A + 4*B)*a^2)*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 = 1.59042, size = 1237, normalized size = 5.47 \begin{align*} \frac{3 \, c^{2} \sqrt{\frac{{\left (4 \, A^{2} - 32 i \, A B - 64 \, B^{2}\right )} a^{5}}{c^{3} f^{2}}} f \log \left (\frac{2 \,{\left ({\left ({\left (4 i \, A + 16 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (4 i \, A + 16 \, B\right )} a^{2}\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 (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{2} f\right )} \sqrt{\frac{{\left (4 \, A^{2} - 32 i \, A B - 64 \, B^{2}\right )} a^{5}}{c^{3} f^{2}}}\right )}}{{\left (i \, A + 4 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A + 4 \, B\right )} a^{2}}\right ) - 3 \, c^{2} \sqrt{\frac{{\left (4 \, A^{2} - 32 i \, A B - 64 \, B^{2}\right )} a^{5}}{c^{3} f^{2}}} f \log \left (\frac{2 \,{\left ({\left ({\left (4 i \, A + 16 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (4 i \, A + 16 \, B\right )} a^{2}\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 (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{2} f\right )} \sqrt{\frac{{\left (4 \, A^{2} - 32 i \, A B - 64 \, B^{2}\right )} a^{5}}{c^{3} f^{2}}}\right )}}{{\left (i \, A + 4 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A + 4 \, B\right )} a^{2}}\right ) + 2 \,{\left ({\left (-4 i \, A - 4 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (8 i \, A + 32 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (12 i \, A + 48 \, B\right )} a^{2}\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 )}}{12 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*c^2*sqrt((4*A^2 - 32*I*A*B - 64*B^2)*a^5/(c^3*f^2))*f*log(2*(((4*I*A + 16*B)*a^2*e^(2*I*f*x + 2*I*e) +
(4*I*A + 16*B)*a^2)*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*e^(2*I*f*x + 2*I*e) - c^2*f)*sqrt((4*A^2 - 32*I*A*B - 64*B^2)*a^5/(c^3*f^2)))/((I*A + 4*B)*a^2*e^(2*I*f*x
+ 2*I*e) + (I*A + 4*B)*a^2)) - 3*c^2*sqrt((4*A^2 - 32*I*A*B - 64*B^2)*a^5/(c^3*f^2))*f*log(2*(((4*I*A + 16*B)*
a^2*e^(2*I*f*x + 2*I*e) + (4*I*A + 16*B)*a^2)*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*e^(2*I*f*x + 2*I*e) - c^2*f)*sqrt((4*A^2 - 32*I*A*B - 64*B^2)*a^5/(c^3*f^2)))/((I
*A + 4*B)*a^2*e^(2*I*f*x + 2*I*e) + (I*A + 4*B)*a^2)) + 2*((-4*I*A - 4*B)*a^2*e^(4*I*f*x + 4*I*e) + (8*I*A + 3
2*B)*a^2*e^(2*I*f*x + 2*I*e) + (12*I*A + 48*B)*a^2)*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))**(5/2)*(A+B*tan(f*x+e))/(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 (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{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))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(5/2)/(-I*c*tan(f*x + e) + c)^(3/2), x)