\(\int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx\) [797]

Optimal result

Integrand size = 45, antiderivative size = 157 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {i a^{3/2} A c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a A c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \]

[Out]

-I*a^(3/2)*A*c^(3/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/f+1/2*a*A*c*(a+
I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)*tan(f*x+e)/f+1/3*B*(a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^
(3/2)/f

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3669, 81, 38, 65, 223, 209} \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {i a^{3/2} A c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a A c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \]

[In]

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

[Out]

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

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x*(a + b*x)^m*((c + d*x)^m/(2*m + 1))
, x] + Dist[2*a*c*(m/(2*m + 1)), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 65

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 81

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

Rule 209

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

Rule 223

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 3669

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \sqrt {a+i a x} (A+B x) \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac {(a A c) \text {Subst}\left (\int \sqrt {a+i a x} \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {a A c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac {\left (a^2 A c^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f} \\ & = \frac {a A c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {\left (i a A c^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{f} \\ & = \frac {a A c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {\left (i a A c^2\right ) \text {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 )}{f} \\ & = -\frac {i a^{3/2} A c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a A c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.46 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.15 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {a^{3/2} c^2 (i+\tan (e+f x)) \left (-6 A \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (e+f x)}+\sqrt {a} \sqrt {1-i \tan (e+f x)} (-i+\tan (e+f x)) \left (2 B+3 A \tan (e+f x)+2 B \tan ^2(e+f x)\right )\right )}{6 f \sqrt {1-i \tan (e+f x)} \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(a^(3/2)*c^2*(I + Tan[e + f*x])*(-6*A*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Sqrt[a + I*a*Tan[e
+ f*x]] + Sqrt[a]*Sqrt[1 - I*Tan[e + f*x]]*(-I + Tan[e + f*x])*(2*B + 3*A*Tan[e + f*x] + 2*B*Tan[e + f*x]^2)))
/(6*f*Sqrt[1 - I*Tan[e + f*x]]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.14

method	result	size
parts	\(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \left (\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}-\frac {B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \left (1+\tan \left (f x +e \right )^{2}\right )}{3 f}\)	\(179\)
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \left (2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+3 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +3 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{6 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(186\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \left (2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+3 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +3 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{6 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(186\)

[In]

int((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*A/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a*c*(tan(f*x+e)*(a*c*(1+tan(f*x+e)^2))^(1/2)*(
a*c)^(1/2)+a*c*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2)))/(a*c*(1+tan(f*x+e)^2
))^(1/2)/(a*c)^(1/2)-1/3*B/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a*c*(1+tan(f*x+e)^2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (119) = 238\).

Time = 0.28 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.76 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {3 \, \sqrt {\frac {A^{2} a^{3} c^{3}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (A a c e^{\left (3 i \, f x + 3 i \, e\right )} + A a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {A^{2} a^{3} c^{3}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{A a c e^{\left (2 i \, f x + 2 i \, e\right )} + A a c}\right ) - 3 \, \sqrt {\frac {A^{2} a^{3} c^{3}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (A a c e^{\left (3 i \, f x + 3 i \, e\right )} + A a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {A^{2} a^{3} c^{3}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{A a c e^{\left (2 i \, f x + 2 i \, e\right )} + A a c}\right ) - 4 \, {\left (3 i \, A a c e^{\left (5 i \, f x + 5 i \, e\right )} - 8 \, B a c e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, A a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

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

[Out]

1/12*(3*sqrt(A^2*a^3*c^3/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(A*a*c*e^(3*I*f*x
+ 3*I*e) + A*a*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(
A^2*a^3*c^3/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I*f))/(A*a*c*e^(2*I*f*x + 2*I*e) + A*a*c)) - 3*sqrt(A^2*a^3*c^3/f^
2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(A*a*c*e^(3*I*f*x + 3*I*e) + A*a*c*e^(I*f*x
+ I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(A^2*a^3*c^3/f^2)*(-I*f*e^(2
*I*f*x + 2*I*e) + I*f))/(A*a*c*e^(2*I*f*x + 2*I*e) + A*a*c)) - 4*(3*I*A*a*c*e^(5*I*f*x + 5*I*e) - 8*B*a*c*e^(3
*I*f*x + 3*I*e) - 3*I*A*a*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1
)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

Sympy [F]

\[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 853 vs. \(2 (119) = 238\).

Time = 0.45 (sec) , antiderivative size = 853, normalized size of antiderivative = 5.43 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-(12*A*a*c*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 32*I*B*a*c*cos(3/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))) - 12*A*a*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*A*a*c*sin(5/2*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 32*B*a*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*I
*A*a*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 6*(A*a*c*cos(6*f*x + 6*e) + 3*A*a*c*cos(4*f*x +
4*e) + 3*A*a*c*cos(2*f*x + 2*e) + I*A*a*c*sin(6*f*x + 6*e) + 3*I*A*a*c*sin(4*f*x + 4*e) + 3*I*A*a*c*sin(2*f*x
+ 2*e) + A*a*c)*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) + 6*(A*a*c*cos(6*f*x + 6*e) + 3*A*a*c*cos(4*f*x + 4*e) + 3*A*a*c*cos(2*f*x + 2*e) +
I*A*a*c*sin(6*f*x + 6*e) + 3*I*A*a*c*sin(4*f*x + 4*e) + 3*I*A*a*c*sin(2*f*x + 2*e) + A*a*c)*arctan2(cos(1/2*ar
ctan2(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) + 3*(I*
A*a*c*cos(6*f*x + 6*e) + 3*I*A*a*c*cos(4*f*x + 4*e) + 3*I*A*a*c*cos(2*f*x + 2*e) - A*a*c*sin(6*f*x + 6*e) - 3*
A*a*c*sin(4*f*x + 4*e) - 3*A*a*c*sin(2*f*x + 2*e) + I*A*a*c)*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) + 3*(-I*A*a*c*cos(6*f*x + 6*e) - 3*I*A*a*c*cos(4*f*x + 4*e) - 3*I*A*a*c*cos(2*f*x + 2*e) +
A*a*c*sin(6*f*x + 6*e) + 3*A*a*c*sin(4*f*x + 4*e) + 3*A*a*c*sin(2*f*x + 2*e) - I*A*a*c)*log(cos(1/2*arctan2(si
n(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*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)*sqrt(c)/(f*(-12*I*cos(6*f*x + 6*e) - 36*I*cos(4*f*x + 4
*e) - 36*I*cos(2*f*x + 2*e) + 12*sin(6*f*x + 6*e) + 36*sin(4*f*x + 4*e) + 36*sin(2*f*x + 2*e) - 12*I))

Giac [F]

\[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int { {\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(3/2),x)

[Out]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(3/2), x)