\(\int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx\) [798]

Optimal result

Integrand size = 45, antiderivative size = 160 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {a^{3/2} (2 i A+B) \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 160, 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, 52, 65, 223, 209} \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {a^{3/2} \sqrt {c} (B+2 i A) \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a (B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f} \]

[In]

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

[Out]

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

Rule 52

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

Mathematica [A] (verified)

Time = 4.37 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.14 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {a^{3/2} c (i+\tan (e+f x)) \left (-2 (2 A-i B) \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (e+f x)}+\sqrt {a} \sqrt {1-i \tan (e+f x)} (1+i \tan (e+f x)) (2 A-2 i B+B \tan (e+f x))\right )}{2 f \sqrt {1-i \tan (e+f x)} \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(a^(3/2)*c*(I + Tan[e + f*x])*(-2*(2*A - I*B)*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]]*(1 + I*Tan[e + f*x])*(2*A - (2*I)*B + B*Tan[e + f*x])))/(2*
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 = 0.38 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.39

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +2 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+2 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(223\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +2 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+2 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(223\)
parts	\(\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )\right )}{f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}-\frac {B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-2 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(277\)

[In]

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

[Out]

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

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. 459 vs. \(2 (120) = 240\).

Time = 0.28 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.87 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {\sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-2 i \, A - B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + B\right )} a}\right ) - \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-2 i \, A - B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + B\right )} a}\right ) + 4 \, {\left ({\left (-2 i \, A - 3 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((I*a*(tan(e + f*x) - I))**(3/2)*sqrt(-I*c*(tan(e + f*x) + I))*(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. 771 vs. \(2 (120) = 240\).

Time = 0.47 (sec) , antiderivative size = 771, normalized size of antiderivative = 4.82 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\text {Too large to display} \]

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

4*(4*(2*A - 3*I*B)*a*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(2*A - I*B)*a*cos(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(2*I*A + 3*B)*a*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
4*(2*I*A + B)*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*((2*A - I*B)*a*cos(4*f*x + 4*e) + 2*(
2*A - I*B)*a*cos(2*f*x + 2*e) - (-2*I*A - B)*a*sin(4*f*x + 4*e) - 2*(-2*I*A - B)*a*sin(2*f*x + 2*e) + (2*A - I
*B)*a)*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) - 2*((2*A - I*B)*a*cos(4*f*x + 4*e) + 2*(2*A - I*B)*a*cos(2*f*x + 2*e) - (-2*I*A - B)*a*sin(4
*f*x + 4*e) - 2*(-2*I*A - B)*a*sin(2*f*x + 2*e) + (2*A - I*B)*a)*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) + ((-2*I*A - B)*a*cos(4*f*x + 4*e)
+ 2*(-2*I*A - B)*a*cos(2*f*x + 2*e) + (2*A - I*B)*a*sin(4*f*x + 4*e) + 2*(2*A - I*B)*a*sin(2*f*x + 2*e) + (-2
*I*A - B)*a)*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), co
s(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((2*I*A + B)*a*cos(4*f*x +
4*e) + 2*(2*I*A + B)*a*cos(2*f*x + 2*e) - (2*A - I*B)*a*sin(4*f*x + 4*e) - 2*(2*A - I*B)*a*sin(2*f*x + 2*e) +
(2*I*A + B)*a)*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))*sqrt(a)*sqrt(c)/(f*(-16*I*
cos(4*f*x + 4*e) - 32*I*cos(2*f*x + 2*e) + 16*sin(4*f*x + 4*e) + 32*sin(2*f*x + 2*e) - 16*I))

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 533 vs. \(2 (120) = 240\).

Time = 1.53 (sec) , antiderivative size = 533, normalized size of antiderivative = 3.33 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {-3 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 18 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) + 3 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 18 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 10 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (7 i \, f x + 7 i \, e\right )} + 26 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (5 i \, f x + 5 i \, e\right )} + 22 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} + 6 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )} - 3 i \, B a^{\frac {3}{2}} \sqrt {c} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) + 3 i \, B a^{\frac {3}{2}} \sqrt {c} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right )}{8 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} - \frac {i \, {\left ({\left (8 \, A a^{\frac {3}{2}} \sqrt {c} - i \, B a^{\frac {3}{2}} \sqrt {c}\right )} \arctan \left (e^{\left (i \, f x + i \, e\right )}\right ) - \frac {8 \, A a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} - 7 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} + 8 \, A a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )} - i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )}}{{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}^{2}}\right )}}{4 \, f} \]

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)),x, algorithm="giac")

[Out]

1/8*(-3*I*B*a^(3/2)*sqrt(c)*e^(8*I*f*x + 8*I*e)*log(e^(I*f*x + I*e) + I) - 12*I*B*a^(3/2)*sqrt(c)*e^(6*I*f*x +
6*I*e)*log(e^(I*f*x + I*e) + I) - 18*I*B*a^(3/2)*sqrt(c)*e^(4*I*f*x + 4*I*e)*log(e^(I*f*x + I*e) + I) - 12*I*
B*a^(3/2)*sqrt(c)*e^(2*I*f*x + 2*I*e)*log(e^(I*f*x + I*e) + I) + 3*I*B*a^(3/2)*sqrt(c)*e^(8*I*f*x + 8*I*e)*log
(e^(I*f*x + I*e) - I) + 12*I*B*a^(3/2)*sqrt(c)*e^(6*I*f*x + 6*I*e)*log(e^(I*f*x + I*e) - I) + 18*I*B*a^(3/2)*s
qrt(c)*e^(4*I*f*x + 4*I*e)*log(e^(I*f*x + I*e) - I) + 12*I*B*a^(3/2)*sqrt(c)*e^(2*I*f*x + 2*I*e)*log(e^(I*f*x
+ I*e) - I) + 10*B*a^(3/2)*sqrt(c)*e^(7*I*f*x + 7*I*e) + 26*B*a^(3/2)*sqrt(c)*e^(5*I*f*x + 5*I*e) + 22*B*a^(3/
2)*sqrt(c)*e^(3*I*f*x + 3*I*e) + 6*B*a^(3/2)*sqrt(c)*e^(I*f*x + I*e) - 3*I*B*a^(3/2)*sqrt(c)*log(e^(I*f*x + I*
e) + I) + 3*I*B*a^(3/2)*sqrt(c)*log(e^(I*f*x + I*e) - I))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6
*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f) - 1/4*I*((8*A*a^(3/2)*sqrt(c) - I*B*a^(3/2)*sqrt(c))*arc
tan(e^(I*f*x + I*e)) - (8*A*a^(3/2)*sqrt(c)*e^(3*I*f*x + 3*I*e) - 7*I*B*a^(3/2)*sqrt(c)*e^(3*I*f*x + 3*I*e) +
8*A*a^(3/2)*sqrt(c)*e^(I*f*x + I*e) - I*B*a^(3/2)*sqrt(c)*e^(I*f*x + I*e))/(e^(2*I*f*x + 2*I*e) + 1)^2)/f

Mupad [F(-1)]

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

[In]

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