Integrand size = 45, antiderivative size = 160 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {a^{3/2} (2 i A+B) \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f} \]
-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
Time = 4.37 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.14 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {a^{3/2} c (i+\tan (e+f x)) \left (-2 (2 A-i B) \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (e+f x)}+\sqrt {a} \sqrt {1-i \tan (e+f x)} (1+i \tan (e+f x)) (2 A-2 i B+B \tan (e+f x))\right )}{2 f \sqrt {1-i \tan (e+f x)} \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
(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*T an[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]])
Time = 0.39 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 4071, 90, 60, 45, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} (A+B \tan (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} (A+B \tan (e+f x))dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {\sqrt {i \tan (e+f x) a+a} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {a c \left (\frac {1}{2} (2 A-i B) \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 a c}\right )}{f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a c \left (\frac {1}{2} (2 A-i B) \left (a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 a c}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (\frac {1}{2} (2 A-i B) \left (2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 a c}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (\frac {1}{2} (2 A-i B) \left (\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}-\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c}}\right )+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 a c}\right )}{f}\) |
(a*c*((B*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(2*a*c) + ((2*A - I*B)*(((-2*I)*Sqrt[a]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]] )/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/Sqrt[c] + (I*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/c))/2))/f
3.8.98.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[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]
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 | \(\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )\right )}{f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}-\frac {B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-2 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\) | \(277\) |
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,m ethod=_RETURNVERBOSE)
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)*tan(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)
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 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {\sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-2 i \, A - B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + B\right )} a}\right ) - \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-2 i \, A - B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + B\right )} a}\right ) + 4 \, {\left ({\left (-2 i \, A - 3 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
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")
-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)
\[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )} \left (A + B \tan {\left (e + f x \right )}\right )\, dx \]
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)
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 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\text {Too large to display} \]
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")
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(sin(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*a rctan2(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), 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) + ( (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*ar ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*...
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} \]
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")
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)*s qrt(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) + 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)*sqr t(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))*arctan(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
Timed out. \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]