\(\int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx\) [1109]

Optimal result

Integrand size = 28, antiderivative size = 98 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=-\frac {2 i a (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {2 a (i c+d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\frac {2 a \left (-3 i (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+(4 i c+3 d+i d \tan (e+f x)) \sqrt {c+d \tan (e+f x)}\right )}{3 f} \] Input:

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

Output:

(2*a*((-3*I)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d 
]] + ((4*I)*c + 3*d + I*d*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]))/(3*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.58 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4011, 3042, 4011, 3042, 4020, 25, 27, 73, 221}

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.

\(\displaystyle \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}dx\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \sqrt {c+d \tan (e+f x)} (a (c-i d)+a (i c+d) \tan (e+f x))dx+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {c+d \tan (e+f x)} (a (c-i d)+a (i c+d) \tan (e+f x))dx+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \frac {a (c-i d)^2+i a \tan (e+f x) (c-i d)^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a (c-i d)^2+i a \tan (e+f x) (c-i d)^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {i a^2 (c-i d)^4 \int -\frac {1}{a (c-i d)^2 \left (a (c-i d)^2-i a (c-i d)^2 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}d\left (i a (c-i d)^2 \tan (e+f x)\right )}{f}+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i a^2 (c-i d)^4 \int \frac {1}{a (c-i d)^2 \left (a (c-i d)^2-i a (c-i d)^2 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}d\left (i a (c-i d)^2 \tan (e+f x)\right )}{f}+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {i a (c-i d)^2 \int \frac {1}{\left (a (c-i d)^2-i a (c-i d)^2 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}d\left (i a (c-i d)^2 \tan (e+f x)\right )}{f}+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 a^2 (c-i d)^4 \int \frac {1}{\frac {i a^3 (c-i d)^6 \tan ^2(e+f x)}{d}-\frac {a (i c+d)^3}{d}}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a (c-i d)^{3/2} \arctan \left (a (c-i d)^{3/2} \tan (e+f x)\right )}{f}+\frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4011

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

rule 4020

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 851 vs. \(2 (81 ) = 162\).

Time = 0.36 (sec) , antiderivative size = 852, normalized size of antiderivative = 8.69

method	result	size
derivativedivides	\(\frac {a \left (\frac {2 i \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i \sqrt {c +d \tan \left (f x +e \right )}\, c +2 \sqrt {c +d \tan \left (f x +e \right )}\, d +\frac {\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}-i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}-\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}-i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {-\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}-i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}-i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )}{f}\)	\(852\)
default	\(\frac {a \left (\frac {2 i \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i \sqrt {c +d \tan \left (f x +e \right )}\, c +2 \sqrt {c +d \tan \left (f x +e \right )}\, d +\frac {\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}-i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}-\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}-i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {-\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}-i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}-i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )}{f}\)	\(852\)
parts	\(\text {Expression too large to display}\)	\(1659\)

Input:

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

Output:

1/f*a*(2/3*I*(c+d*tan(f*x+e))^(3/2)+2*I*(c+d*tan(f*x+e))^(1/2)*c+2*(c+d*ta 
n(f*x+e))^(1/2)*d+1/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*(1/2*(-I 
*c^2*(c^2+d^2)^(1/2)+I*d^2*(c^2+d^2)^(1/2)-I*c^3-I*c*d^2-2*c*d*(c^2+d^2)^( 
1/2)-c^2*d-d^3)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2 
)+2*c)^(1/2)+(c^2+d^2)^(1/2))+2*(-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-I*(2 
*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d^2-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2*d-(2*( 
c^2+d^2)^(1/2)+2*c)^(1/2)*d^3-1/2*(-I*c^2*(c^2+d^2)^(1/2)+I*d^2*(c^2+d^2)^ 
(1/2)-I*c^3-I*c*d^2-2*c*d*(c^2+d^2)^(1/2)-c^2*d-d^3)*(2*(c^2+d^2)^(1/2)+2* 
c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+( 
2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))+1/(2*(c^2+d^ 
2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*(-1/2*(-I*c^2*(c^2+d^2)^(1/2)+I*d^2*(c 
^2+d^2)^(1/2)-I*c^3-I*c*d^2-2*c*d*(c^2+d^2)^(1/2)-c^2*d-d^3)*ln((c+d*tan(f 
*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2)) 
+2*(I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3+I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c* 
d^2+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^3+ 
1/2*(-I*c^2*(c^2+d^2)^(1/2)+I*d^2*(c^2+d^2)^(1/2)-I*c^3-I*c*d^2-2*c*d*(c^2 
+d^2)^(1/2)-c^2*d-d^3)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2 
*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/ 
(2*(c^2+d^2)^(1/2)-2*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. 504 vs. \(2 (76) = 152\).

Time = 0.11 (sec) , antiderivative size = 504, normalized size of antiderivative = 5.14 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\frac {3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {a^{2} c^{3} - 3 i \, a^{2} c^{2} d - 3 \, a^{2} c d^{2} + i \, a^{2} d^{3}}{f^{2}}} \log \left (-\frac {2 \, {\left (-i \, a c^{2} - a c d + {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{2} c^{3} - 3 i \, a^{2} c^{2} d - 3 \, a^{2} c d^{2} + i \, a^{2} d^{3}}{f^{2}}} + {\left (-i \, a c^{2} - 2 \, a c d + i \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{i \, a c + a d}\right ) - 3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {a^{2} c^{3} - 3 i \, a^{2} c^{2} d - 3 \, a^{2} c d^{2} + i \, a^{2} d^{3}}{f^{2}}} \log \left (-\frac {2 \, {\left (-i \, a c^{2} - a c d - {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{2} c^{3} - 3 i \, a^{2} c^{2} d - 3 \, a^{2} c d^{2} + i \, a^{2} d^{3}}{f^{2}}} + {\left (-i \, a c^{2} - 2 \, a c d + i \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{i \, a c + a d}\right ) - 8 \, {\left (-2 i \, a c - a d + 2 \, {\left (-i \, a c - a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{6 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:

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

Output:

1/6*(3*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(a^2*c^3 - 3*I*a^2*c^2*d - 3*a^2* 
c*d^2 + I*a^2*d^3)/f^2)*log(-2*(-I*a*c^2 - a*c*d + (f*e^(2*I*f*x + 2*I*e) 
+ f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 
 1))*sqrt(-(a^2*c^3 - 3*I*a^2*c^2*d - 3*a^2*c*d^2 + I*a^2*d^3)/f^2) + (-I* 
a*c^2 - 2*a*c*d + I*a*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*a* 
c + a*d)) - 3*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(a^2*c^3 - 3*I*a^2*c^2*d - 
 3*a^2*c*d^2 + I*a^2*d^3)/f^2)*log(-2*(-I*a*c^2 - a*c*d - (f*e^(2*I*f*x + 
2*I*e) + f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2 
*I*e) + 1))*sqrt(-(a^2*c^3 - 3*I*a^2*c^2*d - 3*a^2*c*d^2 + I*a^2*d^3)/f^2) 
 + (-I*a*c^2 - 2*a*c*d + I*a*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e 
)/(I*a*c + a*d)) - 8*(-2*I*a*c - a*d + 2*(-I*a*c - a*d)*e^(2*I*f*x + 2*I*e 
))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1 
)))/(f*e^(2*I*f*x + 2*I*e) + f)
 

Sympy [F]

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

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

Output:

I*a*(Integral(-I*c*sqrt(c + d*tan(e + f*x)), x) + Integral(c*sqrt(c + d*ta 
n(e + f*x))*tan(e + f*x), x) + Integral(d*sqrt(c + d*tan(e + f*x))*tan(e + 
 f*x)**2, x) + Integral(-I*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x))
 

Maxima [F(-1)]

Timed out. \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

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. 222 vs. \(2 (76) = 152\).

Time = 0.52 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.27 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=-\frac {2 \, {\left (-i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} - 3 i \, \sqrt {d \tan \left (f x + e\right ) + c} c - 3 \, \sqrt {d \tan \left (f x + e\right ) + c} d + \frac {3 \, \sqrt {2} {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )} a}{3 \, f} \] Input:

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

Output:

-2/3*(-I*(d*tan(f*x + e) + c)^(3/2) - 3*I*sqrt(d*tan(f*x + e) + c)*c - 3*s 
qrt(d*tan(f*x + e) + c)*d + 3*sqrt(2)*(-I*c^2 - 2*c*d + I*d^2)*arctan(2*(s 
qrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(sqr 
t(2)*c*sqrt(-c + sqrt(c^2 + d^2)) - I*sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d 
 - sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2))))/(sqrt(-c + sqrt(c^ 
2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))*a/f
 

Mupad [B] (verification not implemented)

Time = 13.43 (sec) , antiderivative size = 2869, normalized size of antiderivative = 29.28 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

log(((((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^ 
2)/f^4)^(1/2)*((16*c*d^2*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3* 
f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*c^2*1i + a*d^2*1i - f*(((-a^4*d^2*f^4 
*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d 
*tan(e + f*x))^(1/2)))/f - (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d 
^4 - 6*c^2*d^2))/f^2))/2 - (a^3*d^2*(c^2 - d^2)*(c^2*1i + d^2*1i)^2*8i)/f^ 
3)*((6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2)/(4*f^4) - 
(a^2*c^3)/(4*f^2) + (3*a^2*c*d^2)/(4*f^2))^(1/2) - log(((((-a^4*d^2*f^4*(3 
*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*c*d^ 
2*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/ 
f^4)^(1/2)*(a*c^2*1i + a*d^2*1i + f*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) 
 - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/ 
f + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/ 
2 - (a^3*d^2*(c^2 - d^2)*(c^2*1i + d^2*1i)^2*8i)/f^3)*(((6*a^4*c^2*d^4*f^4 
 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2) 
/(4*f^4))^(1/2) - log(((-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f 
^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*c*d^2*(-((-a^4*d^2*f^4*(3*c^2 - d^2) 
^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*c^2*1i + a*d^2*1i 
 + f*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f 
^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*a^2*d^2*(c + d*tan(...
 

Reduce [F]

\[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=a \left (\left (\int \sqrt {d \tan \left (f x +e \right )+c}d x \right ) c +\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}d x \right ) d i +\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )d x \right ) c i +\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )d x \right ) d \right ) \] Input:

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

Output:

a*(int(sqrt(tan(e + f*x)*d + c),x)*c + int(sqrt(tan(e + f*x)*d + c)*tan(e 
+ f*x)**2,x)*d*i + int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x),x)*c*i + int( 
sqrt(tan(e + f*x)*d + c)*tan(e + f*x),x)*d)