\(\int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx\) [1096]

Optimal result

Integrand size = 28, antiderivative size = 125 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx=\frac {2 a^2 x}{(c-i d)^3}-\frac {2 a^2 \log (c \cos (e+f x)+d \sin (e+f x))}{(i c+d)^3 f}+\frac {a^2 (i c-d)}{2 d (i c+d) f (c+d \tan (e+f x))^2}+\frac {2 i a^2}{(c-i d)^2 f (c+d \tan (e+f x))} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(262\) vs. \(2(125)=250\).

Time = 2.16 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.10 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx=-\frac {d (a+i a \tan (e+f x))^2}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}+\frac {(c+i d) \left (-\frac {d (a+i a \tan (e+f x))^2}{\left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {-\frac {i a \left (i a^2 (c+2 i d)+a^2 d\right ) (-2 a \log (i+\tan (e+f x))-i a \tan (e+f x))}{(i a c+a d) f}-\frac {a^2 (a c d-a (c+2 i d) d) \left (-\frac {a (i c-d) \log (c+d \tan (e+f x))}{d^2}+\frac {i a \tan (e+f x)}{d}\right )}{(i a c+a d) f}}{a \left (c^2+d^2\right )}\right )}{c^2+d^2} \] Input:

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

Output:

-1/2*(d*(a + I*a*Tan[e + f*x])^2)/((c^2 + d^2)*f*(c + d*Tan[e + f*x])^2) + 
 ((c + I*d)*(-((d*(a + I*a*Tan[e + f*x])^2)/((c^2 + d^2)*f*(c + d*Tan[e + 
f*x]))) + (((-I)*a*(I*a^2*(c + (2*I)*d) + a^2*d)*(-2*a*Log[I + Tan[e + f*x 
]] - I*a*Tan[e + f*x]))/((I*a*c + a*d)*f) - (a^2*(a*c*d - a*(c + (2*I)*d)* 
d)*(-((a*(I*c - d)*Log[c + d*Tan[e + f*x]])/d^2) + (I*a*Tan[e + f*x])/d))/ 
((I*a*c + a*d)*f))/(a*(c^2 + d^2))))/(c^2 + d^2)
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.36, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4025, 27, 3042, 4012, 3042, 4014, 3042, 4013}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4025

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4013

\(\displaystyle \frac {2 \left (\frac {\frac {a^2 x (c+i d)^3}{c^2+d^2}+\frac {a^2 (c+i d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{f (d+i c)}}{c^2+d^2}+\frac {a^2 (-d+i c)}{f (c-i d) (c+d \tan (e+f x))}\right )}{c^2+d^2}+\frac {a^2 (-d+i c)}{2 d f (d+i c) (c+d \tan (e+f x))^2}\)

Input:

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

Output:

(a^2*(I*c - d))/(2*d*(I*c + d)*f*(c + d*Tan[e + f*x])^2) + (2*(((a^2*(c + 
I*d)^3*x)/(c^2 + d^2) + (a^2*(c + I*d)^2*Log[c*Cos[e + f*x] + d*Sin[e + f* 
x]])/((I*c + d)*f))/(c^2 + d^2) + (a^2*(I*c - d))/((c - I*d)*f*(c + d*Tan[ 
e + f*x]))))/(c^2 + d^2)
 

Defintions of rubi rules used

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 3042

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

rule 4012

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

rule 4013

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* 
(x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si 
n[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] && EqQ[a*c + b*d, 0]
 

rule 4014

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a 
*d)/(a^2 + b^2)   Int[(b - a*Tan[e + f*x])/(a + b*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] && N 
eQ[a*c + b*d, 0]
 

rule 4025

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.73

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\frac {\left (2 i c^{3}-6 i c \,d^{2}-6 c^{2} d +2 d^{3}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}+\left (6 i c^{2} d -2 i d^{3}+2 c^{3}-6 c \,d^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{\left (c^{2}+d^{2}\right )^{3}}-\frac {-2 i c d -c^{2}+d^{2}}{2 \left (c^{2}+d^{2}\right ) d \left (c +d \tan \left (f x +e \right )\right )^{2}}+\frac {2 i c^{2}-2 i d^{2}-4 c d}{\left (c^{2}+d^{2}\right )^{2} \left (c +d \tan \left (f x +e \right )\right )}-\frac {2 \left (i c^{3}-3 i c \,d^{2}-3 c^{2} d +d^{3}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{\left (c^{2}+d^{2}\right )^{3}}\right )}{f}\)	\(216\)
default	\(\frac {a^{2} \left (\frac {\frac {\left (2 i c^{3}-6 i c \,d^{2}-6 c^{2} d +2 d^{3}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}+\left (6 i c^{2} d -2 i d^{3}+2 c^{3}-6 c \,d^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{\left (c^{2}+d^{2}\right )^{3}}-\frac {-2 i c d -c^{2}+d^{2}}{2 \left (c^{2}+d^{2}\right ) d \left (c +d \tan \left (f x +e \right )\right )^{2}}+\frac {2 i c^{2}-2 i d^{2}-4 c d}{\left (c^{2}+d^{2}\right )^{2} \left (c +d \tan \left (f x +e \right )\right )}-\frac {2 \left (i c^{3}-3 i c \,d^{2}-3 c^{2} d +d^{3}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{\left (c^{2}+d^{2}\right )^{3}}\right )}{f}\)	\(216\)
risch	\(-\frac {4 a^{2} x}{3 i c^{2} d -i d^{3}-c^{3}+3 c \,d^{2}}-\frac {4 i a^{2} x}{i c^{3}-3 i c \,d^{2}+3 c^{2} d -d^{3}}-\frac {4 i a^{2} e}{f \left (i c^{3}-3 i c \,d^{2}+3 c^{2} d -d^{3}\right )}-\frac {2 i \left (3 a^{2} d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+2 i a^{2} c d \,{\mathrm e}^{2 i \left (f x +e \right )}-2 a^{2} d^{2}+a^{2} c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+3 i a^{2} c d +a^{2} c^{2}\right )}{\left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+{\mathrm e}^{2 i \left (f x +e \right )} c +i d +c \right )^{2} f \left (-i d +c \right )^{3}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right )}{f \left (i c^{3}-3 i c \,d^{2}+3 c^{2} d -d^{3}\right )}\)	\(290\)
norman	\(\frac {\frac {2 i a^{2} c d +a^{2} c^{2}+a^{2} d^{2}}{2 f \left (-2 i c d +c^{2}-d^{2}\right ) d}+\frac {2 a^{2} c^{2} x}{\left (-i d +c \right ) \left (-2 i c d +c^{2}-d^{2}\right )}+\frac {i d^{2} a^{2} \tan \left (f x +e \right )^{2}}{\left (2 i c d -c^{2}+d^{2}\right ) f c}+\frac {4 c d \,a^{2} x \tan \left (f x +e \right )}{\left (-i d +c \right ) \left (-2 i c d +c^{2}-d^{2}\right )}-\frac {2 i a^{2} d^{2} x \tan \left (f x +e \right )^{2}}{\left (i c +d \right ) \left (2 i c d -c^{2}+d^{2}\right )}}{\left (c +d \tan \left (f x +e \right )\right )^{2}}+\frac {i a^{2} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f \left (-3 i c^{2} d +i d^{3}+c^{3}-3 c \,d^{2}\right )}-\frac {2 i a^{2} \ln \left (c +d \tan \left (f x +e \right )\right )}{f \left (-3 i c^{2} d +i d^{3}+c^{3}-3 c \,d^{2}\right )}\)	\(305\)
parallelrisch	\(\text {Expression too large to display}\)	\(1017\)

Input:

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

Output:

1/f*a^2*(1/(c^2+d^2)^3*(1/2*(2*I*c^3-6*I*c*d^2-6*c^2*d+2*d^3)*ln(1+tan(f*x 
+e)^2)+(6*I*c^2*d-2*I*d^3+2*c^3-6*c*d^2)*arctan(tan(f*x+e)))-1/2*(-2*I*c*d 
-c^2+d^2)/(c^2+d^2)/d/(c+d*tan(f*x+e))^2+2*(I*c^2-I*d^2-2*c*d)/(c^2+d^2)^2 
/(c+d*tan(f*x+e))-2*(I*c^3-3*I*c*d^2-3*c^2*d+d^3)/(c^2+d^2)^3*ln(c+d*tan(f 
*x+e)))
 

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. 310 vs. \(2 (111) = 222\).

Time = 0.12 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.48 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx=\frac {2 \, {\left (a^{2} c^{2} + 3 i \, a^{2} c d - 2 \, a^{2} d^{2} + {\left (a^{2} c^{2} + 2 i \, a^{2} c d + 3 \, a^{2} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2} + {\left (a^{2} c^{2} - 2 i \, a^{2} c d - a^{2} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (a^{2} c^{2} + a^{2} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right )\right )}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left (-i \, c^{5} - 3 \, c^{4} d + 2 i \, c^{3} d^{2} - 2 \, c^{2} d^{3} + 3 i \, c d^{4} + d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{5} + c^{4} d + 2 i \, c^{3} d^{2} + 2 \, c^{2} d^{3} + i \, c d^{4} + d^{5}\right )} f} \] Input:

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

Output:

2*(a^2*c^2 + 3*I*a^2*c*d - 2*a^2*d^2 + (a^2*c^2 + 2*I*a^2*c*d + 3*a^2*d^2) 
*e^(2*I*f*x + 2*I*e) + (a^2*c^2 + 2*I*a^2*c*d - a^2*d^2 + (a^2*c^2 - 2*I*a 
^2*c*d - a^2*d^2)*e^(4*I*f*x + 4*I*e) + 2*(a^2*c^2 + a^2*d^2)*e^(2*I*f*x + 
 2*I*e))*log(((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c - d)/(I*c + d)))/((I*c^5 
 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f*e^(4*I*f*x + 4 
*I*e) - 2*(-I*c^5 - 3*c^4*d + 2*I*c^3*d^2 - 2*c^2*d^3 + 3*I*c*d^4 + d^5)*f 
*e^(2*I*f*x + 2*I*e) + (I*c^5 + c^4*d + 2*I*c^3*d^2 + 2*c^2*d^3 + I*c*d^4 
+ d^5)*f)
 

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (100) = 200\).

Time = 3.79 (sec) , antiderivative size = 391, normalized size of antiderivative = 3.13 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx=- \frac {2 i a^{2} \log {\left (\frac {c + i d}{c e^{2 i e} - i d e^{2 i e}} + e^{2 i f x} \right )}}{f \left (c - i d\right )^{3}} + \frac {- 2 i a^{2} c^{2} + 6 a^{2} c d + 4 i a^{2} d^{2} + \left (- 2 i a^{2} c^{2} e^{2 i e} + 4 a^{2} c d e^{2 i e} - 6 i a^{2} d^{2} e^{2 i e}\right ) e^{2 i f x}}{c^{5} f - i c^{4} d f + 2 c^{3} d^{2} f - 2 i c^{2} d^{3} f + c d^{4} f - i d^{5} f + \left (2 c^{5} f e^{2 i e} - 6 i c^{4} d f e^{2 i e} - 4 c^{3} d^{2} f e^{2 i e} - 4 i c^{2} d^{3} f e^{2 i e} - 6 c d^{4} f e^{2 i e} + 2 i d^{5} f e^{2 i e}\right ) e^{2 i f x} + \left (c^{5} f e^{4 i e} - 5 i c^{4} d f e^{4 i e} - 10 c^{3} d^{2} f e^{4 i e} + 10 i c^{2} d^{3} f e^{4 i e} + 5 c d^{4} f e^{4 i e} - i d^{5} f e^{4 i e}\right ) e^{4 i f x}} \] Input:

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

Output:

-2*I*a**2*log((c + I*d)/(c*exp(2*I*e) - I*d*exp(2*I*e)) + exp(2*I*f*x))/(f 
*(c - I*d)**3) + (-2*I*a**2*c**2 + 6*a**2*c*d + 4*I*a**2*d**2 + (-2*I*a**2 
*c**2*exp(2*I*e) + 4*a**2*c*d*exp(2*I*e) - 6*I*a**2*d**2*exp(2*I*e))*exp(2 
*I*f*x))/(c**5*f - I*c**4*d*f + 2*c**3*d**2*f - 2*I*c**2*d**3*f + c*d**4*f 
 - I*d**5*f + (2*c**5*f*exp(2*I*e) - 6*I*c**4*d*f*exp(2*I*e) - 4*c**3*d**2 
*f*exp(2*I*e) - 4*I*c**2*d**3*f*exp(2*I*e) - 6*c*d**4*f*exp(2*I*e) + 2*I*d 
**5*f*exp(2*I*e))*exp(2*I*f*x) + (c**5*f*exp(4*I*e) - 5*I*c**4*d*f*exp(4*I 
*e) - 10*c**3*d**2*f*exp(4*I*e) + 10*I*c**2*d**3*f*exp(4*I*e) + 5*c*d**4*f 
*exp(4*I*e) - I*d**5*f*exp(4*I*e))*exp(4*I*f*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. 380 vs. \(2 (111) = 222\).

Time = 0.14 (sec) , antiderivative size = 380, normalized size of antiderivative = 3.04 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx=\frac {\frac {4 \, {\left (a^{2} c^{3} + 3 i \, a^{2} c^{2} d - 3 \, a^{2} c d^{2} - i \, a^{2} d^{3}\right )} {\left (f x + e\right )}}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} - \frac {4 \, {\left (i \, a^{2} c^{3} - 3 \, a^{2} c^{2} d - 3 i \, a^{2} c d^{2} + a^{2} d^{3}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} - \frac {2 \, {\left (-i \, a^{2} c^{3} + 3 \, a^{2} c^{2} d + 3 i \, a^{2} c d^{2} - a^{2} d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac {a^{2} c^{4} + 6 i \, a^{2} c^{3} d - 8 \, a^{2} c^{2} d^{2} - 2 i \, a^{2} c d^{3} - a^{2} d^{4} + 4 \, {\left (i \, a^{2} c^{2} d^{2} - 2 \, a^{2} c d^{3} - i \, a^{2} d^{4}\right )} \tan \left (f x + e\right )}{c^{6} d + 2 \, c^{4} d^{3} + c^{2} d^{5} + {\left (c^{4} d^{3} + 2 \, c^{2} d^{5} + d^{7}\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left (c^{5} d^{2} + 2 \, c^{3} d^{4} + c d^{6}\right )} \tan \left (f x + e\right )}}{2 \, f} \] Input:

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

Output:

1/2*(4*(a^2*c^3 + 3*I*a^2*c^2*d - 3*a^2*c*d^2 - I*a^2*d^3)*(f*x + e)/(c^6 
+ 3*c^4*d^2 + 3*c^2*d^4 + d^6) - 4*(I*a^2*c^3 - 3*a^2*c^2*d - 3*I*a^2*c*d^ 
2 + a^2*d^3)*log(d*tan(f*x + e) + c)/(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6) - 
 2*(-I*a^2*c^3 + 3*a^2*c^2*d + 3*I*a^2*c*d^2 - a^2*d^3)*log(tan(f*x + e)^2 
 + 1)/(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6) + (a^2*c^4 + 6*I*a^2*c^3*d - 8*a 
^2*c^2*d^2 - 2*I*a^2*c*d^3 - a^2*d^4 + 4*(I*a^2*c^2*d^2 - 2*a^2*c*d^3 - I* 
a^2*d^4)*tan(f*x + e))/(c^6*d + 2*c^4*d^3 + c^2*d^5 + (c^4*d^3 + 2*c^2*d^5 
 + d^7)*tan(f*x + e)^2 + 2*(c^5*d^2 + 2*c^3*d^4 + c*d^6)*tan(f*x + e)))/f
 

Giac [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.43 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx=\frac {2 \, a^{2} d \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{i \, c^{3} d f + 3 \, c^{2} d^{2} f - 3 i \, c d^{3} f - d^{4} f} + \frac {2 \, a^{2} \log \left (\tan \left (f x + e\right ) + i\right )}{-i \, c^{3} f - 3 \, c^{2} d f + 3 i \, c d^{2} f + d^{3} f} + \frac {a^{2} c^{3} + 3 i \, a^{2} c^{2} d + 5 \, a^{2} c d^{2} - i \, a^{2} d^{3} + 4 i \, {\left (a^{2} c d^{2} - i \, a^{2} d^{3}\right )} \tan \left (f x + e\right )}{2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{2} {\left (c - i \, d\right )}^{3} d f} \] Input:

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

Output:

2*a^2*d*log(abs(d*tan(f*x + e) + c))/(I*c^3*d*f + 3*c^2*d^2*f - 3*I*c*d^3* 
f - d^4*f) + 2*a^2*log(tan(f*x + e) + I)/(-I*c^3*f - 3*c^2*d*f + 3*I*c*d^2 
*f + d^3*f) + 1/2*(a^2*c^3 + 3*I*a^2*c^2*d + 5*a^2*c*d^2 - I*a^2*d^3 + 4*I 
*(a^2*c*d^2 - I*a^2*d^3)*tan(f*x + e))/((d*tan(f*x + e) + c)^2*(c - I*d)^3 
*d*f)
 

Mupad [B] (verification not implemented)

Time = 2.54 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.38 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx=-\frac {\frac {a^2\,c^2+a^2\,c\,d\,4{}\mathrm {i}+a^2\,d^2}{2\,d^3\,\left (-c^2+c\,d\,2{}\mathrm {i}+d^2\right )}+\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}}{d\,\left (-c^2+c\,d\,2{}\mathrm {i}+d^2\right )}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2+\frac {c^2}{d^2}+\frac {2\,c\,\mathrm {tan}\left (e+f\,x\right )}{d}\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {c^3-c^2\,d\,1{}\mathrm {i}+c\,d^2-d^3\,1{}\mathrm {i}}{{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (d+c\,1{}\mathrm {i}\right )}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (2\,c^8\,d^2+8\,c^6\,d^4+12\,c^4\,d^6+8\,c^2\,d^8+2\,d^{10}\right )\,1{}\mathrm {i}}{{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (d+c\,1{}\mathrm {i}\right )\,\left (-c^6\,d\,1{}\mathrm {i}+2\,c^5\,d^2-c^4\,d^3\,1{}\mathrm {i}+4\,c^3\,d^4+c^2\,d^5\,1{}\mathrm {i}+2\,c\,d^6+d^7\,1{}\mathrm {i}\right )}\right )\,4{}\mathrm {i}}{f\,{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (d+c\,1{}\mathrm {i}\right )} \] Input:

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

Output:

(a^2*atan((c*d^2 - c^2*d*1i + c^3 - d^3*1i)/((c - d*1i)^2*(c*1i + d)) - (t 
an(e + f*x)*(2*d^10 + 8*c^2*d^8 + 12*c^4*d^6 + 8*c^6*d^4 + 2*c^8*d^2)*1i)/ 
((c - d*1i)^2*(c*1i + d)*(2*c*d^6 - c^6*d*1i + d^7*1i + c^2*d^5*1i + 4*c^3 
*d^4 - c^4*d^3*1i + 2*c^5*d^2)))*4i)/(f*(c - d*1i)^2*(c*1i + d)) - ((a^2*c 
^2 + a^2*d^2 + a^2*c*d*4i)/(2*d^3*(c*d*2i - c^2 + d^2)) + (a^2*tan(e + f*x 
)*2i)/(d*(c*d*2i - c^2 + d^2)))/(f*(tan(e + f*x)^2 + c^2/d^2 + (2*c*tan(e 
+ f*x))/d))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1007, normalized size of antiderivative = 8.06 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(a**2*(2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2*c**4*d**3*i - 6*log(tan( 
e + f*x)**2 + 1)*tan(e + f*x)**2*c**3*d**4 - 6*log(tan(e + f*x)**2 + 1)*ta 
n(e + f*x)**2*c**2*d**5*i + 2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2*c*d 
**6 + 4*log(tan(e + f*x)**2 + 1)*tan(e + f*x)*c**5*d**2*i - 12*log(tan(e + 
 f*x)**2 + 1)*tan(e + f*x)*c**4*d**3 - 12*log(tan(e + f*x)**2 + 1)*tan(e + 
 f*x)*c**3*d**4*i + 4*log(tan(e + f*x)**2 + 1)*tan(e + f*x)*c**2*d**5 + 2* 
log(tan(e + f*x)**2 + 1)*c**6*d*i - 6*log(tan(e + f*x)**2 + 1)*c**5*d**2 - 
 6*log(tan(e + f*x)**2 + 1)*c**4*d**3*i + 2*log(tan(e + f*x)**2 + 1)*c**3* 
d**4 - 4*log(tan(e + f*x)*d + c)*tan(e + f*x)**2*c**4*d**3*i + 12*log(tan( 
e + f*x)*d + c)*tan(e + f*x)**2*c**3*d**4 + 12*log(tan(e + f*x)*d + c)*tan 
(e + f*x)**2*c**2*d**5*i - 4*log(tan(e + f*x)*d + c)*tan(e + f*x)**2*c*d** 
6 - 8*log(tan(e + f*x)*d + c)*tan(e + f*x)*c**5*d**2*i + 24*log(tan(e + f* 
x)*d + c)*tan(e + f*x)*c**4*d**3 + 24*log(tan(e + f*x)*d + c)*tan(e + f*x) 
*c**3*d**4*i - 8*log(tan(e + f*x)*d + c)*tan(e + f*x)*c**2*d**5 - 4*log(ta 
n(e + f*x)*d + c)*c**6*d*i + 12*log(tan(e + f*x)*d + c)*c**5*d**2 + 12*log 
(tan(e + f*x)*d + c)*c**4*d**3*i - 4*log(tan(e + f*x)*d + c)*c**3*d**4 + 4 
*tan(e + f*x)**2*c**4*d**3*f*x - 2*tan(e + f*x)**2*c**4*d**3*i + 12*tan(e 
+ f*x)**2*c**3*d**4*f*i*x + 4*tan(e + f*x)**2*c**3*d**4 - 12*tan(e + f*x)* 
*2*c**2*d**5*f*x - 4*tan(e + f*x)**2*c*d**6*f*i*x + 4*tan(e + f*x)**2*c*d* 
*6 + 2*tan(e + f*x)**2*d**7*i + 8*tan(e + f*x)*c**5*d**2*f*x + 24*tan(e...