\(\int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx\) [945]

Optimal result

Integrand size = 31, antiderivative size = 169 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {5 x}{16 a^2 c^3}-\frac {i}{24 a^2 f (c-i c \tan (e+f x))^3}-\frac {3 i c^3}{32 a^2 f \left (c^3-i c^3 \tan (e+f x)\right )^2}-\frac {3 i}{16 a^2 f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i c^3}{32 a^2 f \left (c^3+i c^3 \tan (e+f x)\right )^2}+\frac {i}{8 a^2 f \left (c^3+i c^3 \tan (e+f x)\right )} \] Output:

5/16*x/a^2/c^3-1/24*I/a^2/f/(c-I*c*tan(f*x+e))^3-3/32*I*c^3/a^2/f/(c^3-I*c 
^3*tan(f*x+e))^2-3/16*I/a^2/f/(c^3-I*c^3*tan(f*x+e))+1/32*I*c^3/a^2/f/(c^3 
+I*c^3*tan(f*x+e))^2+1/8*I/a^2/f/(c^3+I*c^3*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=-\frac {\sec ^5(e+f x) (-80 \cos (e+f x)+15 \cos (3 (e+f x))+\cos (5 (e+f x))-120 i \arctan (\tan (e+f x)) (\cos (e+f x)-i \sin (e+f x))-40 i \sin (e+f x)-45 i \sin (3 (e+f x))-5 i \sin (5 (e+f x)))}{384 a^2 c^3 f (-i+\tan (e+f x))^2 (i+\tan (e+f x))^3} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^3),x]
 

Output:

-1/384*(Sec[e + f*x]^5*(-80*Cos[e + f*x] + 15*Cos[3*(e + f*x)] + Cos[5*(e 
+ f*x)] - (120*I)*ArcTan[Tan[e + f*x]]*(Cos[e + f*x] - I*Sin[e + f*x]) - ( 
40*I)*Sin[e + f*x] - (45*I)*Sin[3*(e + f*x)] - (5*I)*Sin[5*(e + f*x)]))/(a 
^2*c^3*f*(-I + Tan[e + f*x])^2*(I + Tan[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.83, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4005, 3042, 3968, 54, 2009}

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.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4005

\(\displaystyle \frac {\int \frac {\cos ^4(e+f x)}{c-i c \tan (e+f x)}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\sec (e+f x)^4 (c-i c \tan (e+f x))}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 54

\(\displaystyle \frac {i c^3 \int \left (\frac {3}{16 c^5 (c-i c \tan (e+f x))^2}+\frac {1}{8 c^5 (i \tan (e+f x) c+c)^2}+\frac {3}{16 c^4 (c-i c \tan (e+f x))^3}+\frac {1}{16 c^4 (i \tan (e+f x) c+c)^3}+\frac {1}{8 c^3 (c-i c \tan (e+f x))^4}+\frac {5}{16 c^5 \left (\tan ^2(e+f x) c^2+c^2\right )}\right )d(-i c \tan (e+f x))}{a^2 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i c^3 \left (-\frac {5 i \arctan (\tan (e+f x))}{16 c^6}-\frac {3}{16 c^5 (c-i c \tan (e+f x))}+\frac {1}{8 c^5 (c+i c \tan (e+f x))}-\frac {3}{32 c^4 (c-i c \tan (e+f x))^2}+\frac {1}{32 c^4 (c+i c \tan (e+f x))^2}-\frac {1}{24 c^3 (c-i c \tan (e+f x))^3}\right )}{a^2 f}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^3),x]
 

Output:

(I*c^3*((((-5*I)/16)*ArcTan[Tan[e + f*x]])/c^6 - 1/(24*c^3*(c - I*c*Tan[e 
+ f*x])^3) - 3/(32*c^4*(c - I*c*Tan[e + f*x])^2) - 3/(16*c^5*(c - I*c*Tan[ 
e + f*x])) + 1/(32*c^4*(c + I*c*Tan[e + f*x])^2) + 1/(8*c^5*(c + I*c*Tan[e 
 + f*x]))))/(a^2*f)
 

Defintions of rubi rules used

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[1/(a^(m - 2)*b*f)   Subst[Int[(a - x)^(m/2 - 1)*(a + x 
)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && 
 EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
 

rule 4005

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Sec[e + f*x]^(2*m)*(c + 
 d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[ 
m, 0] || GtQ[m, n]))
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {5 x}{16 a^{2} c^{3}}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )}}{192 a^{2} c^{3} f}-\frac {i \cos \left (4 f x +4 e \right )}{32 a^{2} c^{3} f}+\frac {3 \sin \left (4 f x +4 e \right )}{64 a^{2} c^{3} f}-\frac {5 i \cos \left (2 f x +2 e \right )}{64 a^{2} c^{3} f}+\frac {15 \sin \left (2 f x +2 e \right )}{64 a^{2} c^{3} f}\)	\(114\)
derivativedivides	\(\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}-\frac {i}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {1}{8 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {3 i}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {1}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3}{16 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )}\)	\(132\)
default	\(\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}-\frac {i}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {1}{8 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {3 i}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {1}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3}{16 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )}\)	\(132\)
norman	\(\frac {\frac {5 x}{16 a c}-\frac {i}{6 a c f}+\frac {11 \tan \left (f x +e \right )}{16 a c f}+\frac {5 \tan \left (f x +e \right )^{3}}{6 a c f}+\frac {5 \tan \left (f x +e \right )^{5}}{16 a c f}+\frac {15 x \tan \left (f x +e \right )^{2}}{16 a c}+\frac {15 x \tan \left (f x +e \right )^{4}}{16 a c}+\frac {5 x \tan \left (f x +e \right )^{6}}{16 a c}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} a \,c^{2}}\)	\(148\)

Input:

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

Output:

5/16*x/a^2/c^3-1/192*I/a^2/c^3/f*exp(6*I*(f*x+e))-1/32*I/a^2/c^3/f*cos(4*f 
*x+4*e)+3/64/a^2/c^3/f*sin(4*f*x+4*e)-5/64*I/a^2/c^3/f*cos(2*f*x+2*e)+15/6 
4/a^2/c^3/f*sin(2*f*x+2*e)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.47 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {{\left (120 \, f x e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 15 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 60 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 30 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} c^{3} f} \] Input:

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

Output:

1/384*(120*f*x*e^(4*I*f*x + 4*I*e) - 2*I*e^(10*I*f*x + 10*I*e) - 15*I*e^(8 
*I*f*x + 8*I*e) - 60*I*e^(6*I*f*x + 6*I*e) + 30*I*e^(2*I*f*x + 2*I*e) + 3* 
I)*e^(-4*I*f*x - 4*I*e)/(a^2*c^3*f)
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.53 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (- 33554432 i a^{8} c^{12} f^{4} e^{12 i e} e^{6 i f x} - 251658240 i a^{8} c^{12} f^{4} e^{10 i e} e^{4 i f x} - 1006632960 i a^{8} c^{12} f^{4} e^{8 i e} e^{2 i f x} + 503316480 i a^{8} c^{12} f^{4} e^{4 i e} e^{- 2 i f x} + 50331648 i a^{8} c^{12} f^{4} e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{6442450944 a^{10} c^{15} f^{5}} & \text {for}\: a^{10} c^{15} f^{5} e^{6 i e} \neq 0 \\x \left (\frac {\left (e^{10 i e} + 5 e^{8 i e} + 10 e^{6 i e} + 10 e^{4 i e} + 5 e^{2 i e} + 1\right ) e^{- 4 i e}}{32 a^{2} c^{3}} - \frac {5}{16 a^{2} c^{3}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{16 a^{2} c^{3}} \] Input:

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

Output:

Piecewise(((-33554432*I*a**8*c**12*f**4*exp(12*I*e)*exp(6*I*f*x) - 2516582 
40*I*a**8*c**12*f**4*exp(10*I*e)*exp(4*I*f*x) - 1006632960*I*a**8*c**12*f* 
*4*exp(8*I*e)*exp(2*I*f*x) + 503316480*I*a**8*c**12*f**4*exp(4*I*e)*exp(-2 
*I*f*x) + 50331648*I*a**8*c**12*f**4*exp(2*I*e)*exp(-4*I*f*x))*exp(-6*I*e) 
/(6442450944*a**10*c**15*f**5), Ne(a**10*c**15*f**5*exp(6*I*e), 0)), (x*(( 
exp(10*I*e) + 5*exp(8*I*e) + 10*exp(6*I*e) + 10*exp(4*I*e) + 5*exp(2*I*e) 
+ 1)*exp(-4*I*e)/(32*a**2*c**3) - 5/(16*a**2*c**3)), True)) + 5*x/(16*a**2 
*c**3)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {5 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{32 \, a^{2} c^{3} f} - \frac {5 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{32 \, a^{2} c^{3} f} - \frac {i \, {\left (15 i \, \tan \left (f x + e\right )^{4} - 15 \, \tan \left (f x + e\right )^{3} + 25 i \, \tan \left (f x + e\right )^{2} - 25 \, \tan \left (f x + e\right ) + 8 i\right )}}{48 \, a^{2} c^{3} f {\left (\tan \left (f x + e\right ) + i\right )}^{3} {\left (\tan \left (f x + e\right ) - i\right )}^{2}} \] Input:

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

Output:

5/32*I*log(tan(f*x + e) + I)/(a^2*c^3*f) - 5/32*I*log(tan(f*x + e) - I)/(a 
^2*c^3*f) - 1/48*I*(15*I*tan(f*x + e)^4 - 15*tan(f*x + e)^3 + 25*I*tan(f*x 
 + e)^2 - 25*tan(f*x + e) + 8*I)/(a^2*c^3*f*(tan(f*x + e) + I)^3*(tan(f*x 
+ e) - I)^2)
 

Mupad [B] (verification not implemented)

Time = 2.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.51 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {5\,x}{16\,a^2\,c^3}-\frac {\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^4}{16}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,5{}\mathrm {i}}{16}+\frac {25\,{\mathrm {tan}\left (e+f\,x\right )}^2}{48}+\frac {\mathrm {tan}\left (e+f\,x\right )\,25{}\mathrm {i}}{48}+\frac {1}{6}}{a^2\,c^3\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^3} \] Input:

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

Output:

(5*x)/(16*a^2*c^3) - ((tan(e + f*x)*25i)/48 + (25*tan(e + f*x)^2)/48 + (ta 
n(e + f*x)^3*5i)/16 + (5*tan(e + f*x)^4)/16 + 1/6)/(a^2*c^3*f*(tan(e + f*x 
)*1i + 1)^2*(tan(e + f*x) + 1i)^3)
 

Reduce [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=-\frac {\int \frac {1}{\tan \left (f x +e \right )^{5} i -\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{3} i -2 \tan \left (f x +e \right )^{2}+\tan \left (f x +e \right ) i -1}d x}{a^{2} c^{3}} \] Input:

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

Output:

( - int(1/(tan(e + f*x)**5*i - tan(e + f*x)**4 + 2*tan(e + f*x)**3*i - 2*t 
an(e + f*x)**2 + tan(e + f*x)*i - 1),x))/(a**2*c**3)