3.10.38 \(\int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx\) [938]

3.10.38.1 Optimal result
3.10.38.2 Mathematica [A] (verified)
3.10.38.3 Rubi [A] (verified)
3.10.38.4 Maple [A] (verified)
3.10.38.5 Fricas [A] (verification not implemented)
3.10.38.6 Sympy [A] (verification not implemented)
3.10.38.7 Maxima [F(-2)]
3.10.38.8 Giac [B] (verification not implemented)
3.10.38.9 Mupad [B] (verification not implemented)

3.10.38.1 Optimal result

Integrand size = 31, antiderivative size = 154 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx=-\frac {40 a^6 x}{c^3}+\frac {40 i a^6 \log (\cos (e+f x))}{c^3 f}+\frac {9 a^6 \tan (e+f x)}{c^3 f}+\frac {i a^6 \tan ^2(e+f x)}{2 c^3 f}-\frac {32 i a^6}{3 f (c-i c \tan (e+f x))^3}+\frac {40 i a^6}{c f (c-i c \tan (e+f x))^2}-\frac {80 i a^6}{f \left (c^3-i c^3 \tan (e+f x)\right )} \]

output
-40*a^6*x/c^3+40*I*a^6*ln(cos(f*x+e))/c^3/f+9*a^6*tan(f*x+e)/c^3/f+1/2*I*a 
^6*tan(f*x+e)^2/c^3/f-32/3*I*a^6/f/(c-I*c*tan(f*x+e))^3+40*I*a^6/c/f/(c-I* 
c*tan(f*x+e))^2-80*I*a^6/f/(c^3-I*c^3*tan(f*x+e))
 
3.10.38.2 Mathematica [A] (verified)

Time = 5.98 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.56 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx=\frac {i a^6 \left (-240 \log (i+\tan (e+f x))-54 i \tan (e+f x)+3 \tan ^2(e+f x)+\frac {16 \left (19 i+45 \tan (e+f x)-30 i \tan ^2(e+f x)\right )}{(i+\tan (e+f x))^3}\right )}{6 c^3 f} \]

input
Integrate[(a + I*a*Tan[e + f*x])^6/(c - I*c*Tan[e + f*x])^3,x]
 
output
((I/6)*a^6*(-240*Log[I + Tan[e + f*x]] - (54*I)*Tan[e + f*x] + 3*Tan[e + f 
*x]^2 + (16*(19*I + 45*Tan[e + f*x] - (30*I)*Tan[e + f*x]^2))/(I + Tan[e + 
 f*x])^3))/(c^3*f)
 
3.10.38.3 Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.80, 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, 49, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4005

\(\displaystyle a^6 c^6 \int \frac {\sec ^{12}(e+f x)}{(c-i c \tan (e+f x))^9}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^6 c^6 \int \frac {\sec (e+f x)^{12}}{(c-i c \tan (e+f x))^9}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle \frac {i a^6 \int \frac {(i \tan (e+f x) c+c)^5}{(c-i c \tan (e+f x))^4}d(-i c \tan (e+f x))}{c^5 f}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {i a^6 \int \left (\frac {32 c^5}{(c-i c \tan (e+f x))^4}-\frac {80 c^4}{(c-i c \tan (e+f x))^3}+\frac {80 c^3}{(c-i c \tan (e+f x))^2}-\frac {40 c^2}{c-i c \tan (e+f x)}+i \tan (e+f x) c+9 c\right )d(-i c \tan (e+f x))}{c^5 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i a^6 \left (-\frac {32 c^5}{3 (c-i c \tan (e+f x))^3}+\frac {40 c^4}{(c-i c \tan (e+f x))^2}-\frac {80 c^3}{c-i c \tan (e+f x)}+\frac {1}{2} c^2 \tan ^2(e+f x)-9 i c^2 \tan (e+f x)-40 c^2 \log (c-i c \tan (e+f x))\right )}{c^5 f}\)

input
Int[(a + I*a*Tan[e + f*x])^6/(c - I*c*Tan[e + f*x])^3,x]
 
output
(I*a^6*(-40*c^2*Log[c - I*c*Tan[e + f*x]] - (9*I)*c^2*Tan[e + f*x] + (c^2* 
Tan[e + f*x]^2)/2 - (32*c^5)/(3*(c - I*c*Tan[e + f*x])^3) + (40*c^4)/(c - 
I*c*Tan[e + f*x])^2 - (80*c^3)/(c - I*c*Tan[e + f*x])))/(c^5*f)
 

3.10.38.3.1 Defintions of rubi rules used

rule 49
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[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]))
 
3.10.38.4 Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.90

method result size
risch \(-\frac {4 i a^{6} {\mathrm e}^{6 i \left (f x +e \right )}}{3 c^{3} f}+\frac {6 i a^{6} {\mathrm e}^{4 i \left (f x +e \right )}}{c^{3} f}-\frac {24 i a^{6} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{3} f}+\frac {80 a^{6} e}{f \,c^{3}}+\frac {2 i a^{6} \left (10 \,{\mathrm e}^{2 i \left (f x +e \right )}+9\right )}{f \,c^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {40 i a^{6} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f \,c^{3}}\) \(139\)
derivativedivides \(\frac {9 a^{6} \tan \left (f x +e \right )}{c^{3} f}+\frac {i a^{6} \left (\tan ^{2}\left (f x +e \right )\right )}{2 c^{3} f}-\frac {40 a^{6} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{3}}-\frac {20 i a^{6} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f \,c^{3}}+\frac {80 a^{6}}{f \,c^{3} \left (\tan \left (f x +e \right )+i\right )}-\frac {32 a^{6}}{3 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {40 i a^{6}}{f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{2}}\) \(147\)
default \(\frac {9 a^{6} \tan \left (f x +e \right )}{c^{3} f}+\frac {i a^{6} \left (\tan ^{2}\left (f x +e \right )\right )}{2 c^{3} f}-\frac {40 a^{6} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{3}}-\frac {20 i a^{6} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f \,c^{3}}+\frac {80 a^{6}}{f \,c^{3} \left (\tan \left (f x +e \right )+i\right )}-\frac {32 a^{6}}{3 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {40 i a^{6}}{f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{2}}\) \(147\)
norman \(\frac {-\frac {40 a^{6} x}{c}-\frac {313 i a^{6}}{6 c f}-\frac {120 a^{6} x \left (\tan ^{2}\left (f x +e \right )\right )}{c}-\frac {120 a^{6} x \left (\tan ^{4}\left (f x +e \right )\right )}{c}-\frac {40 a^{6} x \left (\tan ^{6}\left (f x +e \right )\right )}{c}+\frac {41 a^{6} \tan \left (f x +e \right )}{c f}+\frac {289 a^{6} \left (\tan ^{3}\left (f x +e \right )\right )}{3 c f}+\frac {107 a^{6} \left (\tan ^{5}\left (f x +e \right )\right )}{c f}+\frac {9 a^{6} \left (\tan ^{7}\left (f x +e \right )\right )}{c f}-\frac {132 i a^{6} \left (\tan ^{2}\left (f x +e \right )\right )}{c f}-\frac {123 i a^{6} \left (\tan ^{4}\left (f x +e \right )\right )}{c f}+\frac {i a^{6} \left (\tan ^{8}\left (f x +e \right )\right )}{2 c f}}{c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {20 i a^{6} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f \,c^{3}}\) \(248\)

input
int((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 
output
-4/3*I/c^3/f*a^6*exp(6*I*(f*x+e))+6*I/c^3/f*a^6*exp(4*I*(f*x+e))-24*I/c^3/ 
f*a^6*exp(2*I*(f*x+e))+80/f*a^6/c^3*e+2*I*a^6*(10*exp(2*I*(f*x+e))+9)/f/c^ 
3/(exp(2*I*(f*x+e))+1)^2+40*I/f*a^6/c^3*ln(exp(2*I*(f*x+e))+1)
 
3.10.38.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.06 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx=-\frac {2 \, {\left (2 i \, a^{6} e^{\left (10 i \, f x + 10 i \, e\right )} - 5 i \, a^{6} e^{\left (8 i \, f x + 8 i \, e\right )} + 20 i \, a^{6} e^{\left (6 i \, f x + 6 i \, e\right )} + 63 i \, a^{6} e^{\left (4 i \, f x + 4 i \, e\right )} + 6 i \, a^{6} e^{\left (2 i \, f x + 2 i \, e\right )} - 27 i \, a^{6} + 60 \, {\left (-i \, a^{6} e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, a^{6} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{6}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{3 \, {\left (c^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )}} \]

input
integrate((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")
 
output
-2/3*(2*I*a^6*e^(10*I*f*x + 10*I*e) - 5*I*a^6*e^(8*I*f*x + 8*I*e) + 20*I*a 
^6*e^(6*I*f*x + 6*I*e) + 63*I*a^6*e^(4*I*f*x + 4*I*e) + 6*I*a^6*e^(2*I*f*x 
 + 2*I*e) - 27*I*a^6 + 60*(-I*a^6*e^(4*I*f*x + 4*I*e) - 2*I*a^6*e^(2*I*f*x 
 + 2*I*e) - I*a^6)*log(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f*e^(4*I*f*x + 4*I*e 
) + 2*c^3*f*e^(2*I*f*x + 2*I*e) + c^3*f)
 
3.10.38.6 Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.60 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx=\frac {40 i a^{6} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{3} f} + \frac {20 i a^{6} e^{2 i e} e^{2 i f x} + 18 i a^{6}}{c^{3} f e^{4 i e} e^{4 i f x} + 2 c^{3} f e^{2 i e} e^{2 i f x} + c^{3} f} + \begin {cases} \frac {- 4 i a^{6} c^{6} f^{2} e^{6 i e} e^{6 i f x} + 18 i a^{6} c^{6} f^{2} e^{4 i e} e^{4 i f x} - 72 i a^{6} c^{6} f^{2} e^{2 i e} e^{2 i f x}}{3 c^{9} f^{3}} & \text {for}\: c^{9} f^{3} \neq 0 \\\frac {x \left (8 a^{6} e^{6 i e} - 24 a^{6} e^{4 i e} + 48 a^{6} e^{2 i e}\right )}{c^{3}} & \text {otherwise} \end {cases} \]

input
integrate((a+I*a*tan(f*x+e))**6/(c-I*c*tan(f*x+e))**3,x)
 
output
40*I*a**6*log(exp(2*I*f*x) + exp(-2*I*e))/(c**3*f) + (20*I*a**6*exp(2*I*e) 
*exp(2*I*f*x) + 18*I*a**6)/(c**3*f*exp(4*I*e)*exp(4*I*f*x) + 2*c**3*f*exp( 
2*I*e)*exp(2*I*f*x) + c**3*f) + Piecewise(((-4*I*a**6*c**6*f**2*exp(6*I*e) 
*exp(6*I*f*x) + 18*I*a**6*c**6*f**2*exp(4*I*e)*exp(4*I*f*x) - 72*I*a**6*c* 
*6*f**2*exp(2*I*e)*exp(2*I*f*x))/(3*c**9*f**3), Ne(c**9*f**3, 0)), (x*(8*a 
**6*exp(6*I*e) - 24*a**6*exp(4*I*e) + 48*a**6*exp(2*I*e))/c**3, True))
 
3.10.38.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]

input
integrate((a+I*a*tan(f*x+e))^6/(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.
 
3.10.38.8 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. 272 vs. \(2 (134) = 268\).

Time = 1.06 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.77 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx=-\frac {2 \, {\left (-\frac {60 i \, a^{6} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{3}} + \frac {120 i \, a^{6} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{3}} - \frac {60 i \, a^{6} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{3}} - \frac {3 \, {\left (-30 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 9 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 61 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 9 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 30 i \, a^{6}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} c^{3}} + \frac {2 \, {\left (-147 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 930 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 2421 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3340 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2421 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 930 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 147 i \, a^{6}\right )}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{6}}\right )}}{3 \, f} \]

input
integrate((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")
 
output
-2/3*(-60*I*a^6*log(tan(1/2*f*x + 1/2*e) + 1)/c^3 + 120*I*a^6*log(tan(1/2* 
f*x + 1/2*e) + I)/c^3 - 60*I*a^6*log(tan(1/2*f*x + 1/2*e) - 1)/c^3 - 3*(-3 
0*I*a^6*tan(1/2*f*x + 1/2*e)^4 - 9*a^6*tan(1/2*f*x + 1/2*e)^3 + 61*I*a^6*t 
an(1/2*f*x + 1/2*e)^2 + 9*a^6*tan(1/2*f*x + 1/2*e) - 30*I*a^6)/((tan(1/2*f 
*x + 1/2*e)^2 - 1)^2*c^3) + 2*(-147*I*a^6*tan(1/2*f*x + 1/2*e)^6 + 930*a^6 
*tan(1/2*f*x + 1/2*e)^5 + 2421*I*a^6*tan(1/2*f*x + 1/2*e)^4 - 3340*a^6*tan 
(1/2*f*x + 1/2*e)^3 - 2421*I*a^6*tan(1/2*f*x + 1/2*e)^2 + 930*a^6*tan(1/2* 
f*x + 1/2*e) + 147*I*a^6)/(c^3*(tan(1/2*f*x + 1/2*e) + I)^6))/f
 
3.10.38.9 Mupad [B] (verification not implemented)

Time = 6.17 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.90 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^3} \, dx=\frac {9\,a^6\,\mathrm {tan}\left (e+f\,x\right )}{c^3\,f}-\frac {\frac {80\,a^6\,{\mathrm {tan}\left (e+f\,x\right )}^2}{c^3}-\frac {152\,a^6}{3\,c^3}+\frac {a^6\,\mathrm {tan}\left (e+f\,x\right )\,120{}\mathrm {i}}{c^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}+\frac {a^6\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}}{2\,c^3\,f}-\frac {a^6\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,40{}\mathrm {i}}{c^3\,f} \]

input
int((a + a*tan(e + f*x)*1i)^6/(c - c*tan(e + f*x)*1i)^3,x)
 
output
(9*a^6*tan(e + f*x))/(c^3*f) - ((a^6*tan(e + f*x)*120i)/c^3 - (152*a^6)/(3 
*c^3) + (80*a^6*tan(e + f*x)^2)/c^3)/(f*(3*tan(e + f*x) - tan(e + f*x)^2*3 
i - tan(e + f*x)^3 + 1i)) + (a^6*tan(e + f*x)^2*1i)/(2*c^3*f) - (a^6*log(t 
an(e + f*x) + 1i)*40i)/(c^3*f)