Integrand size = 31, antiderivative size = 71 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {4 c^3 x}{a}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}+\frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))} \] Output:
-4*c^3*x/a-4*I*c^3*ln(cos(f*x+e))/a/f+c^3*tan(f*x+e)/a/f+4*I*c^3/f/(a+I*a* tan(f*x+e))
Time = 1.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {i c^3 \left (-4 \log (i-\tan (e+f x))+i \tan (e+f x)+\frac {4 i}{-i+\tan (e+f x)}\right )}{a f} \] Input:
Integrate[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x]),x]
Output:
((-I)*c^3*(-4*Log[I - Tan[e + f*x]] + I*Tan[e + f*x] + (4*I)/(-I + Tan[e + f*x])))/(a*f)
Time = 0.34 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87, 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.
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^3 c^3 \int \frac {\sec ^6(e+f x)}{(i \tan (e+f x) a+a)^4}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \int \frac {\sec (e+f x)^6}{(i \tan (e+f x) a+a)^4}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i c^3 \int \frac {(a-i a \tan (e+f x))^2}{(i \tan (e+f x) a+a)^2}d(i a \tan (e+f x))}{a^2 f}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i c^3 \int \left (\frac {4 a^2}{(i \tan (e+f x) a+a)^2}-\frac {4 a}{i \tan (e+f x) a+a}+1\right )d(i a \tan (e+f x))}{a^2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i c^3 \left (-\frac {4 a^2}{a+i a \tan (e+f x)}+i a \tan (e+f x)-4 a \log (a+i a \tan (e+f x))\right )}{a^2 f}\) |
Input:
Int[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x]),x]
Output:
((-I)*c^3*(-4*a*Log[a + I*a*Tan[e + f*x]] + I*a*Tan[e + f*x] - (4*a^2)/(a + I*a*Tan[e + f*x])))/(a^2*f)
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]
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]
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]))
Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {c^{3} \tan \left (f x +e \right )}{a f}+\frac {4 c^{3}}{f a \left (-i+\tan \left (f x +e \right )\right )}-\frac {4 c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f a}+\frac {2 i c^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f a}\) | \(81\) |
default | \(\frac {c^{3} \tan \left (f x +e \right )}{a f}+\frac {4 c^{3}}{f a \left (-i+\tan \left (f x +e \right )\right )}-\frac {4 c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f a}+\frac {2 i c^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f a}\) | \(81\) |
risch | \(\frac {2 i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )}}{a f}-\frac {8 c^{3} x}{a}-\frac {8 c^{3} e}{a f}+\frac {2 i c^{3}}{f a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {4 i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a f}\) | \(93\) |
norman | \(\frac {\frac {4 i c^{3}}{a f}+\frac {c^{3} \tan \left (f x +e \right )^{3}}{a f}-\frac {4 c^{3} x}{a}-\frac {4 c^{3} x \tan \left (f x +e \right )^{2}}{a}+\frac {5 c^{3} \tan \left (f x +e \right )}{a f}}{1+\tan \left (f x +e \right )^{2}}+\frac {2 i c^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f a}\) | \(112\) |
Input:
int((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),x,method=_RETURNVERBOSE)
Output:
c^3*tan(f*x+e)/a/f+4/f*c^3/a/(-I+tan(f*x+e))-4/f*c^3/a*arctan(tan(f*x+e))+ 2*I/f*c^3/a*ln(1+tan(f*x+e)^2)
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.66 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {2 \, {\left (4 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} - i \, c^{3} + 2 \, {\left (2 \, c^{3} f x - i \, c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}} \] Input:
integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),x, algorithm="fricas")
Output:
-2*(4*c^3*f*x*e^(4*I*f*x + 4*I*e) - I*c^3 + 2*(2*c^3*f*x - I*c^3)*e^(2*I*f *x + 2*I*e) + 2*(I*c^3*e^(4*I*f*x + 4*I*e) + I*c^3*e^(2*I*f*x + 2*I*e))*lo g(e^(2*I*f*x + 2*I*e) + 1))/(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2* I*e))
Time = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.87 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {2 i c^{3}}{a f e^{2 i e} e^{2 i f x} + a f} + \begin {cases} \frac {2 i c^{3} e^{- 2 i e} e^{- 2 i f x}}{a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (\frac {8 c^{3}}{a} + \frac {\left (- 8 c^{3} e^{2 i e} + 4 c^{3}\right ) e^{- 2 i e}}{a}\right ) & \text {otherwise} \end {cases} - \frac {8 c^{3} x}{a} - \frac {4 i c^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} \] Input:
integrate((c-I*c*tan(f*x+e))**3/(a+I*a*tan(f*x+e)),x)
Output:
2*I*c**3/(a*f*exp(2*I*e)*exp(2*I*f*x) + a*f) + Piecewise((2*I*c**3*exp(-2* I*e)*exp(-2*I*f*x)/(a*f), Ne(a*f*exp(2*I*e), 0)), (x*(8*c**3/a + (-8*c**3* exp(2*I*e) + 4*c**3)*exp(-2*I*e)/a), True)) - 8*c**3*x/a - 4*I*c**3*log(ex p(2*I*f*x) + exp(-2*I*e))/(a*f)
Exception generated. \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.52 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {4 i \, c^{3} \log \left (\tan \left (f x + e\right ) - i\right )}{a f} + \frac {c^{3} \tan \left (f x + e\right )}{a f} + \frac {4 \, c^{3}}{a f {\left (\tan \left (f x + e\right ) - i\right )}} \] Input:
integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),x, algorithm="giac")
Output:
4*I*c^3*log(tan(f*x + e) - I)/(a*f) + c^3*tan(f*x + e)/(a*f) + 4*c^3/(a*f* (tan(f*x + e) - I))
Time = 1.75 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {c^3\,\mathrm {tan}\left (e+f\,x\right )}{a\,f}+\frac {c^3\,4{}\mathrm {i}}{a\,f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {c^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,4{}\mathrm {i}}{a\,f} \] Input:
int((c - c*tan(e + f*x)*1i)^3/(a + a*tan(e + f*x)*1i),x)
Output:
(c^3*tan(e + f*x))/(a*f) + (c^3*4i)/(a*f*(tan(e + f*x)*1i + 1)) + (c^3*log (tan(e + f*x) - 1i)*4i)/(a*f)
\[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {c^{3} \left (8 \left (\int \frac {1}{\tan \left (f x +e \right ) i +1}d x \right ) f +2 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) i +\tan \left (f x +e \right )-8 f x \right )}{a f} \] Input:
int((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),x)
Output:
(c**3*(8*int(1/(tan(e + f*x)*i + 1),x)*f + 2*log(tan(e + f*x)**2 + 1)*i + tan(e + f*x) - 8*f*x))/(a*f)