Integrand size = 31, antiderivative size = 83 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^3 x}{a^2}+\frac {i c^3 \log (\cos (e+f x))}{a^2 f}+\frac {2 i c^3}{a^2 f (1+i \tan (e+f x))^2}-\frac {4 i c^3}{a^2 f (1+i \tan (e+f x))} \] Output:
c^3*x/a^2+I*c^3*ln(cos(f*x+e))/a^2/f+2*I*c^3/a^2/f/(1+I*tan(f*x+e))^2-4*I* c^3/a^2/f/(1+I*tan(f*x+e))
Time = 5.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.63 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i c^3 \left (\log (i-\tan (e+f x))+\frac {-2-4 i \tan (e+f x)}{(-i+\tan (e+f x))^2}\right )}{a^2 f} \] Input:
Integrate[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^2,x]
Output:
((-I)*c^3*(Log[I - Tan[e + f*x]] + (-2 - (4*I)*Tan[e + f*x])/(-I + Tan[e + f*x])^2))/(a^2*f)
Time = 0.36 (sec) , antiderivative size = 66, 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.
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^3 c^3 \int \frac {\sec ^6(e+f x)}{(i \tan (e+f x) a+a)^5}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \int \frac {\sec (e+f x)^6}{(i \tan (e+f x) a+a)^5}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i c^3 \int \frac {(a-i a \tan (e+f x))^2}{(i \tan (e+f x) a+a)^3}d(i a \tan (e+f x))}{a^2 f}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i c^3 \int \left (\frac {4 a^2}{(i \tan (e+f x) a+a)^3}-\frac {4 a}{(i \tan (e+f x) a+a)^2}+\frac {1}{i \tan (e+f x) a+a}\right )d(i a \tan (e+f x))}{a^2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i c^3 \left (-\frac {2 a^2}{(a+i a \tan (e+f x))^2}+\frac {4 a}{a+i a \tan (e+f x)}+\log (a+i a \tan (e+f x))\right )}{a^2 f}\) |
Input:
Int[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^2,x]
Output:
((-I)*c^3*(Log[a + I*a*Tan[e + f*x]] - (2*a^2)/(a + I*a*Tan[e + f*x])^2 + (4*a)/(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 = 0.18 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(-\frac {2 i c^{3}}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i c^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}+\frac {c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}-\frac {4 c^{3}}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}\) | \(87\) |
default | \(-\frac {2 i c^{3}}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i c^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}+\frac {c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}-\frac {4 c^{3}}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}\) | \(87\) |
risch | \(-\frac {i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )}}{a^{2} f}+\frac {i c^{3} {\mathrm e}^{-4 i \left (f x +e \right )}}{2 a^{2} f}+\frac {2 c^{3} x}{a^{2}}+\frac {2 c^{3} e}{a^{2} f}+\frac {i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a^{2} f}\) | \(89\) |
norman | \(\frac {\frac {c^{3} x}{a}-\frac {2 i c^{3}}{a f}+\frac {c^{3} x \tan \left (f x +e \right )^{4}}{a}-\frac {4 c^{3} \tan \left (f x +e \right )^{3}}{a f}+\frac {2 c^{3} x \tan \left (f x +e \right )^{2}}{a}-\frac {6 i c^{3} \tan \left (f x +e \right )^{2}}{a f}}{a \left (1+\tan \left (f x +e \right )^{2}\right )^{2}}-\frac {i c^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}\) | \(134\) |
Input:
int((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
-2*I/f*c^3/a^2/(-I+tan(f*x+e))^2-1/2*I/f*c^3/a^2*ln(1+tan(f*x+e)^2)+1/f*c^ 3/a^2*arctan(tan(f*x+e))-4/f*c^3/a^2/(-I+tan(f*x+e))
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (4 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 2 i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{3}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{2 \, a^{2} f} \] Input:
integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")
Output:
1/2*(4*c^3*f*x*e^(4*I*f*x + 4*I*e) + 2*I*c^3*e^(4*I*f*x + 4*I*e)*log(e^(2* I*f*x + 2*I*e) + 1) - 2*I*c^3*e^(2*I*f*x + 2*I*e) + I*c^3)*e^(-4*I*f*x - 4 *I*e)/(a^2*f)
Time = 0.27 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.01 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\begin {cases} \frac {\left (- 2 i a^{2} c^{3} f e^{4 i e} e^{- 2 i f x} + i a^{2} c^{3} f e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{2 a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {2 c^{3}}{a^{2}} + \frac {\left (2 c^{3} e^{4 i e} - 2 c^{3} e^{2 i e} + 2 c^{3}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {2 c^{3} x}{a^{2}} + \frac {i c^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} \] Input:
integrate((c-I*c*tan(f*x+e))**3/(a+I*a*tan(f*x+e))**2,x)
Output:
Piecewise(((-2*I*a**2*c**3*f*exp(4*I*e)*exp(-2*I*f*x) + I*a**2*c**3*f*exp( 2*I*e)*exp(-4*I*f*x))*exp(-6*I*e)/(2*a**4*f**2), Ne(a**4*f**2*exp(6*I*e), 0)), (x*(-2*c**3/a**2 + (2*c**3*exp(4*I*e) - 2*c**3*exp(2*I*e) + 2*c**3)*e xp(-4*I*e)/a**2), True)) + 2*c**3*x/a**2 + I*c**3*log(exp(2*I*f*x) + exp(- 2*I*e))/(a**2*f)
Exception generated. \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \, c^{3} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} f} - \frac {2 \, {\left (2 \, c^{3} \tan \left (f x + e\right ) - i \, c^{3}\right )}}{a^{2} f {\left (\tan \left (f x + e\right ) - i\right )}^{2}} \] Input:
integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")
Output:
-I*c^3*log(tan(f*x + e) - I)/(a^2*f) - 2*(2*c^3*tan(f*x + e) - I*c^3)/(a^2 *f*(tan(f*x + e) - I)^2)
Time = 1.85 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {2\,c^3}{a^2}+\frac {c^3\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}}{a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {c^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{a^2\,f} \] Input:
int((c - c*tan(e + f*x)*1i)^3/(a + a*tan(e + f*x)*1i)^2,x)
Output:
- ((2*c^3)/a^2 + (c^3*tan(e + f*x)*4i)/a^2)/(f*(2*tan(e + f*x) + tan(e + f *x)^2*1i - 1i)) - (c^3*log(tan(e + f*x) - 1i)*1i)/(a^2*f)
\[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^{3} \left (-\left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right ) i +3 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right )+3 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right ) i -\left (\int \frac {1}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right )\right )}{a^{2}} \] Input:
int((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x)
Output:
(c**3*( - int(tan(e + f*x)**3/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x)* i + 3*int(tan(e + f*x)**2/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x) + 3* int(tan(e + f*x)/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x)*i - int(1/(ta n(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x)))/a**2