Integrand size = 31, antiderivative size = 101 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx=\frac {6 c^4 x}{a^2}+\frac {6 i c^4 \log (\cos (e+f x))}{a^2 f}-\frac {c^4 \tan (e+f x)}{a^2 f}+\frac {4 i c^4}{f (a+i a \tan (e+f x))^2}-\frac {12 i c^4}{f \left (a^2+i a^2 \tan (e+f x)\right )} \] Output:
6*c^4*x/a^2+6*I*c^4*ln(cos(f*x+e))/a^2/f-c^4*tan(f*x+e)/a^2/f+4*I*c^4/f/(a +I*a*tan(f*x+e))^2-12*I*c^4/f/(a^2+I*a^2*tan(f*x+e))
Time = 5.56 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i c^4 \left (6 \log (i-\tan (e+f x))-i \tan (e+f x)+\frac {-8-12 i \tan (e+f x)}{(-i+\tan (e+f x))^2}\right )}{a^2 f} \] Input:
Integrate[(c - I*c*Tan[e + f*x])^4/(a + I*a*Tan[e + f*x])^2,x]
Output:
((-I)*c^4*(6*Log[I - Tan[e + f*x]] - I*Tan[e + f*x] + (-8 - (12*I)*Tan[e + f*x])/(-I + Tan[e + f*x])^2))/(a^2*f)
Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81, 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.
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^4 c^4 \int \frac {\sec ^8(e+f x)}{(i \tan (e+f x) a+a)^6}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^4 c^4 \int \frac {\sec (e+f x)^8}{(i \tan (e+f x) a+a)^6}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i c^4 \int \frac {(a-i a \tan (e+f x))^3}{(i \tan (e+f x) a+a)^3}d(i a \tan (e+f x))}{a^3 f}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i c^4 \int \left (\frac {8 a^3}{(i \tan (e+f x) a+a)^3}-\frac {12 a^2}{(i \tan (e+f x) a+a)^2}+\frac {6 a}{i \tan (e+f x) a+a}-1\right )d(i a \tan (e+f x))}{a^3 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i c^4 \left (-\frac {4 a^3}{(a+i a \tan (e+f x))^2}+\frac {12 a^2}{a+i a \tan (e+f x)}-i a \tan (e+f x)+6 a \log (a+i a \tan (e+f x))\right )}{a^3 f}\) |
Input:
Int[(c - I*c*Tan[e + f*x])^4/(a + I*a*Tan[e + f*x])^2,x]
Output:
((-I)*c^4*(6*a*Log[a + I*a*Tan[e + f*x]] - I*a*Tan[e + f*x] - (4*a^3)/(a + I*a*Tan[e + f*x])^2 + (12*a^2)/(a + I*a*Tan[e + f*x])))/(a^3*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 = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(-\frac {c^{4} \tan \left (f x +e \right )}{a^{2} f}-\frac {12 c^{4}}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}+\frac {6 c^{4} \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}-\frac {3 i c^{4} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f \,a^{2}}-\frac {4 i c^{4}}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}\) | \(105\) |
default | \(-\frac {c^{4} \tan \left (f x +e \right )}{a^{2} f}-\frac {12 c^{4}}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}+\frac {6 c^{4} \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}-\frac {3 i c^{4} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f \,a^{2}}-\frac {4 i c^{4}}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}\) | \(105\) |
risch | \(-\frac {4 i c^{4} {\mathrm e}^{-2 i \left (f x +e \right )}}{a^{2} f}+\frac {i c^{4} {\mathrm e}^{-4 i \left (f x +e \right )}}{a^{2} f}+\frac {12 c^{4} x}{a^{2}}+\frac {12 c^{4} e}{a^{2} f}-\frac {2 i c^{4}}{f \,a^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {6 i c^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a^{2} f}\) | \(114\) |
norman | \(\frac {-\frac {8 i c^{4}}{a f}+\frac {6 c^{4} x}{a}-\frac {14 c^{4} \tan \left (f x +e \right )^{3}}{a f}-\frac {c^{4} \tan \left (f x +e \right )^{5}}{a f}+\frac {12 c^{4} x \tan \left (f x +e \right )^{2}}{a}+\frac {6 c^{4} x \tan \left (f x +e \right )^{4}}{a}-\frac {5 c^{4} \tan \left (f x +e \right )}{a f}-\frac {16 i c^{4} \tan \left (f x +e \right )^{2}}{a f}}{a \left (1+\tan \left (f x +e \right )^{2}\right )^{2}}-\frac {3 i c^{4} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f \,a^{2}}\) | \(172\) |
Input:
int((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
-c^4*tan(f*x+e)/a^2/f-12/f*c^4/a^2/(-I+tan(f*x+e))+6/f*c^4/a^2*arctan(tan( f*x+e))-3*I/f*c^4/a^2*ln(1+tan(f*x+e)^2)-4*I/f*c^4/a^2/(-I+tan(f*x+e))^2
Time = 0.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.34 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx=\frac {12 \, c^{4} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{4} + 6 \, {\left (2 \, c^{4} f x - i \, c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 6 \, {\left (-i \, c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} - i \, c^{4} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}} \] Input:
integrate((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")
Output:
(12*c^4*f*x*e^(6*I*f*x + 6*I*e) - 3*I*c^4*e^(2*I*f*x + 2*I*e) + I*c^4 + 6* (2*c^4*f*x - I*c^4)*e^(4*I*f*x + 4*I*e) - 6*(-I*c^4*e^(6*I*f*x + 6*I*e) - I*c^4*e^(4*I*f*x + 4*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))
Time = 0.41 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.97 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx=- \frac {2 i c^{4}}{a^{2} f e^{2 i e} e^{2 i f x} + a^{2} f} + \begin {cases} \frac {\left (- 4 i a^{2} c^{4} f e^{4 i e} e^{- 2 i f x} + i a^{2} c^{4} f e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {12 c^{4}}{a^{2}} + \frac {\left (12 c^{4} e^{4 i e} - 8 c^{4} e^{2 i e} + 4 c^{4}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {12 c^{4} x}{a^{2}} + \frac {6 i c^{4} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} \] Input:
integrate((c-I*c*tan(f*x+e))**4/(a+I*a*tan(f*x+e))**2,x)
Output:
-2*I*c**4/(a**2*f*exp(2*I*e)*exp(2*I*f*x) + a**2*f) + Piecewise(((-4*I*a** 2*c**4*f*exp(4*I*e)*exp(-2*I*f*x) + I*a**2*c**4*f*exp(2*I*e)*exp(-4*I*f*x) )*exp(-6*I*e)/(a**4*f**2), Ne(a**4*f**2*exp(6*I*e), 0)), (x*(-12*c**4/a**2 + (12*c**4*exp(4*I*e) - 8*c**4*exp(2*I*e) + 4*c**4)*exp(-4*I*e)/a**2), Tr ue)) + 12*c**4*x/a**2 + 6*I*c**4*log(exp(2*I*f*x) + exp(-2*I*e))/(a**2*f)
Exception generated. \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((c-I*c*tan(f*x+e))^4/(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 = 0.64 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx=-\frac {6 i \, c^{4} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} f} - \frac {c^{4} \tan \left (f x + e\right )}{a^{2} f} - \frac {4 \, {\left (3 \, c^{4} \tan \left (f x + e\right ) - 2 i \, c^{4}\right )}}{a^{2} f {\left (\tan \left (f x + e\right ) - i\right )}^{2}} \] Input:
integrate((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")
Output:
-6*I*c^4*log(tan(f*x + e) - I)/(a^2*f) - c^4*tan(f*x + e)/(a^2*f) - 4*(3*c ^4*tan(f*x + e) - 2*I*c^4)/(a^2*f*(tan(f*x + e) - I)^2)
Time = 1.78 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {8\,c^4}{a^2}+\frac {c^4\,\mathrm {tan}\left (e+f\,x\right )\,12{}\mathrm {i}}{a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {c^4\,\mathrm {tan}\left (e+f\,x\right )}{a^2\,f}-\frac {c^4\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,6{}\mathrm {i}}{a^2\,f} \] Input:
int((c - c*tan(e + f*x)*1i)^4/(a + a*tan(e + f*x)*1i)^2,x)
Output:
- ((8*c^4)/a^2 + (c^4*tan(e + f*x)*12i)/a^2)/(f*(2*tan(e + f*x) + tan(e + f*x)^2*1i - 1i)) - (c^4*tan(e + f*x))/(a^2*f) - (c^4*log(tan(e + f*x) - 1i )*6i)/(a^2*f)
\[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^{4} \left (-\left (\int \frac {\tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right )-4 \left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right ) i +6 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right )+4 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right ) i -\left (\int \frac {1}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right )\right )}{a^{2}} \] Input:
int((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x)
Output:
(c**4*( - int(tan(e + f*x)**4/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x) - 4*int(tan(e + f*x)**3/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x)*i + 6* int(tan(e + f*x)**2/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x) + 4*int(ta n(e + f*x)/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x)*i - int(1/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x)))/a**2