Integrand size = 31, antiderivative size = 131 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {x}{4 a c^3}-\frac {i}{12 a f (c-i c \tan (e+f x))^3}-\frac {i}{8 a c f (c-i c \tan (e+f x))^2}-\frac {3 i}{16 a f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{16 a f \left (c^3+i c^3 \tan (e+f x)\right )} \] Output:
1/4*x/a/c^3-1/12*I/a/f/(c-I*c*tan(f*x+e))^3-1/8*I/a/c/f/(c-I*c*tan(f*x+e)) ^2-3/16*I/a/f/(c^3-I*c^3*tan(f*x+e))+1/16*I/a/f/(c^3+I*c^3*tan(f*x+e))
Time = 0.59 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {4 i-\tan (e+f x)+6 i \tan ^2(e+f x)+3 \tan ^3(e+f x)+3 \arctan (\tan (e+f x)) (-i+\tan (e+f x)) (i+\tan (e+f x))^3}{12 a c^3 f (-i+\tan (e+f x)) (i+\tan (e+f x))^3} \] Input:
Integrate[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^3),x]
Output:
(4*I - Tan[e + f*x] + (6*I)*Tan[e + f*x]^2 + 3*Tan[e + f*x]^3 + 3*ArcTan[T an[e + f*x]]*(-I + Tan[e + f*x])*(I + Tan[e + f*x])^3)/(12*a*c^3*f*(-I + T an[e + f*x])*(I + Tan[e + f*x])^3)
Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.90, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle \frac {\int \frac {\cos ^2(e+f x)}{(c-i c \tan (e+f x))^2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\sec (e+f x)^2 (c-i c \tan (e+f x))^2}dx}{a c}\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i c^2 \int \frac {1}{(c-i c \tan (e+f x))^4 (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{a f}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {i c^2 \int \left (\frac {3}{16 c^4 (c-i c \tan (e+f x))^2}+\frac {1}{16 c^4 (i \tan (e+f x) c+c)^2}+\frac {1}{4 c^3 (c-i c \tan (e+f x))^3}+\frac {1}{4 c^2 (c-i c \tan (e+f x))^4}+\frac {1}{4 c^4 \left (\tan ^2(e+f x) c^2+c^2\right )}\right )d(-i c \tan (e+f x))}{a f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i c^2 \left (-\frac {i \arctan (\tan (e+f x))}{4 c^5}-\frac {3}{16 c^4 (c-i c \tan (e+f x))}+\frac {1}{16 c^4 (c+i c \tan (e+f x))}-\frac {1}{8 c^3 (c-i c \tan (e+f x))^2}-\frac {1}{12 c^2 (c-i c \tan (e+f x))^3}\right )}{a f}\) |
Input:
Int[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^3),x]
Output:
(I*c^2*(((-1/4*I)*ArcTan[Tan[e + f*x]])/c^5 - 1/(12*c^2*(c - I*c*Tan[e + f *x])^3) - 1/(8*c^3*(c - I*c*Tan[e + f*x])^2) - 3/(16*c^4*(c - I*c*Tan[e + f*x])) + 1/(16*c^4*(c + I*c*Tan[e + f*x]))))/(a*f)
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])
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 = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {x}{4 a \,c^{3}}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )}}{96 a \,c^{3} f}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )}}{16 a \,c^{3} f}-\frac {5 i \cos \left (2 f x +2 e \right )}{32 a \,c^{3} f}+\frac {7 \sin \left (2 f x +2 e \right )}{32 a \,c^{3} f}\) | \(94\) |
derivativedivides | \(\frac {\arctan \left (\tan \left (f x +e \right )\right )}{4 f a \,c^{3}}+\frac {1}{16 f a \,c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i}{8 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {1}{12 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3}{16 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )}\) | \(109\) |
default | \(\frac {\arctan \left (\tan \left (f x +e \right )\right )}{4 f a \,c^{3}}+\frac {1}{16 f a \,c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i}{8 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {1}{12 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3}{16 f a \,c^{3} \left (i+\tan \left (f x +e \right )\right )}\) | \(109\) |
norman | \(\frac {\frac {x}{4 a c}+\frac {3 \tan \left (f x +e \right )}{4 a c f}+\frac {2 \tan \left (f x +e \right )^{3}}{3 a c f}+\frac {\tan \left (f x +e \right )^{5}}{4 a c f}+\frac {3 x \tan \left (f x +e \right )^{2}}{4 a c}+\frac {3 x \tan \left (f x +e \right )^{4}}{4 a c}+\frac {x \tan \left (f x +e \right )^{6}}{4 a c}-\frac {i}{3 a c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} c^{2}}\) | \(145\) |
Input:
int(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
1/4*x/a/c^3-1/96*I/a/c^3/f*exp(6*I*(f*x+e))-1/16*I/a/c^3/f*exp(4*I*(f*x+e) )-5/32*I/a/c^3/f*cos(2*f*x+2*e)+7/32/a/c^3/f*sin(2*f*x+2*e)
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {{\left (24 \, f x e^{\left (2 i \, f x + 2 i \, e\right )} - i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 6 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 18 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{96 \, a c^{3} f} \] Input:
integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")
Output:
1/96*(24*f*x*e^(2*I*f*x + 2*I*e) - I*e^(8*I*f*x + 8*I*e) - 6*I*e^(6*I*f*x + 6*I*e) - 18*I*e^(4*I*f*x + 4*I*e) + 3*I)*e^(-2*I*f*x - 2*I*e)/(a*c^3*f)
Time = 0.24 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (- 8192 i a^{3} c^{9} f^{3} e^{8 i e} e^{6 i f x} - 49152 i a^{3} c^{9} f^{3} e^{6 i e} e^{4 i f x} - 147456 i a^{3} c^{9} f^{3} e^{4 i e} e^{2 i f x} + 24576 i a^{3} c^{9} f^{3} e^{- 2 i f x}\right ) e^{- 2 i e}}{786432 a^{4} c^{12} f^{4}} & \text {for}\: a^{4} c^{12} f^{4} e^{2 i e} \neq 0 \\x \left (\frac {\left (e^{8 i e} + 4 e^{6 i e} + 6 e^{4 i e} + 4 e^{2 i e} + 1\right ) e^{- 2 i e}}{16 a c^{3}} - \frac {1}{4 a c^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x}{4 a c^{3}} \] Input:
integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**3,x)
Output:
Piecewise(((-8192*I*a**3*c**9*f**3*exp(8*I*e)*exp(6*I*f*x) - 49152*I*a**3* c**9*f**3*exp(6*I*e)*exp(4*I*f*x) - 147456*I*a**3*c**9*f**3*exp(4*I*e)*exp (2*I*f*x) + 24576*I*a**3*c**9*f**3*exp(-2*I*f*x))*exp(-2*I*e)/(786432*a**4 *c**12*f**4), Ne(a**4*c**12*f**4*exp(2*I*e), 0)), (x*((exp(8*I*e) + 4*exp( 6*I*e) + 6*exp(4*I*e) + 4*exp(2*I*e) + 1)*exp(-2*I*e)/(16*a*c**3) - 1/(4*a *c**3)), True)) + x/(4*a*c**3)
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(a+I*a*tan(f*x+e))/(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.
Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {i \, \log \left (\tan \left (f x + e\right ) + i\right )}{8 \, a c^{3} f} - \frac {i \, \log \left (\tan \left (f x + e\right ) - i\right )}{8 \, a c^{3} f} + \frac {3 \, \tan \left (f x + e\right )^{3} + 6 i \, \tan \left (f x + e\right )^{2} - \tan \left (f x + e\right ) + 4 i}{12 \, a c^{3} f {\left (\tan \left (f x + e\right ) + i\right )}^{3} {\left (\tan \left (f x + e\right ) - i\right )}} \] Input:
integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")
Output:
1/8*I*log(tan(f*x + e) + I)/(a*c^3*f) - 1/8*I*log(tan(f*x + e) - I)/(a*c^3 *f) + 1/12*(3*tan(f*x + e)^3 + 6*I*tan(f*x + e)^2 - tan(f*x + e) + 4*I)/(a *c^3*f*(tan(f*x + e) + I)^3*(tan(f*x + e) - I))
Time = 2.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.59 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {x}{4\,a\,c^3}-\frac {-\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}}{4}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2}+\frac {\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{12}+\frac {1}{3}}{a\,c^3\,f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^3} \] Input:
int(1/((a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i)^3),x)
Output:
x/(4*a*c^3) - ((tan(e + f*x)*1i)/12 + tan(e + f*x)^2/2 - (tan(e + f*x)^3*1 i)/4 + 1/3)/(a*c^3*f*(tan(e + f*x)*1i + 1)*(tan(e + f*x) + 1i)^3)
\[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=-\frac {\int \frac {1}{\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{3} i +2 \tan \left (f x +e \right ) i -1}d x}{a \,c^{3}} \] Input:
int(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^3,x)
Output:
( - int(1/(tan(e + f*x)**4 + 2*tan(e + f*x)**3*i + 2*tan(e + f*x)*i - 1),x ))/(a*c**3)