Integrand size = 28, antiderivative size = 129 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {(c-i d)^2 x}{8 a^3}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}+\frac {(c+i d) (i c+3 d)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c-i d)^2}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
1/8*(c-I*d)^2*x/a^3+1/6*I*(c+I*d)^2/f/(a+I*a*tan(f*x+e))^3+1/8*(c+I*d)*(I* c+3*d)/a/f/(a+I*a*tan(f*x+e))^2+1/8*I*(c-I*d)^2/f/(a^3+I*a^3*tan(f*x+e))
Time = 2.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.38 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \left (\frac {3 \sec ^2(e+f x) \left (2 c^2+2 d^2+2 c^2 \cos (2 (e+f x))+i c^2 \sin (2 (e+f x))+2 c d \sin (2 (e+f x))+i d^2 \sin (2 (e+f x))+2 (c-i d)^2 \arctan (\tan (e+f x)) (-i \cos (2 (e+f x))+\sin (2 (e+f x)))\right )}{a^3 (-i+\tan (e+f x))^2}-\frac {8 (c+d \tan (e+f x))^3}{(c+i d) (a+i a \tan (e+f x))^3}\right )}{48 f} \]
((-1/48*I)*((3*Sec[e + f*x]^2*(2*c^2 + 2*d^2 + 2*c^2*Cos[2*(e + f*x)] + I* c^2*Sin[2*(e + f*x)] + 2*c*d*Sin[2*(e + f*x)] + I*d^2*Sin[2*(e + f*x)] + 2 *(c - I*d)^2*ArcTan[Tan[e + f*x]]*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x) ])))/(a^3*(-I + Tan[e + f*x])^2) - (8*(c + d*Tan[e + f*x])^3)/((c + I*d)*( a + I*a*Tan[e + f*x])^3)))/f
Time = 0.50 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4023, 3042, 4009, 3042, 3960, 24}
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.
| \(\displaystyle \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3}dx\) |
| \(\Big \downarrow \) 4023 |
| \(\displaystyle \frac {\int \frac {a \left (c^2-2 i d c+d^2\right )-2 i a d^2 \tan (e+f x)}{(i \tan (e+f x) a+a)^2}dx}{2 a^2}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a \left (c^2-2 i d c+d^2\right )-2 i a d^2 \tan (e+f x)}{(i \tan (e+f x) a+a)^2}dx}{2 a^2}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 4009 |
| \(\displaystyle \frac {\frac {1}{2} (c-i d)^2 \int \frac {1}{i \tan (e+f x) a+a}dx+\frac {a (c+i d) (3 d+i c)}{4 f (a+i a \tan (e+f x))^2}}{2 a^2}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} (c-i d)^2 \int \frac {1}{i \tan (e+f x) a+a}dx+\frac {a (c+i d) (3 d+i c)}{4 f (a+i a \tan (e+f x))^2}}{2 a^2}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3960 |
| \(\displaystyle \frac {\frac {1}{2} (c-i d)^2 \left (\frac {\int 1dx}{2 a}+\frac {i}{2 f (a+i a \tan (e+f x))}\right )+\frac {a (c+i d) (3 d+i c)}{4 f (a+i a \tan (e+f x))^2}}{2 a^2}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{2} (c-i d)^2 \left (\frac {x}{2 a}+\frac {i}{2 f (a+i a \tan (e+f x))}\right )+\frac {a (c+i d) (3 d+i c)}{4 f (a+i a \tan (e+f x))^2}}{2 a^2}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}\) |
((I/6)*(c + I*d)^2)/(f*(a + I*a*Tan[e + f*x])^3) + ((a*(c + I*d)*(I*c + 3* d))/(4*f*(a + I*a*Tan[e + f*x])^2) + ((c - I*d)^2*(x/(2*a) + (I/2)/(f*(a + I*a*Tan[e + f*x]))))/2)/(2*a^2)
3.11.76.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a) Int[(a + b*Tan[c + d*x])^ (n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a *f*m)), x] + Simp[(b*c + a*d)/(2*a*b) Int[(a + b*Tan[e + f*x])^(m + 1), x ], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 , 0] && LtQ[m, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^ m/(2*a^3*f*m)), x] + Simp[1/(2*a^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp [a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]
Time = 0.37 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.64
| method | result | size |
| risch | \(-\frac {i x c d}{4 a^{3}}+\frac {x \,c^{2}}{8 a^{3}}-\frac {x \,d^{2}}{8 a^{3}}+\frac {{\mathrm e}^{-2 i \left (f x +e \right )} c d}{8 a^{3} f}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} c^{2}}{16 a^{3} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} d^{2}}{16 a^{3} f}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} c d}{16 a^{3} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )} c^{2}}{32 a^{3} f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} d^{2}}{32 a^{3} f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} c d}{24 a^{3} f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} c^{2}}{48 a^{3} f}-\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} d^{2}}{48 a^{3} f}\) | \(212\) |
| derivativedivides | \(-\frac {i c d \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3}}+\frac {c^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c d}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {d^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {c d}{4 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {3 i d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i c d}{4 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {c^{2}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {d^{2}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {c^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}\) | \(255\) |
| default | \(-\frac {i c d \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3}}+\frac {c^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c d}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {d^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {c d}{4 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {3 i d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i c d}{4 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {c^{2}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {d^{2}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {c^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}\) | \(255\) |
-1/4*I*x/a^3*c*d+1/8*x/a^3*c^2-1/8*x/a^3*d^2+1/8/a^3/f*exp(-2*I*(f*x+e))*c *d+3/16*I/a^3/f*exp(-2*I*(f*x+e))*c^2+1/16*I/a^3/f*exp(-2*I*(f*x+e))*d^2-1 /16/a^3/f*exp(-4*I*(f*x+e))*c*d+3/32*I/a^3/f*exp(-4*I*(f*x+e))*c^2+1/32*I/ a^3/f*exp(-4*I*(f*x+e))*d^2-1/24/a^3/f*exp(-6*I*(f*x+e))*c*d+1/48*I/a^3/f* exp(-6*I*(f*x+e))*c^2-1/48*I/a^3/f*exp(-6*I*(f*x+e))*d^2
Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (12 \, {\left (c^{2} - 2 i \, c d - d^{2}\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 2 i \, c^{2} - 4 \, c d - 2 i \, d^{2} - 6 \, {\left (-3 i \, c^{2} - 2 \, c d - i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-3 i \, c^{2} + 2 \, c d - i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \]
1/96*(12*(c^2 - 2*I*c*d - d^2)*f*x*e^(6*I*f*x + 6*I*e) + 2*I*c^2 - 4*c*d - 2*I*d^2 - 6*(-3*I*c^2 - 2*c*d - I*d^2)*e^(4*I*f*x + 4*I*e) - 3*(-3*I*c^2 + 2*c*d - I*d^2)*e^(2*I*f*x + 2*I*e))*e^(-6*I*f*x - 6*I*e)/(a^3*f)
Time = 0.35 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.11 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (512 i a^{6} c^{2} f^{2} e^{6 i e} - 1024 a^{6} c d f^{2} e^{6 i e} - 512 i a^{6} d^{2} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (2304 i a^{6} c^{2} f^{2} e^{8 i e} - 1536 a^{6} c d f^{2} e^{8 i e} + 768 i a^{6} d^{2} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (4608 i a^{6} c^{2} f^{2} e^{10 i e} + 3072 a^{6} c d f^{2} e^{10 i e} + 1536 i a^{6} d^{2} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text {for}\: a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {c^{2} - 2 i c d - d^{2}}{8 a^{3}} + \frac {\left (c^{2} e^{6 i e} + 3 c^{2} e^{4 i e} + 3 c^{2} e^{2 i e} + c^{2} - 2 i c d e^{6 i e} - 2 i c d e^{4 i e} + 2 i c d e^{2 i e} + 2 i c d - d^{2} e^{6 i e} + d^{2} e^{4 i e} + d^{2} e^{2 i e} - d^{2}\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (c^{2} - 2 i c d - d^{2}\right )}{8 a^{3}} \]
Piecewise((((512*I*a**6*c**2*f**2*exp(6*I*e) - 1024*a**6*c*d*f**2*exp(6*I* e) - 512*I*a**6*d**2*f**2*exp(6*I*e))*exp(-6*I*f*x) + (2304*I*a**6*c**2*f* *2*exp(8*I*e) - 1536*a**6*c*d*f**2*exp(8*I*e) + 768*I*a**6*d**2*f**2*exp(8 *I*e))*exp(-4*I*f*x) + (4608*I*a**6*c**2*f**2*exp(10*I*e) + 3072*a**6*c*d* f**2*exp(10*I*e) + 1536*I*a**6*d**2*f**2*exp(10*I*e))*exp(-2*I*f*x))*exp(- 12*I*e)/(24576*a**9*f**3), Ne(a**9*f**3*exp(12*I*e), 0)), (x*(-(c**2 - 2*I *c*d - d**2)/(8*a**3) + (c**2*exp(6*I*e) + 3*c**2*exp(4*I*e) + 3*c**2*exp( 2*I*e) + c**2 - 2*I*c*d*exp(6*I*e) - 2*I*c*d*exp(4*I*e) + 2*I*c*d*exp(2*I* e) + 2*I*c*d - d**2*exp(6*I*e) + d**2*exp(4*I*e) + d**2*exp(2*I*e) - d**2) *exp(-6*I*e)/(8*a**3)), True)) + x*(c**2 - 2*I*c*d - d**2)/(8*a**3)
Exception generated. \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.68 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.56 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=-\frac {\frac {6 \, {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3}} + \frac {6 \, {\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3}} + \frac {-11 i \, c^{2} \tan \left (f x + e\right )^{3} - 22 \, c d \tan \left (f x + e\right )^{3} + 11 i \, d^{2} \tan \left (f x + e\right )^{3} - 45 \, c^{2} \tan \left (f x + e\right )^{2} + 90 i \, c d \tan \left (f x + e\right )^{2} + 45 \, d^{2} \tan \left (f x + e\right )^{2} + 69 i \, c^{2} \tan \left (f x + e\right ) + 138 \, c d \tan \left (f x + e\right ) - 21 i \, d^{2} \tan \left (f x + e\right ) + 51 \, c^{2} - 38 i \, c d - 3 \, d^{2}}{a^{3} {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \]
-1/96*(6*(-I*c^2 - 2*c*d + I*d^2)*log(tan(f*x + e) + I)/a^3 + 6*(I*c^2 + 2 *c*d - I*d^2)*log(tan(f*x + e) - I)/a^3 + (-11*I*c^2*tan(f*x + e)^3 - 22*c *d*tan(f*x + e)^3 + 11*I*d^2*tan(f*x + e)^3 - 45*c^2*tan(f*x + e)^2 + 90*I *c*d*tan(f*x + e)^2 + 45*d^2*tan(f*x + e)^2 + 69*I*c^2*tan(f*x + e) + 138* c*d*tan(f*x + e) - 21*I*d^2*tan(f*x + e) + 51*c^2 - 38*I*c*d - 3*d^2)/(a^3 *(tan(f*x + e) - I)^3))/f
Time = 6.00 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.15 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {c\,d}{6\,a^3}-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,c^2}{8\,a^3}+\frac {d^2}{8\,a^3}-\frac {c\,d\,3{}\mathrm {i}}{4\,a^3}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {c\,d}{4\,a^3}+\frac {c^2\,1{}\mathrm {i}}{8\,a^3}-\frac {d^2\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {c^2\,5{}\mathrm {i}}{12\,a^3}+\frac {d^2\,1{}\mathrm {i}}{12\,a^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {x\,{\left (d+c\,1{}\mathrm {i}\right )}^2}{8\,a^3} \]