Integrand size = 23, antiderivative size = 202 \[ \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i d^2 e^{-2 i e-2 i f x}}{8 a^2 f^3}-\frac {i d^2 e^{-4 i e-4 i f x}}{128 a^2 f^3}+\frac {d e^{-2 i e-2 i f x} (c+d x)}{4 a^2 f^2}+\frac {d e^{-4 i e-4 i f x} (c+d x)}{32 a^2 f^2}+\frac {i e^{-2 i e-2 i f x} (c+d x)^2}{4 a^2 f}+\frac {i e^{-4 i e-4 i f x} (c+d x)^2}{16 a^2 f}+\frac {(c+d x)^3}{12 a^2 d} \]
-1/8*I*d^2*exp(-2*I*e-2*I*f*x)/a^2/f^3-1/128*I*d^2*exp(-4*I*e-4*I*f*x)/a^2 /f^3+1/4*d*exp(-2*I*e-2*I*f*x)*(d*x+c)/a^2/f^2+1/32*d*exp(-4*I*e-4*I*f*x)* (d*x+c)/a^2/f^2+1/4*I*exp(-2*I*e-2*I*f*x)*(d*x+c)^2/a^2/f+1/16*I*exp(-4*I* e-4*I*f*x)*(d*x+c)^2/a^2/f+1/12*(d*x+c)^3/a^2/d
Time = 1.61 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.40 \[ \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left ((d+(1+i) c f+(1+i) d f x) ((1+i) c f+d (-i+(1+i) f x)) \cos (2 f x)+\frac {1}{16} (d+(2+2 i) c f+(2+2 i) d f x) ((2+2 i) c f+d (-i+(2+2 i) f x)) \cos (4 f x) (\cos (2 e)-i \sin (2 e))+\frac {2}{3} f^3 x \left (3 c^2+3 c d x+d^2 x^2\right ) (\cos (2 e)+i \sin (2 e))-i (d+(1+i) c f+(1+i) d f x) ((1+i) c f+d (-i+(1+i) f x)) \sin (2 f x)-\frac {1}{16} i (d+(2+2 i) c f+(2+2 i) d f x) ((2+2 i) c f+d (-i+(2+2 i) f x)) (\cos (2 e)-i \sin (2 e)) \sin (4 f x)\right )}{8 f^3 (a+i a \tan (e+f x))^2} \]
(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*((d + (1 + I)*c*f + (1 + I)*d*f* x)*((1 + I)*c*f + d*(-I + (1 + I)*f*x))*Cos[2*f*x] + ((d + (2 + 2*I)*c*f + (2 + 2*I)*d*f*x)*((2 + 2*I)*c*f + d*(-I + (2 + 2*I)*f*x))*Cos[4*f*x]*(Cos [2*e] - I*Sin[2*e]))/16 + (2*f^3*x*(3*c^2 + 3*c*d*x + d^2*x^2)*(Cos[2*e] + I*Sin[2*e]))/3 - I*(d + (1 + I)*c*f + (1 + I)*d*f*x)*((1 + I)*c*f + d*(-I + (1 + I)*f*x))*Sin[2*f*x] - (I/16)*(d + (2 + 2*I)*c*f + (2 + 2*I)*d*f*x) *((2 + 2*I)*c*f + d*(-I + (2 + 2*I)*f*x))*(Cos[2*e] - I*Sin[2*e])*Sin[4*f* x]))/(8*f^3*(a + I*a*Tan[e + f*x])^2)
Time = 0.44 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4212, 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.
\(\displaystyle \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 4212 |
\(\displaystyle \int \left (\frac {(c+d x)^2 e^{-2 i e-2 i f x}}{2 a^2}+\frac {(c+d x)^2 e^{-4 i e-4 i f x}}{4 a^2}+\frac {(c+d x)^2}{4 a^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d (c+d x) e^{-2 i e-2 i f x}}{4 a^2 f^2}+\frac {d (c+d x) e^{-4 i e-4 i f x}}{32 a^2 f^2}+\frac {i (c+d x)^2 e^{-2 i e-2 i f x}}{4 a^2 f}+\frac {i (c+d x)^2 e^{-4 i e-4 i f x}}{16 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}-\frac {i d^2 e^{-2 i e-2 i f x}}{8 a^2 f^3}-\frac {i d^2 e^{-4 i e-4 i f x}}{128 a^2 f^3}\) |
((-1/8*I)*d^2*E^((-2*I)*e - (2*I)*f*x))/(a^2*f^3) - ((I/128)*d^2*E^((-4*I) *e - (4*I)*f*x))/(a^2*f^3) + (d*E^((-2*I)*e - (2*I)*f*x)*(c + d*x))/(4*a^2 *f^2) + (d*E^((-4*I)*e - (4*I)*f*x)*(c + d*x))/(32*a^2*f^2) + ((I/4)*E^((- 2*I)*e - (2*I)*f*x)*(c + d*x)^2)/(a^2*f) + ((I/16)*E^((-4*I)*e - (4*I)*f*x )*(c + d*x)^2)/(a^2*f) + (c + d*x)^3/(12*a^2*d)
3.1.25.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (1/(2*a) + E^(2*(a/b)*(e + f* x))/(2*a))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 + b^2 , 0] && ILtQ[n, 0]
Time = 0.78 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {d^{2} x^{3}}{12 a^{2}}+\frac {d c \,x^{2}}{4 a^{2}}+\frac {c^{2} x}{4 a^{2}}+\frac {c^{3}}{12 a^{2} d}+\frac {i \left (2 d^{2} x^{2} f^{2}+4 c d \,f^{2} x -2 i d^{2} f x +2 c^{2} f^{2}-2 i c d f -d^{2}\right ) {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{2} f^{3}}+\frac {i \left (8 d^{2} x^{2} f^{2}+16 c d \,f^{2} x -4 i d^{2} f x +8 c^{2} f^{2}-4 i c d f -d^{2}\right ) {\mathrm e}^{-4 i \left (f x +e \right )}}{128 a^{2} f^{3}}\) | \(173\) |
1/12/a^2*d^2*x^3+1/4/a^2*d*c*x^2+1/4/a^2*c^2*x+1/12/a^2/d*c^3+1/8*I*(2*d^2 *x^2*f^2-2*I*d^2*f*x+4*c*d*f^2*x-2*I*c*d*f+2*c^2*f^2-d^2)/a^2/f^3*exp(-2*I *(f*x+e))+1/128*I*(8*d^2*x^2*f^2-4*I*d^2*f*x+16*c*d*f^2*x-4*I*c*d*f+8*c^2* f^2-d^2)/a^2/f^3*exp(-4*I*(f*x+e))
Time = 0.25 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.81 \[ \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (24 i \, d^{2} f^{2} x^{2} + 24 i \, c^{2} f^{2} + 12 \, c d f - 3 i \, d^{2} - 12 \, {\left (-4 i \, c d f^{2} - d^{2} f\right )} x + 32 \, {\left (d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2} + 3 \, c^{2} f^{3} x\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 48 \, {\left (-2 i \, d^{2} f^{2} x^{2} - 2 i \, c^{2} f^{2} - 2 \, c d f + i \, d^{2} + 2 \, {\left (-2 i \, c d f^{2} - d^{2} f\right )} x\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} f^{3}} \]
1/384*(24*I*d^2*f^2*x^2 + 24*I*c^2*f^2 + 12*c*d*f - 3*I*d^2 - 12*(-4*I*c*d *f^2 - d^2*f)*x + 32*(d^2*f^3*x^3 + 3*c*d*f^3*x^2 + 3*c^2*f^3*x)*e^(4*I*f* x + 4*I*e) - 48*(-2*I*d^2*f^2*x^2 - 2*I*c^2*f^2 - 2*c*d*f + I*d^2 + 2*(-2* I*c*d*f^2 - d^2*f)*x)*e^(2*I*f*x + 2*I*e))*e^(-4*I*f*x - 4*I*e)/(a^2*f^3)
Time = 0.33 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.07 \[ \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx=\begin {cases} \frac {\left (\left (64 i a^{2} c^{2} f^{5} e^{2 i e} + 128 i a^{2} c d f^{5} x e^{2 i e} + 32 a^{2} c d f^{4} e^{2 i e} + 64 i a^{2} d^{2} f^{5} x^{2} e^{2 i e} + 32 a^{2} d^{2} f^{4} x e^{2 i e} - 8 i a^{2} d^{2} f^{3} e^{2 i e}\right ) e^{- 4 i f x} + \left (256 i a^{2} c^{2} f^{5} e^{4 i e} + 512 i a^{2} c d f^{5} x e^{4 i e} + 256 a^{2} c d f^{4} e^{4 i e} + 256 i a^{2} d^{2} f^{5} x^{2} e^{4 i e} + 256 a^{2} d^{2} f^{4} x e^{4 i e} - 128 i a^{2} d^{2} f^{3} e^{4 i e}\right ) e^{- 2 i f x}\right ) e^{- 6 i e}}{1024 a^{4} f^{6}} & \text {for}\: a^{4} f^{6} e^{6 i e} \neq 0 \\\frac {x^{3} \cdot \left (2 d^{2} e^{2 i e} + d^{2}\right ) e^{- 4 i e}}{12 a^{2}} + \frac {x^{2} \cdot \left (2 c d e^{2 i e} + c d\right ) e^{- 4 i e}}{4 a^{2}} + \frac {x \left (2 c^{2} e^{2 i e} + c^{2}\right ) e^{- 4 i e}}{4 a^{2}} & \text {otherwise} \end {cases} + \frac {c^{2} x}{4 a^{2}} + \frac {c d x^{2}}{4 a^{2}} + \frac {d^{2} x^{3}}{12 a^{2}} \]
Piecewise((((64*I*a**2*c**2*f**5*exp(2*I*e) + 128*I*a**2*c*d*f**5*x*exp(2* I*e) + 32*a**2*c*d*f**4*exp(2*I*e) + 64*I*a**2*d**2*f**5*x**2*exp(2*I*e) + 32*a**2*d**2*f**4*x*exp(2*I*e) - 8*I*a**2*d**2*f**3*exp(2*I*e))*exp(-4*I* f*x) + (256*I*a**2*c**2*f**5*exp(4*I*e) + 512*I*a**2*c*d*f**5*x*exp(4*I*e) + 256*a**2*c*d*f**4*exp(4*I*e) + 256*I*a**2*d**2*f**5*x**2*exp(4*I*e) + 2 56*a**2*d**2*f**4*x*exp(4*I*e) - 128*I*a**2*d**2*f**3*exp(4*I*e))*exp(-2*I *f*x))*exp(-6*I*e)/(1024*a**4*f**6), Ne(a**4*f**6*exp(6*I*e), 0)), (x**3*( 2*d**2*exp(2*I*e) + d**2)*exp(-4*I*e)/(12*a**2) + x**2*(2*c*d*exp(2*I*e) + c*d)*exp(-4*I*e)/(4*a**2) + x*(2*c**2*exp(2*I*e) + c**2)*exp(-4*I*e)/(4*a **2), True)) + c**2*x/(4*a**2) + c*d*x**2/(4*a**2) + d**2*x**3/(12*a**2)
Exception generated. \[ \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.45 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (32 \, d^{2} f^{3} x^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 96 \, c d f^{3} x^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 96 \, c^{2} f^{3} x e^{\left (4 i \, f x + 4 i \, e\right )} + 96 i \, d^{2} f^{2} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, d^{2} f^{2} x^{2} + 192 i \, c d f^{2} x e^{\left (2 i \, f x + 2 i \, e\right )} + 48 i \, c d f^{2} x + 96 i \, c^{2} f^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 96 \, d^{2} f x e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, c^{2} f^{2} + 12 \, d^{2} f x + 96 \, c d f e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, c d f - 48 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, d^{2}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} f^{3}} \]
1/384*(32*d^2*f^3*x^3*e^(4*I*f*x + 4*I*e) + 96*c*d*f^3*x^2*e^(4*I*f*x + 4* I*e) + 96*c^2*f^3*x*e^(4*I*f*x + 4*I*e) + 96*I*d^2*f^2*x^2*e^(2*I*f*x + 2* I*e) + 24*I*d^2*f^2*x^2 + 192*I*c*d*f^2*x*e^(2*I*f*x + 2*I*e) + 48*I*c*d*f ^2*x + 96*I*c^2*f^2*e^(2*I*f*x + 2*I*e) + 96*d^2*f*x*e^(2*I*f*x + 2*I*e) + 24*I*c^2*f^2 + 12*d^2*f*x + 96*c*d*f*e^(2*I*f*x + 2*I*e) + 12*c*d*f - 48* I*d^2*e^(2*I*f*x + 2*I*e) - 3*I*d^2)*e^(-4*I*f*x - 4*I*e)/(a^2*f^3)
Time = 3.65 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.91 \[ \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^2\,x}{4\,a^2}-{\mathrm {e}}^{-e\,4{}\mathrm {i}-f\,x\,4{}\mathrm {i}}\,\left (\frac {\left (-8\,c^2\,f^2+c\,d\,f\,4{}\mathrm {i}+d^2\right )\,1{}\mathrm {i}}{128\,a^2\,f^3}-\frac {d^2\,x^2\,1{}\mathrm {i}}{16\,a^2\,f}+\frac {d\,x\,\left (-4\,c\,f+d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^2\,f^2}\right )-{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,\left (\frac {\left (-2\,c^2\,f^2+c\,d\,f\,2{}\mathrm {i}+d^2\right )\,1{}\mathrm {i}}{8\,a^2\,f^3}-\frac {d^2\,x^2\,1{}\mathrm {i}}{4\,a^2\,f}+\frac {d\,x\,\left (-2\,c\,f+d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a^2\,f^2}\right )+\frac {d^2\,x^3}{12\,a^2}+\frac {c\,d\,x^2}{4\,a^2} \]
(c^2*x)/(4*a^2) - exp(- e*4i - f*x*4i)*(((d^2 - 8*c^2*f^2 + c*d*f*4i)*1i)/ (128*a^2*f^3) - (d^2*x^2*1i)/(16*a^2*f) + (d*x*(d*1i - 4*c*f)*1i)/(32*a^2* f^2)) - exp(- e*2i - f*x*2i)*(((d^2 - 2*c^2*f^2 + c*d*f*2i)*1i)/(8*a^2*f^3 ) - (d^2*x^2*1i)/(4*a^2*f) + (d*x*(d*1i - 2*c*f)*1i)/(4*a^2*f^2)) + (d^2*x ^3)/(12*a^2) + (c*d*x^2)/(4*a^2)