Integrand size = 18, antiderivative size = 131 \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2}+\frac {2 i a b d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a b d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f} \] Output:
1/2*a^2*(d*x+c)^2/d-4*I*a*b*(d*x+c)*arctan(exp(I*(f*x+e)))/f+b^2*d*ln(cos( f*x+e))/f^2+2*I*a*b*d*polylog(2,-I*exp(I*(f*x+e)))/f^2-2*I*a*b*d*polylog(2 ,I*exp(I*(f*x+e)))/f^2+b^2*(d*x+c)*tan(f*x+e)/f
Time = 0.38 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.15 \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\frac {2 a^2 c f^2 x+a^2 d f^2 x^2+4 a b c f \coth ^{-1}(\sin (e+f x))-8 i a b d f x \arctan \left (e^{i (e+f x)}\right )+2 b^2 d \log (\cos (e+f x))+4 i a b d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )-4 i a b d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )+2 b^2 c f \tan (e+f x)+2 b^2 d f x \tan (e+f x)}{2 f^2} \] Input:
Integrate[(c + d*x)*(a + b*Sec[e + f*x])^2,x]
Output:
(2*a^2*c*f^2*x + a^2*d*f^2*x^2 + 4*a*b*c*f*ArcCoth[Sin[e + f*x]] - (8*I)*a *b*d*f*x*ArcTan[E^(I*(e + f*x))] + 2*b^2*d*Log[Cos[e + f*x]] + (4*I)*a*b*d *PolyLog[2, (-I)*E^(I*(e + f*x))] - (4*I)*a*b*d*PolyLog[2, I*E^(I*(e + f*x ))] + 2*b^2*c*f*Tan[e + f*x] + 2*b^2*d*f*x*Tan[e + f*x])/(2*f^2)
Time = 0.34 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 4678, 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.
| \(\displaystyle \int (c+d x) (a+b \sec (e+f x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d x) \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 4678 |
| \(\displaystyle \int \left (a^2 (c+d x)+2 a b (c+d x) \sec (e+f x)+b^2 (c+d x) \sec ^2(e+f x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {2 i a b d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a b d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2}\) |
Input:
Int[(c + d*x)*(a + b*Sec[e + f*x])^2,x]
Output:
(a^2*(c + d*x)^2)/(2*d) - ((4*I)*a*b*(c + d*x)*ArcTan[E^(I*(e + f*x))])/f + (b^2*d*Log[Cos[e + f*x]])/f^2 + ((2*I)*a*b*d*PolyLog[2, (-I)*E^(I*(e + f *x))])/f^2 - ((2*I)*a*b*d*PolyLog[2, I*E^(I*(e + f*x))])/f^2 + (b^2*(c + d *x)*Tan[e + f*x])/f
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Time = 0.10 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.45
| method | result | size |
| parts | \(a^{2} \left (\frac {1}{2} d \,x^{2}+c x \right )+\frac {b^{2} d \tan \left (f x +e \right ) x}{f}+\frac {b^{2} d \ln \left (\cos \left (f x +e \right )\right )}{f^{2}}+\frac {b^{2} c \tan \left (f x +e \right )}{f}+\frac {2 a b \left (\frac {d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {e d \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\right )}{f}\) | \(190\) |
| derivativedivides | \(\frac {a^{2} c \left (f x +e \right )-\frac {a^{2} d e \left (f x +e \right )}{f}+\frac {a^{2} d \left (f x +e \right )^{2}}{2 f}+2 a b c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {2 a b d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {2 a b d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+b^{2} c \tan \left (f x +e \right )-\frac {b^{2} d e \tan \left (f x +e \right )}{f}+\frac {b^{2} d \left (\left (f x +e \right ) \tan \left (f x +e \right )+\ln \left (\cos \left (f x +e \right )\right )\right )}{f}}{f}\) | \(232\) |
| default | \(\frac {a^{2} c \left (f x +e \right )-\frac {a^{2} d e \left (f x +e \right )}{f}+\frac {a^{2} d \left (f x +e \right )^{2}}{2 f}+2 a b c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {2 a b d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {2 a b d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+b^{2} c \tan \left (f x +e \right )-\frac {b^{2} d e \tan \left (f x +e \right )}{f}+\frac {b^{2} d \left (\left (f x +e \right ) \tan \left (f x +e \right )+\ln \left (\cos \left (f x +e \right )\right )\right )}{f}}{f}\) | \(232\) |
| risch | \(\frac {a^{2} d \,x^{2}}{2}+a^{2} c x +\frac {2 i b^{2} \left (d x +c \right )}{f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}+\frac {b^{2} d \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b^{2} d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {4 i b a c \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {4 i b a d e \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b a d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 b a d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 b a d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}+\frac {2 b a d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 i b a d \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 i b a d \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}\) | \(266\) |
Input:
int((d*x+c)*(a+b*sec(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
a^2*(1/2*d*x^2+c*x)+b^2/f*d*tan(f*x+e)*x+b^2*d*ln(cos(f*x+e))/f^2+b^2/f*c* tan(f*x+e)+2*a*b/f*(1/f*d*(-(f*x+e)*ln(1+I*exp(I*(f*x+e)))+(f*x+e)*ln(1-I* exp(I*(f*x+e)))+I*dilog(1+I*exp(I*(f*x+e)))-I*dilog(1-I*exp(I*(f*x+e))))+c *ln(sec(f*x+e)+tan(f*x+e))-e/f*d*ln(sec(f*x+e)+tan(f*x+e)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (111) = 222\).
Time = 0.11 (sec) , antiderivative size = 505, normalized size of antiderivative = 3.85 \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\frac {-2 i \, a b d \cos \left (f x + e\right ) {\rm Li}_2\left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) - 2 i \, a b d \cos \left (f x + e\right ) {\rm Li}_2\left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + 2 i \, a b d \cos \left (f x + e\right ) {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + 2 i \, a b d \cos \left (f x + e\right ) {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) - {\left (2 \, a b d e - 2 \, a b c f - b^{2} d\right )} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) + {\left (2 \, a b d e - 2 \, a b c f + b^{2} d\right )} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + 2 \, {\left (a b d f x + a b d e\right )} \cos \left (f x + e\right ) \log \left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (a b d f x + a b d e\right )} \cos \left (f x + e\right ) \log \left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (a b d f x + a b d e\right )} \cos \left (f x + e\right ) \log \left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (a b d f x + a b d e\right )} \cos \left (f x + e\right ) \log \left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) - {\left (2 \, a b d e - 2 \, a b c f - b^{2} d\right )} \cos \left (f x + e\right ) \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) + {\left (2 \, a b d e - 2 \, a b c f + b^{2} d\right )} \cos \left (f x + e\right ) \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + {\left (a^{2} d f^{2} x^{2} + 2 \, a^{2} c f^{2} x\right )} \cos \left (f x + e\right ) + 2 \, {\left (b^{2} d f x + b^{2} c f\right )} \sin \left (f x + e\right )}{2 \, f^{2} \cos \left (f x + e\right )} \] Input:
integrate((d*x+c)*(a+b*sec(f*x+e))^2,x, algorithm="fricas")
Output:
1/2*(-2*I*a*b*d*cos(f*x + e)*dilog(I*cos(f*x + e) + sin(f*x + e)) - 2*I*a* b*d*cos(f*x + e)*dilog(I*cos(f*x + e) - sin(f*x + e)) + 2*I*a*b*d*cos(f*x + e)*dilog(-I*cos(f*x + e) + sin(f*x + e)) + 2*I*a*b*d*cos(f*x + e)*dilog( -I*cos(f*x + e) - sin(f*x + e)) - (2*a*b*d*e - 2*a*b*c*f - b^2*d)*cos(f*x + e)*log(cos(f*x + e) + I*sin(f*x + e) + I) + (2*a*b*d*e - 2*a*b*c*f + b^2 *d)*cos(f*x + e)*log(cos(f*x + e) - I*sin(f*x + e) + I) + 2*(a*b*d*f*x + a *b*d*e)*cos(f*x + e)*log(I*cos(f*x + e) + sin(f*x + e) + 1) - 2*(a*b*d*f*x + a*b*d*e)*cos(f*x + e)*log(I*cos(f*x + e) - sin(f*x + e) + 1) + 2*(a*b*d *f*x + a*b*d*e)*cos(f*x + e)*log(-I*cos(f*x + e) + sin(f*x + e) + 1) - 2*( a*b*d*f*x + a*b*d*e)*cos(f*x + e)*log(-I*cos(f*x + e) - sin(f*x + e) + 1) - (2*a*b*d*e - 2*a*b*c*f - b^2*d)*cos(f*x + e)*log(-cos(f*x + e) + I*sin(f *x + e) + I) + (2*a*b*d*e - 2*a*b*c*f + b^2*d)*cos(f*x + e)*log(-cos(f*x + e) - I*sin(f*x + e) + I) + (a^2*d*f^2*x^2 + 2*a^2*c*f^2*x)*cos(f*x + e) + 2*(b^2*d*f*x + b^2*c*f)*sin(f*x + e))/(f^2*cos(f*x + e))
\[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\int \left (a + b \sec {\left (e + f x \right )}\right )^{2} \left (c + d x\right )\, dx \] Input:
integrate((d*x+c)*(a+b*sec(f*x+e))**2,x)
Output:
Integral((a + b*sec(e + f*x))**2*(c + d*x), x)
\[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\int { {\left (d x + c\right )} {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \,d x } \] Input:
integrate((d*x+c)*(a+b*sec(f*x+e))^2,x, algorithm="maxima")
Output:
1/2*(a^2*d*f^2*x^2 + 2*a^2*c*f^2*x + (a^2*d*f^2*x^2 + 2*a^2*c*f^2*x)*cos(2 *f*x + 2*e)^2 + (a^2*d*f^2*x^2 + 2*a^2*c*f^2*x)*sin(2*f*x + 2*e)^2 + 2*(a^ 2*d*f^2*x^2 + 2*a^2*c*f^2*x)*cos(2*f*x + 2*e) + 8*(a*b*d*f^3*cos(2*f*x + 2 *e)^2 + a*b*d*f^3*sin(2*f*x + 2*e)^2 + 2*a*b*d*f^3*cos(2*f*x + 2*e) + a*b* d*f^3)*integrate((x*cos(2*f*x + 2*e)*cos(f*x + e) + x*sin(2*f*x + 2*e)*sin (f*x + e) + x*cos(f*x + e))/(f*cos(2*f*x + 2*e)^2 + f*sin(2*f*x + 2*e)^2 + 2*f*cos(2*f*x + 2*e) + f), x) + (b^2*d*cos(2*f*x + 2*e)^2 + b^2*d*sin(2*f *x + 2*e)^2 + 2*b^2*d*cos(2*f*x + 2*e) + b^2*d)*log(cos(2*f*x + 2*e)^2 + s in(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1) + 2*(a*b*c*f*cos(2*f*x + 2*e)^ 2 + a*b*c*f*sin(2*f*x + 2*e)^2 + 2*a*b*c*f*cos(2*f*x + 2*e) + a*b*c*f)*log (cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - 2*(a*b*c*f*cos(2* f*x + 2*e)^2 + a*b*c*f*sin(2*f*x + 2*e)^2 + 2*a*b*c*f*cos(2*f*x + 2*e) + a *b*c*f)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) + 4*(b^2 *d*f*x + b^2*c*f)*sin(2*f*x + 2*e))/(f^2*cos(2*f*x + 2*e)^2 + f^2*sin(2*f* x + 2*e)^2 + 2*f^2*cos(2*f*x + 2*e) + f^2)
\[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\int { {\left (d x + c\right )} {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \,d x } \] Input:
integrate((d*x+c)*(a+b*sec(f*x+e))^2,x, algorithm="giac")
Output:
integrate((d*x + c)*(b*sec(f*x + e) + a)^2, x)
Timed out. \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\int {\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^2\,\left (c+d\,x\right ) \,d x \] Input:
int((a + b/cos(e + f*x))^2*(c + d*x),x)
Output:
int((a + b/cos(e + f*x))^2*(c + d*x), x)
\[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\frac {8 \cos \left (f x +e \right ) \left (\int \frac {x}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1}d x \right ) a b d \,f^{2}+2 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a b d -2 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) b^{2} d -4 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) a b c f -2 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) a b d +2 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) b^{2} d +4 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a b c f -2 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a b d +2 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) b^{2} d +2 \cos \left (f x +e \right ) a^{2} c \,f^{2} x +\cos \left (f x +e \right ) a^{2} d \,f^{2} x^{2}-\cos \left (f x +e \right ) a b d \,f^{2} x^{2}-2 \sin \left (f x +e \right ) a b d f x +2 \sin \left (f x +e \right ) b^{2} c f +2 \sin \left (f x +e \right ) b^{2} d f x}{2 \cos \left (f x +e \right ) f^{2}} \] Input:
int((d*x+c)*(a+b*sec(f*x+e))^2,x)
Output:
(8*cos(e + f*x)*int(x/(tan((e + f*x)/2)**4 - 2*tan((e + f*x)/2)**2 + 1),x) *a*b*d*f**2 + 2*cos(e + f*x)*log(tan((e + f*x)/2)**2 + 1)*a*b*d - 2*cos(e + f*x)*log(tan((e + f*x)/2)**2 + 1)*b**2*d - 4*cos(e + f*x)*log(tan((e + f *x)/2) - 1)*a*b*c*f - 2*cos(e + f*x)*log(tan((e + f*x)/2) - 1)*a*b*d + 2*c os(e + f*x)*log(tan((e + f*x)/2) - 1)*b**2*d + 4*cos(e + f*x)*log(tan((e + f*x)/2) + 1)*a*b*c*f - 2*cos(e + f*x)*log(tan((e + f*x)/2) + 1)*a*b*d + 2 *cos(e + f*x)*log(tan((e + f*x)/2) + 1)*b**2*d + 2*cos(e + f*x)*a**2*c*f** 2*x + cos(e + f*x)*a**2*d*f**2*x**2 - cos(e + f*x)*a*b*d*f**2*x**2 - 2*sin (e + f*x)*a*b*d*f*x + 2*sin(e + f*x)*b**2*c*f + 2*sin(e + f*x)*b**2*d*f*x) /(2*cos(e + f*x)*f**2)