Integrand size = 29, antiderivative size = 124 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^2 \, dx=\frac {F^{c (a+b x)} \left (f^2+g^2\right )}{2 b c \log (F)}+\frac {e^{-2 i (d+e x)} F^{c (a+b x)} (f-i g)^2}{4 (2 i e-b c \log (F))}-\frac {e^{2 i (d+e x)} F^{c (a+b x)} (f+i g)^2}{4 (2 i e+b c \log (F))} \] Output:
1/2*F^(c*(b*x+a))*(f^2+g^2)/b/c/ln(F)+1/4*F^(c*(b*x+a))*(f-I*g)^2/exp(2*I* (e*x+d))/(2*I*e-b*c*ln(F))-exp(2*I*(e*x+d))*F^(c*(b*x+a))*(f+I*g)^2/(8*I*e +4*b*c*ln(F))
Time = 1.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^2 \, dx=\frac {F^{c (a+b x)} \left (-b c \cos (2 (d+e x)) \log (F) \left (4 e f g+b c \left (f^2-g^2\right ) \log (F)\right )+\left (f^2+g^2\right ) \left (4 e^2+b^2 c^2 \log ^2(F)\right )+2 b c \log (F) \left (-e f^2+e g^2+b c f g \log (F)\right ) \sin (2 (d+e x))\right )}{8 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)} \] Input:
Integrate[F^(c*(a + b*x))*(g*Cos[d + e*x] + f*Sin[d + e*x])^2,x]
Output:
(F^(c*(a + b*x))*(-(b*c*Cos[2*(d + e*x)]*Log[F]*(4*e*f*g + b*c*(f^2 - g^2) *Log[F])) + (f^2 + g^2)*(4*e^2 + b^2*c^2*Log[F]^2) + 2*b*c*Log[F]*(-(e*f^2 ) + e*g^2 + b*c*f*g*Log[F])*Sin[2*(d + e*x)]))/(8*b*c*e^2*Log[F] + 2*b^3*c ^3*Log[F]^3)
Leaf count is larger than twice the leaf count of optimal. \(367\) vs. \(2(124)=248\).
Time = 0.73 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {7292, 7293, 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.
| \(\displaystyle \int F^{c (a+b x)} (f \sin (d+e x)+g \cos (d+e x))^2 \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int F^{a c+b c x} (f \sin (d+e x)+g \cos (d+e x))^2dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (f^2 \sin ^2(d+e x) F^{a c+b c x}+f g \sin (2 d+2 e x) F^{a c+b c x}+g^2 \cos ^2(d+e x) F^{a c+b c x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b c f^2 \log (F) \sin ^2(d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac {2 e f^2 \sin (d+e x) \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {b c f g \log (F) \sin (2 d+2 e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac {2 e f g \cos (2 d+2 e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {b c g^2 \log (F) \cos ^2(d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e g^2 \sin (d+e x) \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e^2 f^2 F^{a c+b c x}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )}+\frac {2 e^2 g^2 F^{a c+b c x}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )}\) |
Input:
Int[F^(c*(a + b*x))*(g*Cos[d + e*x] + f*Sin[d + e*x])^2,x]
Output:
(-2*e*f*F^(a*c + b*c*x)*g*Cos[2*d + 2*e*x])/(4*e^2 + b^2*c^2*Log[F]^2) + ( 2*e^2*f^2*F^(a*c + b*c*x))/(b*c*Log[F]*(4*e^2 + b^2*c^2*Log[F]^2)) + (2*e^ 2*F^(a*c + b*c*x)*g^2)/(b*c*Log[F]*(4*e^2 + b^2*c^2*Log[F]^2)) + (b*c*F^(a *c + b*c*x)*g^2*Cos[d + e*x]^2*Log[F])/(4*e^2 + b^2*c^2*Log[F]^2) - (2*e*f ^2*F^(a*c + b*c*x)*Cos[d + e*x]*Sin[d + e*x])/(4*e^2 + b^2*c^2*Log[F]^2) + (2*e*F^(a*c + b*c*x)*g^2*Cos[d + e*x]*Sin[d + e*x])/(4*e^2 + b^2*c^2*Log[ F]^2) + (b*c*f^2*F^(a*c + b*c*x)*Log[F]*Sin[d + e*x]^2)/(4*e^2 + b^2*c^2*L og[F]^2) + (b*c*f*F^(a*c + b*c*x)*g*Log[F]*Sin[2*d + 2*e*x])/(4*e^2 + b^2* c^2*Log[F]^2)
Time = 1.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09
| method | result | size |
| parallelrisch | \(-\frac {F^{c \left (b x +a \right )} \left (\ln \left (F \right ) \left (c b \left (f -g \right ) \left (f +g \right ) \ln \left (F \right )+4 f g e \right ) c b \cos \left (2 e x +2 d \right )-2 \ln \left (F \right ) b c \left (\ln \left (F \right ) b c f g -e \,f^{2}+e \,g^{2}\right ) \sin \left (2 e x +2 d \right )-\left (f^{2}+g^{2}\right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )\right )}{2 b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\) | \(135\) |
| risch | \(\frac {f^{2} F^{c \left (b x +a \right )}}{2 b c \ln \left (F \right )}+\frac {g^{2} F^{c \left (b x +a \right )}}{2 b c \ln \left (F \right )}-\frac {\left (2 i f g +f^{2}-g^{2}\right ) F^{c \left (b x +a \right )} {\mathrm e}^{2 i \left (e x +d \right )}}{4 \left (2 i e +b c \ln \left (F \right )\right )}-\frac {\left (-2 i f g +f^{2}-g^{2}\right ) F^{c \left (b x +a \right )} {\mathrm e}^{-2 i \left (e x +d \right )}}{4 \left (b c \ln \left (F \right )-2 i e \right )}\) | \(142\) |
| orering | \(\frac {\left (3 b^{2} c^{2} \ln \left (F \right )^{2}+4 e^{2}\right ) F^{c \left (b x +a \right )} \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )^{2}}{\left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}-\frac {3 \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )^{2}+2 F^{c \left (b x +a \right )} \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right ) \left (-g e \sin \left (e x +d \right )+f e \cos \left (e x +d \right )\right )\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )^{2}+4 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right ) \left (-g e \sin \left (e x +d \right )+f e \cos \left (e x +d \right )\right )+2 F^{c \left (b x +a \right )} \left (-g e \sin \left (e x +d \right )+f e \cos \left (e x +d \right )\right )^{2}+2 F^{c \left (b x +a \right )} \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right ) \left (-g \,e^{2} \cos \left (e x +d \right )-f \,e^{2} \sin \left (e x +d \right )\right )}{\left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}\) | \(389\) |
| norman | \(\frac {\frac {\left (b^{2} c^{2} \ln \left (F \right )^{2} g^{2}-2 \ln \left (F \right ) b c e f g +2 e^{2} f^{2}+2 e^{2} g^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}+\frac {\left (b^{2} c^{2} \ln \left (F \right )^{2} g^{2}-2 \ln \left (F \right ) b c e f g +2 e^{2} f^{2}+2 e^{2} g^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{\left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}+\frac {4 \left (\ln \left (F \right ) b c f g -e \,f^{2}+e \,g^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {4 \left (\ln \left (F \right ) b c f g -e \,f^{2}+e \,g^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 \left (2 \ln \left (F \right )^{2} b^{2} c^{2} f^{2}-b^{2} c^{2} \ln \left (F \right )^{2} g^{2}+6 \ln \left (F \right ) b c e f g +2 e^{2} f^{2}+2 e^{2} g^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{\left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\) | \(417\) |
| parts | \(\text {Expression too large to display}\) | \(829\) |
Input:
int(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d))^2,x,method=_RETURNVERBOSE)
Output:
-1/2*F^(c*(b*x+a))*(ln(F)*(c*b*(f-g)*(f+g)*ln(F)+4*f*g*e)*c*b*cos(2*e*x+2* d)-2*ln(F)*b*c*(ln(F)*b*c*f*g-e*f^2+e*g^2)*sin(2*e*x+2*d)-(f^2+g^2)*(4*e^2 +b^2*c^2*ln(F)^2))/b/c/ln(F)/(4*e^2+b^2*c^2*ln(F)^2)
Time = 0.08 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.45 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^2 \, dx=\frac {{\left (2 \, e^{2} f^{2} + 2 \, e^{2} g^{2} + {\left (b^{2} c^{2} f^{2} - {\left (b^{2} c^{2} f^{2} - b^{2} c^{2} g^{2}\right )} \cos \left (e x + d\right )^{2}\right )} \log \left (F\right )^{2} - 2 \, {\left (2 \, b c e f g \cos \left (e x + d\right )^{2} - b c e f g\right )} \log \left (F\right ) + 2 \, {\left (b^{2} c^{2} f g \cos \left (e x + d\right ) \log \left (F\right )^{2} - {\left (b c e f^{2} - b c e g^{2}\right )} \cos \left (e x + d\right ) \log \left (F\right )\right )} \sin \left (e x + d\right )\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + 4 \, b c e^{2} \log \left (F\right )} \] Input:
integrate(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d))^2,x, algorithm="fricas ")
Output:
(2*e^2*f^2 + 2*e^2*g^2 + (b^2*c^2*f^2 - (b^2*c^2*f^2 - b^2*c^2*g^2)*cos(e* x + d)^2)*log(F)^2 - 2*(2*b*c*e*f*g*cos(e*x + d)^2 - b*c*e*f*g)*log(F) + 2 *(b^2*c^2*f*g*cos(e*x + d)*log(F)^2 - (b*c*e*f^2 - b*c*e*g^2)*cos(e*x + d) *log(F))*sin(e*x + d))*F^(b*c*x + a*c)/(b^3*c^3*log(F)^3 + 4*b*c*e^2*log(F ))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2174 vs. \(2 (100) = 200\).
Time = 1.29 (sec) , antiderivative size = 2174, normalized size of antiderivative = 17.53 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^2 \, dx=\text {Too large to display} \] Input:
integrate(F**(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d))**2,x)
Output:
Piecewise((x*(f*sin(d) + g*cos(d))**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq (e, 0)), (f**2*x*sin(d + e*x)**2/2 + f**2*x*cos(d + e*x)**2/2 + g**2*x*sin (d + e*x)**2/2 + g**2*x*cos(d + e*x)**2/2 - f**2*sin(d + e*x)*cos(d + e*x) /(2*e) - f*g*cos(d + e*x)**2/e + g**2*sin(d + e*x)*cos(d + e*x)/(2*e), Eq( F, 1)), (F**(a*c)*(f**2*x*sin(d + e*x)**2/2 + f**2*x*cos(d + e*x)**2/2 + g **2*x*sin(d + e*x)**2/2 + g**2*x*cos(d + e*x)**2/2 - f**2*sin(d + e*x)*cos (d + e*x)/(2*e) - f*g*cos(d + e*x)**2/e + g**2*sin(d + e*x)*cos(d + e*x)/( 2*e)), Eq(b, 0)), (f**2*x*sin(d + e*x)**2/2 + f**2*x*cos(d + e*x)**2/2 + g **2*x*sin(d + e*x)**2/2 + g**2*x*cos(d + e*x)**2/2 - f**2*sin(d + e*x)*cos (d + e*x)/(2*e) - f*g*cos(d + e*x)**2/e + g**2*sin(d + e*x)*cos(d + e*x)/( 2*e), Eq(c, 0)), (F**(a*c + b*c*x)*f**2*x*sin(I*b*c*x*log(F)/2 - d)**2/4 - I*F**(a*c + b*c*x)*f**2*x*sin(I*b*c*x*log(F)/2 - d)*cos(I*b*c*x*log(F)/2 - d)/2 - F**(a*c + b*c*x)*f**2*x*cos(I*b*c*x*log(F)/2 - d)**2/4 - I*F**(a* c + b*c*x)*f*g*x*sin(I*b*c*x*log(F)/2 - d)**2/2 - F**(a*c + b*c*x)*f*g*x*s in(I*b*c*x*log(F)/2 - d)*cos(I*b*c*x*log(F)/2 - d) + I*F**(a*c + b*c*x)*f* g*x*cos(I*b*c*x*log(F)/2 - d)**2/2 - F**(a*c + b*c*x)*g**2*x*sin(I*b*c*x*l og(F)/2 - d)**2/4 + I*F**(a*c + b*c*x)*g**2*x*sin(I*b*c*x*log(F)/2 - d)*co s(I*b*c*x*log(F)/2 - d)/2 + F**(a*c + b*c*x)*g**2*x*cos(I*b*c*x*log(F)/2 - d)**2/4 + F**(a*c + b*c*x)*f**2*sin(I*b*c*x*log(F)/2 - d)**2/(b*c*log(F)) - I*F**(a*c + b*c*x)*f**2*sin(I*b*c*x*log(F)/2 - d)*cos(I*b*c*x*log(F)...
Leaf count of result is larger than twice the leaf count of optimal. 944 vs. \(2 (105) = 210\).
Time = 0.08 (sec) , antiderivative size = 944, normalized size of antiderivative = 7.61 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^2 \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d))^2,x, algorithm="maxima ")
Output:
-1/4*((F^(a*c)*b^2*c^2*cos(2*d)*log(F)^2 + 2*F^(a*c)*b*c*e*log(F)*sin(2*d) )*F^(b*c*x)*cos(2*e*x) + (F^(a*c)*b^2*c^2*cos(2*d)*log(F)^2 - 2*F^(a*c)*b* c*e*log(F)*sin(2*d))*F^(b*c*x)*cos(2*e*x + 4*d) - (F^(a*c)*b^2*c^2*log(F)^ 2*sin(2*d) - 2*F^(a*c)*b*c*e*cos(2*d)*log(F))*F^(b*c*x)*sin(2*e*x) + (F^(a *c)*b^2*c^2*log(F)^2*sin(2*d) + 2*F^(a*c)*b*c*e*cos(2*d)*log(F))*F^(b*c*x) *sin(2*e*x + 4*d) - 2*(F^(a*c)*b^2*c^2*cos(2*d)^2*log(F)^2 + F^(a*c)*b^2*c ^2*log(F)^2*sin(2*d)^2 + 4*(F^(a*c)*cos(2*d)^2 + F^(a*c)*sin(2*d)^2)*e^2)* F^(b*c*x))*f^2/(b^3*c^3*cos(2*d)^2*log(F)^3 + b^3*c^3*log(F)^3*sin(2*d)^2 + 4*(b*c*cos(2*d)^2*log(F) + b*c*log(F)*sin(2*d)^2)*e^2) + 1/2*((F^(a*c)*b *c*log(F)*sin(2*d) - 2*F^(a*c)*e*cos(2*d))*F^(b*c*x)*cos(2*e*x) - (F^(a*c) *b*c*log(F)*sin(2*d) + 2*F^(a*c)*e*cos(2*d))*F^(b*c*x)*cos(2*e*x + 4*d) + (F^(a*c)*b*c*cos(2*d)*log(F) + 2*F^(a*c)*e*sin(2*d))*F^(b*c*x)*sin(2*e*x) + (F^(a*c)*b*c*cos(2*d)*log(F) - 2*F^(a*c)*e*sin(2*d))*F^(b*c*x)*sin(2*e*x + 4*d))*f*g/(b^2*c^2*cos(2*d)^2*log(F)^2 + b^2*c^2*log(F)^2*sin(2*d)^2 + 4*(cos(2*d)^2 + sin(2*d)^2)*e^2) + 1/4*((F^(a*c)*b^2*c^2*cos(2*d)*log(F)^2 + 2*F^(a*c)*b*c*e*log(F)*sin(2*d))*F^(b*c*x)*cos(2*e*x) + (F^(a*c)*b^2*c^ 2*cos(2*d)*log(F)^2 - 2*F^(a*c)*b*c*e*log(F)*sin(2*d))*F^(b*c*x)*cos(2*e*x + 4*d) - (F^(a*c)*b^2*c^2*log(F)^2*sin(2*d) - 2*F^(a*c)*b*c*e*cos(2*d)*lo g(F))*F^(b*c*x)*sin(2*e*x) + (F^(a*c)*b^2*c^2*log(F)^2*sin(2*d) + 2*F^(a*c )*b*c*e*cos(2*d)*log(F))*F^(b*c*x)*sin(2*e*x + 4*d) + 2*(F^(a*c)*b^2*c^...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 986 vs. \(2 (105) = 210\).
Time = 0.24 (sec) , antiderivative size = 986, normalized size of antiderivative = 7.95 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^2 \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d))^2,x, algorithm="giac")
Output:
(2*(f^2 + g^2)*b*c*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sg n(F) + 1/2*pi*a*c)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2) - (pi*b*c*sgn(F) - pi*b*c)*(f^2 + g^2)*sin(-1/2*pi*b*c*x*sgn(F ) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)/(4*b^2*c^2*log(abs(F))^ 2 + (pi*b*c*sgn(F) - pi*b*c)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - ((f^2 - 2*I*f*g - g^2)*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I* pi*a*c*sgn(F) - 1/2*I*pi*a*c + 2*I*e*x + 2*I*d)/(4*I*pi*b*c*sgn(F) - 4*I*p i*b*c + 8*b*c*log(abs(F)) + 16*I*e) + (f^2 - 2*I*f*g - g^2)*e^(-1/2*I*pi*b *c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c - 2*I*e* x - 2*I*d)/(-4*I*pi*b*c*sgn(F) + 4*I*pi*b*c + 8*b*c*log(abs(F)) - 16*I*e)) *e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + I*((-I*f^2 - 2*f*g + I*g^2)*e^( 1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a* c + 2*I*e*x + 2*I*d)/(4*I*pi*b*c*sgn(F) - 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*I*e) - (-I*f^2 - 2*f*g + I*g^2)*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b *c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c - 2*I*e*x - 2*I*d)/(-4*I*pi*b*c* sgn(F) + 4*I*pi*b*c + 8*b*c*log(abs(F)) - 16*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - ((f^2 + 2*I*f*g - g^2)*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I *pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c - 2*I*e*x - 2*I*d)/(4*I*pi* b*c*sgn(F) - 4*I*pi*b*c + 8*b*c*log(abs(F)) - 16*I*e) + (f^2 + 2*I*f*g - g ^2)*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + ...
Time = 16.99 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.94 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^2 \, dx=\frac {F^{b\,c\,x}\,F^{a\,c}\,{\cos \left (d+e\,x\right )}^2\,\left (b^2\,c^2\,g^2\,{\ln \left (F\right )}^2-2\,b\,c\,e\,f\,g\,\ln \left (F\right )+2\,e^2\,f^2+2\,e^2\,g^2\right )}{b\,\left (b^2\,c^3\,{\ln \left (F\right )}^3+4\,c\,e^2\,\ln \left (F\right )\right )}-\frac {F^{b\,c\,x}\,F^{a\,c}\,\cos \left (d+e\,x\right )\,\sin \left (d+e\,x\right )\,\left (e\,\left (2\,f^2-2\,g^2\right )-2\,b\,c\,f\,g\,\ln \left (F\right )\right )}{b^2\,c^2\,{\ln \left (F\right )}^2+4\,e^2}+\frac {F^{b\,c\,x}\,F^{a\,c}\,{\sin \left (d+e\,x\right )}^2\,\left (b^2\,c^2\,f^2\,{\ln \left (F\right )}^2+2\,b\,c\,e\,f\,g\,\ln \left (F\right )+2\,e^2\,f^2+2\,e^2\,g^2\right )}{b\,c\,\ln \left (F\right )\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+4\,e^2\right )} \] Input:
int(F^(c*(a + b*x))*(g*cos(d + e*x) + f*sin(d + e*x))^2,x)
Output:
(F^(b*c*x)*F^(a*c)*cos(d + e*x)^2*(2*e^2*f^2 + 2*e^2*g^2 + b^2*c^2*g^2*log (F)^2 - 2*b*c*e*f*g*log(F)))/(b*(4*c*e^2*log(F) + b^2*c^3*log(F)^3)) - (F^ (b*c*x)*F^(a*c)*cos(d + e*x)*sin(d + e*x)*(e*(2*f^2 - 2*g^2) - 2*b*c*f*g*l og(F)))/(4*e^2 + b^2*c^2*log(F)^2) + (F^(b*c*x)*F^(a*c)*sin(d + e*x)^2*(2* e^2*f^2 + 2*e^2*g^2 + b^2*c^2*f^2*log(F)^2 + 2*b*c*e*f*g*log(F)))/(b*c*log (F)*(4*e^2 + b^2*c^2*log(F)^2))
Time = 0.16 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.04 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^2 \, dx=\frac {f^{b c x +a c} \left (\cos \left (e x +d \right )^{2} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} g^{2}-2 \cos \left (e x +d \right )^{2} \mathrm {log}\left (f \right ) b c e f g +2 \cos \left (e x +d \right )^{2} e^{2} f^{2}+2 \cos \left (e x +d \right )^{2} e^{2} g^{2}+2 \cos \left (e x +d \right ) \mathrm {log}\left (f \right )^{2} \sin \left (e x +d \right ) b^{2} c^{2} f g -2 \cos \left (e x +d \right ) \mathrm {log}\left (f \right ) \sin \left (e x +d \right ) b c e \,f^{2}+2 \cos \left (e x +d \right ) \mathrm {log}\left (f \right ) \sin \left (e x +d \right ) b c e \,g^{2}+\mathrm {log}\left (f \right )^{2} \sin \left (e x +d \right )^{2} b^{2} c^{2} f^{2}+2 \,\mathrm {log}\left (f \right ) \sin \left (e x +d \right )^{2} b c e f g +2 \sin \left (e x +d \right )^{2} e^{2} f^{2}+2 \sin \left (e x +d \right )^{2} e^{2} g^{2}\right )}{\mathrm {log}\left (f \right ) b c \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+4 e^{2}\right )} \] Input:
int(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d))^2,x)
Output:
(f**(a*c + b*c*x)*(cos(d + e*x)**2*log(f)**2*b**2*c**2*g**2 - 2*cos(d + e* x)**2*log(f)*b*c*e*f*g + 2*cos(d + e*x)**2*e**2*f**2 + 2*cos(d + e*x)**2*e **2*g**2 + 2*cos(d + e*x)*log(f)**2*sin(d + e*x)*b**2*c**2*f*g - 2*cos(d + e*x)*log(f)*sin(d + e*x)*b*c*e*f**2 + 2*cos(d + e*x)*log(f)*sin(d + e*x)* b*c*e*g**2 + log(f)**2*sin(d + e*x)**2*b**2*c**2*f**2 + 2*log(f)*sin(d + e *x)**2*b*c*e*f*g + 2*sin(d + e*x)**2*e**2*f**2 + 2*sin(d + e*x)**2*e**2*g* *2))/(log(f)*b*c*(log(f)**2*b**2*c**2 + 4*e**2))