Integrand size = 27, antiderivative size = 87 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x)) \, dx=-\frac {e^{i (d+e x)} F^{c (a+b x)} (f+i g)}{2 (e-i b c \log (F))}-\frac {e^{-i (d+e x)} F^{c (a+b x)} (f-i g)}{2 (e+i b c \log (F))} \] Output:
-1/2*exp(I*(e*x+d))*F^(c*(b*x+a))*(f+I*g)/(e-I*b*c*ln(F))-1/2*F^(c*(b*x+a) )*(f-I*g)/exp(I*(e*x+d))/(e+I*b*c*ln(F))
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x)) \, dx=\frac {F^{c (a+b x)} (\cos (d+e x) (-e f+b c g \log (F))+(e g+b c f \log (F)) \sin (d+e x))}{e^2+b^2 c^2 \log ^2(F)} \] Input:
Integrate[F^(c*(a + b*x))*(g*Cos[d + e*x] + f*Sin[d + e*x]),x]
Output:
(F^(c*(a + b*x))*(Cos[d + e*x]*(-(e*f) + b*c*g*Log[F]) + (e*g + b*c*f*Log[ F])*Sin[d + e*x]))/(e^2 + b^2*c^2*Log[F]^2)
Time = 0.44 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.75, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int F^{c (a+b x)} (f \sin (d+e x)+g \cos (d+e x)) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int F^{a c+b c x} (f \sin (d+e x)+g \cos (d+e x))dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (f \sin (d+e x) F^{a c+b c x}+g \cos (d+e x) F^{a c+b c x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b c f \log (F) \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}-\frac {e f \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {e g \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {b c g \log (F) \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}\) |
Input:
Int[F^(c*(a + b*x))*(g*Cos[d + e*x] + f*Sin[d + e*x]),x]
Output:
-((e*f*F^(a*c + b*c*x)*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)) + (b*c*F^(a *c + b*c*x)*g*Cos[d + e*x]*Log[F])/(e^2 + b^2*c^2*Log[F]^2) + (e*F^(a*c + b*c*x)*g*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2) + (b*c*f*F^(a*c + b*c*x)*L og[F]*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)
Time = 0.34 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83
method | result | size |
parallelrisch | \(\frac {\left (\cos \left (e x +d \right ) \ln \left (F \right ) b c g +\ln \left (F \right ) \sin \left (e x +d \right ) b c f -f e \cos \left (e x +d \right )+g e \sin \left (e x +d \right )\right ) F^{c \left (b x +a \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(72\) |
risch | \(\frac {\left (-i f +g \right ) F^{c \left (b x +a \right )} {\mathrm e}^{i \left (e x +d \right )}}{2 i e +2 b c \ln \left (F \right )}+\frac {\left (i f +g \right ) F^{c \left (b x +a \right )} {\mathrm e}^{-i \left (e x +d \right )}}{2 b c \ln \left (F \right )-2 i e}\) | \(78\) |
orering | \(\frac {2 b c \ln \left (F \right ) F^{c \left (b x +a \right )} \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {F^{c \left (b x +a \right )} b c \ln \left (F \right ) \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )+F^{c \left (b x +a \right )} \left (-g e \sin \left (e x +d \right )+f e \cos \left (e x +d \right )\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(132\) |
norman | \(\frac {\frac {\left (b c g \ln \left (F \right )-e f \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {\left (b c g \ln \left (F \right )-e f \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 \left (\ln \left (F \right ) b c f +e g \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}\) | \(156\) |
parts | \(\frac {\frac {e f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 b c \ln \left (F \right ) f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}+\frac {\frac {g b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 g e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {g b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}\) | \(269\) |
Input:
int(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d)),x,method=_RETURNVERBOSE)
Output:
1/(e^2+b^2*c^2*ln(F)^2)*(cos(e*x+d)*ln(F)*b*c*g+ln(F)*sin(e*x+d)*b*c*f-f*e *cos(e*x+d)+g*e*sin(e*x+d))*F^(c*(b*x+a))
Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.78 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x)) \, dx=\frac {{\left (b c g \cos \left (e x + d\right ) \log \left (F\right ) - e f \cos \left (e x + d\right ) + {\left (b c f \log \left (F\right ) + e g\right )} \sin \left (e x + d\right )\right )} F^{b c x + a c}}{b^{2} c^{2} \log \left (F\right )^{2} + e^{2}} \] Input:
integrate(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d)),x, algorithm="fricas")
Output:
(b*c*g*cos(e*x + d)*log(F) - e*f*cos(e*x + d) + (b*c*f*log(F) + e*g)*sin(e *x + d))*F^(b*c*x + a*c)/(b^2*c^2*log(F)^2 + e^2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 651 vs. \(2 (70) = 140\).
Time = 0.60 (sec) , antiderivative size = 651, normalized size of antiderivative = 7.48 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x)) \, dx =\text {Too large to display} \] Input:
integrate(F**(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d)),x)
Output:
Piecewise((x*(f*sin(d) + g*cos(d)), Eq(F, 1) & Eq(e, 0)), (F**(a*c)*x*(f*s in(d) + g*cos(d)), Eq(b, 0) & Eq(e, 0)), (x*(f*sin(d) + g*cos(d)), Eq(c, 0 ) & Eq(e, 0)), (-F**(a*c + b*c*x)*f*x*sin(I*b*c*x*log(F) - d)/2 + I*F**(a* c + b*c*x)*f*x*cos(I*b*c*x*log(F) - d)/2 + I*F**(a*c + b*c*x)*g*x*sin(I*b* c*x*log(F) - d)/2 + F**(a*c + b*c*x)*g*x*cos(I*b*c*x*log(F) - d)/2 - I*F** (a*c + b*c*x)*f*cos(I*b*c*x*log(F) - d)/(2*b*c*log(F)) - I*F**(a*c + b*c*x )*g*sin(I*b*c*x*log(F) - d)/(b*c*log(F)) - F**(a*c + b*c*x)*g*cos(I*b*c*x* log(F) - d)/(2*b*c*log(F)), Eq(e, -I*b*c*log(F))), (F**(a*c + b*c*x)*f*x*s in(I*b*c*x*log(F) + d)/2 - I*F**(a*c + b*c*x)*f*x*cos(I*b*c*x*log(F) + d)/ 2 + I*F**(a*c + b*c*x)*g*x*sin(I*b*c*x*log(F) + d)/2 + F**(a*c + b*c*x)*g* x*cos(I*b*c*x*log(F) + d)/2 + I*F**(a*c + b*c*x)*f*cos(I*b*c*x*log(F) + d) /(2*b*c*log(F)) - I*F**(a*c + b*c*x)*g*sin(I*b*c*x*log(F) + d)/(b*c*log(F) ) - F**(a*c + b*c*x)*g*cos(I*b*c*x*log(F) + d)/(2*b*c*log(F)), Eq(e, I*b*c *log(F))), (F**(a*c + b*c*x)*b*c*f*log(F)*sin(d + e*x)/(b**2*c**2*log(F)** 2 + e**2) + F**(a*c + b*c*x)*b*c*g*log(F)*cos(d + e*x)/(b**2*c**2*log(F)** 2 + e**2) - F**(a*c + b*c*x)*e*f*cos(d + e*x)/(b**2*c**2*log(F)**2 + e**2) + F**(a*c + b*c*x)*e*g*sin(d + e*x)/(b**2*c**2*log(F)**2 + e**2), True))
Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (71) = 142\).
Time = 0.06 (sec) , antiderivative size = 389, normalized size of antiderivative = 4.47 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x)) \, dx=-\frac {{\left ({\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) + F^{a c} e \cos \left (d\right )\right )} F^{b c x} \cos \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) - F^{a c} e \cos \left (d\right )\right )} F^{b c x} \cos \left (e x\right ) - {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) - F^{a c} e \sin \left (d\right )\right )} F^{b c x} \sin \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) + F^{a c} e \sin \left (d\right )\right )} F^{b c x} \sin \left (e x\right )\right )} f}{2 \, {\left (b^{2} c^{2} \cos \left (d\right )^{2} \log \left (F\right )^{2} + b^{2} c^{2} \log \left (F\right )^{2} \sin \left (d\right )^{2} + {\left (\cos \left (d\right )^{2} + \sin \left (d\right )^{2}\right )} e^{2}\right )}} + \frac {{\left ({\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) - F^{a c} e \sin \left (d\right )\right )} F^{b c x} \cos \left (e x + 2 \, d\right ) + {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) + F^{a c} e \sin \left (d\right )\right )} F^{b c x} \cos \left (e x\right ) + {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) + F^{a c} e \cos \left (d\right )\right )} F^{b c x} \sin \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) - F^{a c} e \cos \left (d\right )\right )} F^{b c x} \sin \left (e x\right )\right )} g}{2 \, {\left (b^{2} c^{2} \cos \left (d\right )^{2} \log \left (F\right )^{2} + b^{2} c^{2} \log \left (F\right )^{2} \sin \left (d\right )^{2} + {\left (\cos \left (d\right )^{2} + \sin \left (d\right )^{2}\right )} e^{2}\right )}} \] Input:
integrate(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d)),x, algorithm="maxima")
Output:
-1/2*((F^(a*c)*b*c*log(F)*sin(d) + F^(a*c)*e*cos(d))*F^(b*c*x)*cos(e*x + 2 *d) - (F^(a*c)*b*c*log(F)*sin(d) - F^(a*c)*e*cos(d))*F^(b*c*x)*cos(e*x) - (F^(a*c)*b*c*cos(d)*log(F) - F^(a*c)*e*sin(d))*F^(b*c*x)*sin(e*x + 2*d) - (F^(a*c)*b*c*cos(d)*log(F) + F^(a*c)*e*sin(d))*F^(b*c*x)*sin(e*x))*f/(b^2* c^2*cos(d)^2*log(F)^2 + b^2*c^2*log(F)^2*sin(d)^2 + (cos(d)^2 + sin(d)^2)* e^2) + 1/2*((F^(a*c)*b*c*cos(d)*log(F) - F^(a*c)*e*sin(d))*F^(b*c*x)*cos(e *x + 2*d) + (F^(a*c)*b*c*cos(d)*log(F) + F^(a*c)*e*sin(d))*F^(b*c*x)*cos(e *x) + (F^(a*c)*b*c*log(F)*sin(d) + F^(a*c)*e*cos(d))*F^(b*c*x)*sin(e*x + 2 *d) - (F^(a*c)*b*c*log(F)*sin(d) - F^(a*c)*e*cos(d))*F^(b*c*x)*sin(e*x))*g /(b^2*c^2*cos(d)^2*log(F)^2 + b^2*c^2*log(F)^2*sin(d)^2 + (cos(d)^2 + sin( d)^2)*e^2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 611 vs. \(2 (71) = 142\).
Time = 0.16 (sec) , antiderivative size = 611, normalized size of antiderivative = 7.02 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x)) \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d)),x, algorithm="giac")
Output:
-I*((f - I*g)*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 + I*e*x + I*d)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c* log(abs(F)) + 4*I*e) - (f - I*g)*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 - I*e*x - I*d)/(-2*I*pi*b*c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F)) - 4*I*e))*e^(b*c*x*log(abs(F)) + a*c*log( abs(F))) - ((-I*f - g)*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*p i*a*c*sgn(F) - 1/2*I*pi*a*c + I*e*x + I*d)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F)) + 4*I*e) + (-I*f - g)*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 - I*e*x - I*d)/(-2*I*pi*b *c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F)) - 4*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + I*((f + I*g)*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 - I*e*x - I*d)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F)) - 4*I*e) - (f + I*g)*e^(-1/2*I*pi*b*c*x*sg n(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c + I*e*x + I*d)/ (-2*I*pi*b*c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F)) + 4*I*e))*e^(b*c*x*lo g(abs(F)) + a*c*log(abs(F))) - ((I*f - g)*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 - I*e*x - I*d)/(2*I*pi*b*c* sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F)) - 4*I*e) + (I*f - g)*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 + I*e*x + I*d)/(-2*I*pi*b*c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F)) + 4*I*e))*...
Time = 15.48 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x)) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (e\,g\,\sin \left (d+e\,x\right )-e\,f\,\cos \left (d+e\,x\right )+b\,c\,g\,\cos \left (d+e\,x\right )\,\ln \left (F\right )+b\,c\,f\,\sin \left (d+e\,x\right )\,\ln \left (F\right )\right )}{b^2\,c^2\,{\ln \left (F\right )}^2+e^2} \] Input:
int(F^(c*(a + b*x))*(g*cos(d + e*x) + f*sin(d + e*x)),x)
Output:
(F^(a*c + b*c*x)*(e*g*sin(d + e*x) - e*f*cos(d + e*x) + b*c*g*cos(d + e*x) *log(F) + b*c*f*sin(d + e*x)*log(F)))/(e^2 + b^2*c^2*log(F)^2)
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x)) \, dx=\frac {f^{b c x +a c} \left (\cos \left (e x +d \right ) \mathrm {log}\left (f \right ) b c g -\cos \left (e x +d \right ) e f +\mathrm {log}\left (f \right ) \sin \left (e x +d \right ) b c f +\sin \left (e x +d \right ) e g \right )}{\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+e^{2}} \] Input:
int(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d)),x)
Output:
(f**(a*c + b*c*x)*(cos(d + e*x)*log(f)*b*c*g - cos(d + e*x)*e*f + log(f)*s in(d + e*x)*b*c*f + sin(d + e*x)*e*g))/(log(f)**2*b**2*c**2 + e**2)