Integrand size = 42, antiderivative size = 23 \[ \int f F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(1+m+b c x \log (F)) \sin (d+e x)) \, dx=f F^{c (a+b x)} x (f x)^m \sin (d+e x) \]
Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int f F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(1+m+b c x \log (F)) \sin (d+e x)) \, dx=f F^{a c+b c x} x (f x)^m \sin (d+e x) \]
Integrate[f*F^(c*(a + b*x))*(f*x)^m*(e*x*Cos[d + e*x] + (1 + m + b*c*x*Log [F])*Sin[d + e*x]),x]
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.
\(\displaystyle \int f (f x)^m F^{c (a+b x)} (\sin (d+e x) (b c x \log (F)+m+1)+e x \cos (d+e x)) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle f \int F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)+1) \sin (d+e x))dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle f \int F^{a c+b x c} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)+1) \sin (d+e x))dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle f \int \left (F^{a c+b x c} (m+b c x \log (F)+1) \sin (d+e x) (f x)^m+\frac {e F^{a c+b x c} \cos (d+e x) (f x)^{m+1}}{f}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle f \left ((m+1) \int F^{a c+b x c} (f x)^m \sin (d+e x)dx+\frac {e \int F^{a c+b x c} (f x)^{m+1} \cos (d+e x)dx}{f}+\frac {b c \log (F) \int F^{a c+b x c} (f x)^{m+1} \sin (d+e x)dx}{f}\right )\) |
3.1.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 5.87 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(f \,F^{c \left (x b +a \right )} x \left (f x \right )^{m} \sin \left (e x +d \right )\) | \(24\) |
risch | \(-\frac {i x^{m} f^{m} F^{c \left (x b +a \right )} x f \left ({\mathrm e}^{i e x} {\mathrm e}^{i d} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i f x \right )^{3} m}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right ) m}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right ) m}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right ) m}{2}}-{\mathrm e}^{-i e x} {\mathrm e}^{-i d} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i f x \right )^{3} m}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right ) m}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right ) m}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right ) m}{2}}\right )}{2}\) | \(195\) |
int(f*F^(c*(b*x+a))*(f*x)^m*(e*x*cos(e*x+d)+(1+m+b*c*x*ln(F))*sin(e*x+d)), x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int f F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(1+m+b c x \log (F)) \sin (d+e x)) \, dx=\left (f x\right )^{m} F^{b c x + a c} f x \sin \left (e x + d\right ) \]
integrate(f*F^(c*(b*x+a))*(f*x)^m*(e*x*cos(e*x+d)+(1+m+b*c*x*log(F))*sin(e *x+d)),x, algorithm="fricas")
\[ \int f F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(1+m+b c x \log (F)) \sin (d+e x)) \, dx=f \left (\int F^{a c + b c x} \left (f x\right )^{m} \sin {\left (d + e x \right )}\, dx + \int F^{a c + b c x} m \left (f x\right )^{m} \sin {\left (d + e x \right )}\, dx + \int F^{a c + b c x} e x \left (f x\right )^{m} \cos {\left (d + e x \right )}\, dx + \int F^{a c + b c x} b c x \left (f x\right )^{m} \log {\left (F \right )} \sin {\left (d + e x \right )}\, dx\right ) \]
f*(Integral(F**(a*c + b*c*x)*(f*x)**m*sin(d + e*x), x) + Integral(F**(a*c + b*c*x)*m*(f*x)**m*sin(d + e*x), x) + Integral(F**(a*c + b*c*x)*e*x*(f*x) **m*cos(d + e*x), x) + Integral(F**(a*c + b*c*x)*b*c*x*(f*x)**m*log(F)*sin (d + e*x), x))
Time = 0.48 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int f F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(1+m+b c x \log (F)) \sin (d+e x)) \, dx=F^{a c} f^{m + 1} x e^{\left (b c x \log \left (F\right ) + m \log \left (x\right )\right )} \sin \left (e x + d\right ) \]
integrate(f*F^(c*(b*x+a))*(f*x)^m*(e*x*cos(e*x+d)+(1+m+b*c*x*log(F))*sin(e *x+d)),x, algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 4746 vs. \(2 (23) = 46\).
Time = 0.53 (sec) , antiderivative size = 4746, normalized size of antiderivative = 206.35 \[ \int f F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(1+m+b c x \log (F)) \sin (d+e x)) \, dx=\text {Too large to display} \]
integrate(f*F^(c*(b*x+a))*(f*x)^m*(e*x*cos(e*x+d)+(1+m+b*c*x*log(F))*sin(e *x+d)),x, algorithm="giac")
(x*abs(F)^(a*c)*e^(b*c*x*log(abs(F)) + m*log(abs(f)*abs(x)))*tan(1/4*pi*b* c*x*sgn(F) - 1/4*pi*b*c*x + pi*m*floor(-1/4*sgn(f) - 1/4*sgn(x) + 1) + 1/4 *pi*m*sgn(f) + 1/4*pi*m*sgn(x) - 1/2*pi*m + 1/2*e*x)^2*tan(1/4*pi*b*c*x*sg n(F) - 1/4*pi*b*c*x + pi*m*floor(-1/4*sgn(f) - 1/4*sgn(x) + 1) + 1/4*pi*m* sgn(f) + 1/4*pi*m*sgn(x) - 1/2*pi*m - 1/2*e*x)^2*tan(1/4*pi*a*c*sgn(F) - 1 /4*pi*a*c + 1/2*d)^2*tan(1/4*pi*a*c*sgn(F) - 1/4*pi*a*c - 1/2*d) - x*abs(F )^(a*c)*e^(b*c*x*log(abs(F)) + m*log(abs(f)*abs(x)))*tan(1/4*pi*b*c*x*sgn( F) - 1/4*pi*b*c*x + pi*m*floor(-1/4*sgn(f) - 1/4*sgn(x) + 1) + 1/4*pi*m*sg n(f) + 1/4*pi*m*sgn(x) - 1/2*pi*m + 1/2*e*x)^2*tan(1/4*pi*b*c*x*sgn(F) - 1 /4*pi*b*c*x + pi*m*floor(-1/4*sgn(f) - 1/4*sgn(x) + 1) + 1/4*pi*m*sgn(f) + 1/4*pi*m*sgn(x) - 1/2*pi*m - 1/2*e*x)^2*tan(1/4*pi*a*c*sgn(F) - 1/4*pi*a* c + 1/2*d)*tan(1/4*pi*a*c*sgn(F) - 1/4*pi*a*c - 1/2*d)^2 + x*abs(F)^(a*c)* e^(b*c*x*log(abs(F)) + m*log(abs(f)*abs(x)))*tan(1/4*pi*b*c*x*sgn(F) - 1/4 *pi*b*c*x + pi*m*floor(-1/4*sgn(f) - 1/4*sgn(x) + 1) + 1/4*pi*m*sgn(f) + 1 /4*pi*m*sgn(x) - 1/2*pi*m + 1/2*e*x)^2*tan(1/4*pi*b*c*x*sgn(F) - 1/4*pi*b* c*x + pi*m*floor(-1/4*sgn(f) - 1/4*sgn(x) + 1) + 1/4*pi*m*sgn(f) + 1/4*pi* m*sgn(x) - 1/2*pi*m - 1/2*e*x)*tan(1/4*pi*a*c*sgn(F) - 1/4*pi*a*c + 1/2*d) ^2*tan(1/4*pi*a*c*sgn(F) - 1/4*pi*a*c - 1/2*d)^2 - x*abs(F)^(a*c)*e^(b*c*x *log(abs(F)) + m*log(abs(f)*abs(x)))*tan(1/4*pi*b*c*x*sgn(F) - 1/4*pi*b*c* x + pi*m*floor(-1/4*sgn(f) - 1/4*sgn(x) + 1) + 1/4*pi*m*sgn(f) + 1/4*pi...
Time = 28.99 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int f F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(1+m+b c x \log (F)) \sin (d+e x)) \, dx=F^{c\,\left (a+b\,x\right )}\,f\,x\,\sin \left (d+e\,x\right )\,{\left (f\,x\right )}^m \]