Integrand size = 35, antiderivative size = 17 \[ \int F^{c (a+b x)} (e x \cos (d+e x)+(1+b c x \log (F)) \sin (d+e x)) \, dx=F^{c (a+b x)} x \sin (d+e x) \] Output:
F^(c*(b*x+a))*x*sin(e*x+d)
Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int F^{c (a+b x)} (e x \cos (d+e x)+(1+b c x \log (F)) \sin (d+e x)) \, dx=F^{a c+b c x} x \sin (d+e x) \] Input:
Integrate[F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (1 + b*c*x*Log[F])*Sin[d + e *x]),x]
Output:
F^(a*c + b*c*x)*x*Sin[d + e*x]
Leaf count is larger than twice the leaf count of optimal. \(327\) vs. \(2(17)=34\).
Time = 0.99 (sec) , antiderivative size = 327, normalized size of antiderivative = 19.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, 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.
\(\displaystyle \int F^{c (a+b x)} (\sin (d+e x) (b c x \log (F)+1)+e x \cos (d+e x)) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int F^{a c+b c x} (\sin (d+e x) (b c x \log (F)+1)+e x \cos (d+e x))dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e x \cos (d+e x) F^{a c+b c x}+\sin (d+e x) F^{a c+b c x} (b c x \log (F)+1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2 c^2 x \log ^2(F) \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {e^2 x \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {b c \log (F) \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}-\frac {b c e^2 \log (F) \sin (d+e x) F^{a c+b c x}}{\left (b^2 c^2 \log ^2(F)+e^2\right )^2}-\frac {e \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {b^2 c^2 e \log ^2(F) \cos (d+e x) F^{a c+b c x}}{\left (b^2 c^2 \log ^2(F)+e^2\right )^2}+\frac {e^3 \cos (d+e x) F^{a c+b c x}}{\left (b^2 c^2 \log ^2(F)+e^2\right )^2}-\frac {b^3 c^3 \log ^3(F) \sin (d+e x) F^{a c+b c x}}{\left (b^2 c^2 \log ^2(F)+e^2\right )^2}\) |
Input:
Int[F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (1 + b*c*x*Log[F])*Sin[d + e*x]),x ]
Output:
(e^3*F^(a*c + b*c*x)*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)^2 + (b^2*c^2*e *F^(a*c + b*c*x)*Cos[d + e*x]*Log[F]^2)/(e^2 + b^2*c^2*Log[F]^2)^2 - (e*F^ (a*c + b*c*x)*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2) - (b*c*e^2*F^(a*c + b *c*x)*Log[F]*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)^2 - (b^3*c^3*F^(a*c + b*c*x)*Log[F]^3*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)^2 + (e^2*F^(a*c + b *c*x)*x*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2) + (b*c*F^(a*c + b*c*x)*Log[ F]*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2) + (b^2*c^2*F^(a*c + b*c*x)*x*Log [F]^2*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)
Time = 1.90 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06
method | result | size |
risch | \(F^{c \left (b x +a \right )} x \sin \left (e x +d \right )\) | \(18\) |
parallelrisch | \(F^{c \left (b x +a \right )} x \sin \left (e x +d \right )\) | \(18\) |
norman | \(\frac {2 x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\) | \(38\) |
parts | \(\frac {\frac {e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e \,{\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 ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}+\frac {b c \ln \left (F \right ) \left (\frac {e x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e x \,{\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 \left (b^{2} c^{2} \ln \left (F \right )^{2}-e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )^{2}}+\frac {2 e b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )^{2}}+\frac {2 b c \ln \left (F \right ) x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {2 e b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )^{2}}\right )}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}+\frac {\frac {e \left (b^{2} c^{2} \ln \left (F \right )^{2}-e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )^{2}}+\frac {e b c \ln \left (F \right ) x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e \left (b^{2} c^{2} \ln \left (F \right )^{2}-e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )^{2}}+\frac {2 e^{2} x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {4 e^{2} b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )^{2}}-\frac {e b c \ln \left (F \right ) x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\) | \(689\) |
Input:
int(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(1+b*c*x*ln(F))*sin(e*x+d)),x,method=_RE TURNVERBOSE)
Output:
F^(c*(b*x+a))*x*sin(e*x+d)
Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int F^{c (a+b x)} (e x \cos (d+e x)+(1+b c x \log (F)) \sin (d+e x)) \, dx=F^{b c x + a c} x \sin \left (e x + d\right ) \] Input:
integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(1+b*c*x*log(F))*sin(e*x+d)),x, al gorithm="fricas")
Output:
F^(b*c*x + a*c)*x*sin(e*x + d)
Time = 0.36 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} (e x \cos (d+e x)+(1+b c x \log (F)) \sin (d+e x)) \, dx=F^{a c + b c x} x \sin {\left (d + e x \right )} \] Input:
integrate(F**(c*(b*x+a))*(e*x*cos(e*x+d)+(1+b*c*x*ln(F))*sin(e*x+d)),x)
Output:
F**(a*c + b*c*x)*x*sin(d + e*x)
Leaf count of result is larger than twice the leaf count of optimal. 1382 vs. \(2 (17) = 34\).
Time = 0.25 (sec) , antiderivative size = 1382, normalized size of antiderivative = 81.29 \[ \int F^{c (a+b x)} (e x \cos (d+e x)+(1+b c x \log (F)) \sin (d+e x)) \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(1+b*c*x*log(F))*sin(e*x+d)),x, al gorithm="maxima")
Output:
1/2*((F^(a*c)*b^2*c^2*log(F)^2*sin(d) + 2*F^(a*c)*b*c*e*cos(d)*log(F) - F^ (a*c)*e^2*sin(d) - (F^(a*c)*b^3*c^3*log(F)^3*sin(d) + F^(a*c)*b^2*c^2*e*co s(d)*log(F)^2 + F^(a*c)*b*c*e^2*log(F)*sin(d) + F^(a*c)*e^3*cos(d))*x)*F^( b*c*x)*cos(e*x + 2*d) - (F^(a*c)*b^2*c^2*log(F)^2*sin(d) - 2*F^(a*c)*b*c*e *cos(d)*log(F) - F^(a*c)*e^2*sin(d) - (F^(a*c)*b^3*c^3*log(F)^3*sin(d) - F ^(a*c)*b^2*c^2*e*cos(d)*log(F)^2 + F^(a*c)*b*c*e^2*log(F)*sin(d) - F^(a*c) *e^3*cos(d))*x)*F^(b*c*x)*cos(e*x) - (F^(a*c)*b^2*c^2*cos(d)*log(F)^2 - 2* F^(a*c)*b*c*e*log(F)*sin(d) - F^(a*c)*e^2*cos(d) - (F^(a*c)*b^3*c^3*cos(d) *log(F)^3 - F^(a*c)*b^2*c^2*e*log(F)^2*sin(d) + F^(a*c)*b*c*e^2*cos(d)*log (F) - F^(a*c)*e^3*sin(d))*x)*F^(b*c*x)*sin(e*x + 2*d) - (F^(a*c)*b^2*c^2*c os(d)*log(F)^2 + 2*F^(a*c)*b*c*e*log(F)*sin(d) - F^(a*c)*e^2*cos(d) - (F^( a*c)*b^3*c^3*cos(d)*log(F)^3 + F^(a*c)*b^2*c^2*e*log(F)^2*sin(d) + F^(a*c) *b*c*e^2*cos(d)*log(F) + F^(a*c)*e^3*sin(d))*x)*F^(b*c*x)*sin(e*x))*b*c*lo g(F)/(b^4*c^4*cos(d)^2*log(F)^4 + b^4*c^4*log(F)^4*sin(d)^2 + (cos(d)^2 + sin(d)^2)*e^4 + 2*(b^2*c^2*cos(d)^2*log(F)^2 + b^2*c^2*log(F)^2*sin(d)^2)* e^2) - 1/2*((F^(a*c)*b^2*c^2*cos(d)*log(F)^2 - 2*F^(a*c)*b*c*e*log(F)*sin( d) - F^(a*c)*e^2*cos(d) - (F^(a*c)*b^3*c^3*cos(d)*log(F)^3 - F^(a*c)*b^2*c ^2*e*log(F)^2*sin(d) + F^(a*c)*b*c*e^2*cos(d)*log(F) - F^(a*c)*e^3*sin(d)) *x)*F^(b*c*x)*cos(e*x + 2*d) + (F^(a*c)*b^2*c^2*cos(d)*log(F)^2 + 2*F^(a*c )*b*c*e*log(F)*sin(d) - F^(a*c)*e^2*cos(d) - (F^(a*c)*b^3*c^3*cos(d)*lo...
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 1941, normalized size of antiderivative = 114.18 \[ \int F^{c (a+b x)} (e x \cos (d+e x)+(1+b c x \log (F)) \sin (d+e x)) \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(1+b*c*x*log(F))*sin(e*x+d)),x, al gorithm="giac")
Output:
-1/4*((pi*b^2*c^2*x*log(F)*sgn(F) - pi*b^2*c^2*x*log(F) - 2*I*b^2*c^2*x*lo g(F)*log(abs(F)) - I*pi*b*c*e*x*sgn(F) + I*pi*b*c*e*x + 2*b*c*e*x*log(F) - 2*b*c*e*x*log(abs(F)) + pi*b*c*sgn(F) - pi*b*c - 2*I*e^2*x + 2*I*b*c*log( F) - 2*I*b*c*log(abs(F)) + 4*e)*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)/(pi^2*b^2*c^2*sgn(F) + 2*I*pi*b^2*c^2*log(abs(F))*sgn(F) - pi^2*b^2*c^2 - 2*I*pi*b^2*c^2*log(abs (F)) + 2*b^2*c^2*log(abs(F))^2 - 2*pi*b*c*e*sgn(F) + 2*pi*b*c*e + 4*I*b*c* e*log(abs(F)) - 2*e^2) - (pi*b^2*c^2*x*log(F)*sgn(F) - pi*b^2*c^2*x*log(F) + 2*I*b^2*c^2*x*log(F)*log(abs(F)) - I*pi*b*c*e*x*sgn(F) + I*pi*b*c*e*x + 2*b*c*e*x*log(F) + 2*b*c*e*x*log(abs(F)) + pi*b*c*sgn(F) - pi*b*c - 2*I*e ^2*x - 2*I*b*c*log(F) + 2*I*b*c*log(abs(F)))*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)/(pi^2*b^ 2*c^2*sgn(F) - 2*I*pi*b^2*c^2*log(abs(F))*sgn(F) - pi^2*b^2*c^2 + 2*I*pi*b ^2*c^2*log(abs(F)) + 2*b^2*c^2*log(abs(F))^2 - 2*pi*b*c*e*sgn(F) + 2*pi*b* c*e - 4*I*b*c*e*log(abs(F)) - 2*e^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F ))) + 1/4*I*((-I*pi*b^2*c^2*x*log(F)*sgn(F) + I*pi*b^2*c^2*x*log(F) - 2*b^ 2*c^2*x*log(F)*log(abs(F)) - pi*b*c*e*x*sgn(F) + pi*b*c*e*x - 2*I*b*c*e*x* log(F) + 2*I*b*c*e*x*log(abs(F)) - I*pi*b*c*sgn(F) + I*pi*b*c - 2*e^2*x + 2*b*c*log(F) - 2*b*c*log(abs(F)) - 4*I*e)*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)/(pi^2*b^2...
Time = 21.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} (e x \cos (d+e x)+(1+b c x \log (F)) \sin (d+e x)) \, dx=F^{c\,\left (a+b\,x\right )}\,x\,\sin \left (d+e\,x\right ) \] Input:
int(F^(c*(a + b*x))*(sin(d + e*x)*(b*c*x*log(F) + 1) + e*x*cos(d + e*x)),x )
Output:
F^(c*(a + b*x))*x*sin(d + e*x)
Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int F^{c (a+b x)} (e x \cos (d+e x)+(1+b c x \log (F)) \sin (d+e x)) \, dx=f^{b c x +a c} \sin \left (e x +d \right ) x \] Input:
int(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(1+b*c*x*log(F))*sin(e*x+d)),x)
Output:
f**(a*c + b*c*x)*sin(d + e*x)*x