Integrand size = 20, antiderivative size = 98 \[ \int F^{c (a+b x)} (f+f \cos (d+e x)) \, dx=\frac {f F^{a c+b c x}}{b c \log (F)}+\frac {b c f F^{a c+b c x} \cos (d+e x) \log (F)}{e^2+b^2 c^2 \log ^2(F)}+\frac {e f F^{a c+b c x} \sin (d+e x)}{e^2+b^2 c^2 \log ^2(F)} \] Output:
f*F^(b*c*x+a*c)/b/c/ln(F)+b*c*f*F^(b*c*x+a*c)*cos(e*x+d)*ln(F)/(e^2+b^2*c^ 2*ln(F)^2)+e*f*F^(b*c*x+a*c)*sin(e*x+d)/(e^2+b^2*c^2*ln(F)^2)
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.84 \[ \int F^{c (a+b x)} (f+f \cos (d+e x)) \, dx=\frac {f F^{c (a+b x)} \left (e^2+b^2 c^2 \log ^2(F)+b^2 c^2 \cos (d+e x) \log ^2(F)+b c e \log (F) \sin (d+e x)\right )}{b c \log (F) \left (e^2+b^2 c^2 \log ^2(F)\right )} \] Input:
Integrate[F^(c*(a + b*x))*(f + f*Cos[d + e*x]),x]
Output:
(f*F^(c*(a + b*x))*(e^2 + b^2*c^2*Log[F]^2 + b^2*c^2*Cos[d + e*x]*Log[F]^2 + b*c*e*Log[F]*Sin[d + e*x]))/(b*c*Log[F]*(e^2 + b^2*c^2*Log[F]^2))
Time = 0.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7292, 27, 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.
| \(\displaystyle \int F^{c (a+b x)} (f \cos (d+e x)+f) \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int f (\cos (d+e x)+1) F^{a c+b c x}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle f \int F^{a c+b x c} (\cos (d+e x)+1)dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle f \int \left (\cos (d+e x) F^{a c+b x c}+F^{a c+b x c}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle f \left (\frac {e \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {b c \log (F) \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {F^{a c+b c x}}{b c \log (F)}\right )\) |
Input:
Int[F^(c*(a + b*x))*(f + f*Cos[d + e*x]),x]
Output:
f*(F^(a*c + b*c*x)/(b*c*Log[F]) + (b*c*F^(a*c + b*c*x)*Cos[d + e*x]*Log[F] )/(e^2 + b^2*c^2*Log[F]^2) + (e*F^(a*c + b*c*x)*Sin[d + e*x])/(e^2 + b^2*c ^2*Log[F]^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.66 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.85
| method | result | size |
| parallelrisch | \(\frac {f \,F^{c \left (b x +a \right )} \left (\cos \left (e x +d \right ) b^{2} c^{2} \ln \left (F \right )^{2}+b^{2} c^{2} \ln \left (F \right )^{2}+\sin \left (e x +d \right ) b c e \ln \left (F \right )+e^{2}\right )}{b c \ln \left (F \right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\) | \(83\) |
| risch | \(\frac {f \,F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}+\frac {\ln \left (F \right ) c b f \,F^{c \left (b x +a \right )} \cos \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {e f \,F^{c \left (b x +a \right )} \sin \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(96\) |
| parts | \(\frac {f \,F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}+\frac {\frac {f b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {f b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\) | \(158\) |
| norman | \(\frac {\frac {\left (2 b^{2} c^{2} \ln \left (F \right )^{2}+e^{2}\right ) f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{b c \ln \left (F \right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {e^{2} f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}+\frac {2 e f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\) | \(166\) |
| orering | \(\frac {\left (e^{2}+3 b^{2} c^{2} \ln \left (F \right )^{2}\right ) F^{c \left (b x +a \right )} \left (f +f \cos \left (e x +d \right )\right )}{b c \ln \left (F \right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {3 \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \left (f +f \cos \left (e x +d \right )\right )-F^{c \left (b x +a \right )} f e \sin \left (e x +d \right )\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \left (f +f \cos \left (e x +d \right )\right )-2 F^{c \left (b x +a \right )} b c \ln \left (F \right ) f e \sin \left (e x +d \right )-F^{c \left (b x +a \right )} f \,e^{2} \cos \left (e x +d \right )}{b c \ln \left (F \right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\) | \(231\) |
Input:
int(F^(c*(b*x+a))*(f+f*cos(e*x+d)),x,method=_RETURNVERBOSE)
Output:
f*F^(c*(b*x+a))*(cos(e*x+d)*b^2*c^2*ln(F)^2+b^2*c^2*ln(F)^2+sin(e*x+d)*b*c *e*ln(F)+e^2)/b/c/ln(F)/(e^2+b^2*c^2*ln(F)^2)
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int F^{c (a+b x)} (f+f \cos (d+e x)) \, dx=\frac {{\left (b c e f \log \left (F\right ) \sin \left (e x + d\right ) + e^{2} f + {\left (b^{2} c^{2} f \cos \left (e x + d\right ) + b^{2} c^{2} f\right )} \log \left (F\right )^{2}\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + b c e^{2} \log \left (F\right )} \] Input:
integrate(F^(c*(b*x+a))*(f+f*cos(e*x+d)),x, algorithm="fricas")
Output:
(b*c*e*f*log(F)*sin(e*x + d) + e^2*f + (b^2*c^2*f*cos(e*x + d) + b^2*c^2*f )*log(F)^2)*F^(b*c*x + a*c)/(b^3*c^3*log(F)^3 + b*c*e^2*log(F))
Result contains complex when optimal does not.
Time = 0.62 (sec) , antiderivative size = 510, normalized size of antiderivative = 5.20 \[ \int F^{c (a+b x)} (f+f \cos (d+e x)) \, dx=\begin {cases} x \left (f \cos {\left (d \right )} + f\right ) & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge e = 0 \\f x + \frac {f \sin {\left (d + e x \right )}}{e} & \text {for}\: F = 1 \\F^{a c} \left (f x + \frac {f \sin {\left (d + e x \right )}}{e}\right ) & \text {for}\: b = 0 \\f x + \frac {f \sin {\left (d + e x \right )}}{e} & \text {for}\: c = 0 \\\frac {i F^{a c + b c x} f x \sin {\left (i b c x \log {\left (F \right )} - d \right )}}{2} + \frac {F^{a c + b c x} f x \cos {\left (i b c x \log {\left (F \right )} - d \right )}}{2} - \frac {i F^{a c + b c x} f \sin {\left (i b c x \log {\left (F \right )} - d \right )}}{b c \log {\left (F \right )}} - \frac {F^{a c + b c x} f \cos {\left (i b c x \log {\left (F \right )} - d \right )}}{2 b c \log {\left (F \right )}} + \frac {F^{a c + b c x} f}{b c \log {\left (F \right )}} & \text {for}\: e = - i b c \log {\left (F \right )} \\\frac {i F^{a c + b c x} f x \sin {\left (i b c x \log {\left (F \right )} + d \right )}}{2} + \frac {F^{a c + b c x} f x \cos {\left (i b c x \log {\left (F \right )} + d \right )}}{2} - \frac {i F^{a c + b c x} f \sin {\left (i b c x \log {\left (F \right )} + d \right )}}{2 b c \log {\left (F \right )}} + \frac {F^{a c + b c x} f}{b c \log {\left (F \right )}} & \text {for}\: e = i b c \log {\left (F \right )} \\\frac {F^{a c + b c x} b^{2} c^{2} f \log {\left (F \right )}^{2} \cos {\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + b c e^{2} \log {\left (F \right )}} + \frac {F^{a c + b c x} b^{2} c^{2} f \log {\left (F \right )}^{2}}{b^{3} c^{3} \log {\left (F \right )}^{3} + b c e^{2} \log {\left (F \right )}} + \frac {F^{a c + b c x} b c e f \log {\left (F \right )} \sin {\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + b c e^{2} \log {\left (F \right )}} + \frac {F^{a c + b c x} e^{2} f}{b^{3} c^{3} \log {\left (F \right )}^{3} + b c e^{2} \log {\left (F \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(F**(c*(b*x+a))*(f+f*cos(e*x+d)),x)
Output:
Piecewise((x*(f*cos(d) + f), Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0)), ( f*x + f*sin(d + e*x)/e, Eq(F, 1)), (F**(a*c)*(f*x + f*sin(d + e*x)/e), Eq( b, 0)), (f*x + f*sin(d + e*x)/e, Eq(c, 0)), (I*F**(a*c + b*c*x)*f*x*sin(I* b*c*x*log(F) - d)/2 + F**(a*c + b*c*x)*f*x*cos(I*b*c*x*log(F) - d)/2 - I*F **(a*c + b*c*x)*f*sin(I*b*c*x*log(F) - d)/(b*c*log(F)) - F**(a*c + b*c*x)* f*cos(I*b*c*x*log(F) - d)/(2*b*c*log(F)) + F**(a*c + b*c*x)*f/(b*c*log(F)) , Eq(e, -I*b*c*log(F))), (I*F**(a*c + b*c*x)*f*x*sin(I*b*c*x*log(F) + d)/2 + F**(a*c + b*c*x)*f*x*cos(I*b*c*x*log(F) + d)/2 - I*F**(a*c + b*c*x)*f*s in(I*b*c*x*log(F) + d)/(2*b*c*log(F)) + F**(a*c + b*c*x)*f/(b*c*log(F)), E q(e, I*b*c*log(F))), (F**(a*c + b*c*x)*b**2*c**2*f*log(F)**2*cos(d + e*x)/ (b**3*c**3*log(F)**3 + b*c*e**2*log(F)) + F**(a*c + b*c*x)*b**2*c**2*f*log (F)**2/(b**3*c**3*log(F)**3 + b*c*e**2*log(F)) + F**(a*c + b*c*x)*b*c*e*f* log(F)*sin(d + e*x)/(b**3*c**3*log(F)**3 + b*c*e**2*log(F)) + F**(a*c + b* c*x)*e**2*f/(b**3*c**3*log(F)**3 + b*c*e**2*log(F)), True))
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (98) = 196\).
Time = 0.07 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.20 \[ \int F^{c (a+b x)} (f+f \cos (d+e x)) \, dx=\frac {{\left ({\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) - F^{a c} e \sin \left (d\right )\right )} F^{b c x} \cos \left (e x + 2 \, d\right ) + {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) + F^{a c} e \sin \left (d\right )\right )} F^{b c x} \cos \left (e x\right ) + {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) + F^{a c} e \cos \left (d\right )\right )} F^{b c x} \sin \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) - F^{a c} e \cos \left (d\right )\right )} F^{b c x} \sin \left (e x\right )\right )} f}{2 \, {\left (b^{2} c^{2} \cos \left (d\right )^{2} \log \left (F\right )^{2} + b^{2} c^{2} \log \left (F\right )^{2} \sin \left (d\right )^{2} + {\left (\cos \left (d\right )^{2} + \sin \left (d\right )^{2}\right )} e^{2}\right )}} + \frac {F^{b c x + a c} f}{b c \log \left (F\right )} \] Input:
integrate(F^(c*(b*x+a))*(f+f*cos(e*x+d)),x, algorithm="maxima")
Output:
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))*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) + F^(b*c*x + a*c)*f/(b*c*log(F))
Result contains complex when optimal does not.
Time = 0.15 (sec) , antiderivative size = 920, normalized size of antiderivative = 9.39 \[ \int F^{c (a+b x)} (f+f \cos (d+e x)) \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*(f+f*cos(e*x+d)),x, algorithm="giac")
Output:
(2*b*c*f*cos(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 + e*x + d)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 2*e)^2) + (pi*b*c*sgn(F) - pi*b*c + 2*e)*f*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 + e*x + d)/(4*b^2*c^2*lo g(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 2*e)^2))*e^(b*c*x*log(abs(F)) + a* c*log(abs(F))) + (2*b*c*f*cos(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 - e*x - d)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 2*e)^2) + (pi*b*c*sgn(F) - pi*b*c - 2*e)*f*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 - e*x - d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 2*e)^2))*e^(b*c*x *log(abs(F)) + a*c*log(abs(F))) + 2*(2*b*c*f*cos(-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)*log(abs(F))/(4*b^2*c^2*log(ab s(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2) - (pi*b*c*sgn(F) - pi*b*c)*f*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))) + I*(I*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)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b* c + 4*b*c*log(abs(F)) + 4*I*e) - I*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)/(-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...
Time = 20.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.85 \[ \int F^{c (a+b x)} (f+f \cos (d+e x)) \, dx=\frac {F^{a\,c+b\,c\,x}\,f\,\left (e^2+b^2\,c^2\,{\ln \left (F\right )}^2+b^2\,c^2\,\cos \left (d+e\,x\right )\,{\ln \left (F\right )}^2+b\,c\,e\,\sin \left (d+e\,x\right )\,\ln \left (F\right )\right )}{b\,c\,\ln \left (F\right )\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+e^2\right )} \] Input:
int(F^(c*(a + b*x))*(f + f*cos(d + e*x)),x)
Output:
(F^(a*c + b*c*x)*f*(e^2 + b^2*c^2*log(F)^2 + b^2*c^2*cos(d + e*x)*log(F)^2 + b*c*e*sin(d + e*x)*log(F)))/(b*c*log(F)*(e^2 + b^2*c^2*log(F)^2))
Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.85 \[ \int F^{c (a+b x)} (f+f \cos (d+e x)) \, dx=\frac {f^{b c x +a c} f \left (\cos \left (e x +d \right ) \mathrm {log}\left (f \right )^{2} b^{2} c^{2}+\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+\mathrm {log}\left (f \right ) \sin \left (e x +d \right ) b c e +e^{2}\right )}{\mathrm {log}\left (f \right ) b c \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+e^{2}\right )} \] Input:
int(F^(c*(b*x+a))*(f+f*cos(e*x+d)),x)
Output:
(f**(a*c + b*c*x)*f*(cos(d + e*x)*log(f)**2*b**2*c**2 + log(f)**2*b**2*c** 2 + log(f)*sin(d + e*x)*b*c*e + e**2))/(log(f)*b*c*(log(f)**2*b**2*c**2 + e**2))