\(\int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx\) [29]

Optimal result

Integrand size = 20, antiderivative size = 99 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\frac {f F^{a c+b c x}}{b c \log (F)}-\frac {e f F^{a c+b c x} \cos (d+e x)}{e^2+b^2 c^2 \log ^2(F)}+\frac {b c f F^{a c+b c x} \log (F) \sin (d+e x)}{e^2+b^2 c^2 \log ^2(F)} \] Output:

f*F^(b*c*x+a*c)/b/c/ln(F)-e*f*F^(b*c*x+a*c)*cos(e*x+d)/(e^2+b^2*c^2*ln(F)^ 
2)+b*c*f*F^(b*c*x+a*c)*ln(F)*sin(e*x+d)/(e^2+b^2*c^2*ln(F)^2)
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\frac {f F^{c (a+b x)} \left (e^2-b c e \cos (d+e x) \log (F)+b^2 c^2 \log ^2(F)+b^2 c^2 \log ^2(F) \sin (d+e x)\right )}{b c \log (F) \left (e^2+b^2 c^2 \log ^2(F)\right )} \] Input:

Integrate[F^(c*(a + b*x))*(f + f*Sin[d + e*x]),x]
 

Output:

(f*F^(c*(a + b*x))*(e^2 - b*c*e*Cos[d + e*x]*Log[F] + b^2*c^2*Log[F]^2 + b 
^2*c^2*Log[F]^2*Sin[d + e*x]))/(b*c*Log[F]*(e^2 + b^2*c^2*Log[F]^2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 98, 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.

\(\displaystyle \int F^{c (a+b x)} (f \sin (d+e x)+f) \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int f (\sin (d+e x)+1) F^{a c+b c x}dx\)

\(\Big \downarrow \) 27

\(\displaystyle f \int F^{a c+b x c} (\sin (d+e x)+1)dx\)

\(\Big \downarrow \) 7293

\(\displaystyle f \int \left (\sin (d+e x) F^{a c+b x c}+F^{a c+b x c}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle f \left (\frac {b c \log (F) \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}-\frac {e \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {F^{a c+b c x}}{b c \log (F)}\right )\)

Input:

Int[F^(c*(a + b*x))*(f + f*Sin[d + e*x]),x]
 

Output:

f*(F^(a*c + b*c*x)/(b*c*Log[F]) - (e*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)*Log[F]*Sin[d + e*x])/(e^2 + b^2*c 
^2*Log[F]^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85

method	result	size
parallelrisch	\(\frac {f \,F^{c \left (b x +a \right )} \left (\sin \left (e x +d \right ) b^{2} c^{2} \ln \left (F \right )^{2}+b^{2} c^{2} \ln \left (F \right )^{2}-\cos \left (e x +d \right ) b c e \ln \left (F \right )+e^{2}\right )}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}\)	\(84\)
risch	\(\frac {f \,F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}-\frac {e \,F^{c \left (b x +a \right )} f \cos \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {\ln \left (F \right ) c b \,F^{c \left (b x +a \right )} f \sin \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(97\)
parts	\(\frac {f \,F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}+\frac {\frac {e f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e f \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 f b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\)	\(155\)
norman	\(\frac {\frac {f \left (b^{2} c^{2} \ln \left (F \right )^{2}-\ln \left (F \right ) b c e +e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}+\frac {f \left (b^{2} c^{2} \ln \left (F \right )^{2}+\ln \left (F \right ) b c e +e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}+\frac {2 f b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\)	\(193\)
orering	\(\frac {\left (e^{2}+3 b^{2} c^{2} \ln \left (F \right )^{2}\right ) F^{c \left (b x +a \right )} \left (f +f \sin \left (e x +d \right )\right )}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}-\frac {3 \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \left (f +f \sin \left (e x +d \right )\right )+F^{c \left (b x +a \right )} f e \cos \left (e x +d \right )\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \left (f +f \sin \left (e x +d \right )\right )+2 F^{c \left (b x +a \right )} b c \ln \left (F \right ) f e \cos \left (e x +d \right )-F^{c \left (b x +a \right )} f \,e^{2} \sin \left (e x +d \right )}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}\)	\(230\)

Input:

int(F^(c*(b*x+a))*(f+f*sin(e*x+d)),x,method=_RETURNVERBOSE)
 

Output:

f*F^(c*(b*x+a))*(sin(e*x+d)*b^2*c^2*ln(F)^2+b^2*c^2*ln(F)^2-cos(e*x+d)*b*c 
*e*ln(F)+e^2)/(e^2+b^2*c^2*ln(F)^2)/b/c/ln(F)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\frac {{\left (b^{2} c^{2} f \log \left (F\right )^{2} \sin \left (e x + d\right ) + b^{2} c^{2} f \log \left (F\right )^{2} - b c e f \cos \left (e x + d\right ) \log \left (F\right ) + e^{2} f\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + b c e^{2} \log \left (F\right )} \] Input:

integrate(F^(c*(b*x+a))*(f+f*sin(e*x+d)),x, algorithm="fricas")
 

Output:

(b^2*c^2*f*log(F)^2*sin(e*x + d) + b^2*c^2*f*log(F)^2 - b*c*e*f*cos(e*x + 
d)*log(F) + e^2*f)*F^(b*c*x + a*c)/(b^3*c^3*log(F)^3 + b*c*e^2*log(F))
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 542, normalized size of antiderivative = 5.47 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\begin {cases} x \left (f \sin {\left (d \right )} + f\right ) & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge e = 0 \\f x - \frac {f \cos {\left (d + e x \right )}}{e} & \text {for}\: F = 1 \\F^{a c} \left (f x - \frac {f \cos {\left (d + e x \right )}}{e}\right ) & \text {for}\: b = 0 \\f x - \frac {f \cos {\left (d + e x \right )}}{e} & \text {for}\: c = 0 \\- \frac {F^{a c + b c x} f x \sin {\left (i b c x \log {\left (F \right )} - d \right )}}{2} + \frac {i F^{a c + b c x} f x \cos {\left (i b c x \log {\left (F \right )} - d \right )}}{2} + \frac {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 {i F^{a c + b c x} f \cos {\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}{b c \log {\left (F \right )}} & \text {for}\: e = - i b c \log {\left (F \right )} \\\frac {F^{a c + b c x} f x \sin {\left (i b c x \log {\left (F \right )} + d \right )}}{2} - \frac {i F^{a c + b c x} f x \cos {\left (i b c x \log {\left (F \right )} + d \right )}}{2} - \frac {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 {i F^{a c + b c x} f \cos {\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}{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} \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} 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 )} \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} 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*sin(e*x+d)),x)
 

Output:

Piecewise((x*(f*sin(d) + f), Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0)), ( 
f*x - f*cos(d + e*x)/e, Eq(F, 1)), (F**(a*c)*(f*x - f*cos(d + e*x)/e), Eq( 
b, 0)), (f*x - f*cos(d + e*x)/e, Eq(c, 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 + F* 
*(a*c + b*c*x)*f*sin(I*b*c*x*log(F) - d)/(2*b*c*log(F)) - I*F**(a*c + b*c* 
x)*f*cos(I*b*c*x*log(F) - d)/(b*c*log(F)) + F**(a*c + b*c*x)*f/(b*c*log(F) 
), Eq(e, -I*b*c*log(F))), (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 - F**(a*c + b*c*x)*f*si 
n(I*b*c*x*log(F) + d)/(2*b*c*log(F)) + I*F**(a*c + b*c*x)*f*cos(I*b*c*x*lo 
g(F) + d)/(b*c*log(F)) + F**(a*c + b*c*x)*f/(b*c*log(F)), Eq(e, I*b*c*log( 
F))), (F**(a*c + b*c*x)*b**2*c**2*f*log(F)**2*sin(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)*cos(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))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (99) = 198\).

Time = 0.06 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.20 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=-\frac {{\left ({\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) + F^{a c} e \cos \left (d\right )\right )} F^{b c x} \cos \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) - F^{a c} e \cos \left (d\right )\right )} F^{b c x} \cos \left (e x\right ) - {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) - F^{a c} e \sin \left (d\right )\right )} F^{b c x} \sin \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) + F^{a c} e \sin \left (d\right )\right )} F^{b c x} \sin \left (e x\right )\right )} f}{2 \, {\left (b^{2} c^{2} \cos \left (d\right )^{2} \log \left (F\right )^{2} + b^{2} c^{2} \log \left (F\right )^{2} \sin \left (d\right )^{2} + {\left (\cos \left (d\right )^{2} + \sin \left (d\right )^{2}\right )} e^{2}\right )}} + \frac {F^{b c x + a c} f}{b c \log \left (F\right )} \] Input:

integrate(F^(c*(b*x+a))*(f+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) + F^(b*c*x + a*c)*f/(b*c*log(F))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 923, normalized size of antiderivative = 9.32 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\text {Too large to display} \] Input:

integrate(F^(c*(b*x+a))*(f+f*sin(e*x+d)),x, algorithm="giac")
 

Output:

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(abs(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))) + (2*b*c*f*log(abs( 
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) 
- (pi*b*c*sgn(F) - pi*b*c + 2*e)*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)/(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*b*c*f*log(abs(F))*sin(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*s 
gn(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) - (pi*b*c*sgn(F) - pi*b*c - 2*e)*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)/(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))) - (-I*f*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*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*...
 

Mupad [B] (verification not implemented)

Time = 20.49 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\frac {F^{a\,c+b\,c\,x}\,f\,\left (e^2+b^2\,c^2\,{\ln \left (F\right )}^2+b^2\,c^2\,\sin \left (d+e\,x\right )\,{\ln \left (F\right )}^2-b\,c\,e\,\cos \left (d+e\,x\right )\,\ln \left (F\right )\right )}{b\,c\,\ln \left (F\right )\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+e^2\right )} \] Input:

int(F^(c*(a + b*x))*(f + f*sin(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*sin(d + e*x)*log(F)^2 
 - b*c*e*cos(d + e*x)*log(F)))/(b*c*log(F)*(e^2 + b^2*c^2*log(F)^2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\frac {f^{b c x +a c} f \left (-\cos \left (e x +d \right ) \mathrm {log}\left (f \right ) b c e +\mathrm {log}\left (f \right )^{2} \sin \left (e x +d \right ) b^{2} c^{2}+\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+e^{2}\right )}{\mathrm {log}\left (f \right ) b c \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+e^{2}\right )} \] Input:

int(F^(c*(b*x+a))*(f+f*sin(e*x+d)),x)
 

Output:

(f**(a*c + b*c*x)*f*( - cos(d + e*x)*log(f)*b*c*e + log(f)**2*sin(d + e*x) 
*b**2*c**2 + log(f)**2*b**2*c**2 + e**2))/(log(f)*b*c*(log(f)**2*b**2*c**2 
 + e**2))