\(\int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx\) [136]

Optimal result

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

[Out]

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)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6873, 12, 6874, 2225, 4517} \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\frac {b c f \log (F) \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}-\frac {e f \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac {f F^{a c+b c x}}{b c \log (F)} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 4517

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(S
in[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] - Simp[e*F^(c*(a + b*x))*(Cos[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int f F^{a c+b c x} (1+\sin (d+e x)) \, dx \\ & = f \int F^{a c+b c x} (1+\sin (d+e x)) \, dx \\ & = f \int \left (F^{a c+b c x}+F^{a c+b c x} \sin (d+e x)\right ) \, dx \\ & = f \int F^{a c+b c x} \, dx+f \int F^{a c+b c x} \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)} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(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))

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85

method	result	size
parallelrisch	\(\frac {f \,F^{c \left (x b +a \right )} \left (\sin \left (e x +d \right ) b^{2} c^{2} \ln \left (F \right )^{2}+b^{2} c^{2} \ln \left (F \right )^{2}-b c \ln \left (F \right ) e \cos \left (e x +d \right )+e^{2}\right )}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}\)	\(84\)
risch	\(\frac {f \,F^{c \left (x b +a \right )}}{b c \ln \left (F \right )}+\frac {e \,F^{c \left (x b +a \right )} f \cos \left (e x +d \right )}{-e^{2}-b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {\ln \left (F \right ) c b \,F^{c \left (x b +a \right )} f \sin \left (e x +d \right )}{-e^{2}-b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(103\)
parts	\(\frac {f \,F^{c \left (x b +a \right )}}{b c \ln \left (F \right )}+\frac {\frac {e f \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e f \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 f b c \ln \left (F \right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}\)	\(155\)
norman	\(\frac {\frac {f \left (b^{2} c^{2} \ln \left (F \right )^{2}-\ln \left (F \right ) b c e +e^{2}\right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}+\frac {f \left (b^{2} c^{2} \ln \left (F \right )^{2}+\ln \left (F \right ) b c e +e^{2}\right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{\left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) b c \ln \left (F \right )}+\frac {2 f b c \ln \left (F \right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}\)	\(193\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\frac {{\left (b^{2} c^{2} f \log \left (F\right )^{2} \sin \left (e x + d\right ) + b^{2} c^{2} f \log \left (F\right )^{2} - b c e f \cos \left (e x + d\right ) \log \left (F\right ) + e^{2} f\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + b c e^{2} \log \left (F\right )} \]

[In]

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

[Out]

(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 = 0.55 (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} \]

[In]

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

[Out]

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*lo
g(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*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)*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 = 0.22 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.20 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=-\frac {{\left ({\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) + F^{a c} e \cos \left (d\right )\right )} F^{b c x} \cos \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) - F^{a c} e \cos \left (d\right )\right )} F^{b c x} \cos \left (e x\right ) - {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) - F^{a c} e \sin \left (d\right )\right )} F^{b c x} \sin \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) + F^{a c} e \sin \left (d\right )\right )} F^{b c x} \sin \left (e x\right )\right )} f}{2 \, {\left (b^{2} c^{2} \cos \left (d\right )^{2} \log \left (F\right )^{2} + b^{2} c^{2} \log \left (F\right )^{2} \sin \left (d\right )^{2} + {\left (\cos \left (d\right )^{2} + \sin \left (d\right )^{2}\right )} e^{2}\right )}} + \frac {F^{b c x + a c} f}{b c \log \left (F\right )} \]

[In]

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

[Out]

-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*lo
g(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 = 0.31 (sec) , antiderivative size = 923, normalized size of antiderivative = 9.32 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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*lo
g(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*p
i*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*p
i*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*s
gn(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(ab
s(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*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*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*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*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*pi*b*c*sgn(F) - I*pi*b*c + 2*b
*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*pi*a*c*sgn(F) + 1/2*I*pi*a*c)/(-I*pi*
b*c*sgn(F) + I*pi*b*c + 2*b*c*log(abs(F))))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F)))

Mupad [B] (verification not implemented)

Time = 25.80 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx=\frac {F^{a\,c+b\,c\,x}\,f\,\left (e^2+b^2\,c^2\,{\ln \left (F\right )}^2+b^2\,c^2\,\sin \left (d+e\,x\right )\,{\ln \left (F\right )}^2-b\,c\,e\,\cos \left (d+e\,x\right )\,\ln \left (F\right )\right )}{b\,c\,\ln \left (F\right )\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+e^2\right )} \]

[In]

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

[Out]

(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))