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

Optimal. Leaf size=99 \[ \frac{b c f \log (F) \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}-\frac{e f \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac{f F^{a c+b c x}}{b c \log (F)} \]

[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)

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Rubi [A]  time = 0.158877, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6741, 12, 6742, 2194, 4432} \[ \frac{b c f \log (F) \sin (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}-\frac{e f \cos (d+e x) F^{a c+b c x}}{b^2 c^2 \log ^2(F)+e^2}+\frac{f F^{a c+b c x}}{b c \log (F)} \]

Antiderivative was successfully verified.

[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 6741

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

Rule 12

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

Rule 6742

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

Rule 2194

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 4432

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]

Rubi steps

\begin{align*} \int F^{c (a+b x)} (f+f \sin (d+e x)) \, dx &=\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]  time = 0.577816, size = 83, normalized size = 0.84 \[ \frac{f F^{c (a+b x)} \left (b^2 c^2 \log ^2(F) \sin (d+e x)+b^2 c^2 \log ^2(F)-b c e \log (F) \cos (d+e x)+e^2\right )}{b c \log (F) \left (b^2 c^2 \log ^2(F)+e^2\right )} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.014, size = 183, normalized size = 1.9 \begin{align*}{\frac{f{F}^{c \left ( bx+a \right ) }}{bc\ln \left ( F \right ) }}+{\frac{ef{{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}}{{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \left ( 1+ \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \right ) ^{-1}}-{\frac{ef{{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}}{{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}} \left ( 1+ \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \right ) ^{-1}}+2\,{\frac{f\ln \left ( F \right ) bc{{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}\tan \left ( d/2+1/2\,ex \right ) }{ \left ( 1+ \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2} \right ) \left ({e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f/b/c/ln(F)*F^(c*(b*x+a))+f/(1+tan(1/2*d+1/2*e*x)^2)/(e^2+b^2*c^2*ln(F)^2)*e*exp(c*(b*x+a)*ln(F))*tan(1/2*d+1/
2*e*x)^2-f/(1+tan(1/2*d+1/2*e*x)^2)/(e^2+b^2*c^2*ln(F)^2)*e*exp(c*(b*x+a)*ln(F))+2*f/(1+tan(1/2*d+1/2*e*x)^2)*
ln(F)*b*c/(e^2+b^2*c^2*ln(F)^2)*exp(c*(b*x+a)*ln(F))*tan(1/2*d+1/2*e*x)

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Maxima [B]  time = 1.05947, size = 294, normalized size = 2.97 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]  time = 0.487338, size = 197, normalized size = 1.99 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [A]  time = 12.888, size = 408, normalized size = 4.12 \begin{align*} \begin{cases} f x - \frac{f \cos{\left (d + e x \right )}}{e} & \text{for}\: F = 1 \\\tilde{\infty } e^{2} f \left (e^{- \frac{i e}{b c}}\right )^{a c} \left (e^{- \frac{i e}{b c}}\right )^{b c x} \sin{\left (d + e x \right )} + \tilde{\infty } e^{2} f \left (e^{- \frac{i e}{b c}}\right )^{a c} \left (e^{- \frac{i e}{b c}}\right )^{b c x} \cos{\left (d + e x \right )} & \text{for}\: F = e^{- \frac{i e}{b c}} \\\tilde{\infty } e^{2} f \left (e^{\frac{i e}{b c}}\right )^{a c} \left (e^{\frac{i e}{b c}}\right )^{b c x} \sin{\left (d + e x \right )} + \tilde{\infty } e^{2} f \left (e^{\frac{i e}{b c}}\right )^{a c} \left (e^{\frac{i e}{b c}}\right )^{b c x} \cos{\left (d + e x \right )} & \text{for}\: F = e^{\frac{i e}{b c}} \\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} F^{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} F^{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} F^{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} F^{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((f*x - f*cos(d + e*x)/e, Eq(F, 1)), (zoo*e**2*f*exp(-I*e/(b*c))**(a*c)*exp(-I*e/(b*c))**(b*c*x)*sin(
d + e*x) + zoo*e**2*f*exp(-I*e/(b*c))**(a*c)*exp(-I*e/(b*c))**(b*c*x)*cos(d + e*x), Eq(F, exp(-I*e/(b*c)))), (
zoo*e**2*f*exp(I*e/(b*c))**(a*c)*exp(I*e/(b*c))**(b*c*x)*sin(d + e*x) + zoo*e**2*f*exp(I*e/(b*c))**(a*c)*exp(I
*e/(b*c))**(b*c*x)*cos(d + e*x), Eq(F, exp(I*e/(b*c)))), (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)*F**(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)*F**(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)
*F**(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)*F**(b*c*x)*e**2*f/(
b**3*c**3*log(F)**3 + b*c*e**2*log(F)), True))

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Giac [C]  time = 1.26293, size = 1270, normalized size = 12.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 + x*e + 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 + x*e + 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 - x*e - 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 - x*e - 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))) + 1/2*(2*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*x*e + I*d)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F)) + 4*I*e) + 2*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*x*e - 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))) + 1/2*(-2*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*x*e - I*d)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log
(abs(F)) - 4*I*e) - 2*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*
x*e + 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
))) - 1/2*I*(-2*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))) + 2*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)))