3.78 \(\int \frac{F^{c+d x} x}{a+b F^{c+d x}} \, dx\)

Optimal. Leaf size=54 \[ \frac{\text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac{x \log \left (\frac{b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

[Out]

(x*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) + PolyLog[2, -((b*F^(c + d*x))/a)]/(b*d^2*Log[F]^2)

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Rubi [A]  time = 0.0642339, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac{x \log \left (\frac{b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[(F^(c + d*x)*x)/(a + b*F^(c + d*x)),x]

[Out]

(x*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) + PolyLog[2, -((b*F^(c + d*x))/a)]/(b*d^2*Log[F]^2)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{F^{c+d x} x}{a+b F^{c+d x}} \, dx &=\frac{x \log \left (1+\frac{b F^{c+d x}}{a}\right )}{b d \log (F)}-\frac{\int \log \left (1+\frac{b F^{c+d x}}{a}\right ) \, dx}{b d \log (F)}\\ &=\frac{x \log \left (1+\frac{b F^{c+d x}}{a}\right )}{b d \log (F)}-\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{b x}{a}\right )}{x} \, dx,x,F^{c+d x}\right )}{b d^2 \log ^2(F)}\\ &=\frac{x \log \left (1+\frac{b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac{\text{Li}_2\left (-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}\\ \end{align*}

Mathematica [A]  time = 0.0056287, size = 54, normalized size = 1. \[ \frac{\text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac{x \log \left (\frac{b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(c + d*x)*x)/(a + b*F^(c + d*x)),x]

[Out]

(x*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) + PolyLog[2, -((b*F^(c + d*x))/a)]/(b*d^2*Log[F]^2)

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Maple [B]  time = 0.044, size = 154, normalized size = 2.9 \begin{align*} -{\frac{cx}{bd}}-{\frac{{c}^{2}}{2\,b{d}^{2}}}+{\frac{x}{bd\ln \left ( F \right ) }\ln \left ( 1+{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }+{\frac{c}{b\ln \left ( F \right ){d}^{2}}\ln \left ( 1+{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }+{\frac{1}{b \left ( \ln \left ( F \right ) \right ) ^{2}{d}^{2}}{\it polylog} \left ( 2,-{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }+{\frac{c\ln \left ({F}^{dx}{F}^{c} \right ) }{b\ln \left ( F \right ){d}^{2}}}-{\frac{c\ln \left ( a+b{F}^{dx}{F}^{c} \right ) }{b\ln \left ( F \right ){d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(d*x+c)*x/(a+b*F^(d*x+c)),x)

[Out]

-1/b/d*c*x-1/2*c^2/b/d^2+1/b/ln(F)/d*ln(1+b*F^(d*x)*F^c/a)*x+1/b/ln(F)/d^2*ln(1+b*F^(d*x)*F^c/a)*c+1/b/ln(F)^2
/d^2*polylog(2,-b*F^(d*x)*F^c/a)+1/b/ln(F)/d^2*c*ln(F^(d*x)*F^c)-1/b/ln(F)/d^2*c*ln(a+b*F^(d*x)*F^c)

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Maxima [A]  time = 1.13147, size = 107, normalized size = 1.98 \begin{align*} \frac{x^{2}}{2 \, b} - \frac{\log \left (F^{d x}\right )^{2}}{2 \, b d^{2} \log \left (F\right )^{2}} + \frac{\log \left (\frac{F^{d x} F^{c} b}{a} + 1\right ) \log \left (F^{d x}\right ) +{\rm Li}_2\left (-\frac{F^{d x} F^{c} b}{a}\right )}{b d^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2/b - 1/2*log(F^(d*x))^2/(b*d^2*log(F)^2) + (log(F^(d*x)*F^c*b/a + 1)*log(F^(d*x)) + dilog(-F^(d*x)*F^c*
b/a))/(b*d^2*log(F)^2)

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Fricas [A]  time = 1.54216, size = 182, normalized size = 3.37 \begin{align*} -\frac{c \log \left (F^{d x + c} b + a\right ) \log \left (F\right ) -{\left (d x + c\right )} \log \left (F\right ) \log \left (\frac{F^{d x + c} b + a}{a}\right ) -{\rm Li}_2\left (-\frac{F^{d x + c} b + a}{a} + 1\right )}{b d^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*log(F^(d*x + c)*b + a)*log(F) - (d*x + c)*log(F)*log((F^(d*x + c)*b + a)/a) - dilog(-(F^(d*x + c)*b + a)/a
+ 1))/(b*d^2*log(F)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{c + d x} x}{F^{c} F^{d x} b + a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(d*x+c)*x/(a+b*F**(d*x+c)),x)

[Out]

Integral(F**(c + d*x)*x/(F**c*F**(d*x)*b + a), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{d x + c} x}{F^{d x + c} b + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(F^(d*x + c)*x/(F^(d*x + c)*b + a), x)