3.77 \(\int \frac{F^{c+d x} x^2}{a+b F^{c+d x}} \, dx\)

Optimal. Leaf size=85 \[ \frac{2 x \text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac{2 \text{PolyLog}\left (3,-\frac{b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac{x^2 \log \left (\frac{b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

[Out]

(x^2*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) + (2*x*PolyLog[2, -((b*F^(c + d*x))/a)])/(b*d^2*Log[F]^2) - (2*P
olyLog[3, -((b*F^(c + d*x))/a)])/(b*d^3*Log[F]^3)

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Rubi [A]  time = 0.109308, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2190, 2531, 2282, 6589} \[ \frac{2 x \text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac{2 \text{PolyLog}\left (3,-\frac{b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac{x^2 \log \left (\frac{b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[(F^(c + d*x)*x^2)/(a + b*F^(c + d*x)),x]

[Out]

(x^2*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) + (2*x*PolyLog[2, -((b*F^(c + d*x))/a)])/(b*d^2*Log[F]^2) - (2*P
olyLog[3, -((b*F^(c + d*x))/a)])/(b*d^3*Log[F]^3)

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{F^{c+d x} x^2}{a+b F^{c+d x}} \, dx &=\frac{x^2 \log \left (1+\frac{b F^{c+d x}}{a}\right )}{b d \log (F)}-\frac{2 \int x \log \left (1+\frac{b F^{c+d x}}{a}\right ) \, dx}{b d \log (F)}\\ &=\frac{x^2 \log \left (1+\frac{b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac{2 x \text{Li}_2\left (-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac{2 \int \text{Li}_2\left (-\frac{b F^{c+d x}}{a}\right ) \, dx}{b d^2 \log ^2(F)}\\ &=\frac{x^2 \log \left (1+\frac{b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac{2 x \text{Li}_2\left (-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac{2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{b x}{a}\right )}{x} \, dx,x,F^{c+d x}\right )}{b d^3 \log ^3(F)}\\ &=\frac{x^2 \log \left (1+\frac{b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac{2 x \text{Li}_2\left (-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac{2 \text{Li}_3\left (-\frac{b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}\\ \end{align*}

Mathematica [A]  time = 0.0089126, size = 85, normalized size = 1. \[ \frac{2 x \text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac{2 \text{PolyLog}\left (3,-\frac{b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac{x^2 \log \left (\frac{b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(c + d*x)*x^2)/(a + b*F^(c + d*x)),x]

[Out]

(x^2*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) + (2*x*PolyLog[2, -((b*F^(c + d*x))/a)])/(b*d^2*Log[F]^2) - (2*P
olyLog[3, -((b*F^(c + d*x))/a)])/(b*d^3*Log[F]^3)

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Maple [B]  time = 0.047, size = 194, normalized size = 2.3 \begin{align*}{\frac{{c}^{2}x}{b{d}^{2}}}+{\frac{2\,{c}^{3}}{3\,b{d}^{3}}}+{\frac{{x}^{2}}{bd\ln \left ( F \right ) }\ln \left ( 1+{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }-{\frac{{c}^{2}}{b\ln \left ( F \right ){d}^{3}}\ln \left ( 1+{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }+2\,{\frac{x}{b \left ( \ln \left ( F \right ) \right ) ^{2}{d}^{2}}{\it polylog} \left ( 2,-{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }-2\,{\frac{1}{b \left ( \ln \left ( F \right ) \right ) ^{3}{d}^{3}}{\it polylog} \left ( 3,-{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }+{\frac{{c}^{2}\ln \left ( a+b{F}^{dx}{F}^{c} \right ) }{b\ln \left ( F \right ){d}^{3}}}-{\frac{{c}^{2}\ln \left ({F}^{dx}{F}^{c} \right ) }{b\ln \left ( F \right ){d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(d*x+c)*x^2/(a+b*F^(d*x+c)),x)

[Out]

1/b/d^2*c^2*x+2/3/b/d^3*c^3+1/b/ln(F)/d*ln(1+b*F^(d*x)*F^c/a)*x^2-1/b/ln(F)/d^3*ln(1+b*F^(d*x)*F^c/a)*c^2+2/b/
ln(F)^2/d^2*polylog(2,-b*F^(d*x)*F^c/a)*x-2/b/ln(F)^3/d^3*polylog(3,-b*F^(d*x)*F^c/a)+1/b/ln(F)/d^3*c^2*ln(a+b
*F^(d*x)*F^c)-1/b/ln(F)/d^3*c^2*ln(F^(d*x)*F^c)

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Maxima [A]  time = 1.11639, size = 144, normalized size = 1.69 \begin{align*} \frac{x^{3}}{3 \, b} - \frac{\log \left (F^{d x}\right )^{3}}{3 \, b d^{3} \log \left (F\right )^{3}} + \frac{\log \left (\frac{F^{d x} F^{c} b}{a} + 1\right ) \log \left (F^{d x}\right )^{2} + 2 \,{\rm Li}_2\left (-\frac{F^{d x} F^{c} b}{a}\right ) \log \left (F^{d x}\right ) - 2 \,{\rm Li}_{3}(-\frac{F^{d x} F^{c} b}{a})}{b d^{3} \log \left (F\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(d*x+c)*x^2/(a+b*F^(d*x+c)),x, algorithm="maxima")

[Out]

1/3*x^3/b - 1/3*log(F^(d*x))^3/(b*d^3*log(F)^3) + (log(F^(d*x)*F^c*b/a + 1)*log(F^(d*x))^2 + 2*dilog(-F^(d*x)*
F^c*b/a)*log(F^(d*x)) - 2*polylog(3, -F^(d*x)*F^c*b/a))/(b*d^3*log(F)^3)

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Fricas [C]  time = 1.53408, size = 259, normalized size = 3.05 \begin{align*} \frac{c^{2} \log \left (F^{d x + c} b + a\right ) \log \left (F\right )^{2} + 2 \, d x{\rm Li}_2\left (-\frac{F^{d x + c} b + a}{a} + 1\right ) \log \left (F\right ) +{\left (d^{2} x^{2} - c^{2}\right )} \log \left (F\right )^{2} \log \left (\frac{F^{d x + c} b + a}{a}\right ) - 2 \,{\rm polylog}\left (3, -\frac{F^{d x + c} b}{a}\right )}{b d^{3} \log \left (F\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(d*x+c)*x^2/(a+b*F^(d*x+c)),x, algorithm="fricas")

[Out]

(c^2*log(F^(d*x + c)*b + a)*log(F)^2 + 2*d*x*dilog(-(F^(d*x + c)*b + a)/a + 1)*log(F) + (d^2*x^2 - c^2)*log(F)
^2*log((F^(d*x + c)*b + a)/a) - 2*polylog(3, -F^(d*x + c)*b/a))/(b*d^3*log(F)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{c + d x} x^{2}}{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**2/(a+b*F**(d*x+c)),x)

[Out]

Integral(F**(c + d*x)*x**2/(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^{2}}{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^2/(a+b*F^(d*x+c)),x, algorithm="giac")

[Out]

integrate(F^(d*x + c)*x^2/(F^(d*x + c)*b + a), x)