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

Optimal. Leaf size=107 \[ -\frac{2 \text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}-\frac{2 x \log \left (\frac{b F^{c+d x}}{a}+1\right )}{a b d^2 \log ^2(F)}-\frac{x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}+\frac{x^2}{a b d \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.181147, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {2191, 2184, 2190, 2279, 2391} \[ -\frac{2 \text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}-\frac{2 x \log \left (\frac{b F^{c+d x}}{a}+1\right )}{a b d^2 \log ^2(F)}-\frac{x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}+\frac{x^2}{a b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2191

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

Rule 2184

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

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

Mathematica [A]  time = 0.0914333, size = 103, normalized size = 0.96 \[ \frac{d x \log (F) \left (b d x \log (F) F^{c+d x}-2 \left (a+b F^{c+d x}\right ) \log \left (\frac{b F^{c+d x}}{a}+1\right )\right )-2 \left (a+b F^{c+d x}\right ) \text{PolyLog}\left (2,-\frac{b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F) \left (a+b F^{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.027, size = 231, normalized size = 2.2 \begin{align*} -{\frac{{x}^{2}}{bd \left ( a+b{F}^{dx+c} \right ) \ln \left ( F \right ) }}+{\frac{{x}^{2}}{\ln \left ( F \right ) abd}}+2\,{\frac{cx}{b{d}^{2}\ln \left ( F \right ) a}}+{\frac{{c}^{2}}{b{d}^{3}\ln \left ( F \right ) a}}-2\,{\frac{x}{b{d}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}a}\ln \left ( 1+{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }-2\,{\frac{c}{b{d}^{3} \left ( \ln \left ( F \right ) \right ) ^{2}a}\ln \left ( 1+{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }-2\,{\frac{1}{b{d}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}a}{\it polylog} \left ( 2,-{\frac{b{F}^{dx}{F}^{c}}{a}} \right ) }-2\,{\frac{c\ln \left ({F}^{dx}{F}^{c} \right ) }{b{d}^{3} \left ( \ln \left ( F \right ) \right ) ^{2}a}}+2\,{\frac{c\ln \left ( a+b{F}^{dx}{F}^{c} \right ) }{b{d}^{3} \left ( \ln \left ( F \right ) \right ) ^{2}a}} \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.09789, size = 143, normalized size = 1.34 \begin{align*} -\frac{x^{2}}{F^{d x} F^{c} b^{2} d \log \left (F\right ) + a b d \log \left (F\right )} + \frac{\log \left (F^{d x}\right )^{2}}{a b d^{3} \log \left (F\right )^{3}} - \frac{2 \,{\left (\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 )\right )}}{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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.55778, size = 443, normalized size = 4.14 \begin{align*} -\frac{a c^{2} \log \left (F\right )^{2} -{\left (b d^{2} x^{2} - b c^{2}\right )} F^{d x + c} \log \left (F\right )^{2} + 2 \,{\left (F^{d x + c} b + a\right )}{\rm Li}_2\left (-\frac{F^{d x + c} b + a}{a} + 1\right ) - 2 \,{\left (F^{d x + c} b c \log \left (F\right ) + a c \log \left (F\right )\right )} \log \left (F^{d x + c} b + a\right ) + 2 \,{\left ({\left (b d x + b c\right )} F^{d x + c} \log \left (F\right ) +{\left (a d x + a c\right )} \log \left (F\right )\right )} \log \left (\frac{F^{d x + c} b + a}{a}\right )}{F^{d x + c} a b^{2} d^{3} \log \left (F\right )^{3} + a^{2} 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))^2,x, algorithm="fricas")

[Out]

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

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

[Out]

-x**2/(F**(c + d*x)*b**2*d*log(F) + a*b*d*log(F)) + 2*Integral(x/(a + b*exp(c*log(F))*exp(d*x*log(F))), x)/(b*
d*log(F))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{d x + c} x^{2}}{{\left (F^{d x + c} b + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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