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

Optimal. Leaf size=338 \[ -\frac{2 c \text{PolyLog}\left (2,-\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}\right )}{d^2 \log ^2(f) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}+\frac{2 c \text{PolyLog}\left (2,-\frac{2 c f^{c+d x}}{\sqrt{b^2-4 a c}+b}\right )}{d^2 \log ^2(f) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )}-\frac{2 c x \log \left (\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}+1\right )}{d \log (f) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}+\frac{2 c x \log \left (\frac{2 c f^{c+d x}}{\sqrt{b^2-4 a c}+b}+1\right )}{d \log (f) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )}-\frac{c x^2}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{c x^2}{b \sqrt{b^2-4 a c}-4 a c+b^2} \]

[Out]

-((c*x^2)/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])) - (c*x^2)/(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c]) - (2*c*x*Log[1 +
(2*c*f^(c + d*x))/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*Log[f]) + (2*c*x*Log[
1 + (2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*Log[f]) - (2*c*Po
lyLog[2, (-2*c*f^(c + d*x))/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d^2*Log[f]^2)
+ (2*c*PolyLog[2, (-2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d^2
*Log[f]^2)

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Rubi [A]  time = 0.685755, antiderivative size = 338, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2263, 2184, 2190, 2279, 2391} \[ -\frac{2 c \text{PolyLog}\left (2,-\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}\right )}{d^2 \log ^2(f) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}+\frac{2 c \text{PolyLog}\left (2,-\frac{2 c f^{c+d x}}{\sqrt{b^2-4 a c}+b}\right )}{d^2 \log ^2(f) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )}-\frac{2 c x \log \left (\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}+1\right )}{d \log (f) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}+\frac{2 c x \log \left (\frac{2 c f^{c+d x}}{\sqrt{b^2-4 a c}+b}+1\right )}{d \log (f) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )}-\frac{c x^2}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{c x^2}{b \sqrt{b^2-4 a c}-4 a c+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*x^2)/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])) - (c*x^2)/(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c]) - (2*c*x*Log[1 +
(2*c*f^(c + d*x))/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*Log[f]) + (2*c*x*Log[
1 + (2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*Log[f]) - (2*c*Po
lyLog[2, (-2*c*f^(c + d*x))/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d^2*Log[f]^2)
+ (2*c*PolyLog[2, (-2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d^2
*Log[f]^2)

Rule 2263

Int[((f_.) + (g_.)*(x_))^(m_.)/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*
a*c, 2]}, Dist[(2*c)/q, Int[(f + g*x)^m/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[(f + g*x)^m/(b + q + 2*c
*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[
m, 0]

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{x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx &=\frac{(2 c) \int \frac{x}{b-\sqrt{b^2-4 a c}+2 c f^{c+d x}} \, dx}{\sqrt{b^2-4 a c}}-\frac{(2 c) \int \frac{x}{b+\sqrt{b^2-4 a c}+2 c f^{c+d x}} \, dx}{\sqrt{b^2-4 a c}}\\ &=-\frac{c x^2}{b^2-4 a c-b \sqrt{b^2-4 a c}}-\frac{c x^2}{b^2-4 a c+b \sqrt{b^2-4 a c}}+\frac{\left (4 c^2\right ) \int \frac{f^{c+d x} x}{b-\sqrt{b^2-4 a c}+2 c f^{c+d x}} \, dx}{b^2-4 a c-b \sqrt{b^2-4 a c}}+\frac{\left (4 c^2\right ) \int \frac{f^{c+d x} x}{b+\sqrt{b^2-4 a c}+2 c f^{c+d x}} \, dx}{b^2-4 a c+b \sqrt{b^2-4 a c}}\\ &=-\frac{c x^2}{b^2-4 a c-b \sqrt{b^2-4 a c}}-\frac{c x^2}{b^2-4 a c+b \sqrt{b^2-4 a c}}+\frac{2 c x \log \left (1+\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt{b^2-4 a c}\right ) d \log (f)}+\frac{2 c x \log \left (1+\frac{2 c f^{c+d x}}{b+\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt{b^2-4 a c}\right ) d \log (f)}-\frac{(2 c) \int \log \left (1+\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}\right ) \, dx}{\left (b^2-4 a c-b \sqrt{b^2-4 a c}\right ) d \log (f)}-\frac{(2 c) \int \log \left (1+\frac{2 c f^{c+d x}}{b+\sqrt{b^2-4 a c}}\right ) \, dx}{\left (b^2-4 a c+b \sqrt{b^2-4 a c}\right ) d \log (f)}\\ &=-\frac{c x^2}{b^2-4 a c-b \sqrt{b^2-4 a c}}-\frac{c x^2}{b^2-4 a c+b \sqrt{b^2-4 a c}}+\frac{2 c x \log \left (1+\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt{b^2-4 a c}\right ) d \log (f)}+\frac{2 c x \log \left (1+\frac{2 c f^{c+d x}}{b+\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt{b^2-4 a c}\right ) d \log (f)}-\frac{(2 c) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{x} \, dx,x,f^{c+d x}\right )}{\left (b^2-4 a c-b \sqrt{b^2-4 a c}\right ) d^2 \log ^2(f)}-\frac{(2 c) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{x} \, dx,x,f^{c+d x}\right )}{\left (b^2-4 a c+b \sqrt{b^2-4 a c}\right ) d^2 \log ^2(f)}\\ &=-\frac{c x^2}{b^2-4 a c-b \sqrt{b^2-4 a c}}-\frac{c x^2}{b^2-4 a c+b \sqrt{b^2-4 a c}}+\frac{2 c x \log \left (1+\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt{b^2-4 a c}\right ) d \log (f)}+\frac{2 c x \log \left (1+\frac{2 c f^{c+d x}}{b+\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt{b^2-4 a c}\right ) d \log (f)}+\frac{2 c \text{Li}_2\left (-\frac{2 c f^{c+d x}}{b-\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt{b^2-4 a c}\right ) d^2 \log ^2(f)}+\frac{2 c \text{Li}_2\left (-\frac{2 c f^{c+d x}}{b+\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt{b^2-4 a c}\right ) d^2 \log ^2(f)}\\ \end{align*}

Mathematica [F]  time = 5.06031, size = 0, normalized size = 0. \[ \int \frac{x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [B]  time = 0.078, size = 855, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/ln(f)/d/a*ln((-2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*x-1/2/ln(f)/d^2/a*ln((-2*c*
f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*c-1/2/ln(f)/d/a/(-4*a*c+b^2)^(1/2)*ln((-2*c*f^(d*x)
*f^c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*b*x-1/2/ln(f)/d^2/a/(-4*a*c+b^2)^(1/2)*ln((-2*c*f^(d*x)*f^
c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*b*c-1/2/ln(f)/d/a*ln((2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)+b)/(
b+(-4*a*c+b^2)^(1/2)))*x-1/2/ln(f)/d^2/a*ln((2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*c+1
/2/ln(f)/d/a/(-4*a*c+b^2)^(1/2)*ln((2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*b*x+1/2/ln(f
)/d^2/a/(-4*a*c+b^2)^(1/2)*ln((2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*b*c-1/2/ln(f)^2/d
^2/a*dilog((-2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))-1/2/ln(f)^2/d^2/a/(-4*a*c+b^2)^(1/
2)*dilog((-2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*b-1/2/ln(f)^2/d^2/a*dilog((2*c*f^(d*
x)*f^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))+1/2/ln(f)^2/d^2/a/(-4*a*c+b^2)^(1/2)*dilog((2*c*f^(d*x)*f
^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*b+1/2*x^2/a+1/d/a*x*c+1/2/d^2/a*c^2+1/2/ln(f)/d^2*c/a*ln(a+b*
f^(d*x)*f^c+c*(f^(d*x))^2*(f^c)^2)+1/ln(f)/d^2*c/a*b/(4*a*c-b^2)^(1/2)*arctan((2*c*f^(d*x)*f^c+b)/(4*a*c-b^2)^
(1/2))-1/ln(f)/d^2*c/a*ln(f^(d*x)*f^c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51207, size = 1176, normalized size = 3.48 \begin{align*} \frac{{\left (b^{2} - 4 \, a c\right )} d^{2} x^{2} \log \left (f\right )^{2} -{\left (a b \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} + b^{2} - 4 \, a c\right )}{\rm Li}_2\left (-\frac{{\left (a \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} + b\right )} f^{d x + c} + 2 \, a}{2 \, a} + 1\right ) +{\left (a b \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} - b^{2} + 4 \, a c\right )}{\rm Li}_2\left (\frac{{\left (a \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} - b\right )} f^{d x + c} - 2 \, a}{2 \, a} + 1\right ) -{\left (a b c \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) -{\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )\right )} \log \left (2 \, c f^{d x + c} + a \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} + b\right ) +{\left (a b c \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) +{\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )\right )} \log \left (2 \, c f^{d x + c} - a \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} + b\right ) -{\left ({\left (a b d x + a b c\right )} \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) +{\left (b^{2} c - 4 \, a c^{2} +{\left (b^{2} - 4 \, a c\right )} d x\right )} \log \left (f\right )\right )} \log \left (\frac{{\left (a \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} + b\right )} f^{d x + c} + 2 \, a}{2 \, a}\right ) +{\left ({\left (a b d x + a b c\right )} \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) -{\left (b^{2} c - 4 \, a c^{2} +{\left (b^{2} - 4 \, a c\right )} d x\right )} \log \left (f\right )\right )} \log \left (-\frac{{\left (a \sqrt{\frac{b^{2} - 4 \, a c}{a^{2}}} - b\right )} f^{d x + c} - 2 \, a}{2 \, a}\right )}{2 \,{\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b^2 - 4*a*c)*d^2*x^2*log(f)^2 - (a*b*sqrt((b^2 - 4*a*c)/a^2) + b^2 - 4*a*c)*dilog(-1/2*((a*sqrt((b^2 - 4
*a*c)/a^2) + b)*f^(d*x + c) + 2*a)/a + 1) + (a*b*sqrt((b^2 - 4*a*c)/a^2) - b^2 + 4*a*c)*dilog(1/2*((a*sqrt((b^
2 - 4*a*c)/a^2) - b)*f^(d*x + c) - 2*a)/a + 1) - (a*b*c*sqrt((b^2 - 4*a*c)/a^2)*log(f) - (b^2*c - 4*a*c^2)*log
(f))*log(2*c*f^(d*x + c) + a*sqrt((b^2 - 4*a*c)/a^2) + b) + (a*b*c*sqrt((b^2 - 4*a*c)/a^2)*log(f) + (b^2*c - 4
*a*c^2)*log(f))*log(2*c*f^(d*x + c) - a*sqrt((b^2 - 4*a*c)/a^2) + b) - ((a*b*d*x + a*b*c)*sqrt((b^2 - 4*a*c)/a
^2)*log(f) + (b^2*c - 4*a*c^2 + (b^2 - 4*a*c)*d*x)*log(f))*log(1/2*((a*sqrt((b^2 - 4*a*c)/a^2) + b)*f^(d*x + c
) + 2*a)/a) + ((a*b*d*x + a*b*c)*sqrt((b^2 - 4*a*c)/a^2)*log(f) - (b^2*c - 4*a*c^2 + (b^2 - 4*a*c)*d*x)*log(f)
)*log(-1/2*((a*sqrt((b^2 - 4*a*c)/a^2) - b)*f^(d*x + c) - 2*a)/a))/((a*b^2 - 4*a^2*c)*d^2*log(f)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(a + b*f**c*f**(d*x) + c*f**(2*c)*f**(2*d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{c f^{2 \, d x + 2 \, c} + b f^{d x + c} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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