3.6.42 \(\int \frac {x^2}{a+b f^{-c-d x}+c f^{c+d x}} \, dx\) [542]

Optimal. Leaf size=310 \[ \frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}+\frac {2 x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {2 x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {2 \text {Li}_3\left (-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^3 \log ^3(f)}+\frac {2 \text {Li}_3\left (-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^3 \log ^3(f)} \]

[Out]

x^2*ln(1+2*c*f^(d*x+c)/(a-(a^2-4*b*c)^(1/2)))/d/ln(f)/(a^2-4*b*c)^(1/2)-x^2*ln(1+2*c*f^(d*x+c)/(a+(a^2-4*b*c)^
(1/2)))/d/ln(f)/(a^2-4*b*c)^(1/2)+2*x*polylog(2,-2*c*f^(d*x+c)/(a-(a^2-4*b*c)^(1/2)))/d^2/ln(f)^2/(a^2-4*b*c)^
(1/2)-2*x*polylog(2,-2*c*f^(d*x+c)/(a+(a^2-4*b*c)^(1/2)))/d^2/ln(f)^2/(a^2-4*b*c)^(1/2)-2*polylog(3,-2*c*f^(d*
x+c)/(a-(a^2-4*b*c)^(1/2)))/d^3/ln(f)^3/(a^2-4*b*c)^(1/2)+2*polylog(3,-2*c*f^(d*x+c)/(a+(a^2-4*b*c)^(1/2)))/d^
3/ln(f)^3/(a^2-4*b*c)^(1/2)

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Rubi [A]
time = 0.46, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2299, 2296, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {2 \text {PolyLog}\left (3,-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{d^3 \log ^3(f) \sqrt {a^2-4 b c}}+\frac {2 \text {PolyLog}\left (3,-\frac {2 c f^{c+d x}}{\sqrt {a^2-4 b c}+a}\right )}{d^3 \log ^3(f) \sqrt {a^2-4 b c}}+\frac {2 x \text {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{d^2 \log ^2(f) \sqrt {a^2-4 b c}}-\frac {2 x \text {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{\sqrt {a^2-4 b c}+a}\right )}{d^2 \log ^2(f) \sqrt {a^2-4 b c}}+\frac {x^2 \log \left (\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}+1\right )}{d \log (f) \sqrt {a^2-4 b c}}-\frac {x^2 \log \left (\frac {2 c f^{c+d x}}{\sqrt {a^2-4 b c}+a}+1\right )}{d \log (f) \sqrt {a^2-4 b c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Log[1 + (2*c*f^(c + d*x))/(a - Sqrt[a^2 - 4*b*c])])/(Sqrt[a^2 - 4*b*c]*d*Log[f]) - (x^2*Log[1 + (2*c*f^(c
+ d*x))/(a + Sqrt[a^2 - 4*b*c])])/(Sqrt[a^2 - 4*b*c]*d*Log[f]) + (2*x*PolyLog[2, (-2*c*f^(c + d*x))/(a - Sqrt
[a^2 - 4*b*c])])/(Sqrt[a^2 - 4*b*c]*d^2*Log[f]^2) - (2*x*PolyLog[2, (-2*c*f^(c + d*x))/(a + Sqrt[a^2 - 4*b*c])
])/(Sqrt[a^2 - 4*b*c]*d^2*Log[f]^2) - (2*PolyLog[3, (-2*c*f^(c + d*x))/(a - Sqrt[a^2 - 4*b*c])])/(Sqrt[a^2 - 4
*b*c]*d^3*Log[f]^3) + (2*PolyLog[3, (-2*c*f^(c + d*x))/(a + Sqrt[a^2 - 4*b*c])])/(Sqrt[a^2 - 4*b*c]*d^3*Log[f]
^3)

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

Int[((F_)^(u_)*((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*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(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 2299

Int[(u_)/((a_) + (b_.)*(F_)^(v_) + (c_.)*(F_)^(w_)), x_Symbol] :> Int[u*(F^v/(c + a*F^v + b*F^(2*v))), x] /; F
reeQ[{F, a, b, c}, x] && EqQ[w, -v] && LinearQ[v, x] && If[RationalQ[Coefficient[v, x, 1]], GtQ[Coefficient[v,
x, 1], 0], LtQ[LeafCount[v], LeafCount[w]]]

Rule 2320

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 2611

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 6724

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 {x^2}{a+b f^{-c-d x}+c f^{c+d x}} \, dx &=\int \frac {f^{c+d x} x^2}{b+a f^{c+d x}+c f^{2 (c+d x)}} \, dx\\ &=\frac {(2 c) \int \frac {f^{c+d x} x^2}{a-\sqrt {a^2-4 b c}+2 c f^{c+d x}} \, dx}{\sqrt {a^2-4 b c}}-\frac {(2 c) \int \frac {f^{c+d x} x^2}{a+\sqrt {a^2-4 b c}+2 c f^{c+d x}} \, dx}{\sqrt {a^2-4 b c}}\\ &=\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {2 \int x \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right ) \, dx}{\sqrt {a^2-4 b c} d \log (f)}+\frac {2 \int x \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right ) \, dx}{\sqrt {a^2-4 b c} d \log (f)}\\ &=\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}+\frac {2 x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {2 x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {2 \int \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right ) \, dx}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}+\frac {2 \int \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right ) \, dx}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}\\ &=\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}+\frac {2 x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {2 x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {2 \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {2 c x}{-a+\sqrt {a^2-4 b c}}\right )}{x} \, dx,x,f^{c+d x}\right )}{\sqrt {a^2-4 b c} d^3 \log ^3(f)}+\frac {2 \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {2 c x}{a+\sqrt {a^2-4 b c}}\right )}{x} \, dx,x,f^{c+d x}\right )}{\sqrt {a^2-4 b c} d^3 \log ^3(f)}\\ &=\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x^2 \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}+\frac {2 x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {2 x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {2 \text {Li}_3\left (-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^3 \log ^3(f)}+\frac {2 \text {Li}_3\left (-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^3 \log ^3(f)}\\ \end {align*}

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Mathematica [F]
time = 0.22, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2}{a+b f^{-c-d x}+c f^{c+d x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{a +b \,f^{-d x -c}+c \,f^{d x +c}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a^2-4*b*c>0)', see `assume?` f
or more deta

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Fricas [A]
time = 0.38, size = 489, normalized size = 1.58 \begin {gather*} -\frac {b c^{2} \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (2 \, c f^{d x + c} + b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right ) \log \left (f\right )^{2} - b c^{2} \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (2 \, c f^{d x + c} - b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right ) \log \left (f\right )^{2} - 2 \, b d x \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} {\rm Li}_2\left (-\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right )} f^{d x + c} + 2 \, b}{2 \, b} + 1\right ) \log \left (f\right ) + 2 \, b d x \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} {\rm Li}_2\left (\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} - a\right )} f^{d x + c} - 2 \, b}{2 \, b} + 1\right ) \log \left (f\right ) - {\left (b d^{2} x^{2} - b c^{2}\right )} \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (f\right )^{2} \log \left (\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right )} f^{d x + c} + 2 \, b}{2 \, b}\right ) + {\left (b d^{2} x^{2} - b c^{2}\right )} \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (f\right )^{2} \log \left (-\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} - a\right )} f^{d x + c} - 2 \, b}{2 \, b}\right ) + 2 \, b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} {\rm polylog}\left (3, -\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right )} f^{d x + c}}{2 \, b}\right ) - 2 \, b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} {\rm polylog}\left (3, \frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} - a\right )} f^{d x + c}}{2 \, b}\right )}{{\left (a^{2} - 4 \, b c\right )} d^{3} \log \left (f\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*c^2*sqrt((a^2 - 4*b*c)/b^2)*log(2*c*f^(d*x + c) + b*sqrt((a^2 - 4*b*c)/b^2) + a)*log(f)^2 - b*c^2*sqrt((a^
2 - 4*b*c)/b^2)*log(2*c*f^(d*x + c) - b*sqrt((a^2 - 4*b*c)/b^2) + a)*log(f)^2 - 2*b*d*x*sqrt((a^2 - 4*b*c)/b^2
)*dilog(-1/2*((b*sqrt((a^2 - 4*b*c)/b^2) + a)*f^(d*x + c) + 2*b)/b + 1)*log(f) + 2*b*d*x*sqrt((a^2 - 4*b*c)/b^
2)*dilog(1/2*((b*sqrt((a^2 - 4*b*c)/b^2) - a)*f^(d*x + c) - 2*b)/b + 1)*log(f) - (b*d^2*x^2 - b*c^2)*sqrt((a^2
- 4*b*c)/b^2)*log(f)^2*log(1/2*((b*sqrt((a^2 - 4*b*c)/b^2) + a)*f^(d*x + c) + 2*b)/b) + (b*d^2*x^2 - b*c^2)*s
qrt((a^2 - 4*b*c)/b^2)*log(f)^2*log(-1/2*((b*sqrt((a^2 - 4*b*c)/b^2) - a)*f^(d*x + c) - 2*b)/b) + 2*b*sqrt((a^
2 - 4*b*c)/b^2)*polylog(3, -1/2*(b*sqrt((a^2 - 4*b*c)/b^2) + a)*f^(d*x + c)/b) - 2*b*sqrt((a^2 - 4*b*c)/b^2)*p
olylog(3, 1/2*(b*sqrt((a^2 - 4*b*c)/b^2) - a)*f^(d*x + c)/b))/((a^2 - 4*b*c)*d^3*log(f)^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2}{a+c\,f^{c+d\,x}+\frac {b}{f^{c+d\,x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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