3.1.2 \(\int \frac {x^2}{a+b e^{c+d x}} \, dx\) [2]

Optimal. Leaf size=84 \[ \frac {x^3}{3 a}-\frac {x^2 \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a d}-\frac {2 x \text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a d^2}+\frac {2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a}\right )}{a d^3} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2215, 2221, 2611, 2320, 6724} \begin {gather*} \frac {2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{a d^3}-\frac {2 x \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a d^2}-\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{a d}+\frac {x^3}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*E^(c + d*x)),x]

[Out]

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

Rule 2215

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

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Mathematica [A]
time = 0.04, size = 83, normalized size = 0.99 \begin {gather*} -\frac {x^2 \log \left (1+\frac {a e^{-c-d x}}{b}\right )}{a d}+\frac {2 x \text {Li}_2\left (-\frac {a e^{-c-d x}}{b}\right )}{a d^2}+\frac {2 \text {Li}_3\left (-\frac {a e^{-c-d x}}{b}\right )}{a d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*E^(c + d*x)),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(177\) vs. \(2(79)=158\).
time = 0.01, size = 178, normalized size = 2.12

method	result	size
risch	\(\frac {c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{d^{3} a}-\frac {c^{2} \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{d^{3} a}+\frac {x^{3}}{3 a}-\frac {c^{2} x}{d^{2} a}-\frac {2 c^{3}}{3 d^{3} a}-\frac {x^{2} \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a d}+\frac {\ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right ) c^{2}}{d^{3} a}-\frac {2 x \polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a \,d^{2}}+\frac {2 \polylog \left (3, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a \,d^{3}}\)	\(166\)
derivativedivides	\(\frac {\frac {c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a}-\frac {c^{2} \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a}+\frac {\left (d x +c \right )^{3}}{3 a}-\frac {\left (d x +c \right )^{2} \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}-\frac {2 \left (d x +c \right ) \polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}+\frac {2 \polylog \left (3, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}-\frac {c \left (d x +c \right )^{2}}{a}+\frac {2 c \left (d x +c \right ) \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}+\frac {2 c \polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}}{d^{3}}\)	\(178\)
default	\(\frac {\frac {c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a}-\frac {c^{2} \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a}+\frac {\left (d x +c \right )^{3}}{3 a}-\frac {\left (d x +c \right )^{2} \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}-\frac {2 \left (d x +c \right ) \polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}+\frac {2 \polylog \left (3, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}-\frac {c \left (d x +c \right )^{2}}{a}+\frac {2 c \left (d x +c \right ) \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}+\frac {2 c \polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}}{d^{3}}\)	\(178\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+b*exp(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 72, normalized size = 0.86 \begin {gather*} \frac {x^{3}}{3 \, a} - \frac {d^{2} x^{2} \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + 2 \, d x {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right ) - 2 \, {\rm Li}_{3}(-\frac {b e^{\left (d x + c\right )}}{a})}{a d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.42, size = 100, normalized size = 1.19 \begin {gather*} \frac {d^{3} x^{3} - 6 \, d x {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )} + a}{a} + 1\right ) - 3 \, c^{2} \log \left (b e^{\left (d x + c\right )} + a\right ) - 3 \, {\left (d^{2} x^{2} - c^{2}\right )} \log \left (\frac {b e^{\left (d x + c\right )} + a}{a}\right ) + 6 \, {\rm polylog}\left (3, -\frac {b e^{\left (d x + c\right )}}{a}\right )}{3 \, a d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(d^3*x^3 - 6*d*x*dilog(-(b*e^(d*x + c) + a)/a + 1) - 3*c^2*log(b*e^(d*x + c) + a) - 3*(d^2*x^2 - c^2)*log(
(b*e^(d*x + c) + a)/a) + 6*polylog(3, -b*e^(d*x + c)/a))/(a*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*exp(d*x+c)),x, algorithm="giac")

[Out]

integrate(x^2/(b*e^(d*x + c) + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{a+b\,{\mathrm {e}}^{c+d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*exp(c + d*x)),x)

[Out]

int(x^2/(a + b*exp(c + d*x)), x)

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