Integrand size = 17, antiderivative size = 84 \[ \int \frac {x^2}{a+b e^{c+d x}} \, dx=\frac {x^3}{3 a}-\frac {x^2 \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a d}-\frac {2 x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a d^2}+\frac {2 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{a d^3} \]
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
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int \frac {x^2}{a+b e^{c+d x}} \, dx=-\frac {x^2 \log \left (1+\frac {a e^{-c-d x}}{b}\right )}{a d}+\frac {2 x \operatorname {PolyLog}\left (2,-\frac {a e^{-c-d x}}{b}\right )}{a d^2}+\frac {2 \operatorname {PolyLog}\left (3,-\frac {a e^{-c-d x}}{b}\right )}{a d^3} \]
-((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)
Time = 0.50 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2615, 2620, 3011, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^2}{a+b e^{c+d x}} \, dx\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {x^3}{3 a}-\frac {b \int \frac {e^{c+d x} x^2}{a+b e^{c+d x}}dx}{a}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \int x \log \left (\frac {e^{c+d x} b}{a}+1\right )dx}{b d}\right )}{a}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\int \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )dx}{d}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\int e^{-c-d x} \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )de^{c+d x}}{d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\) |
x^3/(3*a) - (b*((x^2*Log[1 + (b*E^(c + d*x))/a])/(b*d) - (2*(-((x*PolyLog[ 2, -((b*E^(c + d*x))/a)])/d) + PolyLog[3, -((b*E^(c + d*x))/a)]/d^2))/(b*d )))/a
3.1.2.3.1 Defintions of rubi rules used
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] - Simp[ 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]
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] - Si mp[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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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] + Simp[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]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
Leaf count of result is larger than twice the leaf count of optimal. \(161\) vs. \(2(79)=158\).
Time = 0.01 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.93
method | result | size |
derivativedivides | \(\frac {c^{2} \left (-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a}+\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a}\right )-\frac {-\frac {\left (d x +c \right )^{3}}{3}+\left (d x +c \right )^{2} \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )+2 \left (d x +c \right ) \operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )-2 \,\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}+\frac {2 c \left (-\frac {\left (d x +c \right )^{2}}{2}+\left (d x +c \right ) \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )+\operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )\right )}{a}}{d^{3}}\) | \(162\) |
default | \(\frac {c^{2} \left (-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a}+\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a}\right )-\frac {-\frac {\left (d x +c \right )^{3}}{3}+\left (d x +c \right )^{2} \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )+2 \left (d x +c \right ) \operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )-2 \,\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}+\frac {2 c \left (-\frac {\left (d x +c \right )^{2}}{2}+\left (d x +c \right ) \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )+\operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )\right )}{a}}{d^{3}}\) | \(162\) |
risch | \(-\frac {c^{2} \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{d^{3} a}+\frac {c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{d^{3} a}+\frac {x^{3}}{3 a}-\frac {c^{2} x}{d^{2} a}-\frac {2 c^{3}}{3 d^{3} a}-\frac {x^{2} \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a d}+\frac {\ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right ) c^{2}}{d^{3} a}-\frac {2 x \,\operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a \,d^{2}}+\frac {2 \,\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a \,d^{3}}\) | \(166\) |
1/d^3*(c^2*(-1/a*ln(a+b*exp(d*x+c))+1/a*ln(exp(d*x+c)))-1/a*(-1/3*(d*x+c)^ 3+(d*x+c)^2*ln(1+b*exp(d*x+c)/a)+2*(d*x+c)*polylog(2,-b*exp(d*x+c)/a)-2*po lylog(3,-b*exp(d*x+c)/a))+2*c/a*(-1/2*(d*x+c)^2+(d*x+c)*ln(1+b*exp(d*x+c)/ a)+polylog(2,-b*exp(d*x+c)/a)))
Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19 \[ \int \frac {x^2}{a+b e^{c+d x}} \, dx=\frac {d^{3} x^{3} - 6 \, d x {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )} + a}{a} + 1\right ) - 3 \, c^{2} \log \left (b e^{\left (d x + c\right )} + a\right ) - 3 \, {\left (d^{2} x^{2} - c^{2}\right )} \log \left (\frac {b e^{\left (d x + c\right )} + a}{a}\right ) + 6 \, {\rm polylog}\left (3, -\frac {b e^{\left (d x + c\right )}}{a}\right )}{3 \, a d^{3}} \]
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)
\[ \int \frac {x^2}{a+b e^{c+d x}} \, dx=\int \frac {x^{2}}{a + b e^{c} e^{d x}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{a+b e^{c+d x}} \, dx=\frac {x^{3}}{3 \, a} - \frac {d^{2} x^{2} \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + 2 \, d x {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right ) - 2 \, {\rm Li}_{3}(-\frac {b e^{\left (d x + c\right )}}{a})}{a d^{3}} \]
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)
\[ \int \frac {x^2}{a+b e^{c+d x}} \, dx=\int { \frac {x^{2}}{b e^{\left (d x + c\right )} + a} \,d x } \]
Timed out. \[ \int \frac {x^2}{a+b e^{c+d x}} \, dx=\int \frac {x^2}{a+b\,{\mathrm {e}}^{c+d\,x}} \,d x \]