Integrand size = 12, antiderivative size = 93 \[ \int x^3 \log \left (a+b e^x\right ) \, dx=\frac {1}{4} x^4 \log \left (a+b e^x\right )-\frac {1}{4} x^4 \log \left (1+\frac {b e^x}{a}\right )-x^3 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+3 x^2 \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-6 x \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )+6 \operatorname {PolyLog}\left (5,-\frac {b e^x}{a}\right ) \]
1/4*x^4*ln(a+b*exp(x))-1/4*x^4*ln(1+b*exp(x)/a)-x^3*polylog(2,-b*exp(x)/a) +3*x^2*polylog(3,-b*exp(x)/a)-6*x*polylog(4,-b*exp(x)/a)+6*polylog(5,-b*ex p(x)/a)
Time = 0.01 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int x^3 \log \left (a+b e^x\right ) \, dx=\frac {1}{4} x^4 \log \left (a+b e^x\right )-\frac {1}{4} x^4 \log \left (1+\frac {b e^x}{a}\right )-x^3 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+3 x^2 \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-6 x \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )+6 \operatorname {PolyLog}\left (5,-\frac {b e^x}{a}\right ) \]
(x^4*Log[a + b*E^x])/4 - (x^4*Log[1 + (b*E^x)/a])/4 - x^3*PolyLog[2, -((b* E^x)/a)] + 3*x^2*PolyLog[3, -((b*E^x)/a)] - 6*x*PolyLog[4, -((b*E^x)/a)] + 6*PolyLog[5, -((b*E^x)/a)]
Time = 0.53 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3012, 3011, 7163, 7163, 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.
\(\displaystyle \int x^3 \log \left (a+b e^x\right ) \, dx\) |
\(\Big \downarrow \) 3012 |
\(\displaystyle \int x^3 \log \left (\frac {e^x b}{a}+1\right )dx+\frac {1}{4} x^4 \log \left (a+b e^x\right )-\frac {1}{4} x^4 \log \left (\frac {b e^x}{a}+1\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle 3 \int x^2 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )dx-x^3 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\frac {1}{4} x^4 \log \left (a+b e^x\right )-\frac {1}{4} x^4 \log \left (\frac {b e^x}{a}+1\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle 3 \left (x^2 \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-2 \int x \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )dx\right )-x^3 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\frac {1}{4} x^4 \log \left (a+b e^x\right )-\frac {1}{4} x^4 \log \left (\frac {b e^x}{a}+1\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle 3 \left (x^2 \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-2 \left (x \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )-\int \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )dx\right )\right )-x^3 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\frac {1}{4} x^4 \log \left (a+b e^x\right )-\frac {1}{4} x^4 \log \left (\frac {b e^x}{a}+1\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle 3 \left (x^2 \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-2 \left (x \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )-\int e^{-x} \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )de^x\right )\right )-x^3 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\frac {1}{4} x^4 \log \left (a+b e^x\right )-\frac {1}{4} x^4 \log \left (\frac {b e^x}{a}+1\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle -x^3 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+3 \left (x^2 \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-2 \left (x \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )-\operatorname {PolyLog}\left (5,-\frac {b e^x}{a}\right )\right )\right )+\frac {1}{4} x^4 \log \left (a+b e^x\right )-\frac {1}{4} x^4 \log \left (\frac {b e^x}{a}+1\right )\) |
(x^4*Log[a + b*E^x])/4 - (x^4*Log[1 + (b*E^x)/a])/4 - x^3*PolyLog[2, -((b* E^x)/a)] + 3*(x^2*PolyLog[3, -((b*E^x)/a)] - 2*(x*PolyLog[4, -((b*E^x)/a)] - PolyLog[5, -((b*E^x)/a)]))
3.2.13.3.1 Defintions of rubi rules used
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[Log[(d_) + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g _.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*(Log[d + e*(F^(c*(a + b*x)))^n]/(g*(m + 1))), x] + (Int[(f + g*x)^m*Log[1 + (e/d)*(F^(c*(a + b*x) ))^n], x] - Simp[(f + g*x)^(m + 1)*(Log[1 + (e/d)*(F^(c*(a + b*x)))^n]/(g*( m + 1))), x]) /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[ d, 1]
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]
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. )*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F])) Int[(e + f*x) ^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c , d, e, f, n, p}, x] && GtQ[m, 0]
Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {x^{4} \ln \left (a +b \,{\mathrm e}^{x}\right )}{4}-\frac {x^{4} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{4}-x^{3} \operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+3 x^{2} \operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )-6 x \,\operatorname {Li}_{4}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+6 \,\operatorname {Li}_{5}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\) | \(84\) |
risch | \(\frac {x^{4} \ln \left (a +b \,{\mathrm e}^{x}\right )}{4}-\frac {x^{4} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{4}-x^{3} \operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+3 x^{2} \operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )-6 x \,\operatorname {Li}_{4}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+6 \,\operatorname {Li}_{5}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\) | \(84\) |
parts | \(\frac {x^{4} \ln \left (a +b \,{\mathrm e}^{x}\right )}{4}-\frac {x^{4} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{4}-x^{3} \operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+3 x^{2} \operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )-6 x \,\operatorname {Li}_{4}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+6 \,\operatorname {Li}_{5}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\) | \(84\) |
1/4*x^4*ln(a+b*exp(x))-1/4*x^4*ln(1+b*exp(x)/a)-x^3*polylog(2,-b*exp(x)/a) +3*x^2*polylog(3,-b*exp(x)/a)-6*x*polylog(4,-b*exp(x)/a)+6*polylog(5,-b*ex p(x)/a)
Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95 \[ \int x^3 \log \left (a+b e^x\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (b e^{x} + a\right ) - \frac {1}{4} \, x^{4} \log \left (\frac {b e^{x} + a}{a}\right ) - x^{3} {\rm Li}_2\left (-\frac {b e^{x} + a}{a} + 1\right ) + 3 \, x^{2} {\rm polylog}\left (3, -\frac {b e^{x}}{a}\right ) - 6 \, x {\rm polylog}\left (4, -\frac {b e^{x}}{a}\right ) + 6 \, {\rm polylog}\left (5, -\frac {b e^{x}}{a}\right ) \]
1/4*x^4*log(b*e^x + a) - 1/4*x^4*log((b*e^x + a)/a) - x^3*dilog(-(b*e^x + a)/a + 1) + 3*x^2*polylog(3, -b*e^x/a) - 6*x*polylog(4, -b*e^x/a) + 6*poly log(5, -b*e^x/a)
\[ \int x^3 \log \left (a+b e^x\right ) \, dx=- \frac {b \int \frac {x^{4} e^{x}}{a + b e^{x}}\, dx}{4} + \frac {x^{4} \log {\left (a + b e^{x} \right )}}{4} \]
Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.88 \[ \int x^3 \log \left (a+b e^x\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (b e^{x} + a\right ) - \frac {1}{4} \, x^{4} \log \left (\frac {b e^{x}}{a} + 1\right ) - x^{3} {\rm Li}_2\left (-\frac {b e^{x}}{a}\right ) + 3 \, x^{2} {\rm Li}_{3}(-\frac {b e^{x}}{a}) - 6 \, x {\rm Li}_{4}(-\frac {b e^{x}}{a}) + 6 \, {\rm Li}_{5}(-\frac {b e^{x}}{a}) \]
1/4*x^4*log(b*e^x + a) - 1/4*x^4*log(b*e^x/a + 1) - x^3*dilog(-b*e^x/a) + 3*x^2*polylog(3, -b*e^x/a) - 6*x*polylog(4, -b*e^x/a) + 6*polylog(5, -b*e^ x/a)
\[ \int x^3 \log \left (a+b e^x\right ) \, dx=\int { x^{3} \log \left (b e^{x} + a\right ) \,d x } \]
Timed out. \[ \int x^3 \log \left (a+b e^x\right ) \, dx=\int x^3\,\ln \left (a+b\,{\mathrm {e}}^x\right ) \,d x \]