Integrand size = 101, antiderivative size = 23 \begin {dmath*} \int \frac {e^x \left (-9 x^3-3 x^4\right )+\left (-3 x^4+e^x \left (36 x^3+21 x^4+3 x^5\right )\right ) \log (x)+\left (18 x^2+6 x^3\right ) \log ^2(x)+\left (9 x^3+3 x^4+\left (-36 x^3-12 x^4\right ) \log (x)\right ) \log (3+x)}{(3+x) \log ^2(x)} \, dx=x^3 \left (2+\frac {3 x \left (e^x-\log (3+x)\right )}{\log (x)}\right ) \end {dmath*}
Time = 0.57 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \begin {dmath*} \int \frac {e^x \left (-9 x^3-3 x^4\right )+\left (-3 x^4+e^x \left (36 x^3+21 x^4+3 x^5\right )\right ) \log (x)+\left (18 x^2+6 x^3\right ) \log ^2(x)+\left (9 x^3+3 x^4+\left (-36 x^3-12 x^4\right ) \log (x)\right ) \log (3+x)}{(3+x) \log ^2(x)} \, dx=3 \left (\frac {2 x^3}{3}+\frac {x^4 \left (e^x-\log (3+x)\right )}{\log (x)}\right ) \end {dmath*}
Integrate[(E^x*(-9*x^3 - 3*x^4) + (-3*x^4 + E^x*(36*x^3 + 21*x^4 + 3*x^5)) *Log[x] + (18*x^2 + 6*x^3)*Log[x]^2 + (9*x^3 + 3*x^4 + (-36*x^3 - 12*x^4)* Log[x])*Log[3 + x])/((3 + x)*Log[x]^2),x]
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.
\(\displaystyle \int \frac {e^x \left (-3 x^4-9 x^3\right )+\left (3 x^4+9 x^3+\left (-12 x^4-36 x^3\right ) \log (x)\right ) \log (x+3)+\left (6 x^3+18 x^2\right ) \log ^2(x)+\left (e^x \left (3 x^5+21 x^4+36 x^3\right )-3 x^4\right ) \log (x)}{(x+3) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int 3 x^2 \left (\frac {x \left (e^x \left (x^2+7 x+12\right )-x-4 (x+3) \log (x+3)\right )}{(x+3) \log (x)}+\frac {x \left (\log (x+3)-e^x\right )}{\log ^2(x)}+2\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int x^2 \left (-\frac {x \left (e^x-\log (x+3)\right )}{\log ^2(x)}-\frac {x \left (x-e^x \left (x^2+7 x+12\right )+4 (x+3) \log (x+3)\right )}{(x+3) \log (x)}+2\right )dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle 3 \int \left (\frac {e^x x^3 (x \log (x)+4 \log (x)-1)}{\log ^2(x)}-\frac {x^2 \left (\log (x) x^2+4 \log (x) \log (x+3) x^2-\log (x+3) x^2-2 \log ^2(x) x+12 \log (x) \log (x+3) x-3 \log (x+3) x-6 \log ^2(x)\right )}{(x+3) \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (-\int \frac {x^4}{(x+3) \log (x)}dx+\int \frac {x^3 \log (x+3)}{\log ^2(x)}dx-4 \int \frac {x^3 \log (x+3)}{\log (x)}dx+\frac {e^x x^4}{\log (x)}+\frac {2 x^3}{3}\right )\) |
Int[(E^x*(-9*x^3 - 3*x^4) + (-3*x^4 + E^x*(36*x^3 + 21*x^4 + 3*x^5))*Log[x ] + (18*x^2 + 6*x^3)*Log[x]^2 + (9*x^3 + 3*x^4 + (-36*x^3 - 12*x^4)*Log[x] )*Log[3 + x])/((3 + x)*Log[x]^2),x]
3.2.82.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.80 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35
method | result | size |
parallelrisch | \(-\frac {-18 \,{\mathrm e}^{x} x^{4}+18 \ln \left (3+x \right ) x^{4}-12 x^{3} \ln \left (x \right )}{6 \ln \left (x \right )}\) | \(31\) |
risch | \(-\frac {3 x^{4} \ln \left (3+x \right )}{\ln \left (x \right )}+\frac {x^{3} \left (3 \,{\mathrm e}^{x} x +2 \ln \left (x \right )\right )}{\ln \left (x \right )}\) | \(33\) |
int((((-12*x^4-36*x^3)*ln(x)+3*x^4+9*x^3)*ln(3+x)+(6*x^3+18*x^2)*ln(x)^2+( (3*x^5+21*x^4+36*x^3)*exp(x)-3*x^4)*ln(x)+(-3*x^4-9*x^3)*exp(x))/(3+x)/ln( x)^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {e^x \left (-9 x^3-3 x^4\right )+\left (-3 x^4+e^x \left (36 x^3+21 x^4+3 x^5\right )\right ) \log (x)+\left (18 x^2+6 x^3\right ) \log ^2(x)+\left (9 x^3+3 x^4+\left (-36 x^3-12 x^4\right ) \log (x)\right ) \log (3+x)}{(3+x) \log ^2(x)} \, dx=\frac {3 \, x^{4} e^{x} - 3 \, x^{4} \log \left (x + 3\right ) + 2 \, x^{3} \log \left (x\right )}{\log \left (x\right )} \end {dmath*}
integrate((((-12*x^4-36*x^3)*log(x)+3*x^4+9*x^3)*log(3+x)+(6*x^3+18*x^2)*l og(x)^2+((3*x^5+21*x^4+36*x^3)*exp(x)-3*x^4)*log(x)+(-3*x^4-9*x^3)*exp(x)) /(3+x)/log(x)^2,x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {e^x \left (-9 x^3-3 x^4\right )+\left (-3 x^4+e^x \left (36 x^3+21 x^4+3 x^5\right )\right ) \log (x)+\left (18 x^2+6 x^3\right ) \log ^2(x)+\left (9 x^3+3 x^4+\left (-36 x^3-12 x^4\right ) \log (x)\right ) \log (3+x)}{(3+x) \log ^2(x)} \, dx=\frac {3 x^{4} e^{x}}{\log {\left (x \right )}} - \frac {3 x^{4} \log {\left (x + 3 \right )}}{\log {\left (x \right )}} + 2 x^{3} \end {dmath*}
integrate((((-12*x**4-36*x**3)*ln(x)+3*x**4+9*x**3)*ln(3+x)+(6*x**3+18*x** 2)*ln(x)**2+((3*x**5+21*x**4+36*x**3)*exp(x)-3*x**4)*ln(x)+(-3*x**4-9*x**3 )*exp(x))/(3+x)/ln(x)**2,x)
Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {e^x \left (-9 x^3-3 x^4\right )+\left (-3 x^4+e^x \left (36 x^3+21 x^4+3 x^5\right )\right ) \log (x)+\left (18 x^2+6 x^3\right ) \log ^2(x)+\left (9 x^3+3 x^4+\left (-36 x^3-12 x^4\right ) \log (x)\right ) \log (3+x)}{(3+x) \log ^2(x)} \, dx=\frac {3 \, x^{4} e^{x} - 3 \, x^{4} \log \left (x + 3\right ) + 2 \, x^{3} \log \left (x\right )}{\log \left (x\right )} \end {dmath*}
integrate((((-12*x^4-36*x^3)*log(x)+3*x^4+9*x^3)*log(3+x)+(6*x^3+18*x^2)*l og(x)^2+((3*x^5+21*x^4+36*x^3)*exp(x)-3*x^4)*log(x)+(-3*x^4-9*x^3)*exp(x)) /(3+x)/log(x)^2,x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {e^x \left (-9 x^3-3 x^4\right )+\left (-3 x^4+e^x \left (36 x^3+21 x^4+3 x^5\right )\right ) \log (x)+\left (18 x^2+6 x^3\right ) \log ^2(x)+\left (9 x^3+3 x^4+\left (-36 x^3-12 x^4\right ) \log (x)\right ) \log (3+x)}{(3+x) \log ^2(x)} \, dx=\frac {3 \, x^{4} e^{x} - 3 \, x^{4} \log \left (x + 3\right ) + 2 \, x^{3} \log \left (x\right )}{\log \left (x\right )} \end {dmath*}
integrate((((-12*x^4-36*x^3)*log(x)+3*x^4+9*x^3)*log(3+x)+(6*x^3+18*x^2)*l og(x)^2+((3*x^5+21*x^4+36*x^3)*exp(x)-3*x^4)*log(x)+(-3*x^4-9*x^3)*exp(x)) /(3+x)/log(x)^2,x, algorithm=\
Time = 13.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.91 \begin {dmath*} \int \frac {e^x \left (-9 x^3-3 x^4\right )+\left (-3 x^4+e^x \left (36 x^3+21 x^4+3 x^5\right )\right ) \log (x)+\left (18 x^2+6 x^3\right ) \log ^2(x)+\left (9 x^3+3 x^4+\left (-36 x^3-12 x^4\right ) \log (x)\right ) \log (3+x)}{(3+x) \log ^2(x)} \, dx={\mathrm {e}}^x\,\left (3\,x^5+12\,x^4\right )+2\,x^3+\frac {3\,x^4\,{\mathrm {e}}^x-3\,x^4\,{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x+4\right )}{\ln \left (x\right )}+\frac {\ln \left (x+3\right )\,\left (\ln \left (x\right )\,\left (12\,x^4-\frac {12\,x^6+36\,x^5}{x\,\left (x+3\right )}\right )-3\,x^4\right )}{\ln \left (x\right )} \end {dmath*}
int((log(x)*(exp(x)*(36*x^3 + 21*x^4 + 3*x^5) - 3*x^4) - exp(x)*(9*x^3 + 3 *x^4) + log(x)^2*(18*x^2 + 6*x^3) + log(x + 3)*(9*x^3 - log(x)*(36*x^3 + 1 2*x^4) + 3*x^4))/(log(x)^2*(x + 3)),x)