Integrand size = 81, antiderivative size = 29 \begin {dmath*} \int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx=\log (x)+\frac {-\frac {2}{3}-x+\frac {-2+\frac {1}{5} \log (1+x)}{e^4}}{\log (x)} \end {dmath*}
Time = 0.84 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \begin {dmath*} \int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx=-\frac {6+2 e^4+3 e^4 x}{3 e^4 \log (x)}+\log (x)+\frac {\log (1+x)}{5 e^4 \log (x)} \end {dmath*}
Integrate[(30 + 30*x + E^4*(10 + 25*x + 15*x^2) + (3*x + E^4*(-15*x - 15*x ^2))*Log[x] + E^4*(15 + 15*x)*Log[x]^2 + (-3 - 3*x)*Log[1 + x])/(E^4*(15*x + 15*x^2)*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.
\(\displaystyle \int \frac {e^4 \left (15 x^2+25 x+10\right )+\left (e^4 \left (-15 x^2-15 x\right )+3 x\right ) \log (x)+30 x+e^4 (15 x+15) \log ^2(x)+(-3 x-3) \log (x+1)+30}{e^4 \left (15 x^2+15 x\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {15 e^4 (x+1) \log ^2(x)+3 \left (x-5 e^4 \left (x^2+x\right )\right ) \log (x)+30 x+5 e^4 \left (3 x^2+5 x+2\right )-3 (x+1) \log (x+1)+30}{15 \left (x^2+x\right ) \log ^2(x)}dx}{e^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {15 e^4 (x+1) \log ^2(x)+3 \left (x-5 e^4 \left (x^2+x\right )\right ) \log (x)+30 x+5 e^4 \left (3 x^2+5 x+2\right )-3 (x+1) \log (x+1)+30}{\left (x^2+x\right ) \log ^2(x)}dx}{15 e^4}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \frac {\int \frac {15 e^4 (x+1) \log ^2(x)+3 \left (x-5 e^4 \left (x^2+x\right )\right ) \log (x)+30 x+5 e^4 \left (3 x^2+5 x+2\right )-3 (x+1) \log (x+1)+30}{x (x+1) \log ^2(x)}dx}{15 e^4}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {-15 e^4 \log (x) x^2+15 e^4 x^2+15 e^4 \log ^2(x) x+3 \left (1-5 e^4\right ) \log (x) x+30 \left (1+\frac {5 e^4}{6}\right ) x+15 e^4 \log ^2(x)+30 \left (1+\frac {e^4}{3}\right )}{x (x+1) \log ^2(x)}-\frac {3 \log (x+1)}{x \log ^2(x)}\right )dx}{15 e^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-3 \int \frac {\log (x+1)}{x \log ^2(x)}dx-3 \int \frac {5 e^4 x+5 e^4-1}{(x+1) \log (x)}dx+15 e^4 \operatorname {LogIntegral}(x)-\frac {15 e^4 x}{\log (x)}+15 e^4 \log (x)-\frac {10 \left (3+e^4\right )}{\log (x)}}{15 e^4}\) |
Int[(30 + 30*x + E^4*(10 + 25*x + 15*x^2) + (3*x + E^4*(-15*x - 15*x^2))*L og[x] + E^4*(15 + 15*x)*Log[x]^2 + (-3 - 3*x)*Log[1 + x])/(E^4*(15*x + 15* x^2)*Log[x]^2),x]
3.1.95.3.1 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[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 3.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{-4} \left (-15 \,{\mathrm e}^{4} \ln \left (x \right )^{2}+15 x \,{\mathrm e}^{4}+30+10 \,{\mathrm e}^{4}-3 \ln \left (1+x \right )\right )}{15 \ln \left (x \right )}\) | \(36\) |
risch | \(\frac {{\mathrm e}^{-4} \ln \left (1+x \right )}{5 \ln \left (x \right )}+\frac {{\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (x \right )^{2}-3 x \,{\mathrm e}^{4}-2 \,{\mathrm e}^{4}-6\right )}{3 \ln \left (x \right )}\) | \(41\) |
default | \(\frac {{\mathrm e}^{-4} \left (\frac {3 \ln \left (1+x \right )}{\ln \left (x \right )}-\frac {10 \,{\mathrm e}^{4}+30}{\ln \left (x \right )}+15 \,{\mathrm e}^{4} \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )+15 \,{\mathrm e}^{4} \ln \left (x \right )+15 \,{\mathrm e}^{4} \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )}{15}\) | \(66\) |
int(((-3*x-3)*ln(1+x)+(15*x+15)*exp(4)*ln(x)^2+((-15*x^2-15*x)*exp(4)+3*x) *ln(x)+(15*x^2+25*x+10)*exp(4)+30*x+30)/(15*x^2+15*x)/exp(4)/ln(x)^2,x,met hod=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \begin {dmath*} \int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {{\left (15 \, e^{4} \log \left (x\right )^{2} - 5 \, {\left (3 \, x + 2\right )} e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \left (x\right )} \end {dmath*}
integrate(((-3*x-3)*log(1+x)+(15*x+15)*exp(4)*log(x)^2+((-15*x^2-15*x)*exp (4)+3*x)*log(x)+(15*x^2+25*x+10)*exp(4)+30*x+30)/(15*x^2+15*x)/exp(4)/log( x)^2,x, algorithm=\
Exception generated. \begin {dmath*} \int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \end {dmath*}
integrate(((-3*x-3)*ln(1+x)+(15*x+15)*exp(4)*ln(x)**2+((-15*x**2-15*x)*exp (4)+3*x)*ln(x)+(15*x**2+25*x+10)*exp(4)+30*x+30)/(15*x**2+15*x)/exp(4)/ln( x)**2,x)
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \begin {dmath*} \int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {{\left (15 \, e^{4} \log \left (x\right )^{2} - 15 \, x e^{4} - 10 \, e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \left (x\right )} \end {dmath*}
integrate(((-3*x-3)*log(1+x)+(15*x+15)*exp(4)*log(x)^2+((-15*x^2-15*x)*exp (4)+3*x)*log(x)+(15*x^2+25*x+10)*exp(4)+30*x+30)/(15*x^2+15*x)/exp(4)/log( x)^2,x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \begin {dmath*} \int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {{\left (15 \, e^{4} \log \left (x\right )^{2} - 15 \, x e^{4} - 10 \, e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \left (x\right )} \end {dmath*}
integrate(((-3*x-3)*log(1+x)+(15*x+15)*exp(4)*log(x)^2+((-15*x^2-15*x)*exp (4)+3*x)*log(x)+(15*x^2+25*x+10)*exp(4)+30*x+30)/(15*x^2+15*x)/exp(4)/log( x)^2,x, algorithm=\
Time = 14.46 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \begin {dmath*} \int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx=\ln \left (x\right )-\frac {{\mathrm {e}}^{-4}\,\left (10\,{\mathrm {e}}^4-3\,\ln \left (x+1\right )+15\,x\,{\mathrm {e}}^4+30\right )}{15\,\ln \left (x\right )} \end {dmath*}