Integrand size = 109, antiderivative size = 35 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=4+e^{x+e^x x}+\frac {5}{-x+\frac {4+x}{4}}-e^x \log (1+x) \]
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=e^{x+e^x x}-\frac {20}{-4+3 x}-e^x \log (1+x) \]
Integrate[(60 + 60*x + E^x*(-16 + 24*x - 9*x^2) + E^(x + E^x*x)*(16 - 8*x - 15*x^2 + 9*x^3 + E^x*(16 + 8*x - 23*x^2 - 6*x^3 + 9*x^4)) + E^x*(-16 + 8 *x + 15*x^2 - 9*x^3)*Log[1 + x])/(16 - 8*x - 15*x^2 + 9*x^3),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.
\(\displaystyle \int \frac {e^x \left (-9 x^2+24 x-16\right )+e^x \left (-9 x^3+15 x^2+8 x-16\right ) \log (x+1)+e^{e^x x+x} \left (9 x^3-15 x^2+e^x \left (9 x^4-6 x^3-23 x^2+8 x+16\right )-8 x+16\right )+60 x+60}{9 x^3-15 x^2-8 x+16} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {e^x \left (-9 x^2+24 x-16\right )+e^x \left (-9 x^3+15 x^2+8 x-16\right ) \log (x+1)+e^{e^x x+x} \left (9 x^3-15 x^2+e^x \left (9 x^4-6 x^3-23 x^2+8 x+16\right )-8 x+16\right )+60 x+60}{49 (x+1)}-\frac {3 \left (e^x \left (-9 x^2+24 x-16\right )+e^x \left (-9 x^3+15 x^2+8 x-16\right ) \log (x+1)+e^{e^x x+x} \left (9 x^3-15 x^2+e^x \left (9 x^4-6 x^3-23 x^2+8 x+16\right )-8 x+16\right )+60 x+60\right )}{49 (3 x-4)}+\frac {3 \left (e^x \left (-9 x^2+24 x-16\right )+e^x \left (-9 x^3+15 x^2+8 x-16\right ) \log (x+1)+e^{e^x x+x} \left (9 x^3-15 x^2+e^x \left (9 x^4-6 x^3-23 x^2+8 x+16\right )-8 x+16\right )+60 x+60\right )}{7 (3 x-4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{e^x x+x}dx+\int e^{e^x x+2 x}dx+\int e^{e^x x+2 x} xdx+\frac {9}{49} e^x x^2 \log (x+1)+\frac {20}{4-3 x}-\frac {1}{49} e^x (4-3 x)^2 \log (x+1)-\frac {6}{49} e^x (4-3 x) \log (x+1)+\frac {12}{49} e^x \log (x+1)-\frac {3}{7} e^x x \log (x+1)-\frac {3}{7} e^x (x+1) \log (x+1)\) |
Int[(60 + 60*x + E^x*(-16 + 24*x - 9*x^2) + E^(x + E^x*x)*(16 - 8*x - 15*x ^2 + 9*x^3 + E^x*(16 + 8*x - 23*x^2 - 6*x^3 + 9*x^4)) + E^x*(-16 + 8*x + 1 5*x^2 - 9*x^3)*Log[1 + x])/(16 - 8*x - 15*x^2 + 9*x^3),x]
3.21.21.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.72 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74
method | result | size |
risch | \(-\ln \left (1+x \right ) {\mathrm e}^{x}-\frac {20}{-4+3 x}+{\mathrm e}^{x \left ({\mathrm e}^{x}+1\right )}\) | \(26\) |
parallelrisch | \(\frac {-6 \,{\mathrm e}^{x} \ln \left (1+x \right ) x +6 \,{\mathrm e}^{x \left ({\mathrm e}^{x}+1\right )} x -120+8 \ln \left (1+x \right ) {\mathrm e}^{x}+60 x -8 \,{\mathrm e}^{x \left ({\mathrm e}^{x}+1\right )}}{-8+6 x}\) | \(51\) |
int((((9*x^4-6*x^3-23*x^2+8*x+16)*exp(x)+9*x^3-15*x^2-8*x+16)*exp(exp(x)*x +x)+(-9*x^3+15*x^2+8*x-16)*exp(x)*ln(1+x)+(-9*x^2+24*x-16)*exp(x)+60*x+60) /(9*x^3-15*x^2-8*x+16),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=-\frac {{\left (3 \, x - 4\right )} e^{x} \log \left (x + 1\right ) - {\left (3 \, x - 4\right )} e^{\left (x e^{x} + x\right )} + 20}{3 \, x - 4} \]
integrate((((9*x^4-6*x^3-23*x^2+8*x+16)*exp(x)+9*x^3-15*x^2-8*x+16)*exp(ex p(x)*x+x)+(-9*x^3+15*x^2+8*x-16)*exp(x)*log(1+x)+(-9*x^2+24*x-16)*exp(x)+6 0*x+60)/(9*x^3-15*x^2-8*x+16),x, algorithm=\
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.63 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=- e^{x} \log {\left (x + 1 \right )} + e^{x e^{x} + x} - \frac {60}{9 x - 12} \]
integrate((((9*x**4-6*x**3-23*x**2+8*x+16)*exp(x)+9*x**3-15*x**2-8*x+16)*e xp(exp(x)*x+x)+(-9*x**3+15*x**2+8*x-16)*exp(x)*ln(1+x)+(-9*x**2+24*x-16)*e xp(x)+60*x+60)/(9*x**3-15*x**2-8*x+16),x)
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=-e^{x} \log \left (x + 1\right ) - \frac {20}{3 \, x - 4} + e^{\left (x e^{x} + x\right )} \]
integrate((((9*x^4-6*x^3-23*x^2+8*x+16)*exp(x)+9*x^3-15*x^2-8*x+16)*exp(ex p(x)*x+x)+(-9*x^3+15*x^2+8*x-16)*exp(x)*log(1+x)+(-9*x^2+24*x-16)*exp(x)+6 0*x+60)/(9*x^3-15*x^2-8*x+16),x, algorithm=\
\[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=\int { -\frac {{\left (9 \, x^{3} - 15 \, x^{2} - 8 \, x + 16\right )} e^{x} \log \left (x + 1\right ) - {\left (9 \, x^{3} - 15 \, x^{2} + {\left (9 \, x^{4} - 6 \, x^{3} - 23 \, x^{2} + 8 \, x + 16\right )} e^{x} - 8 \, x + 16\right )} e^{\left (x e^{x} + x\right )} + {\left (9 \, x^{2} - 24 \, x + 16\right )} e^{x} - 60 \, x - 60}{9 \, x^{3} - 15 \, x^{2} - 8 \, x + 16} \,d x } \]
integrate((((9*x^4-6*x^3-23*x^2+8*x+16)*exp(x)+9*x^3-15*x^2-8*x+16)*exp(ex p(x)*x+x)+(-9*x^3+15*x^2+8*x-16)*exp(x)*log(1+x)+(-9*x^2+24*x-16)*exp(x)+6 0*x+60)/(9*x^3-15*x^2-8*x+16),x, algorithm=\
integrate(-((9*x^3 - 15*x^2 - 8*x + 16)*e^x*log(x + 1) - (9*x^3 - 15*x^2 + (9*x^4 - 6*x^3 - 23*x^2 + 8*x + 16)*e^x - 8*x + 16)*e^(x*e^x + x) + (9*x^ 2 - 24*x + 16)*e^x - 60*x - 60)/(9*x^3 - 15*x^2 - 8*x + 16), x)
Time = 9.62 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^x}\,{\mathrm {e}}^x-\frac {20}{3\,\left (x-\frac {4}{3}\right )}-\ln \left (x+1\right )\,{\mathrm {e}}^x \]