\(\int \frac {60+60 x+e^x (-16+24 x-9 x^2)+e^{x+e^x x} (16-8 x-15 x^2+9 x^3+e^x (16+8 x-23 x^2-6 x^3+9 x^4))+e^x (-16+8 x+15 x^2-9 x^3) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx\) [2021]

Optimal result

Integrand size = 109, antiderivative size = 35 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=4+e^{x+e^x x}+\frac {5}{-x+\frac {4+x}{4}}-e^x \log (1+x) \]

[Out]

4-ln(1+x)*exp(x)+exp(exp(x)*x+x)+5/(1-3/4*x)

Rubi [F]

\[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=\int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx \]

[In]

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 + 15*x^2 - 9*x^3)*Log[1 + x])/(16 - 8*x - 15*x^2 + 9*x^3),x]

[Out]

20/(4 - 3*x) - E^x*Log[1 + x] + Defer[Int][E^(x + E^x*x), x] + Defer[Int][E^(2*x + E^x*x), x] + Defer[Int][E^(
2*x + E^x*x)*x, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (e^{x+e^x x}+\frac {60}{(4-3 x)^2}-\frac {e^x}{1+x}+e^{2 x+e^x x} (1+x)-e^x \log (1+x)\right ) \, dx \\ & = \frac {20}{4-3 x}+\int e^{x+e^x x} \, dx-\int \frac {e^x}{1+x} \, dx+\int e^{2 x+e^x x} (1+x) \, dx-\int e^x \log (1+x) \, dx \\ & = \frac {20}{4-3 x}-\frac {\operatorname {ExpIntegralEi}(1+x)}{e}-e^x \log (1+x)+\int e^{x+e^x x} \, dx+\int \frac {e^x}{1+x} \, dx+\int \left (e^{2 x+e^x x}+e^{2 x+e^x x} x\right ) \, dx \\ & = \frac {20}{4-3 x}-e^x \log (1+x)+\int e^{x+e^x x} \, dx+\int e^{2 x+e^x x} \, dx+\int e^{2 x+e^x x} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=e^{x+e^x x}-\frac {20}{-4+3 x}-e^x \log (1+x) \]

[In]

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 - 2
3*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]

[Out]

E^(x + E^x*x) - 20/(-4 + 3*x) - E^x*Log[1 + x]

Maple [A] (verified)

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	\(\frac {-6 \,{\mathrm e}^{x} \ln \left (1+x \right ) x +6 \,{\mathrm e}^{x \left ({\mathrm e}^{x}+1\right )} x -120+8 \ln \left (1+x \right ) {\mathrm e}^{x}+60 x -8 \,{\mathrm e}^{x \left ({\mathrm e}^{x}+1\right )}}{-8+6 x}\)	\(51\)

[In]

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)

[Out]

-ln(1+x)*exp(x)-20/(-4+3*x)+exp(x*(exp(x)+1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=-\frac {{\left (3 \, x - 4\right )} e^{x} \log \left (x + 1\right ) - {\left (3 \, x - 4\right )} e^{\left (x e^{x} + x\right )} + 20}{3 \, x - 4} \]

[In]

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(exp(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)+60*x+60)/(9*x^3-15*x^2-8*x+16),x, algorithm="fricas")

[Out]

-((3*x - 4)*e^x*log(x + 1) - (3*x - 4)*e^(x*e^x + x) + 20)/(3*x - 4)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.63 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=- e^{x} \log {\left (x + 1 \right )} + e^{x e^{x} + x} - \frac {60}{9 x - 12} \]

[In]

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(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)

[Out]

-exp(x)*log(x + 1) + exp(x*exp(x) + x) - 60/(9*x - 12)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=-e^{x} \log \left (x + 1\right ) - \frac {20}{3 \, x - 4} + e^{\left (x e^{x} + x\right )} \]

[In]

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(exp(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)+60*x+60)/(9*x^3-15*x^2-8*x+16),x, algorithm="maxima")

[Out]

-e^x*log(x + 1) - 20/(3*x - 4) + e^(x*e^x + x)

Giac [F]

\[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx=\int { -\frac {{\left (9 \, x^{3} - 15 \, x^{2} - 8 \, x + 16\right )} e^{x} \log \left (x + 1\right ) - {\left (9 \, x^{3} - 15 \, x^{2} + {\left (9 \, x^{4} - 6 \, x^{3} - 23 \, x^{2} + 8 \, x + 16\right )} e^{x} - 8 \, x + 16\right )} e^{\left (x e^{x} + x\right )} + {\left (9 \, x^{2} - 24 \, x + 16\right )} e^{x} - 60 \, x - 60}{9 \, x^{3} - 15 \, x^{2} - 8 \, x + 16} \,d x } \]

[In]

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(exp(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)+60*x+60)/(9*x^3-15*x^2-8*x+16),x, algorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 9.62 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \frac {60+60 x+e^x \left (-16+24 x-9 x^2\right )+e^{x+e^x x} \left (16-8 x-15 x^2+9 x^3+e^x \left (16+8 x-23 x^2-6 x^3+9 x^4\right )\right )+e^x \left (-16+8 x+15 x^2-9 x^3\right ) \log (1+x)}{16-8 x-15 x^2+9 x^3} \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^x}\,{\mathrm {e}}^x-\frac {20}{3\,\left (x-\frac {4}{3}\right )}-\ln \left (x+1\right )\,{\mathrm {e}}^x \]

[In]

int(-(60*x + exp(x + x*exp(x))*(exp(x)*(8*x - 23*x^2 - 6*x^3 + 9*x^4 + 16) - 8*x - 15*x^2 + 9*x^3 + 16) - exp(
x)*(9*x^2 - 24*x + 16) + log(x + 1)*exp(x)*(8*x + 15*x^2 - 9*x^3 - 16) + 60)/(8*x + 15*x^2 - 9*x^3 - 16),x)

[Out]

exp(x*exp(x))*exp(x) - 20/(3*(x - 4/3)) - log(x + 1)*exp(x)