\(\int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x (8 x-x^2)+(4 e^x x \log (5)+4 x^2 \log (5)) \log (3 x^2)+(x^2 \log (5)+e^x (2 x-x^2) \log (5)) \log ^2(3 x^2)}{e^{2 x}+2 e^x x+x^2} \, dx\) [2969]

Optimal result

Integrand size = 109, antiderivative size = 32 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=-e^x-x+x \left (5+\frac {x \log (5) \log ^2\left (3 x^2\right )}{e^x+x}\right ) \]

[Out]

x*(ln(3*x^2)^2/(exp(x)+x)*x*ln(5)+5)-exp(x)-x

Rubi [F]

\[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx \]

[In]

Int[(-E^(3*x) + E^(2*x)*(4 - 2*x) + 4*x^2 + E^x*(8*x - x^2) + (4*E^x*x*Log[5] + 4*x^2*Log[5])*Log[3*x^2] + (x^
2*Log[5] + E^x*(2*x - x^2)*Log[5])*Log[3*x^2]^2)/(E^(2*x) + 2*E^x*x + x^2),x]

[Out]

-E^x + 4*x + 4*Log[5]*Log[3*x^2]*Defer[Int][x/(E^x + x), x] - Log[5]*Defer[Int][(x^2*Log[3*x^2]^2)/(E^x + x)^2
, x] + Log[5]*Defer[Int][(x^3*Log[3*x^2]^2)/(E^x + x)^2, x] + 2*Log[5]*Defer[Int][(x*Log[3*x^2]^2)/(E^x + x),
x] - Log[5]*Defer[Int][(x^2*Log[3*x^2]^2)/(E^x + x), x] - 8*Log[5]*Defer[Int][Defer[Int][x/(E^x + x), x]/x, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (4-e^x+\frac {4 x \log (5) \log \left (3 x^2\right )}{e^x+x}-\frac {\left (e^x (-2+x)-x\right ) x \log (5) \log ^2\left (3 x^2\right )}{\left (e^x+x\right )^2}\right ) \, dx \\ & = 4 x-\log (5) \int \frac {\left (e^x (-2+x)-x\right ) x \log ^2\left (3 x^2\right )}{\left (e^x+x\right )^2} \, dx+(4 \log (5)) \int \frac {x \log \left (3 x^2\right )}{e^x+x} \, dx-\int e^x \, dx \\ & = -e^x+4 x-\log (5) \int \left (-\frac {(-1+x) x^2 \log ^2\left (3 x^2\right )}{\left (e^x+x\right )^2}+\frac {(-2+x) x \log ^2\left (3 x^2\right )}{e^x+x}\right ) \, dx-(4 \log (5)) \int \frac {2 \int \frac {x}{e^x+x} \, dx}{x} \, dx+\left (4 \log (5) \log \left (3 x^2\right )\right ) \int \frac {x}{e^x+x} \, dx \\ & = -e^x+4 x+\log (5) \int \frac {(-1+x) x^2 \log ^2\left (3 x^2\right )}{\left (e^x+x\right )^2} \, dx-\log (5) \int \frac {(-2+x) x \log ^2\left (3 x^2\right )}{e^x+x} \, dx-(8 \log (5)) \int \frac {\int \frac {x}{e^x+x} \, dx}{x} \, dx+\left (4 \log (5) \log \left (3 x^2\right )\right ) \int \frac {x}{e^x+x} \, dx \\ & = -e^x+4 x+\log (5) \int \left (-\frac {x^2 \log ^2\left (3 x^2\right )}{\left (e^x+x\right )^2}+\frac {x^3 \log ^2\left (3 x^2\right )}{\left (e^x+x\right )^2}\right ) \, dx-\log (5) \int \left (-\frac {2 x \log ^2\left (3 x^2\right )}{e^x+x}+\frac {x^2 \log ^2\left (3 x^2\right )}{e^x+x}\right ) \, dx-(8 \log (5)) \int \frac {\int \frac {x}{e^x+x} \, dx}{x} \, dx+\left (4 \log (5) \log \left (3 x^2\right )\right ) \int \frac {x}{e^x+x} \, dx \\ & = -e^x+4 x-\log (5) \int \frac {x^2 \log ^2\left (3 x^2\right )}{\left (e^x+x\right )^2} \, dx+\log (5) \int \frac {x^3 \log ^2\left (3 x^2\right )}{\left (e^x+x\right )^2} \, dx-\log (5) \int \frac {x^2 \log ^2\left (3 x^2\right )}{e^x+x} \, dx+(2 \log (5)) \int \frac {x \log ^2\left (3 x^2\right )}{e^x+x} \, dx-(8 \log (5)) \int \frac {\int \frac {x}{e^x+x} \, dx}{x} \, dx+\left (4 \log (5) \log \left (3 x^2\right )\right ) \int \frac {x}{e^x+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.50 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=-e^x+4 x+\frac {x^2 \log (5) \log ^2\left (3 x^2\right )}{e^x+x} \]

[In]

Integrate[(-E^(3*x) + E^(2*x)*(4 - 2*x) + 4*x^2 + E^x*(8*x - x^2) + (4*E^x*x*Log[5] + 4*x^2*Log[5])*Log[3*x^2]
+ (x^2*Log[5] + E^x*(2*x - x^2)*Log[5])*Log[3*x^2]^2)/(E^(2*x) + 2*E^x*x + x^2),x]

[Out]

-E^x + 4*x + (x^2*Log[5]*Log[3*x^2]^2)/(E^x + x)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28

method	result	size
parallelrisch	\(\frac {2 \ln \left (5\right ) \ln \left (3 x^{2}\right )^{2} x^{2}+8 x^{2}+6 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{2 x}}{2 \,{\mathrm e}^{x}+2 x}\)	\(41\)
risch	\(\frac {4 x^{2} \ln \left (5\right ) \ln \left (x \right )^{2}}{{\mathrm e}^{x}+x}+\frac {2 \left (-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 \ln \left (3\right )\right ) x^{2} \ln \left (5\right ) \ln \left (x \right )}{{\mathrm e}^{x}+x}-\frac {-12 \,{\mathrm e}^{x} x -16 x^{2}+4 \,{\mathrm e}^{2 x}+\pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}+\pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}-4 \pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}+6 \pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}-4 \pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}-4 \ln \left (3\right )^{2} \ln \left (5\right ) x^{2}+4 i \pi \ln \left (3\right ) \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}-8 i \pi \ln \left (3\right ) \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+4 i \pi \ln \left (3\right ) \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{4 \left ({\mathrm e}^{x}+x \right )}\)	\(313\)

[In]

int((((-x^2+2*x)*ln(5)*exp(x)+x^2*ln(5))*ln(3*x^2)^2+(4*x*exp(x)*ln(5)+4*x^2*ln(5))*ln(3*x^2)-exp(x)^3+(4-2*x)
*exp(x)^2+(-x^2+8*x)*exp(x)+4*x^2)/(exp(x)^2+2*exp(x)*x+x^2),x,method=_RETURNVERBOSE)

[Out]

1/2*(2*ln(5)*ln(3*x^2)^2*x^2+8*x^2+6*exp(x)*x-2*exp(x)^2)/(exp(x)+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\frac {x^{2} \log \left (5\right ) \log \left (3 \, x^{2}\right )^{2} + 4 \, x^{2} + 3 \, x e^{x} - e^{\left (2 \, x\right )}}{x + e^{x}} \]

[In]

integrate((((-x^2+2*x)*log(5)*exp(x)+x^2*log(5))*log(3*x^2)^2+(4*x*exp(x)*log(5)+4*x^2*log(5))*log(3*x^2)-exp(
x)^3+(4-2*x)*exp(x)^2+(-x^2+8*x)*exp(x)+4*x^2)/(exp(x)^2+2*exp(x)*x+x^2),x, algorithm="fricas")

[Out]

(x^2*log(5)*log(3*x^2)^2 + 4*x^2 + 3*x*e^x - e^(2*x))/(x + e^x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\frac {x^{2} \log {\left (5 \right )} \log {\left (3 x^{2} \right )}^{2}}{x + e^{x}} + 4 x - e^{x} \]

[In]

integrate((((-x**2+2*x)*ln(5)*exp(x)+x**2*ln(5))*ln(3*x**2)**2+(4*x*exp(x)*ln(5)+4*x**2*ln(5))*ln(3*x**2)-exp(
x)**3+(4-2*x)*exp(x)**2+(-x**2+8*x)*exp(x)+4*x**2)/(exp(x)**2+2*exp(x)*x+x**2),x)

[Out]

x**2*log(5)*log(3*x**2)**2/(x + exp(x)) + 4*x - exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\frac {4 \, x^{2} \log \left (5\right ) \log \left (3\right ) \log \left (x\right ) + 4 \, x^{2} \log \left (5\right ) \log \left (x\right )^{2} + {\left (\log \left (5\right ) \log \left (3\right )^{2} + 4\right )} x^{2} + 3 \, x e^{x} - e^{\left (2 \, x\right )}}{x + e^{x}} \]

[In]

integrate((((-x^2+2*x)*log(5)*exp(x)+x^2*log(5))*log(3*x^2)^2+(4*x*exp(x)*log(5)+4*x^2*log(5))*log(3*x^2)-exp(
x)^3+(4-2*x)*exp(x)^2+(-x^2+8*x)*exp(x)+4*x^2)/(exp(x)^2+2*exp(x)*x+x^2),x, algorithm="maxima")

[Out]

(4*x^2*log(5)*log(3)*log(x) + 4*x^2*log(5)*log(x)^2 + (log(5)*log(3)^2 + 4)*x^2 + 3*x*e^x - e^(2*x))/(x + e^x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\frac {x^{2} \log \left (5\right ) \log \left (3 \, x^{2}\right )^{2} + 4 \, x^{2} + 3 \, x e^{x} - e^{\left (2 \, x\right )}}{x + e^{x}} \]

[In]

integrate((((-x^2+2*x)*log(5)*exp(x)+x^2*log(5))*log(3*x^2)^2+(4*x*exp(x)*log(5)+4*x^2*log(5))*log(3*x^2)-exp(
x)^3+(4-2*x)*exp(x)^2+(-x^2+8*x)*exp(x)+4*x^2)/(exp(x)^2+2*exp(x)*x+x^2),x, algorithm="giac")

[Out]

(x^2*log(5)*log(3*x^2)^2 + 4*x^2 + 3*x*e^x - e^(2*x))/(x + e^x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\int \frac {\ln \left (3\,x^2\right )\,\left (4\,x^2\,\ln \left (5\right )+4\,x\,{\mathrm {e}}^x\,\ln \left (5\right )\right )-{\mathrm {e}}^{3\,x}+{\ln \left (3\,x^2\right )}^2\,\left (x^2\,\ln \left (5\right )+{\mathrm {e}}^x\,\ln \left (5\right )\,\left (2\,x-x^2\right )\right )+{\mathrm {e}}^x\,\left (8\,x-x^2\right )-{\mathrm {e}}^{2\,x}\,\left (2\,x-4\right )+4\,x^2}{{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^x+x^2} \,d x \]

[In]

int((log(3*x^2)*(4*x^2*log(5) + 4*x*exp(x)*log(5)) - exp(3*x) + log(3*x^2)^2*(x^2*log(5) + exp(x)*log(5)*(2*x
- x^2)) + exp(x)*(8*x - x^2) - exp(2*x)*(2*x - 4) + 4*x^2)/(exp(2*x) + 2*x*exp(x) + x^2),x)

[Out]

int((log(3*x^2)*(4*x^2*log(5) + 4*x*exp(x)*log(5)) - exp(3*x) + log(3*x^2)^2*(x^2*log(5) + exp(x)*log(5)*(2*x
- x^2)) + exp(x)*(8*x - x^2) - exp(2*x)*(2*x - 4) + 4*x^2)/(exp(2*x) + 2*x*exp(x) + x^2), x)