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

Optimal. Leaf size=32 \[ -e^x-x+x \left (5+\frac {x \log (5) \log ^2\left (3 x^2\right )}{e^x+x}\right ) \]

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Rubi [F]  time = 1.74, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 2.80, size = 30, normalized size = 0.94 \begin {gather*} -e^x+4 x+\frac {x^2 \log (5) \log ^2\left (3 x^2\right )}{e^x+x} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.91, size = 38, normalized size = 1.19 \begin {gather*} \frac {x^{2} \log \relax (5) \log \left (3 \, x^{2}\right )^{2} + 4 \, x^{2} + 3 \, x e^{x} - e^{\left (2 \, x\right )}}{x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 1.02, size = 212, normalized size = 6.62 \begin {gather*} \frac {x^{3} \log \relax (5) \log \relax (3)^{2} + 4 \, x^{3} \log \relax (5) \log \relax (3) \log \relax (x) + 4 \, x^{3} \log \relax (5) \log \relax (x)^{2} + 4 \, x^{3} \log \relax (5) \log \relax (3) \log \left (\mathrm {sgn}\relax (x)\right ) + 8 \, x^{3} \log \relax (5) \log \relax (x) \log \left (\mathrm {sgn}\relax (x)\right ) + 4 \, x^{3} \log \relax (5) \log \left (\mathrm {sgn}\relax (x)\right )^{2} - x^{2} \log \relax (5) \log \relax (3)^{2} - 4 \, x^{2} \log \relax (5) \log \relax (3) \log \relax (x) - 4 \, x^{2} \log \relax (5) \log \relax (x)^{2} - 4 \, x^{2} \log \relax (5) \log \relax (3) \log \left (\mathrm {sgn}\relax (x)\right ) - 8 \, x^{2} \log \relax (5) \log \relax (x) \log \left (\mathrm {sgn}\relax (x)\right ) - 4 \, x^{2} \log \relax (5) \log \left (\mathrm {sgn}\relax (x)\right )^{2} + x^{3} - 4 \, x^{2}}{x^{2} + x e^{x} - x - e^{x}} + \frac {3 \, x^{3} + 3 \, x^{2} e^{x} - x e^{\left (2 \, x\right )} - 3 \, x e^{x} + e^{\left (2 \, x\right )}}{x^{2} + x e^{x} - x - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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^3*log(5)*log(3)^2 + 4*x^3*log(5)*log(3)*log(x) + 4*x^3*log(5)*log(x)^2 + 4*x^3*log(5)*log(3)*log(sgn(x)) +
8*x^3*log(5)*log(x)*log(sgn(x)) + 4*x^3*log(5)*log(sgn(x))^2 - x^2*log(5)*log(3)^2 - 4*x^2*log(5)*log(3)*log(x
) - 4*x^2*log(5)*log(x)^2 - 4*x^2*log(5)*log(3)*log(sgn(x)) - 8*x^2*log(5)*log(x)*log(sgn(x)) - 4*x^2*log(5)*l
og(sgn(x))^2 + x^3 - 4*x^2)/(x^2 + x*e^x - x - e^x) + (3*x^3 + 3*x^2*e^x - x*e^(2*x) - 3*x*e^x + e^(2*x))/(x^2
+ x*e^x - x - e^x)

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maple [C]  time = 0.29, size = 313, normalized size = 9.78

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

4*x^2*ln(5)/(exp(x)+x)*ln(x)^2+2*(-I*Pi*csgn(I*x)^2*csgn(I*x^2)+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2-I*Pi*csgn(I*x^2
)^3+2*ln(3))*x^2*ln(5)/(exp(x)+x)*ln(x)-1/4*(4*exp(2*x)-16*x^2+ln(5)*Pi^2*x^2*csgn(I*x)^4*csgn(I*x^2)^2-4*ln(5
)*Pi^2*x^2*csgn(I*x)^3*csgn(I*x^2)^3+6*ln(5)*Pi^2*x^2*csgn(I*x)^2*csgn(I*x^2)^4-4*ln(5)*Pi^2*x^2*csgn(I*x)*csg
n(I*x^2)^5-12*exp(x)*x+ln(5)*Pi^2*x^2*csgn(I*x^2)^6-4*ln(5)*ln(3)^2*x^2-8*I*ln(5)*Pi*ln(3)*x^2*csgn(I*x)*csgn(
I*x^2)^2+4*I*ln(5)*Pi*ln(3)*x^2*csgn(I*x)^2*csgn(I*x^2)+4*I*ln(5)*Pi*ln(3)*x^2*csgn(I*x^2)^3)/(exp(x)+x)

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maxima [A]  time = 1.03, size = 54, normalized size = 1.69 \begin {gather*} \frac {4 \, x^{2} \log \relax (5) \log \relax (3) \log \relax (x) + 4 \, x^{2} \log \relax (5) \log \relax (x)^{2} + {\left (\log \relax (5) \log \relax (3)^{2} + 4\right )} x^{2} + 3 \, x e^{x} - e^{\left (2 \, x\right )}}{x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\ln \left (3\,x^2\right )\,\left (4\,x^2\,\ln \relax (5)+4\,x\,{\mathrm {e}}^x\,\ln \relax (5)\right )-{\mathrm {e}}^{3\,x}+{\ln \left (3\,x^2\right )}^2\,\left (x^2\,\ln \relax (5)+{\mathrm {e}}^x\,\ln \relax (5)\,\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.30, size = 26, normalized size = 0.81 \begin {gather*} \frac {x^{2} \log {\relax (5 )} \log {\left (3 x^{2} \right )}^{2}}{x + e^{x}} + 4 x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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