3.87.5 \(\int \frac {e^x (2 x+2 x^2+(-8 x^4-2 x^5) \log ^2(5))+e^x (-2 x-x^2+(2 x^4+x^5) \log ^2(5)) \log (e^x (-x+x^4 \log ^2(5)))}{(-1+x^3 \log ^2(5)) \log ^3(e^x (-x+x^4 \log ^2(5)))} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^x x^2}{\log ^2\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \]

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Rubi [F]  time = 8.21, 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^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(2*x + 2*x^2 + (-8*x^4 - 2*x^5)*Log[5]^2) + E^x*(-2*x - x^2 + (2*x^4 + x^5)*Log[5]^2)*Log[E^x*(-x + x
^4*Log[5]^2)])/((-1 + x^3*Log[5]^2)*Log[E^x*(-x + x^4*Log[5]^2)]^3),x]

[Out]

-8*Defer[Int][(E^x*x)/Log[E^x*x*(-1 + x^3*Log[5]^2)]^3, x] - 2*Defer[Int][(E^x*x^2)/Log[E^x*x*(-1 + x^3*Log[5]
^2)]^3, x] - (2*(-1)^(1/3)*Defer[Int][E^x/((1 - x*(-Log[5])^(2/3))*Log[E^x*x*(-1 + x^3*Log[5]^2)]^3), x])/Log[
5]^(2/3) + (2*Defer[Int][E^x/((1 - x*Log[5]^(2/3))*Log[E^x*x*(-1 + x^3*Log[5]^2)]^3), x])/Log[5]^(2/3) + 2*(-L
og[5]^(-1))^(2/3)*Defer[Int][E^x/((1 + (-1)^(1/3)*x*Log[5]^(2/3))*Log[E^x*x*(-1 + x^3*Log[5]^2)]^3), x] + 2*De
fer[Int][(E^x*x)/Log[E^x*x*(-1 + x^3*Log[5]^2)]^2, x] + Defer[Int][(E^x*x^2)/Log[E^x*x*(-1 + x^3*Log[5]^2)]^2,
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 e^x x}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )}+\frac {2 e^x x^2}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )}-\frac {8 e^x x^4 \log ^2(5)}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )}-\frac {2 e^x x^5 \log ^2(5)}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )}-\frac {2 e^x x}{\left (-1+x^3 \log ^2(5)\right ) \log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )}-\frac {e^x x^2}{\left (-1+x^3 \log ^2(5)\right ) \log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )}+\frac {2 e^x x^4 \log ^2(5)}{\left (-1+x^3 \log ^2(5)\right ) \log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )}+\frac {e^x x^5 \log ^2(5)}{\left (-1+x^3 \log ^2(5)\right ) \log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )}\right ) \, dx\\ &=2 \int \frac {e^x x}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \, dx+2 \int \frac {e^x x^2}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \, dx-2 \int \frac {e^x x}{\left (-1+x^3 \log ^2(5)\right ) \log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \, dx+\log ^2(5) \int \frac {e^x x^5}{\left (-1+x^3 \log ^2(5)\right ) \log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \, dx-\left (2 \log ^2(5)\right ) \int \frac {e^x x^5}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \, dx+\left (2 \log ^2(5)\right ) \int \frac {e^x x^4}{\left (-1+x^3 \log ^2(5)\right ) \log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \, dx-\left (8 \log ^2(5)\right ) \int \frac {e^x x^4}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \, dx-\int \frac {e^x x^2}{\left (-1+x^3 \log ^2(5)\right ) \log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 25, normalized size = 0.96 \begin {gather*} \frac {e^x x^2}{\log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(2*x + 2*x^2 + (-8*x^4 - 2*x^5)*Log[5]^2) + E^x*(-2*x - x^2 + (2*x^4 + x^5)*Log[5]^2)*Log[E^x*(
-x + x^4*Log[5]^2)])/((-1 + x^3*Log[5]^2)*Log[E^x*(-x + x^4*Log[5]^2)]^3),x]

[Out]

(E^x*x^2)/Log[E^x*x*(-1 + x^3*Log[5]^2)]^2

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fricas [A]  time = 1.22, size = 24, normalized size = 0.92 \begin {gather*} \frac {x^{2} e^{x}}{\log \left ({\left (x^{4} \log \relax (5)^{2} - x\right )} e^{x}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^5+2*x^4)*log(5)^2-x^2-2*x)*exp(x)*log((x^4*log(5)^2-x)*exp(x))+((-2*x^5-8*x^4)*log(5)^2+2*x^2+2
*x)*exp(x))/(x^3*log(5)^2-1)/log((x^4*log(5)^2-x)*exp(x))^3,x, algorithm="fricas")

[Out]

x^2*e^x/log((x^4*log(5)^2 - x)*e^x)^2

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giac [A]  time = 0.32, size = 43, normalized size = 1.65 \begin {gather*} \frac {x^{2} e^{x}}{x^{2} + 2 \, x \log \left (x^{4} \log \relax (5)^{2} - x\right ) + \log \left (x^{4} \log \relax (5)^{2} - x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^5+2*x^4)*log(5)^2-x^2-2*x)*exp(x)*log((x^4*log(5)^2-x)*exp(x))+((-2*x^5-8*x^4)*log(5)^2+2*x^2+2
*x)*exp(x))/(x^3*log(5)^2-1)/log((x^4*log(5)^2-x)*exp(x))^3,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.56, size = 282, normalized size = 10.85

method	result	size
risch	\(-\frac {4 x^{2} {\mathrm e}^{x}}{\left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \relax (5)^{2}-1\right )\right ) \mathrm {csgn}\left (i x \left (x^{3} \ln \relax (5)^{2}-1\right ) {\mathrm e}^{x}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x^{3} \ln \relax (5)^{2}-1\right ) {\mathrm e}^{x}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i \left (x^{3} \ln \relax (5)^{2}-1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \relax (5)^{2}-1\right )\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \relax (5)^{2}-1\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x^{3} \ln \relax (5)^{2}-1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \relax (5)^{2}-1\right )\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \relax (5)^{2}-1\right )\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \relax (5)^{2}-1\right )\right ) \mathrm {csgn}\left (i x \left (x^{3} \ln \relax (5)^{2}-1\right ) {\mathrm e}^{x}\right )^{2}+\pi \mathrm {csgn}\left (i x \left (x^{3} \ln \relax (5)^{2}-1\right ) {\mathrm e}^{x}\right )^{3}+2 i \ln \relax (x )+2 i \ln \left (x^{3} \ln \relax (5)^{2}-1\right )+2 i \ln \left ({\mathrm e}^{x}\right )\right )^{2}}\)	\(282\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^5+2*x^4)*ln(5)^2-x^2-2*x)*exp(x)*ln((x^4*ln(5)^2-x)*exp(x))+((-2*x^5-8*x^4)*ln(5)^2+2*x^2+2*x)*exp(x)
)/(x^3*ln(5)^2-1)/ln((x^4*ln(5)^2-x)*exp(x))^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.54, size = 51, normalized size = 1.96 \begin {gather*} \frac {x^{2} e^{x}}{x^{2} + 2 \, {\left (x + \log \relax (x)\right )} \log \left (x^{3} \log \relax (5)^{2} - 1\right ) + \log \left (x^{3} \log \relax (5)^{2} - 1\right )^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^5+2*x^4)*log(5)^2-x^2-2*x)*exp(x)*log((x^4*log(5)^2-x)*exp(x))+((-2*x^5-8*x^4)*log(5)^2+2*x^2+2
*x)*exp(x))/(x^3*log(5)^2-1)/log((x^4*log(5)^2-x)*exp(x))^3,x, algorithm="maxima")

[Out]

x^2*e^x/(x^2 + 2*(x + log(x))*log(x^3*log(5)^2 - 1) + log(x^3*log(5)^2 - 1)^2 + 2*x*log(x) + log(x)^2)

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mupad [B]  time = 8.32, size = 24, normalized size = 0.92 \begin {gather*} \frac {x^2\,{\mathrm {e}}^x}{{\ln \left (-{\mathrm {e}}^x\,\left (x-x^4\,{\ln \relax (5)}^2\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(2*x + 2*x^2 - log(5)^2*(8*x^4 + 2*x^5)) - exp(x)*log(-exp(x)*(x - x^4*log(5)^2))*(2*x - log(5)^2*
(2*x^4 + x^5) + x^2))/(log(-exp(x)*(x - x^4*log(5)^2))^3*(x^3*log(5)^2 - 1)),x)

[Out]

(x^2*exp(x))/log(-exp(x)*(x - x^4*log(5)^2))^2

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sympy [A]  time = 0.23, size = 22, normalized size = 0.85 \begin {gather*} \frac {x^{2} e^{x}}{\log {\left (\left (x^{4} \log {\relax (5 )}^{2} - x\right ) e^{x} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**5+2*x**4)*ln(5)**2-x**2-2*x)*exp(x)*ln((x**4*ln(5)**2-x)*exp(x))+((-2*x**5-8*x**4)*ln(5)**2+2*
x**2+2*x)*exp(x))/(x**3*ln(5)**2-1)/ln((x**4*ln(5)**2-x)*exp(x))**3,x)

[Out]

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

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