3.48.71 \(\int \frac {-8-18 x-4 \log (5)+e^{1-x} (20 x+4 x^2+8 x \log (5))+x \log (x^2)}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+(-64 x+16 e^{1-x} x) \log (x^2)+2 x \log ^2(x^2)} \, dx\)

Optimal. Leaf size=32 \[ \frac {2+\frac {x}{2}+\log (5)}{4 \left (-4+e^{1-x}+\frac {\log \left (x^2\right )}{4}\right )} \]

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Rubi [F]  time = 9.90, 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 {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 - 18*x - 4*Log[5] + E^(1 - x)*(20*x + 4*x^2 + 8*x*Log[5]) + x*Log[x^2])/(512*x + 32*E^(2 - 2*x)*x - 25
6*E^(1 - x)*x + (-64*x + 16*E^(1 - x)*x)*Log[x^2] + 2*x*Log[x^2]^2),x]

[Out]

4*Defer[Int][E^(1 + x)/((-16 + Log[x^2])*(4*E - 16*E^x + E^x*Log[x^2])^2), x] - 32*(4 + Log[25])*Defer[Int][E^
(1 + x)/((-16 + Log[x^2])*(4*E - 16*E^x + E^x*Log[x^2])^2), x] + 4*(4 + Log[25])*Defer[Int][E^(1 + x)/(x*(-16
+ Log[x^2])*(4*E - 16*E^x + E^x*Log[x^2])^2), x] - 32*Defer[Int][(E^(1 + x)*x)/((-16 + Log[x^2])*(4*E - 16*E^x
+ E^x*Log[x^2])^2), x] + 2*(4 + Log[25])*Defer[Int][(E^(1 + x)*Log[x^2])/((-16 + Log[x^2])*(4*E - 16*E^x + E^
x*Log[x^2])^2), x] + 2*Defer[Int][(E^(1 + x)*x*Log[x^2])/((-16 + Log[x^2])*(4*E - 16*E^x + E^x*Log[x^2])^2), x
] - 9*Defer[Int][E^x/((-16 + Log[x^2])*(4*E - 16*E^x + E^x*Log[x^2])), x] - 2*(2 + Log[5])*Defer[Int][E^x/(x*(
-16 + Log[x^2])*(4*E - 16*E^x + E^x*Log[x^2])), x] + Defer[Int][(E^x*Log[x^2])/((-16 + Log[x^2])*(4*E - 16*E^x
+ E^x*Log[x^2])), x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (4 e x (5+x+\log (25))-2 e^x (4+9 x+\log (25))+e^x x \log \left (x^2\right )\right )}{2 x \left (4 \left (e-4 e^x\right )+e^x \log \left (x^2\right )\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^x \left (4 e x (5+x+\log (25))-2 e^x (4+9 x+\log (25))+e^x x \log \left (x^2\right )\right )}{x \left (4 \left (e-4 e^x\right )+e^x \log \left (x^2\right )\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {e^x \left (18 x+8 \left (1+\frac {\log (5)}{2}\right )-x \log \left (x^2\right )\right )}{x \left (16-\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )}+\frac {4 e^{1+x} (4+x+\log (25)) \left (2-16 x+x \log \left (x^2\right )\right )}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^x \left (18 x+8 \left (1+\frac {\log (5)}{2}\right )-x \log \left (x^2\right )\right )}{x \left (16-\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+2 \int \frac {e^{1+x} (4+x+\log (25)) \left (2-16 x+x \log \left (x^2\right )\right )}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {18 e^x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )}-\frac {4 e^x (2+\log (5))}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )}+\frac {e^x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )}\right ) \, dx+2 \int \left (\frac {e^{1+x} \left (2-16 x+x \log \left (x^2\right )\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}+\frac {e^{1+x} (4+\log (25)) \left (2-16 x+x \log \left (x^2\right )\right )}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+2 \int \frac {e^{1+x} \left (2-16 x+x \log \left (x^2\right )\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx-9 \int \frac {e^x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx-(2 (2+\log (5))) \int \frac {e^x}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+(2 (4+\log (25))) \int \frac {e^{1+x} \left (2-16 x+x \log \left (x^2\right )\right )}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+2 \int \left (\frac {2 e^{1+x}}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}-\frac {16 e^{1+x} x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}+\frac {e^{1+x} x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}\right ) \, dx-9 \int \frac {e^x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx-(2 (2+\log (5))) \int \frac {e^x}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+(2 (4+\log (25))) \int \left (-\frac {16 e^{1+x}}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}+\frac {2 e^{1+x}}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}+\frac {e^{1+x} \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+2 \int \frac {e^{1+x} x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx+4 \int \frac {e^{1+x}}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx-9 \int \frac {e^x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx-32 \int \frac {e^{1+x} x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx-(2 (2+\log (5))) \int \frac {e^x}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+(2 (4+\log (25))) \int \frac {e^{1+x} \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx+(4 (4+\log (25))) \int \frac {e^{1+x}}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx-(32 (4+\log (25))) \int \frac {e^{1+x}}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.05, size = 31, normalized size = 0.97 \begin {gather*} \frac {e^x (4+x+\log (25))}{2 \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 - 18*x - 4*Log[5] + E^(1 - x)*(20*x + 4*x^2 + 8*x*Log[5]) + x*Log[x^2])/(512*x + 32*E^(2 - 2*x)*
x - 256*E^(1 - x)*x + (-64*x + 16*E^(1 - x)*x)*Log[x^2] + 2*x*Log[x^2]^2),x]

[Out]

(E^x*(4 + x + Log[25]))/(2*(4*E - 16*E^x + E^x*Log[x^2]))

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fricas [A]  time = 0.70, size = 25, normalized size = 0.78 \begin {gather*} \frac {x + 2 \, \log \relax (5) + 4}{2 \, {\left (4 \, e^{\left (-x + 1\right )} + \log \left (x^{2}\right ) - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x^2)+(8*x*log(5)+4*x^2+20*x)*exp(-x+1)-4*log(5)-18*x-8)/(2*x*log(x^2)^2+(16*x*exp(-x+1)-64*x)
*log(x^2)+32*x*exp(-x+1)^2-256*x*exp(-x+1)+512*x),x, algorithm="fricas")

[Out]

1/2*(x + 2*log(5) + 4)/(4*e^(-x + 1) + log(x^2) - 16)

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giac [A]  time = 0.25, size = 25, normalized size = 0.78 \begin {gather*} \frac {x + 2 \, \log \relax (5) + 4}{2 \, {\left (4 \, e^{\left (-x + 1\right )} + \log \left (x^{2}\right ) - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x^2)+(8*x*log(5)+4*x^2+20*x)*exp(-x+1)-4*log(5)-18*x-8)/(2*x*log(x^2)^2+(16*x*exp(-x+1)-64*x)
*log(x^2)+32*x*exp(-x+1)^2-256*x*exp(-x+1)+512*x),x, algorithm="giac")

[Out]

1/2*(x + 2*log(5) + 4)/(4*e^(-x + 1) + log(x^2) - 16)

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maple [C]  time = 0.30, size = 74, normalized size = 2.31

method	result	size
risch	\(\frac {i \left (2 \ln \relax (5)+x +4\right )}{\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+8 i {\mathrm e}^{1-x}+4 i \ln \relax (x )-32 i}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(x^2)+(8*x*ln(5)+4*x^2+20*x)*exp(1-x)-4*ln(5)-18*x-8)/(2*x*ln(x^2)^2+(16*x*exp(1-x)-64*x)*ln(x^2)+32*
x*exp(1-x)^2-256*x*exp(1-x)+512*x),x,method=_RETURNVERBOSE)

[Out]

I*(2*ln(5)+x+4)/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+8*I*exp(1-x)+4*I*ln(
x)-32*I)

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maxima [A]  time = 0.51, size = 25, normalized size = 0.78 \begin {gather*} \frac {{\left (x + 2 \, \log \relax (5) + 4\right )} e^{x}}{4 \, {\left ({\left (\log \relax (x) - 8\right )} e^{x} + 2 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x^2)+(8*x*log(5)+4*x^2+20*x)*exp(-x+1)-4*log(5)-18*x-8)/(2*x*log(x^2)^2+(16*x*exp(-x+1)-64*x)
*log(x^2)+32*x*exp(-x+1)^2-256*x*exp(-x+1)+512*x),x, algorithm="maxima")

[Out]

1/4*(x + 2*log(5) + 4)*e^x/((log(x) - 8)*e^x + 2*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {18\,x+4\,\ln \relax (5)-x\,\ln \left (x^2\right )-{\mathrm {e}}^{1-x}\,\left (20\,x+8\,x\,\ln \relax (5)+4\,x^2\right )+8}{2\,x\,{\ln \left (x^2\right )}^2+\left (16\,x\,{\mathrm {e}}^{1-x}-64\,x\right )\,\ln \left (x^2\right )+512\,x-256\,x\,{\mathrm {e}}^{1-x}+32\,x\,{\mathrm {e}}^{2-2\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(18*x + 4*log(5) - x*log(x^2) - exp(1 - x)*(20*x + 8*x*log(5) + 4*x^2) + 8)/(512*x - log(x^2)*(64*x - 16*
x*exp(1 - x)) - 256*x*exp(1 - x) + 32*x*exp(2 - 2*x) + 2*x*log(x^2)^2),x)

[Out]

-int((18*x + 4*log(5) - x*log(x^2) - exp(1 - x)*(20*x + 8*x*log(5) + 4*x^2) + 8)/(512*x - log(x^2)*(64*x - 16*
x*exp(1 - x)) - 256*x*exp(1 - x) + 32*x*exp(2 - 2*x) + 2*x*log(x^2)^2), x)

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sympy [A]  time = 0.30, size = 22, normalized size = 0.69 \begin {gather*} \frac {x + 2 \log {\relax (5 )} + 4}{8 e^{1 - x} + 2 \log {\left (x^{2} \right )} - 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(x**2)+(8*x*ln(5)+4*x**2+20*x)*exp(-x+1)-4*ln(5)-18*x-8)/(2*x*ln(x**2)**2+(16*x*exp(-x+1)-64*x)
*ln(x**2)+32*x*exp(-x+1)**2-256*x*exp(-x+1)+512*x),x)

[Out]

(x + 2*log(5) + 4)/(8*exp(1 - x) + 2*log(x**2) - 32)

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