3.94.10 \(\int \frac {(8+2 x-2 x^2+e (16 x+4 x^2-4 x^3)) \log (x)+(-4+8 x+e (-4 x+8 x^2)) \log ^2(x) \log (\frac {x+e x^2}{e})+(-8-2 x+2 x^2+e (-8 x-2 x^2+2 x^3)+(8+4 x-6 x^2+e (8 x+4 x^2-6 x^3)) \log (x)) \log (\frac {x+e x^2}{e}) \log (\log (\frac {x+e x^2}{e}))}{(1+e x) \log ^2(x) \log (\frac {x+e x^2}{e})} \, dx\)

Optimal. Leaf size=29 \[ \left (4+x-x^2\right ) \left (-4+\frac {2 x \log \left (\log \left (\frac {x}{e}+x^2\right )\right )}{\log (x)}\right ) \]

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Rubi [F]  time = 6.81, 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 {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((8 + 2*x - 2*x^2 + E*(16*x + 4*x^2 - 4*x^3))*Log[x] + (-4 + 8*x + E*(-4*x + 8*x^2))*Log[x]^2*Log[(x + E*x
^2)/E] + (-8 - 2*x + 2*x^2 + E*(-8*x - 2*x^2 + 2*x^3) + (8 + 4*x - 6*x^2 + E*(8*x + 4*x^2 - 6*x^3))*Log[x])*Lo
g[(x + E*x^2)/E]*Log[Log[(x + E*x^2)/E]])/((1 + E*x)*Log[x]^2*Log[(x + E*x^2)/E]),x]

[Out]

-4*x + 4*x^2 - (2*(1 + E - 8*E^2)*Defer[Int][1/(Log[x]*Log[x*(E^(-1) + x)]), x])/E^2 - 4*Defer[Int][x^2/(Log[x
]*Log[x*(E^(-1) + x)]), x] + (2*(1 + E - 4*E^2)*Defer[Int][1/((1 + E*x)*Log[x]*Log[x*(E^(-1) + x)]), x])/E^2 -
(2*(1 + 2*E)*Defer[Int][x/(Log[x]*(1 - Log[x*(1 + E*x)])), x])/E - 8*Defer[Int][Log[Log[x*(E^(-1) + x)]]/Log[
x]^2, x] - 2*Defer[Int][(x*Log[Log[x*(E^(-1) + x)]])/Log[x]^2, x] + 2*Defer[Int][(x^2*Log[Log[x*(E^(-1) + x)]]
)/Log[x]^2, x] + 8*Defer[Int][Log[Log[x*(E^(-1) + x)]]/Log[x], x] + 4*Defer[Int][(x*Log[Log[x*(E^(-1) + x)]])/
Log[x], x] - 6*Defer[Int][(x^2*Log[Log[x*(E^(-1) + x)]])/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+8 x+4 e x (-1+2 x)-\frac {2 (1+2 e x) \left (-4-x+x^2\right )}{\log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )}-\frac {2 (1+e x) \left (4+x-x^2+\left (-4-2 x+3 x^2\right ) \log (x)\right ) \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)}}{1+e x} \, dx\\ &=\int \left (\frac {2 \left (4+(1+8 e) x-(1-2 e) x^2-2 e x^3-2 \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )+4 \left (1-\frac {e}{2}\right ) x \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )+4 e x^2 \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{(1+e x) \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )}-\frac {2 \left (4+x-x^2-4 \log (x)-2 x \log (x)+3 x^2 \log (x)\right ) \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)}\right ) \, dx\\ &=2 \int \frac {4+(1+8 e) x-(1-2 e) x^2-2 e x^3-2 \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )+4 \left (1-\frac {e}{2}\right ) x \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )+4 e x^2 \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )}{(1+e x) \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )} \, dx-2 \int \frac {\left (4+x-x^2-4 \log (x)-2 x \log (x)+3 x^2 \log (x)\right ) \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx\\ &=2 \int \left (-2+4 x-\frac {(1+2 e x) \left (-4-x+x^2\right )}{(1+e x) \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )}\right ) \, dx-2 \int \left (\frac {4 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)}+\frac {x \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)}-\frac {x^2 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)}-\frac {4 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)}-\frac {2 x \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)}+\frac {3 x^2 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)}\right ) \, dx\\ &=-4 x+4 x^2-2 \int \frac {(1+2 e x) \left (-4-x+x^2\right )}{(1+e x) \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )} \, dx-2 \int \frac {x \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx+2 \int \frac {x^2 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx+4 \int \frac {x \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx-6 \int \frac {x^2 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx-8 \int \frac {\log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx+8 \int \frac {\log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx\\ &=-4 x+4 x^2-2 \int \left (\frac {1+e-8 e^2}{e^2 \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )}+\frac {2 x^2}{\log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )}+\frac {-1-e+4 e^2}{e^2 (1+e x) \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )}+\frac {(1+2 e) x}{e \log (x) (1-\log (x (1+e x)))}\right ) \, dx-2 \int \frac {x \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx+2 \int \frac {x^2 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx+4 \int \frac {x \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx-6 \int \frac {x^2 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx-8 \int \frac {\log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx+8 \int \frac {\log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx\\ &=-4 x+4 x^2-2 \int \frac {x \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx+2 \int \frac {x^2 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx-4 \int \frac {x^2}{\log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )} \, dx+4 \int \frac {x \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx-6 \int \frac {x^2 \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx-8 \int \frac {\log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log ^2(x)} \, dx+8 \int \frac {\log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \, dx-\frac {(2 (1+2 e)) \int \frac {x}{\log (x) (1-\log (x (1+e x)))} \, dx}{e}-\frac {\left (2 \left (1+e-8 e^2\right )\right ) \int \frac {1}{\log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )} \, dx}{e^2}+\frac {\left (2 \left (1+e-4 e^2\right )\right ) \int \frac {1}{(1+e x) \log (x) \log \left (x \left (\frac {1}{e}+x\right )\right )} \, dx}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 33, normalized size = 1.14 \begin {gather*} -4 x+4 x^2-\frac {2 x \left (-4-x+x^2\right ) \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((8 + 2*x - 2*x^2 + E*(16*x + 4*x^2 - 4*x^3))*Log[x] + (-4 + 8*x + E*(-4*x + 8*x^2))*Log[x]^2*Log[(x
+ E*x^2)/E] + (-8 - 2*x + 2*x^2 + E*(-8*x - 2*x^2 + 2*x^3) + (8 + 4*x - 6*x^2 + E*(8*x + 4*x^2 - 6*x^3))*Log[
x])*Log[(x + E*x^2)/E]*Log[Log[(x + E*x^2)/E]])/((1 + E*x)*Log[x]^2*Log[(x + E*x^2)/E]),x]

[Out]

-4*x + 4*x^2 - (2*x*(-4 - x + x^2)*Log[Log[x*(E^(-1) + x)]])/Log[x]

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fricas [A]  time = 0.58, size = 45, normalized size = 1.55 \begin {gather*} \frac {2 \, {\left (2 \, {\left (x^{2} - x\right )} \log \relax (x) - {\left (x^{3} - x^{2} - 4 \, x\right )} \log \left (\log \left ({\left (x^{2} e + x\right )} e^{\left (-1\right )}\right )\right )\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8*x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(
1)+x)/exp(1))*log(log((x^2*exp(1)+x)/exp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+(
(-4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(x*exp(1)+1)/log(x)^2/log((x^2*exp(1)+x)/exp(1)),x, algorithm=
"fricas")

[Out]

2*(2*(x^2 - x)*log(x) - (x^3 - x^2 - 4*x)*log(log((x^2*e + x)*e^(-1))))/log(x)

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giac [B]  time = 0.56, size = 67, normalized size = 2.31 \begin {gather*} -\frac {2 \, {\left (x^{3} \log \left (\log \left (x^{2} e + x\right ) - 1\right ) - 2 \, x^{2} \log \relax (x) - x^{2} \log \left (\log \left (x^{2} e + x\right ) - 1\right ) + 2 \, x \log \relax (x) - 4 \, x \log \left (\log \left (x^{2} e + x\right ) - 1\right )\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8*x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(
1)+x)/exp(1))*log(log((x^2*exp(1)+x)/exp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+(
(-4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(x*exp(1)+1)/log(x)^2/log((x^2*exp(1)+x)/exp(1)),x, algorithm=
"giac")

[Out]

-2*(x^3*log(log(x^2*e + x) - 1) - 2*x^2*log(x) - x^2*log(log(x^2*e + x) - 1) + 2*x*log(x) - 4*x*log(log(x^2*e
+ x) - 1))/log(x)

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maple [C]  time = 0.68, size = 95, normalized size = 3.28

method	result	size
risch	\(-\frac {2 x \left (x^{2}-x -4\right ) \ln \left (-1+\ln \relax (x )+\ln \left (x \,{\mathrm e}+1\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right ) \left (-\mathrm {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right )+\mathrm {csgn}\left (i \left (x \,{\mathrm e}+1\right )\right )\right )}{2}\right )}{\ln \relax (x )}+4 x^{2}-4 x\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*ln(x)+(2*x^3-2*x^2-8*x)*exp(1)+2*x^2-2*x-8)*ln((x^2*exp(1)+x)/ex
p(1))*ln(ln((x^2*exp(1)+x)/exp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*ln(x)^2*ln((x^2*exp(1)+x)/exp(1))+((-4*x^3+4*x^
2+16*x)*exp(1)-2*x^2+2*x+8)*ln(x))/(x*exp(1)+1)/ln(x)^2/ln((x^2*exp(1)+x)/exp(1)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 1.98, size = 44, normalized size = 1.52 \begin {gather*} \frac {2 \, {\left (2 \, {\left (x^{2} - x\right )} \log \relax (x) - {\left (x^{3} - x^{2} - 4 \, x\right )} \log \left (\log \left (x e + 1\right ) + \log \relax (x) - 1\right )\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8*x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(
1)+x)/exp(1))*log(log((x^2*exp(1)+x)/exp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+(
(-4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(x*exp(1)+1)/log(x)^2/log((x^2*exp(1)+x)/exp(1)),x, algorithm=
"maxima")

[Out]

2*(2*(x^2 - x)*log(x) - (x^3 - x^2 - 4*x)*log(log(x*e + 1) + log(x) - 1))/log(x)

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mupad [B]  time = 9.72, size = 55, normalized size = 1.90 \begin {gather*} 4\,x^2-4\,x+\frac {2\,x^2\,\ln \left (\ln \left ({\mathrm {e}}^{-1}\,\left (\mathrm {e}\,x^2+x\right )\right )\right )\,\left (x\,\mathrm {e}+1\right )\,\left (-x^2+x+4\right )}{\ln \relax (x)\,\left (\mathrm {e}\,x^2+x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(exp(-1)*(x + x^2*exp(1))))*log(exp(-1)*(x + x^2*exp(1)))*(2*x - log(x)*(4*x + exp(1)*(8*x + 4*x^
2 - 6*x^3) - 6*x^2 + 8) + exp(1)*(8*x + 2*x^2 - 2*x^3) - 2*x^2 + 8) - log(x)*(2*x + exp(1)*(16*x + 4*x^2 - 4*x
^3) - 2*x^2 + 8) + log(exp(-1)*(x + x^2*exp(1)))*log(x)^2*(exp(1)*(4*x - 8*x^2) - 8*x + 4))/(log(exp(-1)*(x +
x^2*exp(1)))*log(x)^2*(x*exp(1) + 1)),x)

[Out]

4*x^2 - 4*x + (2*x^2*log(log(exp(-1)*(x + x^2*exp(1))))*(x*exp(1) + 1)*(x - x^2 + 4))/(log(x)*(x + x^2*exp(1))
)

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sympy [A]  time = 0.90, size = 39, normalized size = 1.34 \begin {gather*} 4 x^{2} - 4 x + \frac {\left (- 2 x^{3} + 2 x^{2} + 8 x\right ) \log {\left (\log {\left (\frac {e x^{2} + x}{e} \right )} \right )}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-6*x**3+4*x**2+8*x)*exp(1)-6*x**2+4*x+8)*ln(x)+(2*x**3-2*x**2-8*x)*exp(1)+2*x**2-2*x-8)*ln((x**2
*exp(1)+x)/exp(1))*ln(ln((x**2*exp(1)+x)/exp(1)))+((8*x**2-4*x)*exp(1)+8*x-4)*ln(x)**2*ln((x**2*exp(1)+x)/exp(
1))+((-4*x**3+4*x**2+16*x)*exp(1)-2*x**2+2*x+8)*ln(x))/(x*exp(1)+1)/ln(x)**2/ln((x**2*exp(1)+x)/exp(1)),x)

[Out]

4*x**2 - 4*x + (-2*x**3 + 2*x**2 + 8*x)*log(log((E*x**2 + x)*exp(-1)))/log(x)

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