3.94.9 \(\int \frac {2+12 e^{2 x}+8 e^{3 x}+e^x (8-4 x)+(1+6 e^x+12 e^{2 x}+8 e^{3 x}) \log (x)}{(4+48 e^{2 x}+32 e^{3 x}+x+e^x (24+2 x)+(x+6 e^x x+12 e^{2 x} x+8 e^{3 x} x) \log (x)) \log (\frac {4+16 e^x+16 e^{2 x}+x+(x+4 e^x x+4 e^{2 x} x) \log (x)}{1+4 e^x+4 e^{2 x}})} \, dx\)

Optimal. Leaf size=21 \[ 4+\log \left (\log \left (4+\frac {x}{\left (1+2 e^x\right )^2}+x \log (x)\right )\right ) \]

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Rubi [F]  time = 17.52, 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 {2+12 e^{2 x}+8 e^{3 x}+e^x (8-4 x)+\left (1+6 e^x+12 e^{2 x}+8 e^{3 x}\right ) \log (x)}{\left (4+48 e^{2 x}+32 e^{3 x}+x+e^x (24+2 x)+\left (x+6 e^x x+12 e^{2 x} x+8 e^{3 x} x\right ) \log (x)\right ) \log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (x+4 e^x x+4 e^{2 x} x\right ) \log (x)}{1+4 e^x+4 e^{2 x}}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 + 12*E^(2*x) + 8*E^(3*x) + E^x*(8 - 4*x) + (1 + 6*E^x + 12*E^(2*x) + 8*E^(3*x))*Log[x])/((4 + 48*E^(2*x
) + 32*E^(3*x) + x + E^x*(24 + 2*x) + (x + 6*E^x*x + 12*E^(2*x)*x + 8*E^(3*x)*x)*Log[x])*Log[(4 + 16*E^x + 16*
E^(2*x) + x + (x + 4*E^x*x + 4*E^(2*x)*x)*Log[x])/(1 + 4*E^x + 4*E^(2*x))]),x]

[Out]

2*Defer[Int][1/((1 + 2*E^x)*Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x] + D
efer[Int][1/((4 + x*Log[x])*Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x] + D
efer[Int][Log[x]/((4 + x*Log[x])*Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x
] - 28*Defer[Int][1/((4 + x*Log[x])*(4 + 16*E^x + 16*E^(2*x) + x + x*Log[x] + 4*E^x*x*Log[x] + 4*E^(2*x)*x*Log
[x])*Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x] - 64*Defer[Int][E^x/((4 +
x*Log[x])*(4 + 16*E^x + 16*E^(2*x) + x + x*Log[x] + 4*E^x*x*Log[x] + 4*E^(2*x)*x*Log[x])*Log[(4 + 16*E^x + 16*
E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x] - 9*Defer[Int][x/((4 + x*Log[x])*(4 + 16*E^x + 16*E^
(2*x) + x + x*Log[x] + 4*E^x*x*Log[x] + 4*E^(2*x)*x*Log[x])*Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x
*Log[x])/(1 + 2*E^x)^2]), x] - 16*Defer[Int][(x*Log[x])/((4 + x*Log[x])*(4 + 16*E^x + 16*E^(2*x) + x + x*Log[x
] + 4*E^x*x*Log[x] + 4*E^(2*x)*x*Log[x])*Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x
)^2]), x] - 32*Defer[Int][(E^x*x*Log[x])/((4 + x*Log[x])*(4 + 16*E^x + 16*E^(2*x) + x + x*Log[x] + 4*E^x*x*Log
[x] + 4*E^(2*x)*x*Log[x])*Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x] - 2*D
efer[Int][(x^2*Log[x])/((4 + x*Log[x])*(4 + 16*E^x + 16*E^(2*x) + x + x*Log[x] + 4*E^x*x*Log[x] + 4*E^(2*x)*x*
Log[x])*Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x] - 2*Defer[Int][(x^2*Log
[x]^2)/((4 + x*Log[x])*(4 + 16*E^x + 16*E^(2*x) + x + x*Log[x] + 4*E^x*x*Log[x] + 4*E^(2*x)*x*Log[x])*Log[(4 +
16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x] - 4*Defer[Int][(E^x*x^2*Log[x]^2)/((4 +
x*Log[x])*(4 + 16*E^x + 16*E^(2*x) + x + x*Log[x] + 4*E^x*x*Log[x] + 4*E^(2*x)*x*Log[x])*Log[(4 + 16*E^x + 16
*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2+12 e^{2 x}+8 e^{3 x}-4 e^x (-2+x)+\left (1+2 e^x\right )^3 \log (x)}{\left (1+2 e^x\right ) \left (4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)\right ) \log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)}{\left (1+2 e^x\right )^2}\right )} \, dx\\ &=\int \left (\frac {2}{\left (1+2 e^x\right ) \log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)}{\left (1+2 e^x\right )^2}\right )}+\frac {1+\log (x)}{(4+x \log (x)) \log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)}{\left (1+2 e^x\right )^2}\right )}-\frac {28+64 e^x+9 x+16 x \log (x)+32 e^x x \log (x)+2 x^2 \log (x)+2 x^2 \log ^2(x)+4 e^x x^2 \log ^2(x)}{(4+x \log (x)) \left (4+16 e^x+16 e^{2 x}+x+x \log (x)+4 e^x x \log (x)+4 e^{2 x} x \log (x)\right ) \log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)}{\left (1+2 e^x\right )^2}\right )}\right ) \, dx\\ &=2 \int \frac {1}{\left (1+2 e^x\right ) \log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)}{\left (1+2 e^x\right )^2}\right )} \, dx+\int \frac {1+\log (x)}{(4+x \log (x)) \log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)}{\left (1+2 e^x\right )^2}\right )} \, dx-\int \frac {28+64 e^x+9 x+16 x \log (x)+32 e^x x \log (x)+2 x^2 \log (x)+2 x^2 \log ^2(x)+4 e^x x^2 \log ^2(x)}{(4+x \log (x)) \left (4+16 e^x+16 e^{2 x}+x+x \log (x)+4 e^x x \log (x)+4 e^{2 x} x \log (x)\right ) \log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)}{\left (1+2 e^x\right )^2}\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 40, normalized size = 1.90 \begin {gather*} \log \left (\log \left (\frac {4+16 e^x+16 e^{2 x}+x+\left (1+2 e^x\right )^2 x \log (x)}{\left (1+2 e^x\right )^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 12*E^(2*x) + 8*E^(3*x) + E^x*(8 - 4*x) + (1 + 6*E^x + 12*E^(2*x) + 8*E^(3*x))*Log[x])/((4 + 48*
E^(2*x) + 32*E^(3*x) + x + E^x*(24 + 2*x) + (x + 6*E^x*x + 12*E^(2*x)*x + 8*E^(3*x)*x)*Log[x])*Log[(4 + 16*E^x
+ 16*E^(2*x) + x + (x + 4*E^x*x + 4*E^(2*x)*x)*Log[x])/(1 + 4*E^x + 4*E^(2*x))]),x]

[Out]

Log[Log[(4 + 16*E^x + 16*E^(2*x) + x + (1 + 2*E^x)^2*x*Log[x])/(1 + 2*E^x)^2]]

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fricas [B]  time = 0.68, size = 47, normalized size = 2.24 \begin {gather*} \log \left (\log \left (\frac {{\left (4 \, x e^{\left (2 \, x\right )} + 4 \, x e^{x} + x\right )} \log \relax (x) + x + 16 \, e^{\left (2 \, x\right )} + 16 \, e^{x} + 4}{4 \, e^{\left (2 \, x\right )} + 4 \, e^{x} + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*exp(x)^3+12*exp(x)^2+6*exp(x)+1)*log(x)+8*exp(x)^3+12*exp(x)^2+(-4*x+8)*exp(x)+2)/((8*x*exp(x)^3
+12*x*exp(x)^2+6*exp(x)*x+x)*log(x)+32*exp(x)^3+48*exp(x)^2+(2*x+24)*exp(x)+4+x)/log(((4*x*exp(x)^2+4*exp(x)*x
+x)*log(x)+16*exp(x)^2+16*exp(x)+4+x)/(4*exp(x)^2+4*exp(x)+1)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.33, size = 47, normalized size = 2.24 \begin {gather*} \log \left (-\log \left (4 \, x e^{\left (2 \, x\right )} \log \relax (x) + 4 \, x e^{x} \log \relax (x) + x \log \relax (x) + x + 16 \, e^{\left (2 \, x\right )} + 16 \, e^{x} + 4\right ) + 2 \, \log \left (2 \, e^{x} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*exp(x)^3+12*exp(x)^2+6*exp(x)+1)*log(x)+8*exp(x)^3+12*exp(x)^2+(-4*x+8)*exp(x)+2)/((8*x*exp(x)^3
+12*x*exp(x)^2+6*exp(x)*x+x)*log(x)+32*exp(x)^3+48*exp(x)^2+(2*x+24)*exp(x)+4+x)/log(((4*x*exp(x)^2+4*exp(x)*x
+x)*log(x)+16*exp(x)^2+16*exp(x)+4+x)/(4*exp(x)^2+4*exp(x)+1)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.30, size = 385, normalized size = 18.33

method	result	size
risch	\(\ln \left (\ln \left (\left (\ln \relax (x ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \ln \relax (x )+\frac {\ln \relax (x )}{4}+\frac {1}{4}\right ) x +4 \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x}+1\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{\left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}}\right ) \mathrm {csgn}\left (i \left (\left (\ln \relax (x ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \ln \relax (x )+\frac {\ln \relax (x )}{4}+\frac {1}{4}\right ) x +4 \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \ln \relax (x )+\frac {\ln \relax (x )}{4}+\frac {1}{4}\right ) x +4 \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x}+1\right )}{\left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{\left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \ln \relax (x )+\frac {\ln \relax (x )}{4}+\frac {1}{4}\right ) x +4 \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x}+1\right )}{\left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}}\right )^{2}-\pi \mathrm {csgn}\left (i \left (\frac {1}{2}+{\mathrm e}^{x}\right )\right )^{2} \mathrm {csgn}\left (i \left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}\right )+2 \pi \,\mathrm {csgn}\left (i \left (\frac {1}{2}+{\mathrm e}^{x}\right )\right ) \mathrm {csgn}\left (i \left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}\right )^{2}-\pi \mathrm {csgn}\left (i \left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i \left (\left (\ln \relax (x ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \ln \relax (x )+\frac {\ln \relax (x )}{4}+\frac {1}{4}\right ) x +4 \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \ln \relax (x )+\frac {\ln \relax (x )}{4}+\frac {1}{4}\right ) x +4 \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x}+1\right )}{\left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \ln \relax (x )+\frac {\ln \relax (x )}{4}+\frac {1}{4}\right ) x +4 \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x}+1\right )}{\left (\frac {1}{2}+{\mathrm e}^{x}\right )^{2}}\right )^{3}-4 i \ln \left (\frac {1}{2}+{\mathrm e}^{x}\right )\right )}{2}\right )\)	\(385\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*exp(x)^3+12*exp(x)^2+6*exp(x)+1)*ln(x)+8*exp(x)^3+12*exp(x)^2+(-4*x+8)*exp(x)+2)/((8*x*exp(x)^3+12*x*e
xp(x)^2+6*exp(x)*x+x)*ln(x)+32*exp(x)^3+48*exp(x)^2+(2*x+24)*exp(x)+4+x)/ln(((4*x*exp(x)^2+4*exp(x)*x+x)*ln(x)
+16*exp(x)^2+16*exp(x)+4+x)/(4*exp(x)^2+4*exp(x)+1)),x,method=_RETURNVERBOSE)

[Out]

ln(ln((ln(x)*exp(2*x)+exp(x)*ln(x)+1/4*ln(x)+1/4)*x+4*exp(2*x)+4*exp(x)+1)-1/2*I*(Pi*csgn(I/(1/2+exp(x))^2)*cs
gn(I*((ln(x)*exp(2*x)+exp(x)*ln(x)+1/4*ln(x)+1/4)*x+4*exp(2*x)+4*exp(x)+1))*csgn(I/(1/2+exp(x))^2*((ln(x)*exp(
2*x)+exp(x)*ln(x)+1/4*ln(x)+1/4)*x+4*exp(2*x)+4*exp(x)+1))-Pi*csgn(I/(1/2+exp(x))^2)*csgn(I/(1/2+exp(x))^2*((l
n(x)*exp(2*x)+exp(x)*ln(x)+1/4*ln(x)+1/4)*x+4*exp(2*x)+4*exp(x)+1))^2-Pi*csgn(I*(1/2+exp(x)))^2*csgn(I*(1/2+ex
p(x))^2)+2*Pi*csgn(I*(1/2+exp(x)))*csgn(I*(1/2+exp(x))^2)^2-Pi*csgn(I*(1/2+exp(x))^2)^3-Pi*csgn(I*((ln(x)*exp(
2*x)+exp(x)*ln(x)+1/4*ln(x)+1/4)*x+4*exp(2*x)+4*exp(x)+1))*csgn(I/(1/2+exp(x))^2*((ln(x)*exp(2*x)+exp(x)*ln(x)
+1/4*ln(x)+1/4)*x+4*exp(2*x)+4*exp(x)+1))^2+Pi*csgn(I/(1/2+exp(x))^2*((ln(x)*exp(2*x)+exp(x)*ln(x)+1/4*ln(x)+1
/4)*x+4*exp(2*x)+4*exp(x)+1))^3-4*I*ln(1/2+exp(x))))

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maxima [B]  time = 17.87, size = 42, normalized size = 2.00 \begin {gather*} \log \left (\log \left ({\left (4 \, x e^{\left (2 \, x\right )} + 4 \, x e^{x} + x\right )} \log \relax (x) + x + 16 \, e^{\left (2 \, x\right )} + 16 \, e^{x} + 4\right ) - 2 \, \log \left (2 \, e^{x} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*exp(x)^3+12*exp(x)^2+6*exp(x)+1)*log(x)+8*exp(x)^3+12*exp(x)^2+(-4*x+8)*exp(x)+2)/((8*x*exp(x)^3
+12*x*exp(x)^2+6*exp(x)*x+x)*log(x)+32*exp(x)^3+48*exp(x)^2+(2*x+24)*exp(x)+4+x)/log(((4*x*exp(x)^2+4*exp(x)*x
+x)*log(x)+16*exp(x)^2+16*exp(x)+4+x)/(4*exp(x)^2+4*exp(x)+1)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.24, size = 51, normalized size = 2.43 \begin {gather*} \ln \left (\ln \left (x+16\,{\mathrm {e}}^{2\,x}+16\,{\mathrm {e}}^x+x\,\ln \relax (x)+4\,x\,{\mathrm {e}}^x\,\ln \relax (x)+4\,x\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)+4\right )-\ln \left (4\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^x+1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*exp(2*x) + 8*exp(3*x) + log(x)*(12*exp(2*x) + 8*exp(3*x) + 6*exp(x) + 1) - exp(x)*(4*x - 8) + 2)/(log(
(x + 16*exp(2*x) + 16*exp(x) + log(x)*(x + 4*x*exp(2*x) + 4*x*exp(x)) + 4)/(4*exp(2*x) + 4*exp(x) + 1))*(x + 4
8*exp(2*x) + 32*exp(3*x) + exp(x)*(2*x + 24) + log(x)*(x + 12*x*exp(2*x) + 8*x*exp(3*x) + 6*x*exp(x)) + 4)),x)

[Out]

log(log(x + 16*exp(2*x) + 16*exp(x) + x*log(x) + 4*x*exp(x)*log(x) + 4*x*exp(2*x)*log(x) + 4) - log(4*exp(2*x)
+ 4*exp(x) + 1))

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sympy [B]  time = 7.69, size = 51, normalized size = 2.43 \begin {gather*} \log {\left (\log {\left (\frac {x + \left (4 x e^{2 x} + 4 x e^{x} + x\right ) \log {\relax (x )} + 16 e^{2 x} + 16 e^{x} + 4}{4 e^{2 x} + 4 e^{x} + 1} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*exp(x)**3+12*exp(x)**2+6*exp(x)+1)*ln(x)+8*exp(x)**3+12*exp(x)**2+(-4*x+8)*exp(x)+2)/((8*x*exp(x
)**3+12*x*exp(x)**2+6*exp(x)*x+x)*ln(x)+32*exp(x)**3+48*exp(x)**2+(2*x+24)*exp(x)+4+x)/ln(((4*x*exp(x)**2+4*ex
p(x)*x+x)*ln(x)+16*exp(x)**2+16*exp(x)+4+x)/(4*exp(x)**2+4*exp(x)+1)),x)

[Out]

log(log((x + (4*x*exp(2*x) + 4*x*exp(x) + x)*log(x) + 16*exp(2*x) + 16*exp(x) + 4)/(4*exp(2*x) + 4*exp(x) + 1)
))

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