3.36.20 \(\int \frac {e^{\frac {1}{x}} (-104 x+26 x^2-2 x^4+2 x^5+(-52 x-x^4) \log (3))+e^{\frac {1}{x}} (52-26 x-2 x^3+x^4+(26-x^3) \log (3)) \log (\frac {52-26 x-2 x^3+x^4+(26-x^3) \log (3)}{x^2})}{(-52 x^2+26 x^3+2 x^5-x^6+(-26 x^2+x^5) \log (3)) \log ^2(\frac {52-26 x-2 x^3+x^4+(26-x^3) \log (3)}{x^2})} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{\frac {1}{x}}}{\log \left (\left (\frac {26}{x^2}-x\right ) (2-x+\log (3))\right )} \]

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Rubi [A]  time = 0.74, antiderivative size = 27, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, integrand size = 161, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6688, 2288} \begin {gather*} \frac {e^{\frac {1}{x}}}{\log \left (\frac {\left (26-x^3\right ) (-x+2+\log (3))}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x^(-1)*(-104*x + 26*x^2 - 2*x^4 + 2*x^5 + (-52*x - x^4)*Log[3]) + E^x^(-1)*(52 - 26*x - 2*x^3 + x^4 + (
26 - x^3)*Log[3])*Log[(52 - 26*x - 2*x^3 + x^4 + (26 - x^3)*Log[3])/x^2])/((-52*x^2 + 26*x^3 + 2*x^5 - x^6 + (
-26*x^2 + x^5)*Log[3])*Log[(52 - 26*x - 2*x^3 + x^4 + (26 - x^3)*Log[3])/x^2]^2),x]

[Out]

E^x^(-1)/Log[((26 - x^3)*(2 - x + Log[3]))/x^2]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{x}} \left (\frac {x \left (-26 x-2 x^4+52 (2+\log (3))+x^3 (2+\log (3))\right )}{\left (-26+x^3\right ) (-2+x-\log (3))}-\log \left (\frac {\left (-26+x^3\right ) (-2+x-\log (3))}{x^2}\right )\right )}{x^2 \log ^2\left (\frac {\left (-26+x^3\right ) (-2+x-\log (3))}{x^2}\right )} \, dx\\ &=\frac {e^{\frac {1}{x}}}{\log \left (\frac {\left (26-x^3\right ) (2-x+\log (3))}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 25, normalized size = 0.96 \begin {gather*} \frac {e^{\frac {1}{x}}}{\log \left (\frac {\left (-26+x^3\right ) (-2+x-\log (3))}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x^(-1)*(-104*x + 26*x^2 - 2*x^4 + 2*x^5 + (-52*x - x^4)*Log[3]) + E^x^(-1)*(52 - 26*x - 2*x^3 + x
^4 + (26 - x^3)*Log[3])*Log[(52 - 26*x - 2*x^3 + x^4 + (26 - x^3)*Log[3])/x^2])/((-52*x^2 + 26*x^3 + 2*x^5 - x
^6 + (-26*x^2 + x^5)*Log[3])*Log[(52 - 26*x - 2*x^3 + x^4 + (26 - x^3)*Log[3])/x^2]^2),x]

[Out]

E^x^(-1)/Log[((-26 + x^3)*(-2 + x - Log[3]))/x^2]

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fricas [A]  time = 1.17, size = 34, normalized size = 1.31 \begin {gather*} \frac {e^{\frac {1}{x}}}{\log \left (\frac {x^{4} - 2 \, x^{3} - {\left (x^{3} - 26\right )} \log \relax (3) - 26 \, x + 52}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+26)*log(3)+x^4-2*x^3-26*x+52)*exp(1/x)*log(((-x^3+26)*log(3)+x^4-2*x^3-26*x+52)/x^2)+((-x^4-
52*x)*log(3)+2*x^5-2*x^4+26*x^2-104*x)*exp(1/x))/((x^5-26*x^2)*log(3)-x^6+2*x^5+26*x^3-52*x^2)/log(((-x^3+26)*
log(3)+x^4-2*x^3-26*x+52)/x^2)^2,x, algorithm="fricas")

[Out]

e^(1/x)/log((x^4 - 2*x^3 - (x^3 - 26)*log(3) - 26*x + 52)/x^2)

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giac [A]  time = 0.54, size = 39, normalized size = 1.50 \begin {gather*} \frac {e^{\frac {1}{x}}}{\log \left (x^{4} - x^{3} \log \relax (3) - 2 \, x^{3} - 26 \, x + 26 \, \log \relax (3) + 52\right ) - \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+26)*log(3)+x^4-2*x^3-26*x+52)*exp(1/x)*log(((-x^3+26)*log(3)+x^4-2*x^3-26*x+52)/x^2)+((-x^4-
52*x)*log(3)+2*x^5-2*x^4+26*x^2-104*x)*exp(1/x))/((x^5-26*x^2)*log(3)-x^6+2*x^5+26*x^3-52*x^2)/log(((-x^3+26)*
log(3)+x^4-2*x^3-26*x+52)/x^2)^2,x, algorithm="giac")

[Out]

e^(1/x)/(log(x^4 - x^3*log(3) - 2*x^3 - 26*x + 26*log(3) + 52) - log(x^2))

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maple [C]  time = 0.17, size = 364, normalized size = 14.00

method	result	size
risch	\(-\frac {2 i {\mathrm e}^{\frac {1}{x}}}{\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 \pi \mathrm {csgn}\left (\frac {i \left (x^{3}-26\right ) \left (\ln \relax (3)-x +2\right )}{x^{2}}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (\ln \relax (3)-x +2\right ) \left (x^{3}-26\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{3}-26\right ) \left (\ln \relax (3)-x +2\right )}{x^{2}}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{3}-26\right ) \left (\ln \relax (3)-x +2\right )}{x^{2}}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (\ln \relax (3)-x +2\right )\right ) \mathrm {csgn}\left (i \left (x^{3}-26\right )\right ) \mathrm {csgn}\left (i \left (\ln \relax (3)-x +2\right ) \left (x^{3}-26\right )\right )+\pi \,\mathrm {csgn}\left (i \left (\ln \relax (3)-x +2\right )\right ) \mathrm {csgn}\left (i \left (\ln \relax (3)-x +2\right ) \left (x^{3}-26\right )\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (x^{3}-26\right )\right ) \mathrm {csgn}\left (i \left (\ln \relax (3)-x +2\right ) \left (x^{3}-26\right )\right )^{2}-\pi \mathrm {csgn}\left (i \left (\ln \relax (3)-x +2\right ) \left (x^{3}-26\right )\right )^{3}+\pi \,\mathrm {csgn}\left (i \left (\ln \relax (3)-x +2\right ) \left (x^{3}-26\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{3}-26\right ) \left (\ln \relax (3)-x +2\right )}{x^{2}}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x^{3}-26\right ) \left (\ln \relax (3)-x +2\right )}{x^{2}}\right )^{3}+2 \pi -2 i \ln \left (x^{3}-26\right )-2 i \ln \left (\ln \relax (3)-x +2\right )+4 i \ln \relax (x )}\)	\(364\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^3+26)*ln(3)+x^4-2*x^3-26*x+52)*exp(1/x)*ln(((-x^3+26)*ln(3)+x^4-2*x^3-26*x+52)/x^2)+((-x^4-52*x)*ln(
3)+2*x^5-2*x^4+26*x^2-104*x)*exp(1/x))/((x^5-26*x^2)*ln(3)-x^6+2*x^5+26*x^3-52*x^2)/ln(((-x^3+26)*ln(3)+x^4-2*
x^3-26*x+52)/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-2*I*exp(1/x)/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3-2*Pi*csgn(I/x^2*(x^3-2
6)*(ln(3)-x+2))^2-Pi*csgn(I/x^2)*csgn(I*(ln(3)-x+2)*(x^3-26))*csgn(I/x^2*(x^3-26)*(ln(3)-x+2))+Pi*csgn(I/x^2)*
csgn(I/x^2*(x^3-26)*(ln(3)-x+2))^2-Pi*csgn(I*(ln(3)-x+2))*csgn(I*(x^3-26))*csgn(I*(ln(3)-x+2)*(x^3-26))+Pi*csg
n(I*(ln(3)-x+2))*csgn(I*(ln(3)-x+2)*(x^3-26))^2+Pi*csgn(I*(x^3-26))*csgn(I*(ln(3)-x+2)*(x^3-26))^2-Pi*csgn(I*(
ln(3)-x+2)*(x^3-26))^3+Pi*csgn(I*(ln(3)-x+2)*(x^3-26))*csgn(I/x^2*(x^3-26)*(ln(3)-x+2))^2+Pi*csgn(I/x^2*(x^3-2
6)*(ln(3)-x+2))^3+2*Pi-2*I*ln(x^3-26)-2*I*ln(ln(3)-x+2)+4*I*ln(x))

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maxima [A]  time = 0.59, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {1}{x}}}{\log \left (x^{3} - 26\right ) + \log \left (x - \log \relax (3) - 2\right ) - 2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+26)*log(3)+x^4-2*x^3-26*x+52)*exp(1/x)*log(((-x^3+26)*log(3)+x^4-2*x^3-26*x+52)/x^2)+((-x^4-
52*x)*log(3)+2*x^5-2*x^4+26*x^2-104*x)*exp(1/x))/((x^5-26*x^2)*log(3)-x^6+2*x^5+26*x^3-52*x^2)/log(((-x^3+26)*
log(3)+x^4-2*x^3-26*x+52)/x^2)^2,x, algorithm="maxima")

[Out]

e^(1/x)/(log(x^3 - 26) + log(x - log(3) - 2) - 2*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{1/x}\,\left (104\,x-26\,x^2+2\,x^4-2\,x^5+\ln \relax (3)\,\left (x^4+52\,x\right )\right )+\ln \left (-\frac {26\,x+\ln \relax (3)\,\left (x^3-26\right )+2\,x^3-x^4-52}{x^2}\right )\,{\mathrm {e}}^{1/x}\,\left (26\,x+\ln \relax (3)\,\left (x^3-26\right )+2\,x^3-x^4-52\right )}{{\ln \left (-\frac {26\,x+\ln \relax (3)\,\left (x^3-26\right )+2\,x^3-x^4-52}{x^2}\right )}^2\,\left (\ln \relax (3)\,\left (26\,x^2-x^5\right )+52\,x^2-26\,x^3-2\,x^5+x^6\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/x)*(104*x - 26*x^2 + 2*x^4 - 2*x^5 + log(3)*(52*x + x^4)) + log(-(26*x + log(3)*(x^3 - 26) + 2*x^3
- x^4 - 52)/x^2)*exp(1/x)*(26*x + log(3)*(x^3 - 26) + 2*x^3 - x^4 - 52))/(log(-(26*x + log(3)*(x^3 - 26) + 2*x
^3 - x^4 - 52)/x^2)^2*(log(3)*(26*x^2 - x^5) + 52*x^2 - 26*x^3 - 2*x^5 + x^6)),x)

[Out]

int((exp(1/x)*(104*x - 26*x^2 + 2*x^4 - 2*x^5 + log(3)*(52*x + x^4)) + log(-(26*x + log(3)*(x^3 - 26) + 2*x^3
- x^4 - 52)/x^2)*exp(1/x)*(26*x + log(3)*(x^3 - 26) + 2*x^3 - x^4 - 52))/(log(-(26*x + log(3)*(x^3 - 26) + 2*x
^3 - x^4 - 52)/x^2)^2*(log(3)*(26*x^2 - x^5) + 52*x^2 - 26*x^3 - 2*x^5 + x^6)), x)

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sympy [A]  time = 0.75, size = 31, normalized size = 1.19 \begin {gather*} \frac {e^{\frac {1}{x}}}{\log {\left (\frac {x^{4} - 2 x^{3} - 26 x + \left (26 - x^{3}\right ) \log {\relax (3 )} + 52}{x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**3+26)*ln(3)+x**4-2*x**3-26*x+52)*exp(1/x)*ln(((-x**3+26)*ln(3)+x**4-2*x**3-26*x+52)/x**2)+((-
x**4-52*x)*ln(3)+2*x**5-2*x**4+26*x**2-104*x)*exp(1/x))/((x**5-26*x**2)*ln(3)-x**6+2*x**5+26*x**3-52*x**2)/ln(
((-x**3+26)*ln(3)+x**4-2*x**3-26*x+52)/x**2)**2,x)

[Out]

exp(1/x)/log((x**4 - 2*x**3 - 26*x + (26 - x**3)*log(3) + 52)/x**2)

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