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

Optimal. Leaf size=27 \[ x \log \left (\log \left (-e^x+e^5 \left (4-e^3-\frac {1}{x}+x\right )\right )\right ) \]

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Rubi [F]  time = 3.72, 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 x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (-e^x+\frac {e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx\\ &=\int \left (\frac {e^5 \left (-1-x+\left (3-e^3\right ) x^2+x^3\right )}{\left (e^5+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )}+\frac {x+\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right ) \log \left (\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )\right )}{\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )}\right ) \, dx\\ &=e^5 \int \frac {-1-x+\left (3-e^3\right ) x^2+x^3}{\left (e^5+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )} \, dx+\int \frac {x+\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right ) \log \left (\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )\right )}{\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )} \, dx\\ &=e^5 \int \left (\frac {\left (3-e^3\right ) x^2}{\left (e^5+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )}+\frac {x^3}{\left (e^5+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )}+\frac {1}{\left (-e^5-e^x x+4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )}+\frac {x}{\left (-e^5-e^x x+4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )}\right ) \, dx+\int \left (\frac {x}{\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )}+\log \left (\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )\right )\right ) \, dx\\ &=e^5 \int \frac {x^3}{\left (e^5+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )} \, dx+e^5 \int \frac {1}{\left (-e^5-e^x x+4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )} \, dx+e^5 \int \frac {x}{\left (-e^5-e^x x+4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )} \, dx+\left (e^5 \left (3-e^3\right )\right ) \int \frac {x^2}{\left (e^5+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5 x^2\right ) \log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )} \, dx+\int \frac {x}{\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )} \, dx+\int \log \left (\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 27, normalized size = 1.00 \begin {gather*} x \log \left (\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*Log[Log[-E^8 - E^x + E^5*(4 - x^(-1) + x)]]

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fricas [A]  time = 1.68, size = 30, normalized size = 1.11 \begin {gather*} x \log \left (\log \left (-\frac {x e^{8} - {\left (x^{2} + 4 \, x - 1\right )} e^{5} + x e^{x}}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {x^{2} e^{x} - {\left ({\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}\right )} \log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right ) \log \left (\log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right )\right ) - {\left (x^{2} + 1\right )} e^{5}}{{\left ({\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}\right )} \log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(x^2*e^x - ((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)*log(((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)/x)*log(l
og(((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)/x)) - (x^2 + 1)*e^5)/(((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)*log(((x^2
- x*e^3 + 4*x - 1)*e^5 - x*e^x)/x)), x)

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maple [C]  time = 0.19, size = 232, normalized size = 8.59

method	result	size
risch	\(x \ln \left (i \pi -\ln \relax (x )+\ln \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right )+\mathrm {csgn}\left (i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )\right )\right )}{2}+i \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right )^{2} \left (\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right )-1\right )\right )\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.62, size = 35, normalized size = 1.30 \begin {gather*} x \log \left (\log \left (x^{2} e^{5} - x {\left (e^{8} - 4 \, e^{5}\right )} - x e^{x} - e^{5}\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.14, size = 30, normalized size = 1.11 \begin {gather*} x\,\ln \left (\ln \left (\frac {{\mathrm {e}}^5\,\left (4\,x-x\,{\mathrm {e}}^3+x^2-1\right )-x\,{\mathrm {e}}^x}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: CoercionFailed

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