3.53.20 \(\int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+(16-4 x-32 x^2+4 x^3) \log (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x})+(-2+4 x^2) \log ^2(\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x})}{32 x-8 x^2+(-16 x+2 x^2) \log (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x})+2 x \log ^2(\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x})} \, dx\)

Optimal. Leaf size=40 \[ x^2-\log \left (x-\frac {x^2}{4-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}\right ) \]

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Rubi [F]  time = 8.22, 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 {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{32 x-8 x^2+\left (-16 x+2 x^2\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-32 + E^((-1 + 9*x)/x) + 14*x + 64*x^2 - 16*x^3 + (16 - 4*x - 32*x^2 + 4*x^3)*Log[E^(E^((-1 + 9*x)/x)/2)/
x] + (-2 + 4*x^2)*Log[E^(E^((-1 + 9*x)/x)/2)/x]^2)/(32*x - 8*x^2 + (-16*x + 2*x^2)*Log[E^(E^((-1 + 9*x)/x)/2)/
x] + 2*x*Log[E^(E^((-1 + 9*x)/x)/2)/x]^2),x]

[Out]

x^2 - Log[x] + 32*Defer[Int][(4 - x - Log[E^(E^(9 - x^(-1))/2)/x])^(-1), x] + 31*Defer[Int][(-4 + x + Log[E^(E
^(9 - x^(-1))/2)/x])^(-1), x] - Defer[Int][E^(9 - x^(-1))/(x^2*(-4 + x + Log[E^(E^(9 - x^(-1))/2)/x])), x]/2 +
Defer[Int][1/(x*(-4 + x + Log[E^(E^(9 - x^(-1))/2)/x])), x] - Defer[Subst][Defer[Int][E^(9 - x)/(-4 + Log[E^(
E^(9 - x)/2)*x]), x], x, x^(-1)]/2 + Defer[Subst][Defer[Int][1/(x*(-4 + Log[E^(E^(9 - x)/2)*x])), x], x, x^(-1
)]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{2 x \left (4-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (4-x-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx\\ &=\frac {1}{2} \int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{x \left (4-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (4-x-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx\\ &=\frac {1}{2} \int \left (\frac {14}{\left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}-\frac {32}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}+\frac {e^{9-\frac {1}{x}}}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}+\frac {64 x}{\left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}-\frac {16 x^2}{\left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}+\frac {4 \left (4-x-8 x^2+x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}+\frac {2 \left (-1+2 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{9-\frac {1}{x}}}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+2 \int \frac {\left (4-x-8 x^2+x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+7 \int \frac {1}{\left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx-8 \int \frac {x^2}{\left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx-16 \int \frac {1}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+32 \int \frac {x}{\left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+\int \frac {\left (-1+2 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx\\ &=\frac {1}{2} \int \left (\frac {e^{9-\frac {1}{x}}}{x^2 \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}-\frac {e^{9-\frac {1}{x}}}{x^2 \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}\right ) \, dx+2 \int \left (\frac {4 \left (4-x-8 x^2+x^3\right )}{x^2 \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}+\frac {(-4+x) \left (4-x-8 x^2+x^3\right )}{x^2 \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}\right ) \, dx+7 \int \left (\frac {1}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}-\frac {1}{x \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}\right ) \, dx-8 \int \left (\frac {x}{-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}-\frac {x}{-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}\right ) \, dx-16 \int \left (\frac {1}{x^2 \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}-\frac {1}{x^2 \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}\right ) \, dx+32 \int \left (\frac {1}{4-x-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}+\frac {1}{-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}\right ) \, dx+\int \left (\frac {-1+2 x^2}{x}+\frac {16 \left (-1+2 x^2\right )}{x^2 \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}-\frac {(-4+x)^2 \left (-1+2 x^2\right )}{x^2 \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{9-\frac {1}{x}}}{x^2 \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx-\frac {1}{2} \int \frac {e^{9-\frac {1}{x}}}{x^2 \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+2 \int \frac {(-4+x) \left (4-x-8 x^2+x^3\right )}{x^2 \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+7 \int \frac {1}{x \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx-7 \int \frac {1}{x \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx-8 \int \frac {x}{-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )} \, dx+8 \int \frac {4-x-8 x^2+x^3}{x^2 \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+8 \int \frac {x}{-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )} \, dx-16 \int \frac {1}{x^2 \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+16 \int \frac {-1+2 x^2}{x^2 \left (-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+16 \int \frac {1}{x^2 \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx+32 \int \frac {1}{4-x-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )} \, dx+32 \int \frac {1}{-4+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )} \, dx+\int \frac {-1+2 x^2}{x} \, dx-\int \frac {(-4+x)^2 \left (-1+2 x^2\right )}{x^2 \left (-4+x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 145, normalized size = 3.62 \begin {gather*} \frac {1}{2} \left (2 x^2-2 \log (x)+2 \log \left (8-e^{9-\frac {1}{x}}-2 \log \left (\frac {1}{x}\right )-2 \left (-\frac {1}{2} e^{9-\frac {1}{x}}-\log \left (\frac {1}{x}\right )+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )\right )-2 \log \left (8-e^{9-\frac {1}{x}}-2 x-2 \log \left (\frac {1}{x}\right )-2 \left (-\frac {1}{2} e^{9-\frac {1}{x}}-\log \left (\frac {1}{x}\right )+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32 + E^((-1 + 9*x)/x) + 14*x + 64*x^2 - 16*x^3 + (16 - 4*x - 32*x^2 + 4*x^3)*Log[E^(E^((-1 + 9*x)/
x)/2)/x] + (-2 + 4*x^2)*Log[E^(E^((-1 + 9*x)/x)/2)/x]^2)/(32*x - 8*x^2 + (-16*x + 2*x^2)*Log[E^(E^((-1 + 9*x)/
x)/2)/x] + 2*x*Log[E^(E^((-1 + 9*x)/x)/2)/x]^2),x]

[Out]

(2*x^2 - 2*Log[x] + 2*Log[8 - E^(9 - x^(-1)) - 2*Log[x^(-1)] - 2*(-1/2*E^(9 - x^(-1)) - Log[x^(-1)] + Log[E^(E
^(9 - x^(-1))/2)/x])] - 2*Log[8 - E^(9 - x^(-1)) - 2*x - 2*Log[x^(-1)] - 2*(-1/2*E^(9 - x^(-1)) - Log[x^(-1)]
+ Log[E^(E^(9 - x^(-1))/2)/x])])/2

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fricas [B]  time = 0.72, size = 79, normalized size = 1.98 \begin {gather*} x^{2} - \frac {1}{2} \, e^{\left (\frac {9 \, x - 1}{x}\right )} - \log \left (x + \log \left (\frac {e^{\left (\frac {1}{2} \, e^{\left (\frac {9 \, x - 1}{x}\right )}\right )}}{x}\right ) - 4\right ) + \log \left (\frac {e^{\left (\frac {1}{2} \, e^{\left (\frac {9 \, x - 1}{x}\right )}\right )}}{x}\right ) + \log \left (\log \left (\frac {e^{\left (\frac {1}{2} \, e^{\left (\frac {9 \, x - 1}{x}\right )}\right )}}{x}\right ) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-2)*log(exp(1/2*exp((9*x-1)/x))/x)^2+(4*x^3-32*x^2-4*x+16)*log(exp(1/2*exp((9*x-1)/x))/x)+exp
((9*x-1)/x)-16*x^3+64*x^2+14*x-32)/(2*x*log(exp(1/2*exp((9*x-1)/x))/x)^2+(2*x^2-16*x)*log(exp(1/2*exp((9*x-1)/
x))/x)-8*x^2+32*x),x, algorithm="fricas")

[Out]

x^2 - 1/2*e^((9*x - 1)/x) - log(x + log(e^(1/2*e^((9*x - 1)/x))/x) - 4) + log(e^(1/2*e^((9*x - 1)/x))/x) + log
(log(e^(1/2*e^((9*x - 1)/x))/x) - 4)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-2)*log(exp(1/2*exp((9*x-1)/x))/x)^2+(4*x^3-32*x^2-4*x+16)*log(exp(1/2*exp((9*x-1)/x))/x)+exp
((9*x-1)/x)-16*x^3+64*x^2+14*x-32)/(2*x*log(exp(1/2*exp((9*x-1)/x))/x)^2+(2*x^2-16*x)*log(exp(1/2*exp((9*x-1)/
x))/x)-8*x^2+32*x),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.20, size = 357, normalized size = 8.92

method	result	size
risch	\(x^{2}-\ln \relax (x )+\ln \left (\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{3}-2 i \ln \relax (x )-8 i\right )}{2}\right )-\ln \left (\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{3}+2 i x -2 i \ln \relax (x )-8 i\right )}{2}\right )\)	\(357\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2-2)*ln(exp(1/2*exp((9*x-1)/x))/x)^2+(4*x^3-32*x^2-4*x+16)*ln(exp(1/2*exp((9*x-1)/x))/x)+exp((9*x-1)
/x)-16*x^3+64*x^2+14*x-32)/(2*x*ln(exp(1/2*exp((9*x-1)/x))/x)^2+(2*x^2-16*x)*ln(exp(1/2*exp((9*x-1)/x))/x)-8*x
^2+32*x),x,method=_RETURNVERBOSE)

[Out]

x^2-ln(x)+ln(ln(exp(1/2*exp((9*x-1)/x)))-1/2*I*(Pi*csgn(I/x)*csgn(I*exp(1/2*exp((9*x-1)/x)))*csgn(I/x*exp(1/2*
exp((9*x-1)/x)))-Pi*csgn(I/x)*csgn(I/x*exp(1/2*exp((9*x-1)/x)))^2-Pi*csgn(I*exp(1/2*exp((9*x-1)/x)))*csgn(I/x*
exp(1/2*exp((9*x-1)/x)))^2+Pi*csgn(I/x*exp(1/2*exp((9*x-1)/x)))^3-2*I*ln(x)-8*I))-ln(ln(exp(1/2*exp((9*x-1)/x)
))-1/2*I*(Pi*csgn(I/x)*csgn(I*exp(1/2*exp((9*x-1)/x)))*csgn(I/x*exp(1/2*exp((9*x-1)/x)))-Pi*csgn(I/x)*csgn(I/x
*exp(1/2*exp((9*x-1)/x)))^2-Pi*csgn(I*exp(1/2*exp((9*x-1)/x)))*csgn(I/x*exp(1/2*exp((9*x-1)/x)))^2+Pi*csgn(I/x
*exp(1/2*exp((9*x-1)/x)))^3+2*I*x-2*I*ln(x)-8*I))

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maxima [B]  time = 0.42, size = 77, normalized size = 1.92 \begin {gather*} x^{2} - \log \relax (x) - \log \left (-x + \log \relax (x) + 4\right ) - \log \left (\frac {2 \, {\left (x - \log \relax (x) - 4\right )} e^{\frac {1}{x}} + e^{9}}{2 \, {\left (x - \log \relax (x) - 4\right )}}\right ) + \log \left (\frac {2 \, {\left (\log \relax (x) + 4\right )} e^{\frac {1}{x}} - e^{9}}{2 \, {\left (\log \relax (x) + 4\right )}}\right ) + \log \left (\log \relax (x) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-2)*log(exp(1/2*exp((9*x-1)/x))/x)^2+(4*x^3-32*x^2-4*x+16)*log(exp(1/2*exp((9*x-1)/x))/x)+exp
((9*x-1)/x)-16*x^3+64*x^2+14*x-32)/(2*x*log(exp(1/2*exp((9*x-1)/x))/x)^2+(2*x^2-16*x)*log(exp(1/2*exp((9*x-1)/
x))/x)-8*x^2+32*x),x, algorithm="maxima")

[Out]

x^2 - log(x) - log(-x + log(x) + 4) - log(1/2*(2*(x - log(x) - 4)*e^(1/x) + e^9)/(x - log(x) - 4)) + log(1/2*(
2*(log(x) + 4)*e^(1/x) - e^9)/(log(x) + 4)) + log(log(x) + 4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {14\,x+{\mathrm {e}}^{\frac {9\,x-1}{x}}-\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {9\,x-1}{x}}}{2}}}{x}\right )\,\left (-4\,x^3+32\,x^2+4\,x-16\right )+{\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {9\,x-1}{x}}}{2}}}{x}\right )}^2\,\left (4\,x^2-2\right )+64\,x^2-16\,x^3-32}{32\,x-\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {9\,x-1}{x}}}{2}}}{x}\right )\,\left (16\,x-2\,x^2\right )+2\,x\,{\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {9\,x-1}{x}}}{2}}}{x}\right )}^2-8\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((14*x + exp((9*x - 1)/x) - log(exp(exp((9*x - 1)/x)/2)/x)*(4*x + 32*x^2 - 4*x^3 - 16) + log(exp(exp((9*x -
1)/x)/2)/x)^2*(4*x^2 - 2) + 64*x^2 - 16*x^3 - 32)/(32*x - log(exp(exp((9*x - 1)/x)/2)/x)*(16*x - 2*x^2) + 2*x
*log(exp(exp((9*x - 1)/x)/2)/x)^2 - 8*x^2),x)

[Out]

int((14*x + exp((9*x - 1)/x) - log(exp(exp((9*x - 1)/x)/2)/x)*(4*x + 32*x^2 - 4*x^3 - 16) + log(exp(exp((9*x -
1)/x)/2)/x)^2*(4*x^2 - 2) + 64*x^2 - 16*x^3 - 32)/(32*x - log(exp(exp((9*x - 1)/x)/2)/x)*(16*x - 2*x^2) + 2*x
*log(exp(exp((9*x - 1)/x)/2)/x)^2 - 8*x^2), x)

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sympy [A]  time = 0.56, size = 44, normalized size = 1.10 \begin {gather*} x^{2} - \log {\relax (x )} + \log {\left (\log {\left (\frac {e^{\frac {e^{\frac {9 x - 1}{x}}}{2}}}{x} \right )} - 4 \right )} - \log {\left (x + \log {\left (\frac {e^{\frac {e^{\frac {9 x - 1}{x}}}{2}}}{x} \right )} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2-2)*ln(exp(1/2*exp((9*x-1)/x))/x)**2+(4*x**3-32*x**2-4*x+16)*ln(exp(1/2*exp((9*x-1)/x))/x)+e
xp((9*x-1)/x)-16*x**3+64*x**2+14*x-32)/(2*x*ln(exp(1/2*exp((9*x-1)/x))/x)**2+(2*x**2-16*x)*ln(exp(1/2*exp((9*x
-1)/x))/x)-8*x**2+32*x),x)

[Out]

x**2 - log(x) + log(log(exp(exp((9*x - 1)/x)/2)/x) - 4) - log(x + log(exp(exp((9*x - 1)/x)/2)/x) - 4)

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