3.29.16 \(\int \frac {(-1+x^2) (\frac {-1+x^2}{x})^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} (-2 x^2+2 x^4+(2 x-2 x^3) \log (2)) \log (\frac {-1+x^2}{x})))}{x (-x+x^3)} \, dx\)

Optimal. Leaf size=24 \[ \left (-\frac {1}{x}+x\right )^{1+e^{e^{15+(x-\log (2))^2}}} \]

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Rubi [F]  time = 18.62, 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 (-1+x^2\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x \left (-x+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-1 + x^2)*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)*(1 + x^2 + E^E^(15 + x^2 - 2*x*Log[2] +
Log[2]^2)*(1 + x^2 + E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)*(-2*x^2 + 2*x^4 + (2*x - 2*x^3)*Log[2])*Log[(-1 + x^
2)/x])))/(x*(-x + x^3)),x]

[Out]

Defer[Int][((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2), x] + Defer[Int][E^(E^(15 + x^2 + Log[2]^2)/4
^x)*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2), x] - Log[-((1 - x^2)/x)]*Defer[Int][2^(1 - 2*x)*E^(
E^(15 + x^2 + Log[2]^2)/2^(2*x) + x^2 + 15*(1 + Log[2]^2/15))*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[
2]^2), x] + Defer[Int][((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)/x^2, x] + Defer[Int][(E^(E^(15 +
x^2 + Log[2]^2)/4^x)*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2))/x^2, x] + Log[2]*Log[-((1 - x^2)/x
)]*Defer[Int][(2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/2^(2*x) + x^2 + 15*(1 + Log[2]^2/15))*((-1 + x^2)/x)^E^E
^(15 + x^2 - 2*x*Log[2] + Log[2]^2))/x, x] - Log[2]*Log[-((1 - x^2)/x)]*Defer[Int][2^(1 - 2*x)*E^(E^(15 + x^2
+ Log[2]^2)/2^(2*x) + x^2 + 15*(1 + Log[2]^2/15))*x*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2), x]
+ Log[-((1 - x^2)/x)]*Defer[Int][2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/2^(2*x) + x^2 + 15*(1 + Log[2]^2/15))*
x^2*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2), x] + Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15 + x
^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*((-1 + x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/(-1 + x),
x] - Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*((-1 + x^2
)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/x, x] + Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/
4^x + x^2 + 15*(1 + Log[2]^2/15))*((-1 + x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/(1 + x), x] + Log[2]*Defe
r[Int][Defer[Int][(2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*((-1 + x^2)/x)^E^(
E^(15 + x^2 + Log[2]^2)/4^x))/x, x]/(-1 - x), x] + Log[2]*Defer[Int][Defer[Int][(2^(1 - 2*x)*E^(E^(15 + x^2 +
Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*((-1 + x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x))/x, x]/(1 - x), x]
+ Log[2]*Defer[Int][Defer[Int][(2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*((-1
+ x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x))/x, x]/x, x] + Log[2]*Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15 + x
^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*x*((-1 + x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/(-1 + x)
, x] - Log[2]*Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*x
*((-1 + x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/x, x] + Log[2]*Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15
+ x^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*x*((-1 + x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/(1 +
x), x] - Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*x^2*((
-1 + x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/(-1 + x), x] + Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15 + x
^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*x^2*((-1 + x^2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/x, x]
- Defer[Int][Defer[Int][2^(1 - 2*x)*E^(E^(15 + x^2 + Log[2]^2)/4^x + x^2 + 15*(1 + Log[2]^2/15))*x^2*((-1 + x^
2)/x)^E^(E^(15 + x^2 + Log[2]^2)/4^x), x]/(1 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x^2} \, dx\\ &=\int \left (\frac {\left (1+e^{4^{-x} e^{15+x^2+\log ^2(2)}}\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2\right )}{x^2}+\frac {2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) (1-x) (1+x) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} (-x+\log (2)) \log \left (\frac {-1+x^2}{x}\right )}{x}\right ) \, dx\\ &=\int \frac {\left (1+e^{4^{-x} e^{15+x^2+\log ^2(2)}}\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2\right )}{x^2} \, dx+\int \frac {2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) (1-x) (1+x) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} (-x+\log (2)) \log \left (\frac {-1+x^2}{x}\right )}{x} \, dx\\ &=-\left (\log \left (\frac {-1+x^2}{x}\right ) \int 2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \, dx\right )+\log \left (\frac {-1+x^2}{x}\right ) \int 2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) x^2 \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \, dx+\left (\log (2) \log \left (\frac {-1+x^2}{x}\right )\right ) \int \frac {2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}}}{x} \, dx-\left (\log (2) \log \left (\frac {-1+x^2}{x}\right )\right ) \int 2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) x \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \, dx+\int \left (\frac {\left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2\right )}{x^2}+\frac {e^{4^{-x} e^{15+x^2+\log ^2(2)}} \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2\right )}{x^2}\right ) \, dx-\int \frac {\left (1+x^2\right ) \left (\int 2^{1-2 x} e^{15+4^{-x} e^{15+x^2+\log ^2(2)}+x^2+\log ^2(2)} \left (-\frac {1}{x}+x\right )^{e^{4^{-x} e^{15+x^2+\log ^2(2)}}} \, dx-\log (2) \int \frac {2^{1-2 x} e^{15+4^{-x} e^{15+x^2+\log ^2(2)}+x^2+\log ^2(2)} \left (-\frac {1}{x}+x\right )^{e^{4^{-x} e^{15+x^2+\log ^2(2)}}}}{x} \, dx+\log (2) \int 2^{1-2 x} e^{15+4^{-x} e^{15+x^2+\log ^2(2)}+x^2+\log ^2(2)} x \left (-\frac {1}{x}+x\right )^{e^{4^{-x} e^{15+x^2+\log ^2(2)}}} \, dx-\int 2^{1-2 x} e^{15+4^{-x} e^{15+x^2+\log ^2(2)}+x^2+\log ^2(2)} x^2 \left (-\frac {1}{x}+x\right )^{e^{4^{-x} e^{15+x^2+\log ^2(2)}}} \, dx\right )}{x \left (1-x^2\right )} \, dx\\ &=-\left (\log \left (\frac {-1+x^2}{x}\right ) \int 2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \, dx\right )+\log \left (\frac {-1+x^2}{x}\right ) \int 2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) x^2 \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \, dx+\left (\log (2) \log \left (\frac {-1+x^2}{x}\right )\right ) \int \frac {2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}}}{x} \, dx-\left (\log (2) \log \left (\frac {-1+x^2}{x}\right )\right ) \int 2^{1-2 x} \exp \left (2^{-2 x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) x \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \, dx+\int \frac {\left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2\right )}{x^2} \, dx+\int \frac {e^{4^{-x} e^{15+x^2+\log ^2(2)}} \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2\right )}{x^2} \, dx-\int \left (-\frac {\left (1+x^2\right ) \left (\int 2^{1-2 x} e^{15+4^{-x} e^{15+x^2+\log ^2(2)}+x^2+\log ^2(2)} \left (-\frac {1}{x}+x\right )^{e^{4^{-x} e^{15+x^2+\log ^2(2)}}} \, dx-\log (2) \int \frac {2^{1-2 x} e^{15+4^{-x} e^{15+x^2+\log ^2(2)}+x^2+\log ^2(2)} \left (-\frac {1}{x}+x\right )^{e^{4^{-x} e^{15+x^2+\log ^2(2)}}}}{x} \, dx+\log (2) \int 2^{1-2 x} e^{15+4^{-x} e^{15+x^2+\log ^2(2)}+x^2+\log ^2(2)} x \left (-\frac {1}{x}+x\right )^{e^{4^{-x} e^{15+x^2+\log ^2(2)}}} \, dx\right )}{x \left (-1+x^2\right )}+\frac {\left (-1-x^2\right ) \int 2^{1-2 x} \exp \left (4^{-x} e^{15+x^2+\log ^2(2)}+x^2+15 \left (1+\frac {\log ^2(2)}{15}\right )\right ) x^2 \left (\frac {-1+x^2}{x}\right )^{e^{4^{-x} e^{15+x^2+\log ^2(2)}}} \, dx}{x \left (1-x^2\right )}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.34, size = 28, normalized size = 1.17 \begin {gather*} \left (-\frac {1}{x}+x\right )^{1+e^{e^{15+x^2+\log ^2(2)-x \log (4)}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^2)*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)*(1 + x^2 + E^E^(15 + x^2 - 2*x*Log
[2] + Log[2]^2)*(1 + x^2 + E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)*(-2*x^2 + 2*x^4 + (2*x - 2*x^3)*Log[2])*Log[(-
1 + x^2)/x])))/(x*(-x + x^3)),x]

[Out]

(-x^(-1) + x)^(1 + E^E^(15 + x^2 + Log[2]^2 - x*Log[4]))

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fricas [A]  time = 1.07, size = 39, normalized size = 1.62 \begin {gather*} e^{\left (e^{\left (e^{\left (x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2} + 15\right )}\right )} \log \left (\frac {x^{2} - 1}{x}\right ) + \log \left (\frac {x^{2} - 1}{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-2*x^3+2*x)*log(2)+2*x^4-2*x^2)*exp(log(2)^2-2*x*log(2)+x^2+15)*log((x^2-1)/x)+x^2+1)*exp(exp(lo
g(2)^2-2*x*log(2)+x^2+15))+x^2+1)*exp(log((x^2-1)/x)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+log((x^2-1)/x))/(x^3
-x),x, algorithm="fricas")

[Out]

e^(e^(e^(x^2 - 2*x*log(2) + log(2)^2 + 15))*log((x^2 - 1)/x) + log((x^2 - 1)/x))

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giac [A]  time = 5.31, size = 35, normalized size = 1.46 \begin {gather*} e^{\left (e^{\left (e^{\left (x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2} + 15\right )}\right )} \log \left (x - \frac {1}{x}\right ) + \log \left (x - \frac {1}{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-2*x^3+2*x)*log(2)+2*x^4-2*x^2)*exp(log(2)^2-2*x*log(2)+x^2+15)*log((x^2-1)/x)+x^2+1)*exp(exp(lo
g(2)^2-2*x*log(2)+x^2+15))+x^2+1)*exp(log((x^2-1)/x)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+log((x^2-1)/x))/(x^3
-x),x, algorithm="giac")

[Out]

e^(e^(e^(x^2 - 2*x*log(2) + log(2)^2 + 15))*log(x - 1/x) + log(x - 1/x))

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maple [C]  time = 0.52, size = 134, normalized size = 5.58

method	result	size
risch	\(\frac {\left (x^{2}-1\right ) x^{-{\mathrm e}^{4^{-x} {\mathrm e}^{\ln \relax (2)^{2}+15+x^{2}}}} \left (x^{2}-1\right )^{{\mathrm e}^{4^{-x} {\mathrm e}^{\ln \relax (2)^{2}+15+x^{2}}}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right )+\mathrm {csgn}\left (i \left (x^{2}-1\right )\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left ({\mathrm e}^{\left (\frac {1}{4}\right )^{x} {\mathrm e}^{\ln \relax (2)^{2}+15+x^{2}}}+1\right )}{2}}}{x}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-2*x^3+2*x)*ln(2)+2*x^4-2*x^2)*exp(ln(2)^2-2*x*ln(2)+x^2+15)*ln((x^2-1)/x)+x^2+1)*exp(exp(ln(2)^2-2*x*
ln(2)+x^2+15))+x^2+1)*exp(ln((x^2-1)/x)*exp(exp(ln(2)^2-2*x*ln(2)+x^2+15))+ln((x^2-1)/x))/(x^3-x),x,method=_RE
TURNVERBOSE)

[Out]

(x^2-1)/x*x^(-exp(4^(-x)*exp(ln(2)^2+15+x^2)))*(x^2-1)^exp(4^(-x)*exp(ln(2)^2+15+x^2))*exp(-1/2*I*Pi*csgn(I/x*
(x^2-1))*(-csgn(I/x*(x^2-1))+csgn(I*(x^2-1)))*(-csgn(I/x*(x^2-1))+csgn(I/x))*(exp((1/4)^x*exp(ln(2)^2+15+x^2))
+1))

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maxima [B]  time = 1.34, size = 73, normalized size = 3.04 \begin {gather*} \frac {{\left (x^{2} - 1\right )} e^{\left (e^{\left (e^{\left (x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2} + 15\right )}\right )} \log \left (x + 1\right ) + e^{\left (e^{\left (x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2} + 15\right )}\right )} \log \left (x - 1\right ) - e^{\left (e^{\left (x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2} + 15\right )}\right )} \log \relax (x)\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-2*x^3+2*x)*log(2)+2*x^4-2*x^2)*exp(log(2)^2-2*x*log(2)+x^2+15)*log((x^2-1)/x)+x^2+1)*exp(exp(lo
g(2)^2-2*x*log(2)+x^2+15))+x^2+1)*exp(log((x^2-1)/x)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+log((x^2-1)/x))/(x^3
-x),x, algorithm="maxima")

[Out]

(x^2 - 1)*e^(e^(e^(x^2 - 2*x*log(2) + log(2)^2 + 15))*log(x + 1) + e^(e^(x^2 - 2*x*log(2) + log(2)^2 + 15))*lo
g(x - 1) - e^(e^(x^2 - 2*x*log(2) + log(2)^2 + 15))*log(x))/x

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mupad [B]  time = 2.03, size = 37, normalized size = 1.54 \begin {gather*} \frac {{\left (x-\frac {1}{x}\right )}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\ln \relax (2)}^2}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{15}}{2^{2\,x}}}}\,\left (x^2-1\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log((x^2 - 1)/x) + exp(exp(log(2)^2 - 2*x*log(2) + x^2 + 15))*log((x^2 - 1)/x))*(exp(exp(log(2)^2 -
2*x*log(2) + x^2 + 15))*(x^2 + exp(log(2)^2 - 2*x*log(2) + x^2 + 15)*log((x^2 - 1)/x)*(log(2)*(2*x - 2*x^3) -
2*x^2 + 2*x^4) + 1) + x^2 + 1))/(x - x^3),x)

[Out]

((x - 1/x)^exp((exp(log(2)^2)*exp(x^2)*exp(15))/2^(2*x))*(x^2 - 1))/x

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sympy [A]  time = 8.93, size = 36, normalized size = 1.50 \begin {gather*} \frac {\left (x^{2} - 1\right ) e^{e^{e^{x^{2} - 2 x \log {\relax (2 )} + \log {\relax (2 )}^{2} + 15}} \log {\left (\frac {x^{2} - 1}{x} \right )}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-2*x**3+2*x)*ln(2)+2*x**4-2*x**2)*exp(ln(2)**2-2*x*ln(2)+x**2+15)*ln((x**2-1)/x)+x**2+1)*exp(exp
(ln(2)**2-2*x*ln(2)+x**2+15))+x**2+1)*exp(ln((x**2-1)/x)*exp(exp(ln(2)**2-2*x*ln(2)+x**2+15))+ln((x**2-1)/x))/
(x**3-x),x)

[Out]

(x**2 - 1)*exp(exp(exp(x**2 - 2*x*log(2) + log(2)**2 + 15))*log((x**2 - 1)/x))/x

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