3.2.17 \(\int \frac {e^{\frac {-x+(-2 x-x^3) \log (4)+(1+x^2 \log (4)) \log (2 x)}{(-5 x^2-x^3) \log (4)+(5 x+x^2) \log (4) \log (2 x)}} (-5 x^2-2 x^3+(10 x-8 x^2-4 x^3+5 x^4) \log (4)+(10 x+4 x^2+(2 x^2-10 x^3) \log (4)) \log (2 x)+(-5-2 x+5 x^2 \log (4)) \log ^2(2 x))}{(25 x^4+10 x^5+x^6) \log (4)+(-50 x^3-20 x^4-2 x^5) \log (4) \log (2 x)+(25 x^2+10 x^3+x^4) \log (4) \log ^2(2 x)} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {x+\frac {1}{x \log (4)}+\frac {2}{x-\log (2 x)}}{5+x}} \]

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Rubi [F]  time = 20.91, 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 {\exp \left (\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{\left (-5 x^2-x^3\right ) \log (4)+\left (5 x+x^2\right ) \log (4) \log (2 x)}\right ) \left (-5 x^2-2 x^3+\left (10 x-8 x^2-4 x^3+5 x^4\right ) \log (4)+\left (10 x+4 x^2+\left (2 x^2-10 x^3\right ) \log (4)\right ) \log (2 x)+\left (-5-2 x+5 x^2 \log (4)\right ) \log ^2(2 x)\right )}{\left (25 x^4+10 x^5+x^6\right ) \log (4)+\left (-50 x^3-20 x^4-2 x^5\right ) \log (4) \log (2 x)+\left (25 x^2+10 x^3+x^4\right ) \log (4) \log ^2(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-x + (-2*x - x^3)*Log[4] + (1 + x^2*Log[4])*Log[2*x])/((-5*x^2 - x^3)*Log[4] + (5*x + x^2)*Log[4]*Log
[2*x]))*(-5*x^2 - 2*x^3 + (10*x - 8*x^2 - 4*x^3 + 5*x^4)*Log[4] + (10*x + 4*x^2 + (2*x^2 - 10*x^3)*Log[4])*Log
[2*x] + (-5 - 2*x + 5*x^2*Log[4])*Log[2*x]^2))/((25*x^4 + 10*x^5 + x^6)*Log[4] + (-50*x^3 - 20*x^4 - 2*x^5)*Lo
g[4]*Log[2*x] + (25*x^2 + 10*x^3 + x^4)*Log[4]*Log[2*x]^2),x]

[Out]

-1/5*Defer[Int][1/(E^((-x + (-2*x - x^3)*Log[4] + (1 + x^2*Log[4])*Log[2*x])/(x*(5 + x)*Log[4]*(x - Log[2*x]))
)*x^2), x]/Log[4] + ((1 + 25*Log[4])*Defer[Int][1/(E^((-x + (-2*x - x^3)*Log[4] + (1 + x^2*Log[4])*Log[2*x])/(
x*(5 + x)*Log[4]*(x - Log[2*x])))*(5 + x)^2), x])/(5*Log[4]) + (2*Defer[Int][1/(E^((-x + (-2*x - x^3)*Log[4] +
(1 + x^2*Log[4])*Log[2*x])/(x*(5 + x)*Log[4]*(x - Log[2*x])))*x*(x - Log[2*x])^2), x])/5 - (12*Defer[Int][1/(
E^((-x + (-2*x - x^3)*Log[4] + (1 + x^2*Log[4])*Log[2*x])/(x*(5 + x)*Log[4]*(x - Log[2*x])))*(5 + x)*(x - Log[
2*x])^2), x])/5 - 2*Defer[Int][1/(E^((-x + (-2*x - x^3)*Log[4] + (1 + x^2*Log[4])*Log[2*x])/(x*(5 + x)*Log[4]*
(x - Log[2*x])))*(5 + x)^2*(x - Log[2*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \left (-5 x^2-2 x^3+\left (10 x-8 x^2-4 x^3+5 x^4\right ) \log (4)+\left (10 x+4 x^2+\left (2 x^2-10 x^3\right ) \log (4)\right ) \log (2 x)+\left (-5-2 x+5 x^2 \log (4)\right ) \log ^2(2 x)\right )}{x^2 (5+x)^2 \log (4) (x-\log (2 x))^2} \, dx\\ &=\frac {\int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \left (-5 x^2-2 x^3+\left (10 x-8 x^2-4 x^3+5 x^4\right ) \log (4)+\left (10 x+4 x^2+\left (2 x^2-10 x^3\right ) \log (4)\right ) \log (2 x)+\left (-5-2 x+5 x^2 \log (4)\right ) \log ^2(2 x)\right )}{x^2 (5+x)^2 (x-\log (2 x))^2} \, dx}{\log (4)}\\ &=\frac {\int \left (\frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \left (-5-2 x+5 x^2 \log (4)\right )}{x^2 (5+x)^2}-\frac {2 \exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) (-1+x) \log (4)}{x (5+x) (x-\log (2 x))^2}-\frac {2 \exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \log (4)}{(5+x)^2 (x-\log (2 x))}\right ) \, dx}{\log (4)}\\ &=-\left (2 \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) (-1+x)}{x (5+x) (x-\log (2 x))^2} \, dx\right )-2 \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x)^2 (x-\log (2 x))} \, dx+\frac {\int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \left (-5-2 x+5 x^2 \log (4)\right )}{x^2 (5+x)^2} \, dx}{\log (4)}\\ &=-\left (2 \int \left (-\frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{5 x (x-\log (2 x))^2}+\frac {6 \exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{5 (5+x) (x-\log (2 x))^2}\right ) \, dx\right )-2 \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x)^2 (x-\log (2 x))} \, dx+\frac {\int \left (-\frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{5 x^2}+\frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) (1+25 \log (4))}{5 (5+x)^2}\right ) \, dx}{\log (4)}\\ &=\frac {2}{5} \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{x (x-\log (2 x))^2} \, dx-2 \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x)^2 (x-\log (2 x))} \, dx-\frac {12}{5} \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x) (x-\log (2 x))^2} \, dx-\frac {\int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{x^2} \, dx}{5 \log (4)}+\frac {(1+25 \log (4)) \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x)^2} \, dx}{5 \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 16.97, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{\left (-5 x^2-x^3\right ) \log (4)+\left (5 x+x^2\right ) \log (4) \log (2 x)}} \left (-5 x^2-2 x^3+\left (10 x-8 x^2-4 x^3+5 x^4\right ) \log (4)+\left (10 x+4 x^2+\left (2 x^2-10 x^3\right ) \log (4)\right ) \log (2 x)+\left (-5-2 x+5 x^2 \log (4)\right ) \log ^2(2 x)\right )}{\left (25 x^4+10 x^5+x^6\right ) \log (4)+\left (-50 x^3-20 x^4-2 x^5\right ) \log (4) \log (2 x)+\left (25 x^2+10 x^3+x^4\right ) \log (4) \log ^2(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((-x + (-2*x - x^3)*Log[4] + (1 + x^2*Log[4])*Log[2*x])/((-5*x^2 - x^3)*Log[4] + (5*x + x^2)*Log[
4]*Log[2*x]))*(-5*x^2 - 2*x^3 + (10*x - 8*x^2 - 4*x^3 + 5*x^4)*Log[4] + (10*x + 4*x^2 + (2*x^2 - 10*x^3)*Log[4
])*Log[2*x] + (-5 - 2*x + 5*x^2*Log[4])*Log[2*x]^2))/((25*x^4 + 10*x^5 + x^6)*Log[4] + (-50*x^3 - 20*x^4 - 2*x
^5)*Log[4]*Log[2*x] + (25*x^2 + 10*x^3 + x^4)*Log[4]*Log[2*x]^2),x]

[Out]

Integrate[(E^((-x + (-2*x - x^3)*Log[4] + (1 + x^2*Log[4])*Log[2*x])/((-5*x^2 - x^3)*Log[4] + (5*x + x^2)*Log[
4]*Log[2*x]))*(-5*x^2 - 2*x^3 + (10*x - 8*x^2 - 4*x^3 + 5*x^4)*Log[4] + (10*x + 4*x^2 + (2*x^2 - 10*x^3)*Log[4
])*Log[2*x] + (-5 - 2*x + 5*x^2*Log[4])*Log[2*x]^2))/((25*x^4 + 10*x^5 + x^6)*Log[4] + (-50*x^3 - 20*x^4 - 2*x
^5)*Log[4]*Log[2*x] + (25*x^2 + 10*x^3 + x^4)*Log[4]*Log[2*x]^2), x]

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fricas [A]  time = 2.37, size = 61, normalized size = 2.03 \begin {gather*} e^{\left (-\frac {2 \, {\left (x^{3} + 2 \, x\right )} \log \relax (2) - {\left (2 \, x^{2} \log \relax (2) + 1\right )} \log \left (2 \, x\right ) + x}{2 \, {\left ({\left (x^{2} + 5 \, x\right )} \log \relax (2) \log \left (2 \, x\right ) - {\left (x^{3} + 5 \, x^{2}\right )} \log \relax (2)\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2*log(2)-2*x-5)*log(2*x)^2+(2*(-10*x^3+2*x^2)*log(2)+4*x^2+10*x)*log(2*x)+2*(5*x^4-4*x^3-8*x^
2+10*x)*log(2)-2*x^3-5*x^2)*exp(((2*x^2*log(2)+1)*log(2*x)+2*(-x^3-2*x)*log(2)-x)/(2*(x^2+5*x)*log(2)*log(2*x)
+2*(-x^3-5*x^2)*log(2)))/(2*(x^4+10*x^3+25*x^2)*log(2)*log(2*x)^2+2*(-2*x^5-20*x^4-50*x^3)*log(2)*log(2*x)+2*(
x^6+10*x^5+25*x^4)*log(2)),x, algorithm="fricas")

[Out]

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

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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(((10*x^2*log(2)-2*x-5)*log(2*x)^2+(2*(-10*x^3+2*x^2)*log(2)+4*x^2+10*x)*log(2*x)+2*(5*x^4-4*x^3-8*x^
2+10*x)*log(2)-2*x^3-5*x^2)*exp(((2*x^2*log(2)+1)*log(2*x)+2*(-x^3-2*x)*log(2)-x)/(2*(x^2+5*x)*log(2)*log(2*x)
+2*(-x^3-5*x^2)*log(2)))/(2*(x^4+10*x^3+25*x^2)*log(2)*log(2*x)^2+2*(-2*x^5-20*x^4-50*x^3)*log(2)*log(2*x)+2*(
x^6+10*x^5+25*x^4)*log(2)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.07, size = 57, normalized size = 1.90

method	result	size
risch	\({\mathrm e}^{\frac {2 \ln \relax (2) \ln \left (2 x \right ) x^{2}-2 x^{3} \ln \relax (2)-4 x \ln \relax (2)+\ln \left (2 x \right )-x}{2 \ln \relax (2) x \left (5+x \right ) \left (\ln \left (2 x \right )-x \right )}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2*ln(2)-2*x-5)*ln(2*x)^2+(2*(-10*x^3+2*x^2)*ln(2)+4*x^2+10*x)*ln(2*x)+2*(5*x^4-4*x^3-8*x^2+10*x)*ln
(2)-2*x^3-5*x^2)*exp(((2*x^2*ln(2)+1)*ln(2*x)+2*(-x^3-2*x)*ln(2)-x)/(2*(x^2+5*x)*ln(2)*ln(2*x)+2*(-x^3-5*x^2)*
ln(2)))/(2*(x^4+10*x^3+25*x^2)*ln(2)*ln(2*x)^2+2*(-2*x^5-20*x^4-50*x^3)*ln(2)*ln(2*x)+2*(x^6+10*x^5+25*x^4)*ln
(2)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 1.79, size = 197, normalized size = 6.57 \begin {gather*} e^{\left (-\frac {5 \, \log \relax (2)}{x {\left (\log \relax (2) + 5\right )} + {\left (x + 5\right )} \log \relax (x) + 5 \, \log \relax (2) + 25} - \frac {\log \relax (x)}{10 \, {\left ({\left (\log \relax (2)^{2} + 5 \, \log \relax (2)\right )} x + 5 \, \log \relax (2)^{2} + {\left (x \log \relax (2) + 5 \, \log \relax (2)\right )} \log \relax (x) + 25 \, \log \relax (2)\right )}} - \frac {5 \, \log \relax (x)}{x {\left (\log \relax (2) + 5\right )} + {\left (x + 5\right )} \log \relax (x) + 5 \, \log \relax (2) + 25} - \frac {1}{2 \, {\left ({\left (\log \relax (2)^{2} + 5 \, \log \relax (2)\right )} x + 5 \, \log \relax (2)^{2} + {\left (x \log \relax (2) + 5 \, \log \relax (2)\right )} \log \relax (x) + 25 \, \log \relax (2)\right )}} + \frac {2}{x {\left (\log \relax (2) + 5\right )} - \log \relax (2)^{2} + {\left (x - 2 \, \log \relax (2) - 5\right )} \log \relax (x) - \log \relax (x)^{2} - 5 \, \log \relax (2)} - \frac {271}{10 \, {\left (x {\left (\log \relax (2) + 5\right )} + {\left (x + 5\right )} \log \relax (x) + 5 \, \log \relax (2) + 25\right )}} + \frac {1}{10 \, x \log \relax (2)} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2*log(2)-2*x-5)*log(2*x)^2+(2*(-10*x^3+2*x^2)*log(2)+4*x^2+10*x)*log(2*x)+2*(5*x^4-4*x^3-8*x^
2+10*x)*log(2)-2*x^3-5*x^2)*exp(((2*x^2*log(2)+1)*log(2*x)+2*(-x^3-2*x)*log(2)-x)/(2*(x^2+5*x)*log(2)*log(2*x)
+2*(-x^3-5*x^2)*log(2)))/(2*(x^4+10*x^3+25*x^2)*log(2)*log(2*x)^2+2*(-2*x^5-20*x^4-50*x^3)*log(2)*log(2*x)+2*(
x^6+10*x^5+25*x^4)*log(2)),x, algorithm="maxima")

[Out]

e^(-5*log(2)/(x*(log(2) + 5) + (x + 5)*log(x) + 5*log(2) + 25) - 1/10*log(x)/((log(2)^2 + 5*log(2))*x + 5*log(
2)^2 + (x*log(2) + 5*log(2))*log(x) + 25*log(2)) - 5*log(x)/(x*(log(2) + 5) + (x + 5)*log(x) + 5*log(2) + 25)
- 1/2/((log(2)^2 + 5*log(2))*x + 5*log(2)^2 + (x*log(2) + 5*log(2))*log(x) + 25*log(2)) + 2/(x*(log(2) + 5) -
log(2)^2 + (x - 2*log(2) - 5)*log(x) - log(x)^2 - 5*log(2)) - 271/10/(x*(log(2) + 5) + (x + 5)*log(x) + 5*log(
2) + 25) + 1/10/(x*log(2)) + 1)

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mupad [B]  time = 1.71, size = 272, normalized size = 9.07 \begin {gather*} 2^{\frac {1}{2\,x^2\,{\ln \relax (2)}^2+10\,x\,{\ln \relax (2)}^2-10\,x^2\,\ln \relax (2)-2\,x^3\,\ln \relax (2)+10\,x\,\ln \relax (2)\,\ln \relax (x)+2\,x^2\,\ln \relax (2)\,\ln \relax (x)}-\frac {x^2+2}{x\,{\ln \relax (2)}^2-5\,x\,\ln \relax (2)-x^2\,\ln \relax (2)+5\,\ln \relax (2)\,\ln \relax (x)+5\,{\ln \relax (2)}^2+x\,\ln \relax (2)\,\ln \relax (x)}}\,x^{\frac {2\,\ln \relax (2)\,x^2+1}{2\,\left (x^2\,{\ln \relax (2)}^2+5\,x\,{\ln \relax (2)}^2-5\,x^2\,\ln \relax (2)-x^3\,\ln \relax (2)+5\,x\,\ln \relax (2)\,\ln \relax (x)+x^2\,\ln \relax (2)\,\ln \relax (x)\right )}}\,{\mathrm {e}}^{-\frac {x}{2\,x^2\,{\ln \relax (2)}^2+10\,x\,{\ln \relax (2)}^2-10\,x^2\,\ln \relax (2)-2\,x^3\,\ln \relax (2)+10\,x\,\ln \relax (2)\,\ln \relax (x)+2\,x^2\,\ln \relax (2)\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {2\,x^2\,{\ln \relax (2)}^2}{2\,x^2\,{\ln \relax (2)}^2+10\,x\,{\ln \relax (2)}^2-10\,x^2\,\ln \relax (2)-2\,x^3\,\ln \relax (2)+10\,x\,\ln \relax (2)\,\ln \relax (x)+2\,x^2\,\ln \relax (2)\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x - log(2*x)*(2*x^2*log(2) + 1) + 2*log(2)*(2*x + x^3))/(2*log(2)*(5*x^2 + x^3) - 2*log(2*x)*log(2)
*(5*x + x^2)))*(5*x^2 - log(2*x)*(10*x + 2*log(2)*(2*x^2 - 10*x^3) + 4*x^2) - 2*log(2)*(10*x - 8*x^2 - 4*x^3 +
5*x^4) + 2*x^3 + log(2*x)^2*(2*x - 10*x^2*log(2) + 5)))/(2*log(2)*(25*x^4 + 10*x^5 + x^6) - 2*log(2*x)*log(2)
*(50*x^3 + 20*x^4 + 2*x^5) + 2*log(2*x)^2*log(2)*(25*x^2 + 10*x^3 + x^4)),x)

[Out]

2^(1/(2*x^2*log(2)^2 + 10*x*log(2)^2 - 10*x^2*log(2) - 2*x^3*log(2) + 10*x*log(2)*log(x) + 2*x^2*log(2)*log(x)
) - (x^2 + 2)/(x*log(2)^2 - 5*x*log(2) - x^2*log(2) + 5*log(2)*log(x) + 5*log(2)^2 + x*log(2)*log(x)))*x^((2*x
^2*log(2) + 1)/(2*(x^2*log(2)^2 + 5*x*log(2)^2 - 5*x^2*log(2) - x^3*log(2) + 5*x*log(2)*log(x) + x^2*log(2)*lo
g(x))))*exp(-x/(2*x^2*log(2)^2 + 10*x*log(2)^2 - 10*x^2*log(2) - 2*x^3*log(2) + 10*x*log(2)*log(x) + 2*x^2*log
(2)*log(x)))*exp((2*x^2*log(2)^2)/(2*x^2*log(2)^2 + 10*x*log(2)^2 - 10*x^2*log(2) - 2*x^3*log(2) + 10*x*log(2)
*log(x) + 2*x^2*log(2)*log(x)))

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sympy [B]  time = 2.17, size = 63, normalized size = 2.10 \begin {gather*} e^{\frac {- x + \left (- 2 x^{3} - 4 x\right ) \log {\relax (2 )} + \left (2 x^{2} \log {\relax (2 )} + 1\right ) \log {\left (2 x \right )}}{\left (2 x^{2} + 10 x\right ) \log {\relax (2 )} \log {\left (2 x \right )} + \left (- 2 x^{3} - 10 x^{2}\right ) \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2*ln(2)-2*x-5)*ln(2*x)**2+(2*(-10*x**3+2*x**2)*ln(2)+4*x**2+10*x)*ln(2*x)+2*(5*x**4-4*x**3-8
*x**2+10*x)*ln(2)-2*x**3-5*x**2)*exp(((2*x**2*ln(2)+1)*ln(2*x)+2*(-x**3-2*x)*ln(2)-x)/(2*(x**2+5*x)*ln(2)*ln(2
*x)+2*(-x**3-5*x**2)*ln(2)))/(2*(x**4+10*x**3+25*x**2)*ln(2)*ln(2*x)**2+2*(-2*x**5-20*x**4-50*x**3)*ln(2)*ln(2
*x)+2*(x**6+10*x**5+25*x**4)*ln(2)),x)

[Out]

exp((-x + (-2*x**3 - 4*x)*log(2) + (2*x**2*log(2) + 1)*log(2*x))/((2*x**2 + 10*x)*log(2)*log(2*x) + (-2*x**3 -
10*x**2)*log(2)))

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