3.7.64 \(\int \frac {-2 x+4 x^2+e^5 (-4+12 x-8 x^2)+e^{10} (4-10 x+4 x^2)+(-2 x^2+e^{10} (-8+8 x-2 x^2)+e^5 (-8 x+4 x^2)+2 e^{10} \log (2)) \log (\frac {x^2+e^5 (4 x-2 x^2)+e^{10} (4-4 x+x^2)-e^{10} \log (2)}{e^{10}})}{(-x^2+e^{10} (-4+4 x-x^2)+e^5 (-4 x+2 x^2)+e^{10} \log (2)) \log ^2(\frac {x^2+e^5 (4 x-2 x^2)+e^{10} (4-4 x+x^2)-e^{10} \log (2)}{e^{10}})} \, dx\)

Optimal. Leaf size=26 \[ \frac {-1+2 x}{\log \left (\left (2-x+\frac {x}{e^5}\right )^2-\log (2)\right )} \]

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Rubi [F]  time = 7.18, 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 {-2 x+4 x^2+e^5 \left (-4+12 x-8 x^2\right )+e^{10} \left (4-10 x+4 x^2\right )+\left (-2 x^2+e^{10} \left (-8+8 x-2 x^2\right )+e^5 \left (-8 x+4 x^2\right )+2 e^{10} \log (2)\right ) \log \left (\frac {x^2+e^5 \left (4 x-2 x^2\right )+e^{10} \left (4-4 x+x^2\right )-e^{10} \log (2)}{e^{10}}\right )}{\left (-x^2+e^{10} \left (-4+4 x-x^2\right )+e^5 \left (-4 x+2 x^2\right )+e^{10} \log (2)\right ) \log ^2\left (\frac {x^2+e^5 \left (4 x-2 x^2\right )+e^{10} \left (4-4 x+x^2\right )-e^{10} \log (2)}{e^{10}}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x + 4*x^2 + E^5*(-4 + 12*x - 8*x^2) + E^10*(4 - 10*x + 4*x^2) + (-2*x^2 + E^10*(-8 + 8*x - 2*x^2) + E^
5*(-8*x + 4*x^2) + 2*E^10*Log[2])*Log[(x^2 + E^5*(4*x - 2*x^2) + E^10*(4 - 4*x + x^2) - E^10*Log[2])/E^10])/((
-x^2 + E^10*(-4 + 4*x - x^2) + E^5*(-4*x + 2*x^2) + E^10*Log[2])*Log[(x^2 + E^5*(4*x - 2*x^2) + E^10*(4 - 4*x
+ x^2) - E^10*Log[2])/E^10]^2),x]

[Out]

(-4*Defer[Int][Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^(-2), x])/(1 - E^5)^2 + (8*E^5*Defe
r[Int][Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^(-2), x])/(1 - E^5)^2 - (4*E^10*Defer[Int][
Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^(-2), x])/(1 - E^5)^2 - (4*E^5*(4 - Log[2])*Defer[
Int][1/((-4*E^5 + 4*E^10 - 2*(1 - E^5)^2*x - 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E
^5)^2*x^2)/E^10 - Log[2]]^2), x])/((1 - E^5)*Sqrt[Log[2]]) + (2*E^5*(2 - 4*E^5 - E^10*(6 - Log[4]))*Defer[Int]
[1/((-4*E^5 + 4*E^10 - 2*(1 - E^5)^2*x - 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^
2*x^2)/E^10 - Log[2]]^2), x])/((1 - E^5)*Sqrt[Log[2]]) - (4*(1 - 2*E^5 - E^10*(7 - Log[4]))*Defer[Int][1/((-4*
E^5 + 4*E^10 - 2*(1 - E^5)^2*x - 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E
^10 - Log[2]]^2), x])/((1 - E^5)*Sqrt[Log[2]]) - (4*E^5*(4 - Log[2])*Defer[Int][1/((4*E^5 - 4*E^10 + 2*(1 - E^
5)^2*x - 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^2), x])/((
1 - E^5)*Sqrt[Log[2]]) + (2*E^5*(2 - 4*E^5 - E^10*(6 - Log[4]))*Defer[Int][1/((4*E^5 - 4*E^10 + 2*(1 - E^5)^2*
x - 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^2), x])/((1 - E
^5)*Sqrt[Log[2]]) - (4*(1 - 2*E^5 - E^10*(7 - Log[4]))*Defer[Int][1/((4*E^5 - 4*E^10 + 2*(1 - E^5)^2*x - 2*E^5
*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^2), x])/((1 - E^5)*Sqrt[
Log[2]]) + 2*(1 + 2/Sqrt[Log[2]])*Defer[Int][1/((4*E^5*(1 - E^5) + 2*(1 - E^5)^2*x - 2*E^5*(-1 + E^5)*Sqrt[Log
[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^2), x] + (16*E^5*(1 + 2/Sqrt[Log[2]])*Defer[
Int][1/((4*E^5*(1 - E^5) + 2*(1 - E^5)^2*x - 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E
^5)^2*x^2)/E^10 - Log[2]]^2), x])/(1 - E^5) + (2*E^10*(5 + 3*E^5)*(2 + Sqrt[Log[2]])*Defer[Int][1/((4*E^5*(1 -
E^5) + 2*(1 - E^5)^2*x - 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - L
og[2]]^2), x])/((1 - E^5)*Sqrt[Log[2]]) - (4*E^5*(3 + 5*E^5)*(2 + Sqrt[Log[2]])*Defer[Int][1/((4*E^5*(1 - E^5)
+ 2*(1 - E^5)^2*x - 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]
]^2), x])/((1 - E^5)*Sqrt[Log[2]]) + 2*(1 - 2/Sqrt[Log[2]])*Defer[Int][1/((4*E^5*(1 - E^5) + 2*(1 - E^5)^2*x +
2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^2), x] + (16*E^5*(
1 - 2/Sqrt[Log[2]])*Defer[Int][1/((4*E^5*(1 - E^5) + 2*(1 - E^5)^2*x + 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 -
4*(1 - E^(-5))*x + ((1 - E^5)^2*x^2)/E^10 - Log[2]]^2), x])/(1 - E^5) - (2*E^10*(5 + 3*E^5)*(2 - Sqrt[Log[2]])
*Defer[Int][1/((4*E^5*(1 - E^5) + 2*(1 - E^5)^2*x + 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x +
((1 - E^5)^2*x^2)/E^10 - Log[2]]^2), x])/((1 - E^5)*Sqrt[Log[2]]) + (4*E^5*(3 + 5*E^5)*(2 - Sqrt[Log[2]])*Defe
r[Int][1/((4*E^5*(1 - E^5) + 2*(1 - E^5)^2*x + 2*E^5*(-1 + E^5)*Sqrt[Log[2]])*Log[4 - 4*(1 - E^(-5))*x + ((1 -
E^5)^2*x^2)/E^10 - Log[2]]^2), x])/((1 - E^5)*Sqrt[Log[2]]) + 2*Defer[Int][Log[4 - 4*(1 - E^(-5))*x + ((1 - E
^5)^2*x^2)/E^10 - Log[2]]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x-4 x^2-e^5 \left (-4+12 x-8 x^2\right )-e^{10} \left (4-10 x+4 x^2\right )-\left (-2 x^2+e^{10} \left (-8+8 x-2 x^2\right )+e^5 \left (-8 x+4 x^2\right )+2 e^{10} \log (2)\right ) \log \left (\frac {x^2+e^5 \left (4 x-2 x^2\right )+e^{10} \left (4-4 x+x^2\right )-e^{10} \log (2)}{e^{10}}\right )}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (\frac {4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))}{e^{10}}\right )} \, dx\\ &=\int \left (\frac {4 x^2}{\left (-4 e^5 \left (1-e^5\right ) x-\left (1-e^5\right )^2 x^2-e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {4 e^5 (1-2 x) (1-x)}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {2 e^{10} (1-2 x) (-2+x)}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {2 x}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {2}{\log \left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx\\ &=2 \int \frac {x}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+2 \int \frac {1}{\log \left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+4 \int \frac {x^2}{\left (-4 e^5 \left (1-e^5\right ) x-\left (1-e^5\right )^2 x^2-e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+\left (4 e^5\right ) \int \frac {(1-2 x) (1-x)}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+\left (2 e^{10}\right ) \int \frac {(1-2 x) (-2+x)}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx\\ &=2 \int \left (\frac {1+\frac {2}{\sqrt {\log (2)}}}{\left (4 e^5 \left (1-e^5\right )+2 \left (1-e^5\right )^2 x-2 e^5 \left (-1+e^5\right ) \sqrt {\log (2)}\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {1-\frac {2}{\sqrt {\log (2)}}}{\left (4 e^5 \left (1-e^5\right )+2 \left (1-e^5\right )^2 x+2 e^5 \left (-1+e^5\right ) \sqrt {\log (2)}\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx+2 \int \frac {1}{\log \left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+4 \int \left (-\frac {1}{\left (1-e^5\right )^2 \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {e^5 \left (4 \left (1-e^5\right ) x+e^5 (4-\log (2))\right )}{\left (1-e^5\right )^2 \left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx+\left (4 e^5\right ) \int \left (\frac {2}{\left (1-e^5\right )^2 \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {1-2 e^5-\left (3+2 e^5-5 e^{10}\right ) x-e^{10} (7-\log (4))}{\left (1-e^5\right )^2 \left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx+\left (2 e^{10}\right ) \int \left (-\frac {2}{\left (1-e^5\right )^2 \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {-2+4 e^5+\left (5-2 e^5-3 e^{10}\right ) x+e^{10} (6-\log (4))}{\left (1-e^5\right )^2 \left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (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 = 0.12, size = 38, normalized size = 1.46 \begin {gather*} \frac {-1+2 x}{\log \left (4+\left (-4+\frac {4}{e^5}\right ) x+\frac {\left (-1+e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + 4*x^2 + E^5*(-4 + 12*x - 8*x^2) + E^10*(4 - 10*x + 4*x^2) + (-2*x^2 + E^10*(-8 + 8*x - 2*x^2
) + E^5*(-8*x + 4*x^2) + 2*E^10*Log[2])*Log[(x^2 + E^5*(4*x - 2*x^2) + E^10*(4 - 4*x + x^2) - E^10*Log[2])/E^1
0])/((-x^2 + E^10*(-4 + 4*x - x^2) + E^5*(-4*x + 2*x^2) + E^10*Log[2])*Log[(x^2 + E^5*(4*x - 2*x^2) + E^10*(4
- 4*x + x^2) - E^10*Log[2])/E^10]^2),x]

[Out]

(-1 + 2*x)/Log[4 + (-4 + 4/E^5)*x + ((-1 + E^5)^2*x^2)/E^10 - Log[2]]

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fricas [A]  time = 1.13, size = 44, normalized size = 1.69 \begin {gather*} \frac {2 \, x - 1}{\log \left ({\left (x^{2} + {\left (x^{2} - 4 \, x + 4\right )} e^{10} - 2 \, {\left (x^{2} - 2 \, x\right )} e^{5} - e^{10} \log \relax (2)\right )} e^{\left (-10\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)^2*log(2)+(-2*x^2+8*x-8)*exp(5)^2+(4*x^2-8*x)*exp(5)-2*x^2)*log((-exp(5)^2*log(2)+(x^2-4*x
+4)*exp(5)^2+(-2*x^2+4*x)*exp(5)+x^2)/exp(5)^2)+(4*x^2-10*x+4)*exp(5)^2+(-8*x^2+12*x-4)*exp(5)+4*x^2-2*x)/(exp
(5)^2*log(2)+(-x^2+4*x-4)*exp(5)^2+(2*x^2-4*x)*exp(5)-x^2)/log((-exp(5)^2*log(2)+(x^2-4*x+4)*exp(5)^2+(-2*x^2+
4*x)*exp(5)+x^2)/exp(5)^2)^2,x, algorithm="fricas")

[Out]

(2*x - 1)/log((x^2 + (x^2 - 4*x + 4)*e^10 - 2*(x^2 - 2*x)*e^5 - e^10*log(2))*e^(-10))

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giac [B]  time = 1.80, size = 51, normalized size = 1.96 \begin {gather*} \frac {2 \, x - 1}{\log \left (x^{2} e^{15} - 2 \, x^{2} e^{10} + x^{2} e^{5} - 4 \, x e^{15} + 4 \, x e^{10} - e^{15} \log \relax (2) + 4 \, e^{15}\right ) - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)^2*log(2)+(-2*x^2+8*x-8)*exp(5)^2+(4*x^2-8*x)*exp(5)-2*x^2)*log((-exp(5)^2*log(2)+(x^2-4*x
+4)*exp(5)^2+(-2*x^2+4*x)*exp(5)+x^2)/exp(5)^2)+(4*x^2-10*x+4)*exp(5)^2+(-8*x^2+12*x-4)*exp(5)+4*x^2-2*x)/(exp
(5)^2*log(2)+(-x^2+4*x-4)*exp(5)^2+(2*x^2-4*x)*exp(5)-x^2)/log((-exp(5)^2*log(2)+(x^2-4*x+4)*exp(5)^2+(-2*x^2+
4*x)*exp(5)+x^2)/exp(5)^2)^2,x, algorithm="giac")

[Out]

(2*x - 1)/(log(x^2*e^15 - 2*x^2*e^10 + x^2*e^5 - 4*x*e^15 + 4*x*e^10 - e^15*log(2) + 4*e^15) - 15)

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maple [A]  time = 0.46, size = 46, normalized size = 1.77

method	result	size
risch	\(\frac {2 x -1}{\ln \left (\left (-{\mathrm e}^{10} \ln \relax (2)+\left (x^{2}-4 x +4\right ) {\mathrm e}^{10}+\left (-2 x^{2}+4 x \right ) {\mathrm e}^{5}+x^{2}\right ) {\mathrm e}^{-10}\right )}\)	\(46\)
norman	\(\frac {2 x -1}{\ln \left (\left (-{\mathrm e}^{10} \ln \relax (2)+\left (x^{2}-4 x +4\right ) {\mathrm e}^{10}+\left (-2 x^{2}+4 x \right ) {\mathrm e}^{5}+x^{2}\right ) {\mathrm e}^{-10}\right )}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(5)^2*ln(2)+(-2*x^2+8*x-8)*exp(5)^2+(4*x^2-8*x)*exp(5)-2*x^2)*ln((-exp(5)^2*ln(2)+(x^2-4*x+4)*exp(5
)^2+(-2*x^2+4*x)*exp(5)+x^2)/exp(5)^2)+(4*x^2-10*x+4)*exp(5)^2+(-8*x^2+12*x-4)*exp(5)+4*x^2-2*x)/(exp(5)^2*ln(
2)+(-x^2+4*x-4)*exp(5)^2+(2*x^2-4*x)*exp(5)-x^2)/ln((-exp(5)^2*ln(2)+(x^2-4*x+4)*exp(5)^2+(-2*x^2+4*x)*exp(5)+
x^2)/exp(5)^2)^2,x,method=_RETURNVERBOSE)

[Out]

(2*x-1)/ln((-exp(10)*ln(2)+(x^2-4*x+4)*exp(10)+(-2*x^2+4*x)*exp(5)+x^2)*exp(-10))

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maxima [A]  time = 0.76, size = 44, normalized size = 1.69 \begin {gather*} \frac {2 \, x - 1}{\log \left (x^{2} {\left (e^{10} - 2 \, e^{5} + 1\right )} - 4 \, x {\left (e^{10} - e^{5}\right )} - e^{10} \log \relax (2) + 4 \, e^{10}\right ) - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)^2*log(2)+(-2*x^2+8*x-8)*exp(5)^2+(4*x^2-8*x)*exp(5)-2*x^2)*log((-exp(5)^2*log(2)+(x^2-4*x
+4)*exp(5)^2+(-2*x^2+4*x)*exp(5)+x^2)/exp(5)^2)+(4*x^2-10*x+4)*exp(5)^2+(-8*x^2+12*x-4)*exp(5)+4*x^2-2*x)/(exp
(5)^2*log(2)+(-x^2+4*x-4)*exp(5)^2+(2*x^2-4*x)*exp(5)-x^2)/log((-exp(5)^2*log(2)+(x^2-4*x+4)*exp(5)^2+(-2*x^2+
4*x)*exp(5)+x^2)/exp(5)^2)^2,x, algorithm="maxima")

[Out]

(2*x - 1)/(log(x^2*(e^10 - 2*e^5 + 1) - 4*x*(e^10 - e^5) - e^10*log(2) + 4*e^10) - 10)

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mupad [B]  time = 1.35, size = 101, normalized size = 3.88 \begin {gather*} \frac {2\,x\,{\mathrm {e}}^5-{\mathrm {e}}^5-2\,x+2\,{\mathrm {e}}^5\,\ln \left ({\mathrm {e}}^{-10}\,\left ({\mathrm {e}}^5\,\left (4\,x-2\,x^2\right )-{\mathrm {e}}^{10}\,\ln \relax (2)+{\mathrm {e}}^{10}\,\left (x^2-4\,x+4\right )+x^2\right )\right )+1}{\ln \left ({\mathrm {e}}^{-10}\,\left ({\mathrm {e}}^5\,\left (4\,x-2\,x^2\right )-{\mathrm {e}}^{10}\,\ln \relax (2)+{\mathrm {e}}^{10}\,\left (x^2-4\,x+4\right )+x^2\right )\right )\,\left ({\mathrm {e}}^5-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + log(exp(-10)*(exp(5)*(4*x - 2*x^2) - exp(10)*log(2) + exp(10)*(x^2 - 4*x + 4) + x^2))*(exp(5)*(8*x
- 4*x^2) - 2*exp(10)*log(2) + exp(10)*(2*x^2 - 8*x + 8) + 2*x^2) - exp(10)*(4*x^2 - 10*x + 4) + exp(5)*(8*x^2
- 12*x + 4) - 4*x^2)/(log(exp(-10)*(exp(5)*(4*x - 2*x^2) - exp(10)*log(2) + exp(10)*(x^2 - 4*x + 4) + x^2))^2*
(exp(5)*(4*x - 2*x^2) - exp(10)*log(2) + exp(10)*(x^2 - 4*x + 4) + x^2)),x)

[Out]

(2*x*exp(5) - exp(5) - 2*x + 2*exp(5)*log(exp(-10)*(exp(5)*(4*x - 2*x^2) - exp(10)*log(2) + exp(10)*(x^2 - 4*x
+ 4) + x^2)) + 1)/(log(exp(-10)*(exp(5)*(4*x - 2*x^2) - exp(10)*log(2) + exp(10)*(x^2 - 4*x + 4) + x^2))*(exp
(5) - 1))

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sympy [B]  time = 0.40, size = 42, normalized size = 1.62 \begin {gather*} \frac {2 x - 1}{\log {\left (\frac {x^{2} + \left (- 2 x^{2} + 4 x\right ) e^{5} + \left (x^{2} - 4 x + 4\right ) e^{10} - e^{10} \log {\relax (2 )}}{e^{10}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)**2*ln(2)+(-2*x**2+8*x-8)*exp(5)**2+(4*x**2-8*x)*exp(5)-2*x**2)*ln((-exp(5)**2*ln(2)+(x**2
-4*x+4)*exp(5)**2+(-2*x**2+4*x)*exp(5)+x**2)/exp(5)**2)+(4*x**2-10*x+4)*exp(5)**2+(-8*x**2+12*x-4)*exp(5)+4*x*
*2-2*x)/(exp(5)**2*ln(2)+(-x**2+4*x-4)*exp(5)**2+(2*x**2-4*x)*exp(5)-x**2)/ln((-exp(5)**2*ln(2)+(x**2-4*x+4)*e
xp(5)**2+(-2*x**2+4*x)*exp(5)+x**2)/exp(5)**2)**2,x)

[Out]

(2*x - 1)/log((x**2 + (-2*x**2 + 4*x)*exp(5) + (x**2 - 4*x + 4)*exp(10) - exp(10)*log(2))*exp(-10))

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