3.56.11 \(\int \frac {-10 x-20 e^{10} x+20 x^2+e^5 (-30 x+20 x^2)+(-4-4 e^{10}+4 x+e^5 (-8+4 x)) \log (3)+(-20 x^2-20 e^5 x^2+5 x^3+(-4 x-4 e^5 x+x^2) \log (3)) \log (4)+(-5 x^3-x^2 \log (3)) \log ^2(4)+(10 x+10 e^5 x+(2+2 e^5) \log (3)+(-20 x-20 e^5 x+10 x^2+(-4-4 e^5+2 x) \log (3)) \log (4)+(-10 x^2-2 x \log (3)) \log ^2(4)) \log (\frac {1}{5} (5 x+\log (3)))+((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)) \log ^2(\frac {1}{5} (5 x+\log (3)))}{20 x+40 e^5 x+20 e^{10} x+(4+8 e^5+4 e^{10}) \log (3)+(20 x^2+20 e^5 x^2+(4 x+4 e^5 x) \log (3)) \log (4)+(5 x^3+x^2 \log (3)) \log ^2(4)+((20 x+20 e^5 x+(4+4 e^5) \log (3)) \log (4)+(10 x^2+2 x \log (3)) \log ^2(4)) \log (\frac {1}{5} (5 x+\log (3)))+(5 x+\log (3)) \log ^2(4) \log ^2(\frac {1}{5} (5 x+\log (3)))} \, dx\)

Optimal. Leaf size=31 \[ -x+\frac {x}{\log (4)+\frac {2 \left (1+e^5\right )}{x+\log \left (x+\frac {\log (3)}{5}\right )}} \]

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Rubi [F]  time = 2.81, 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 {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 x+40 e^5 x+20 e^{10} x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*x - 20*E^10*x + 20*x^2 + E^5*(-30*x + 20*x^2) + (-4 - 4*E^10 + 4*x + E^5*(-8 + 4*x))*Log[3] + (-20*x^
2 - 20*E^5*x^2 + 5*x^3 + (-4*x - 4*E^5*x + x^2)*Log[3])*Log[4] + (-5*x^3 - x^2*Log[3])*Log[4]^2 + (10*x + 10*E
^5*x + (2 + 2*E^5)*Log[3] + (-20*x - 20*E^5*x + 10*x^2 + (-4 - 4*E^5 + 2*x)*Log[3])*Log[4] + (-10*x^2 - 2*x*Lo
g[3])*Log[4]^2)*Log[(5*x + Log[3])/5] + ((5*x + Log[3])*Log[4] + (-5*x - Log[3])*Log[4]^2)*Log[(5*x + Log[3])/
5]^2)/(20*x + 40*E^5*x + 20*E^10*x + (4 + 8*E^5 + 4*E^10)*Log[3] + (20*x^2 + 20*E^5*x^2 + (4*x + 4*E^5*x)*Log[
3])*Log[4] + (5*x^3 + x^2*Log[3])*Log[4]^2 + ((20*x + 20*E^5*x + (4 + 4*E^5)*Log[3])*Log[4] + (10*x^2 + 2*x*Lo
g[3])*Log[4]^2)*Log[(5*x + Log[3])/5] + (5*x + Log[3])*Log[4]^2*Log[(5*x + Log[3])/5]^2),x]

[Out]

(x*(1 - Log[4]))/Log[4] + ((1 + E^5)*(Log[4]*Log[9]*Log[16] - Log[4]^2*Log[81] + Log[1048576])*Defer[Int][(2*(
1 + E^5) + x*Log[4] + Log[4]*Log[x + Log[3]/5])^(-2), x])/(5*Log[4]) + 2*(1 + E^5)*Defer[Int][x/(2*(1 + E^5) +
x*Log[4] + Log[4]*Log[x + Log[3]/5])^2, x] - ((1 + E^5)*Log[3]*(Log[4]*Log[9]*Log[16] - Log[4]^2*Log[81] + Lo
g[1048576])*Defer[Int][1/((5*x + Log[3])*(2*(1 + E^5) + x*Log[4] + Log[4]*Log[x + Log[3]/5])^2), x])/(5*Log[4]
) - (2*(1 + E^5)*Defer[Int][(2*(1 + E^5) + x*Log[4] + Log[4]*Log[x + Log[3]/5])^(-1), x])/Log[4]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{20 e^{10} x+\left (20+40 e^5\right ) x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx\\ &=\int \frac {-10 x-20 e^{10} x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{\left (20+40 e^5+20 e^{10}\right ) x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx\\ &=\int \frac {\left (-10-20 e^{10}\right ) x+20 x^2+e^5 \left (-30 x+20 x^2\right )+\left (-4-4 e^{10}+4 x+e^5 (-8+4 x)\right ) \log (3)+\left (-20 x^2-20 e^5 x^2+5 x^3+\left (-4 x-4 e^5 x+x^2\right ) \log (3)\right ) \log (4)+\left (-5 x^3-x^2 \log (3)\right ) \log ^2(4)+\left (10 x+10 e^5 x+\left (2+2 e^5\right ) \log (3)+\left (-20 x-20 e^5 x+10 x^2+\left (-4-4 e^5+2 x\right ) \log (3)\right ) \log (4)+\left (-10 x^2-2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+\left ((5 x+\log (3)) \log (4)+(-5 x-\log (3)) \log ^2(4)\right ) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )}{\left (20+40 e^5+20 e^{10}\right ) x+\left (4+8 e^5+4 e^{10}\right ) \log (3)+\left (20 x^2+20 e^5 x^2+\left (4 x+4 e^5 x\right ) \log (3)\right ) \log (4)+\left (5 x^3+x^2 \log (3)\right ) \log ^2(4)+\left (\left (20 x+20 e^5 x+\left (4+4 e^5\right ) \log (3)\right ) \log (4)+\left (10 x^2+2 x \log (3)\right ) \log ^2(4)\right ) \log \left (\frac {1}{5} (5 x+\log (3))\right )+(5 x+\log (3)) \log ^2(4) \log ^2\left (\frac {1}{5} (5 x+\log (3))\right )} \, dx\\ &=\int \frac {-4 \left (1+e^5\right )^2 \log (3)-5 x^3 (-1+\log (4)) \log (4)-x^2 (-1+\log (4)) \left (20+20 e^5+\log (3) \log (4)\right )-\left (1+e^5\right ) x \left (10+20 e^5+(-1+\log (4)) \log (81)\right )-2 (5 x+\log (3)) \left (-1+x (-1+\log (4)) \log (4)+e^5 (-1+\log (16))+\log (16)\right ) \log \left (x+\frac {\log (3)}{5}\right )-(5 x+\log (3)) (-1+\log (4)) \log (4) \log ^2\left (x+\frac {\log (3)}{5}\right )}{(5 x+\log (3)) \left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2} \, dx\\ &=\int \left (\frac {1-\log (4)}{\log (4)}+\frac {\left (1+e^5\right ) x \left (10 x \log (4)+\log (4) \log (9)+\log (4) \log (9) \log (16)-\log ^2(4) \log (81)+\log (1048576)\right )}{(5 x+\log (3)) \log (4) \left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2}+\frac {2 \left (-1-e^5\right )}{\log (4) \left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )}\right ) \, dx\\ &=\frac {x (1-\log (4))}{\log (4)}+\frac {\left (1+e^5\right ) \int \frac {x \left (10 x \log (4)+\log (4) \log (9)+\log (4) \log (9) \log (16)-\log ^2(4) \log (81)+\log (1048576)\right )}{(5 x+\log (3)) \left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2} \, dx}{\log (4)}-\frac {\left (2 \left (1+e^5\right )\right ) \int \frac {1}{2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )} \, dx}{\log (4)}\\ &=\frac {x (1-\log (4))}{\log (4)}+\frac {\left (1+e^5\right ) \int \left (\frac {2 x \log (4)}{\left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2}+\frac {\log (3) \left (-\log (4) \log (9) \log (16)+\log ^2(4) \log (81)-\log (1048576)\right )}{5 (5 x+\log (3)) \left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2}+\frac {\log (4) \log (9) \log (16)-\log ^2(4) \log (81)+\log (1048576)}{5 \left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2}\right ) \, dx}{\log (4)}-\frac {\left (2 \left (1+e^5\right )\right ) \int \frac {1}{2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )} \, dx}{\log (4)}\\ &=\frac {x (1-\log (4))}{\log (4)}+\left (2 \left (1+e^5\right )\right ) \int \frac {x}{\left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2} \, dx-\frac {\left (2 \left (1+e^5\right )\right ) \int \frac {1}{2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )} \, dx}{\log (4)}+\frac {\left (\left (1+e^5\right ) \left (\log (4) \log (9) \log (16)-\log ^2(4) \log (81)+\log (1048576)\right )\right ) \int \frac {1}{\left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2} \, dx}{5 \log (4)}-\frac {\left (\left (1+e^5\right ) \log (3) \left (\log (4) \log (9) \log (16)-\log ^2(4) \log (81)+\log (1048576)\right )\right ) \int \frac {1}{(5 x+\log (3)) \left (2 \left (1+e^5\right )+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )^2} \, dx}{5 \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.22, size = 81, normalized size = 2.61 \begin {gather*} -\frac {x (-1+\log (4)) \log (4)+\frac {\left (1+e^5\right ) x \left (10 x \log (4)+\log (4) \log (9) (1+\log (16))-\log ^2(4) \log (81)+\log (1048576)\right )}{(5+5 x+\log (3)) \left (2+2 e^5+x \log (4)+\log (4) \log \left (x+\frac {\log (3)}{5}\right )\right )}}{\log ^2(4)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(-10*x - 20*E^10*x + 20*x^2 + E^5*(-30*x + 20*x^2) + (-4 - 4*E^10 + 4*x + E^5*(-8 + 4*x))*Log[3] + (
-20*x^2 - 20*E^5*x^2 + 5*x^3 + (-4*x - 4*E^5*x + x^2)*Log[3])*Log[4] + (-5*x^3 - x^2*Log[3])*Log[4]^2 + (10*x
+ 10*E^5*x + (2 + 2*E^5)*Log[3] + (-20*x - 20*E^5*x + 10*x^2 + (-4 - 4*E^5 + 2*x)*Log[3])*Log[4] + (-10*x^2 -
2*x*Log[3])*Log[4]^2)*Log[(5*x + Log[3])/5] + ((5*x + Log[3])*Log[4] + (-5*x - Log[3])*Log[4]^2)*Log[(5*x + Lo
g[3])/5]^2)/(20*x + 40*E^5*x + 20*E^10*x + (4 + 8*E^5 + 4*E^10)*Log[3] + (20*x^2 + 20*E^5*x^2 + (4*x + 4*E^5*x
)*Log[3])*Log[4] + (5*x^3 + x^2*Log[3])*Log[4]^2 + ((20*x + 20*E^5*x + (4 + 4*E^5)*Log[3])*Log[4] + (10*x^2 +
2*x*Log[3])*Log[4]^2)*Log[(5*x + Log[3])/5] + (5*x + Log[3])*Log[4]^2*Log[(5*x + Log[3])/5]^2),x]

[Out]

-((x*(-1 + Log[4])*Log[4] + ((1 + E^5)*x*(10*x*Log[4] + Log[4]*Log[9]*(1 + Log[16]) - Log[4]^2*Log[81] + Log[1
048576]))/((5 + 5*x + Log[3])*(2 + 2*E^5 + x*Log[4] + Log[4]*Log[x + Log[3]/5])))/Log[4]^2)

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fricas [B]  time = 0.59, size = 60, normalized size = 1.94 \begin {gather*} -\frac {2 \, x^{2} \log \relax (2) - x^{2} + 2 \, x e^{5} + {\left (2 \, x \log \relax (2) - x\right )} \log \left (x + \frac {1}{5} \, \log \relax (3)\right ) + 2 \, x}{2 \, {\left (x \log \relax (2) + \log \relax (2) \log \left (x + \frac {1}{5} \, \log \relax (3)\right ) + e^{5} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-log(3)-5*x)*log(2)^2+2*(log(3)+5*x)*log(2))*log(1/5*log(3)+x)^2+(4*(-2*x*log(3)-10*x^2)*log(2)
^2+2*((-4*exp(5)+2*x-4)*log(3)-20*x*exp(5)+10*x^2-20*x)*log(2)+(2*exp(5)+2)*log(3)+10*x*exp(5)+10*x)*log(1/5*l
og(3)+x)+4*(-x^2*log(3)-5*x^3)*log(2)^2+2*((-4*x*exp(5)+x^2-4*x)*log(3)-20*x^2*exp(5)+5*x^3-20*x^2)*log(2)+(-4
*exp(5)^2+(4*x-8)*exp(5)+4*x-4)*log(3)-20*x*exp(5)^2+(20*x^2-30*x)*exp(5)+20*x^2-10*x)/(4*(log(3)+5*x)*log(2)^
2*log(1/5*log(3)+x)^2+(4*(2*x*log(3)+10*x^2)*log(2)^2+2*((4*exp(5)+4)*log(3)+20*x*exp(5)+20*x)*log(2))*log(1/5
*log(3)+x)+4*(x^2*log(3)+5*x^3)*log(2)^2+2*((4*x*exp(5)+4*x)*log(3)+20*x^2*exp(5)+20*x^2)*log(2)+(4*exp(5)^2+8
*exp(5)+4)*log(3)+20*x*exp(5)^2+40*x*exp(5)+20*x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.39, size = 82, normalized size = 2.65 \begin {gather*} -\frac {2 \, x^{2} \log \relax (2) - 2 \, x \log \relax (5) \log \relax (2) + 2 \, x \log \relax (2) \log \left (5 \, x + \log \relax (3)\right ) - x^{2} + 2 \, x e^{5} + x \log \relax (5) - x \log \left (5 \, x + \log \relax (3)\right ) + 2 \, x}{2 \, {\left (x \log \relax (2) - \log \relax (5) \log \relax (2) + \log \relax (2) \log \left (5 \, x + \log \relax (3)\right ) + e^{5} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-log(3)-5*x)*log(2)^2+2*(log(3)+5*x)*log(2))*log(1/5*log(3)+x)^2+(4*(-2*x*log(3)-10*x^2)*log(2)
^2+2*((-4*exp(5)+2*x-4)*log(3)-20*x*exp(5)+10*x^2-20*x)*log(2)+(2*exp(5)+2)*log(3)+10*x*exp(5)+10*x)*log(1/5*l
og(3)+x)+4*(-x^2*log(3)-5*x^3)*log(2)^2+2*((-4*x*exp(5)+x^2-4*x)*log(3)-20*x^2*exp(5)+5*x^3-20*x^2)*log(2)+(-4
*exp(5)^2+(4*x-8)*exp(5)+4*x-4)*log(3)-20*x*exp(5)^2+(20*x^2-30*x)*exp(5)+20*x^2-10*x)/(4*(log(3)+5*x)*log(2)^
2*log(1/5*log(3)+x)^2+(4*(2*x*log(3)+10*x^2)*log(2)^2+2*((4*exp(5)+4)*log(3)+20*x*exp(5)+20*x)*log(2))*log(1/5
*log(3)+x)+4*(x^2*log(3)+5*x^3)*log(2)^2+2*((4*x*exp(5)+4*x)*log(3)+20*x^2*exp(5)+20*x^2)*log(2)+(4*exp(5)^2+8
*exp(5)+4)*log(3)+20*x*exp(5)^2+40*x*exp(5)+20*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.45, size = 43, normalized size = 1.39

method	result	size
risch	\(-x +\frac {x}{2 \ln \relax (2)}-\frac {\left ({\mathrm e}^{5}+1\right ) x}{2 \ln \relax (2) \left (\ln \relax (2) \ln \left (\frac {\ln \relax (3)}{5}+x \right )+x \ln \relax (2)+{\mathrm e}^{5}+1\right )}\)	\(43\)
norman	\(\frac {\left (-\ln \relax (2)+\frac {1}{2}\right ) x^{2}+\left ({\mathrm e}^{5}+1\right ) \ln \left (\frac {\ln \relax (3)}{5}+x \right )+\left (-\ln \relax (2)+\frac {1}{2}\right ) x \ln \left (\frac {\ln \relax (3)}{5}+x \right )+\frac {{\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1}{\ln \relax (2)}}{\ln \relax (2) \ln \left (\frac {\ln \relax (3)}{5}+x \right )+x \ln \relax (2)+{\mathrm e}^{5}+1}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*(-ln(3)-5*x)*ln(2)^2+2*(ln(3)+5*x)*ln(2))*ln(1/5*ln(3)+x)^2+(4*(-2*x*ln(3)-10*x^2)*ln(2)^2+2*((-4*exp(
5)+2*x-4)*ln(3)-20*x*exp(5)+10*x^2-20*x)*ln(2)+(2*exp(5)+2)*ln(3)+10*x*exp(5)+10*x)*ln(1/5*ln(3)+x)+4*(-x^2*ln
(3)-5*x^3)*ln(2)^2+2*((-4*x*exp(5)+x^2-4*x)*ln(3)-20*x^2*exp(5)+5*x^3-20*x^2)*ln(2)+(-4*exp(5)^2+(4*x-8)*exp(5
)+4*x-4)*ln(3)-20*x*exp(5)^2+(20*x^2-30*x)*exp(5)+20*x^2-10*x)/(4*(ln(3)+5*x)*ln(2)^2*ln(1/5*ln(3)+x)^2+(4*(2*
x*ln(3)+10*x^2)*ln(2)^2+2*((4*exp(5)+4)*ln(3)+20*x*exp(5)+20*x)*ln(2))*ln(1/5*ln(3)+x)+4*(x^2*ln(3)+5*x^3)*ln(
2)^2+2*((4*x*exp(5)+4*x)*ln(3)+20*x^2*exp(5)+20*x^2)*ln(2)+(4*exp(5)^2+8*exp(5)+4)*ln(3)+20*x*exp(5)^2+40*x*ex
p(5)+20*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.55, size = 73, normalized size = 2.35 \begin {gather*} -\frac {x^{2} {\left (2 \, \log \relax (2) - 1\right )} + x {\left (2 \, \log \relax (2) - 1\right )} \log \left (5 \, x + \log \relax (3)\right ) - {\left (2 \, \log \relax (5) \log \relax (2) - 2 \, e^{5} - \log \relax (5) - 2\right )} x}{2 \, {\left (x \log \relax (2) - \log \relax (5) \log \relax (2) + \log \relax (2) \log \left (5 \, x + \log \relax (3)\right ) + e^{5} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-log(3)-5*x)*log(2)^2+2*(log(3)+5*x)*log(2))*log(1/5*log(3)+x)^2+(4*(-2*x*log(3)-10*x^2)*log(2)
^2+2*((-4*exp(5)+2*x-4)*log(3)-20*x*exp(5)+10*x^2-20*x)*log(2)+(2*exp(5)+2)*log(3)+10*x*exp(5)+10*x)*log(1/5*l
og(3)+x)+4*(-x^2*log(3)-5*x^3)*log(2)^2+2*((-4*x*exp(5)+x^2-4*x)*log(3)-20*x^2*exp(5)+5*x^3-20*x^2)*log(2)+(-4
*exp(5)^2+(4*x-8)*exp(5)+4*x-4)*log(3)-20*x*exp(5)^2+(20*x^2-30*x)*exp(5)+20*x^2-10*x)/(4*(log(3)+5*x)*log(2)^
2*log(1/5*log(3)+x)^2+(4*(2*x*log(3)+10*x^2)*log(2)^2+2*((4*exp(5)+4)*log(3)+20*x*exp(5)+20*x)*log(2))*log(1/5
*log(3)+x)+4*(x^2*log(3)+5*x^3)*log(2)^2+2*((4*x*exp(5)+4*x)*log(3)+20*x^2*exp(5)+20*x^2)*log(2)+(4*exp(5)^2+8
*exp(5)+4)*log(3)+20*x*exp(5)^2+40*x*exp(5)+20*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.55, size = 126, normalized size = 4.06 \begin {gather*} \frac {2\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+x^2\,{\ln \relax (2)}^2-2\,x^2\,{\ln \relax (2)}^3+x\,\ln \relax (2)-2\,x\,{\ln \relax (2)}^2+\ln \left (x+\frac {\ln \relax (3)}{5}\right )\,\ln \relax (2)-2\,x\,{\mathrm {e}}^5\,{\ln \relax (2)}^2+\ln \left (x+\frac {\ln \relax (3)}{5}\right )\,{\mathrm {e}}^5\,\ln \relax (2)+x\,{\mathrm {e}}^5\,\ln \relax (2)+x\,\ln \left (x+\frac {\ln \relax (3)}{5}\right )\,{\ln \relax (2)}^2-2\,x\,\ln \left (x+\frac {\ln \relax (3)}{5}\right )\,{\ln \relax (2)}^3+1}{2\,{\ln \relax (2)}^2\,\left ({\mathrm {e}}^5+x\,\ln \relax (2)+\ln \left (x+\frac {\ln \relax (3)}{5}\right )\,\ln \relax (2)+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x - log(x + log(3)/5)*(10*x - 4*log(2)^2*(2*x*log(3) + 10*x^2) + 10*x*exp(5) - 2*log(2)*(20*x + 20*x*
exp(5) + log(3)*(4*exp(5) - 2*x + 4) - 10*x^2) + log(3)*(2*exp(5) + 2)) + exp(5)*(30*x - 20*x^2) + 20*x*exp(10
) + 4*log(2)^2*(x^2*log(3) + 5*x^3) - log(x + log(3)/5)^2*(2*log(2)*(5*x + log(3)) - 4*log(2)^2*(5*x + log(3))
) + 2*log(2)*(log(3)*(4*x + 4*x*exp(5) - x^2) + 20*x^2*exp(5) + 20*x^2 - 5*x^3) - log(3)*(4*x - 4*exp(10) + ex
p(5)*(4*x - 8) - 4) - 20*x^2)/(20*x + log(x + log(3)/5)*(4*log(2)^2*(2*x*log(3) + 10*x^2) + 2*log(2)*(20*x + 2
0*x*exp(5) + log(3)*(4*exp(5) + 4))) + 40*x*exp(5) + 20*x*exp(10) + 2*log(2)*(log(3)*(4*x + 4*x*exp(5)) + 20*x
^2*exp(5) + 20*x^2) + 4*log(2)^2*(x^2*log(3) + 5*x^3) + log(3)*(8*exp(5) + 4*exp(10) + 4) + 4*log(x + log(3)/5
)^2*log(2)^2*(5*x + log(3))),x)

[Out]

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

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sympy [B]  time = 0.37, size = 58, normalized size = 1.87 \begin {gather*} \frac {x \left (1 - 2 \log {\relax (2 )}\right )}{2 \log {\relax (2 )}} + \frac {- x e^{5} - x}{2 x \log {\relax (2 )}^{2} + 2 \log {\relax (2 )}^{2} \log {\left (x + \frac {\log {\relax (3 )}}{5} \right )} + 2 \log {\relax (2 )} + 2 e^{5} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-ln(3)-5*x)*ln(2)**2+2*(ln(3)+5*x)*ln(2))*ln(1/5*ln(3)+x)**2+(4*(-2*x*ln(3)-10*x**2)*ln(2)**2+2
*((-4*exp(5)+2*x-4)*ln(3)-20*x*exp(5)+10*x**2-20*x)*ln(2)+(2*exp(5)+2)*ln(3)+10*x*exp(5)+10*x)*ln(1/5*ln(3)+x)
+4*(-x**2*ln(3)-5*x**3)*ln(2)**2+2*((-4*x*exp(5)+x**2-4*x)*ln(3)-20*x**2*exp(5)+5*x**3-20*x**2)*ln(2)+(-4*exp(
5)**2+(4*x-8)*exp(5)+4*x-4)*ln(3)-20*x*exp(5)**2+(20*x**2-30*x)*exp(5)+20*x**2-10*x)/(4*(ln(3)+5*x)*ln(2)**2*l
n(1/5*ln(3)+x)**2+(4*(2*x*ln(3)+10*x**2)*ln(2)**2+2*((4*exp(5)+4)*ln(3)+20*x*exp(5)+20*x)*ln(2))*ln(1/5*ln(3)+
x)+4*(x**2*ln(3)+5*x**3)*ln(2)**2+2*((4*x*exp(5)+4*x)*ln(3)+20*x**2*exp(5)+20*x**2)*ln(2)+(4*exp(5)**2+8*exp(5
)+4)*ln(3)+20*x*exp(5)**2+40*x*exp(5)+20*x),x)

[Out]

x*(1 - 2*log(2))/(2*log(2)) + (-x*exp(5) - x)/(2*x*log(2)**2 + 2*log(2)**2*log(x + log(3)/5) + 2*log(2) + 2*ex
p(5)*log(2))

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