3.41.75 \(\int \frac {2 e^6 x+2 e^3 x^2+(-2 e^3 x^2-2 x^3) \log (-\frac {15 x}{16})+(4 e^3 x+(-2 e^3 x-6 x^2) \log (-\frac {15 x}{16})) \log (x)+(2 e^3-6 x \log (-\frac {15 x}{16})) \log ^2(x)-2 \log (-\frac {15 x}{16}) \log ^3(x)+(-4 e^3 x+(2 e^3 x+6 x^2) \log (-\frac {15 x}{16})+(-4 e^3+12 x \log (-\frac {15 x}{16})) \log (x)+6 \log (-\frac {15 x}{16}) \log ^2(x)) \log (2 x)+(2 e^3-6 x \log (-\frac {15 x}{16})-6 \log (-\frac {15 x}{16}) \log (x)) \log ^2(2 x)+2 \log (-\frac {15 x}{16}) \log ^3(2 x)}{-x^4-3 x^3 \log (x)-3 x^2 \log ^2(x)-x \log ^3(x)+(3 x^3+6 x^2 \log (x)+3 x \log ^2(x)) \log (2 x)+(-3 x^2-3 x \log (x)) \log ^2(2 x)+x \log ^3(2 x)} \, dx\)

Optimal. Leaf size=27 \[ \left (-\log \left (-\frac {15 x}{16}\right )+\frac {e^3}{x+\log (x)-\log (2 x)}\right )^2 \]

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Rubi [F]  time = 2.49, 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 e^6 x+2 e^3 x^2+\left (-2 e^3 x^2-2 x^3\right ) \log \left (-\frac {15 x}{16}\right )+\left (4 e^3 x+\left (-2 e^3 x-6 x^2\right ) \log \left (-\frac {15 x}{16}\right )\right ) \log (x)+\left (2 e^3-6 x \log \left (-\frac {15 x}{16}\right )\right ) \log ^2(x)-2 \log \left (-\frac {15 x}{16}\right ) \log ^3(x)+\left (-4 e^3 x+\left (2 e^3 x+6 x^2\right ) \log \left (-\frac {15 x}{16}\right )+\left (-4 e^3+12 x \log \left (-\frac {15 x}{16}\right )\right ) \log (x)+6 \log \left (-\frac {15 x}{16}\right ) \log ^2(x)\right ) \log (2 x)+\left (2 e^3-6 x \log \left (-\frac {15 x}{16}\right )-6 \log \left (-\frac {15 x}{16}\right ) \log (x)\right ) \log ^2(2 x)+2 \log \left (-\frac {15 x}{16}\right ) \log ^3(2 x)}{-x^4-3 x^3 \log (x)-3 x^2 \log ^2(x)-x \log ^3(x)+\left (3 x^3+6 x^2 \log (x)+3 x \log ^2(x)\right ) \log (2 x)+\left (-3 x^2-3 x \log (x)\right ) \log ^2(2 x)+x \log ^3(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*E^6*x + 2*E^3*x^2 + (-2*E^3*x^2 - 2*x^3)*Log[(-15*x)/16] + (4*E^3*x + (-2*E^3*x - 6*x^2)*Log[(-15*x)/16
])*Log[x] + (2*E^3 - 6*x*Log[(-15*x)/16])*Log[x]^2 - 2*Log[(-15*x)/16]*Log[x]^3 + (-4*E^3*x + (2*E^3*x + 6*x^2
)*Log[(-15*x)/16] + (-4*E^3 + 12*x*Log[(-15*x)/16])*Log[x] + 6*Log[(-15*x)/16]*Log[x]^2)*Log[2*x] + (2*E^3 - 6
*x*Log[(-15*x)/16] - 6*Log[(-15*x)/16]*Log[x])*Log[2*x]^2 + 2*Log[(-15*x)/16]*Log[2*x]^3)/(-x^4 - 3*x^3*Log[x]
- 3*x^2*Log[x]^2 - x*Log[x]^3 + (3*x^3 + 6*x^2*Log[x] + 3*x*Log[x]^2)*Log[2*x] + (-3*x^2 - 3*x*Log[x])*Log[2*
x]^2 + x*Log[2*x]^3),x]

[Out]

-2*Log[16/15]*Log[2]^2*Defer[Int][(x + Log[x] - Log[2*x])^(-3), x] - 2*(E^3 - Log[16/15]*Log[2])*(E^3 - Log[4]
)*Defer[Int][(x + Log[x] - Log[2*x])^(-3), x] - 2*Log[2]^2*(E^3 - Log[16/15]*Log[2])*Defer[Int][1/(x*(x + Log[
x] - Log[2*x])^3), x] - 2*(E^3 - Log[16/15]*Log[2])*Defer[Int][x/(x + Log[x] - Log[2*x])^3, x] - 2*Log[16/15]*
(E^3 - Log[4])*Defer[Int][x/(x + Log[x] - Log[2*x])^3, x] - 2*Log[16/15]*Defer[Int][x^2/(x + Log[x] - Log[2*x]
)^3, x] + 2*Log[2]^2*Defer[Int][Log[-x]/(x + Log[x] - Log[2*x])^3, x] - 2*Log[2]*(E^3 - Log[4])*Defer[Int][Log
[-x]/(x + Log[x] - Log[2*x])^3, x] - 2*Log[2]^3*Defer[Int][Log[-x]/(x*(x + Log[x] - Log[2*x])^3), x] - 2*Log[2
]*Defer[Int][(x*Log[-x])/(x + Log[x] - Log[2*x])^3, x] + 2*(E^3 - Log[4])*Defer[Int][(x*Log[-x])/(x + Log[x] -
Log[2*x])^3, x] + 2*Defer[Int][(x^2*Log[-x])/(x + Log[x] - Log[2*x])^3, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (x^2+\log ^2(2)+x \left (e^3-\log (4)\right )\right ) \left (-e^3-\log \left (\frac {16}{15}\right ) (x-\log (2))-(-x+\log (2)) \log (-x)\right )}{x (x+\log (x)-\log (2 x))^3} \, dx\\ &=2 \int \frac {\left (x^2+\log ^2(2)+x \left (e^3-\log (4)\right )\right ) \left (-e^3-\log \left (\frac {16}{15}\right ) (x-\log (2))-(-x+\log (2)) \log (-x)\right )}{x (x+\log (x)-\log (2 x))^3} \, dx\\ &=2 \int \left (\frac {x \left (-x \log \left (\frac {16}{15}\right )-e^3 \left (1-\frac {\log \left (\frac {16}{15}\right ) \log (2)}{e^3}\right )+x \log (-x)-\log (2) \log (-x)\right )}{(x+\log (x)-\log (2 x))^3}+\frac {\log ^2(2) \left (-x \log \left (\frac {16}{15}\right )-e^3 \left (1-\frac {\log \left (\frac {16}{15}\right ) \log (2)}{e^3}\right )+x \log (-x)-\log (2) \log (-x)\right )}{x (x+\log (x)-\log (2 x))^3}+\frac {e^3 \left (1-\frac {\log (4)}{e^3}\right ) \left (-x \log \left (\frac {16}{15}\right )-e^3 \left (1-\frac {\log \left (\frac {16}{15}\right ) \log (2)}{e^3}\right )+x \log (-x)-\log (2) \log (-x)\right )}{(x+\log (x)-\log (2 x))^3}\right ) \, dx\\ &=2 \int \frac {x \left (-x \log \left (\frac {16}{15}\right )-e^3 \left (1-\frac {\log \left (\frac {16}{15}\right ) \log (2)}{e^3}\right )+x \log (-x)-\log (2) \log (-x)\right )}{(x+\log (x)-\log (2 x))^3} \, dx+\left (2 \log ^2(2)\right ) \int \frac {-x \log \left (\frac {16}{15}\right )-e^3 \left (1-\frac {\log \left (\frac {16}{15}\right ) \log (2)}{e^3}\right )+x \log (-x)-\log (2) \log (-x)}{x (x+\log (x)-\log (2 x))^3} \, dx+\left (2 \left (e^3-\log (4)\right )\right ) \int \frac {-x \log \left (\frac {16}{15}\right )-e^3 \left (1-\frac {\log \left (\frac {16}{15}\right ) \log (2)}{e^3}\right )+x \log (-x)-\log (2) \log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx\\ &=2 \int \left (-\frac {x^2 \log \left (\frac {16}{15}\right )}{(x+\log (x)-\log (2 x))^3}-\frac {x \left (e^3-\log \left (\frac {16}{15}\right ) \log (2)\right )}{(x+\log (x)-\log (2 x))^3}+\frac {x^2 \log (-x)}{(x+\log (x)-\log (2 x))^3}-\frac {x \log (2) \log (-x)}{(x+\log (x)-\log (2 x))^3}\right ) \, dx+\left (2 \log ^2(2)\right ) \int \frac {-e^3+\log \left (\frac {16}{15}\right ) (-x+\log (2))+(x-\log (2)) \log (-x)}{x (x+\log (x)-\log (2 x))^3} \, dx+\left (2 \left (e^3-\log (4)\right )\right ) \int \frac {-e^3+\log \left (\frac {16}{15}\right ) (-x+\log (2))+(x-\log (2)) \log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx\\ &=2 \int \frac {x^2 \log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx-\left (2 \log \left (\frac {16}{15}\right )\right ) \int \frac {x^2}{(x+\log (x)-\log (2 x))^3} \, dx-(2 \log (2)) \int \frac {x \log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx+\left (2 \log ^2(2)\right ) \int \left (-\frac {\log \left (\frac {16}{15}\right )}{(x+\log (x)-\log (2 x))^3}-\frac {e^3 \left (1-\frac {\log \left (\frac {16}{15}\right ) \log (2)}{e^3}\right )}{x (x+\log (x)-\log (2 x))^3}+\frac {\log (-x)}{(x+\log (x)-\log (2 x))^3}-\frac {\log (2) \log (-x)}{x (x+\log (x)-\log (2 x))^3}\right ) \, dx-\left (2 \left (e^3-\log \left (\frac {16}{15}\right ) \log (2)\right )\right ) \int \frac {x}{(x+\log (x)-\log (2 x))^3} \, dx+\left (2 \left (e^3-\log (4)\right )\right ) \int \left (-\frac {x \log \left (\frac {16}{15}\right )}{(x+\log (x)-\log (2 x))^3}-\frac {e^3 \left (1-\frac {\log \left (\frac {16}{15}\right ) \log (2)}{e^3}\right )}{(x+\log (x)-\log (2 x))^3}+\frac {x \log (-x)}{(x+\log (x)-\log (2 x))^3}-\frac {\log (2) \log (-x)}{(x+\log (x)-\log (2 x))^3}\right ) \, dx\\ &=2 \int \frac {x^2 \log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx-\left (2 \log \left (\frac {16}{15}\right )\right ) \int \frac {x^2}{(x+\log (x)-\log (2 x))^3} \, dx-(2 \log (2)) \int \frac {x \log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx+\left (2 \log ^2(2)\right ) \int \frac {\log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx-\left (2 \log \left (\frac {16}{15}\right ) \log ^2(2)\right ) \int \frac {1}{(x+\log (x)-\log (2 x))^3} \, dx-\left (2 \log ^3(2)\right ) \int \frac {\log (-x)}{x (x+\log (x)-\log (2 x))^3} \, dx-\left (2 \left (e^3-\log \left (\frac {16}{15}\right ) \log (2)\right )\right ) \int \frac {x}{(x+\log (x)-\log (2 x))^3} \, dx-\left (2 \log ^2(2) \left (e^3-\log \left (\frac {16}{15}\right ) \log (2)\right )\right ) \int \frac {1}{x (x+\log (x)-\log (2 x))^3} \, dx+\left (2 \left (e^3-\log (4)\right )\right ) \int \frac {x \log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx-\left (2 \log \left (\frac {16}{15}\right ) \left (e^3-\log (4)\right )\right ) \int \frac {x}{(x+\log (x)-\log (2 x))^3} \, dx-\left (2 \log (2) \left (e^3-\log (4)\right )\right ) \int \frac {\log (-x)}{(x+\log (x)-\log (2 x))^3} \, dx-\left (2 \left (e^3-\log \left (\frac {16}{15}\right ) \log (2)\right ) \left (e^3-\log (4)\right )\right ) \int \frac {1}{(x+\log (x)-\log (2 x))^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 1.53, size = 295, normalized size = 10.93 \begin {gather*} -\frac {\log ^3(2) \log ^2(-x)-2 \log (x) \left (\log \left (\frac {16}{15}\right ) \log ^3(2)+\left (-\log ^2(4)+\log (2) \log (16)\right ) \log (x)\right )+\frac {e^6 \log (2) (\log (x)-\log (2 x))^2}{(x+\log (x)-\log (2 x))^2}+\frac {2 \log (-x) \left (e^3 \log ^2(2)+\left (-\log ^2(4)+\log (2) \log (16)\right ) \log (x)\right ) (\log (x)-\log (2 x))}{x+\log (x)-\log (2 x)}+\frac {2 (\log (x)-\log (2 x)) \left (\log ^2(2) \left (e^3-\log \left (\frac {16}{15}\right ) \log (2)\right )+2 \log \left (\frac {16}{15}\right ) \log ^3(x)+\left (-e^3 \left (1+\log \left (\frac {16}{15}\right )\right )+\log \left (\frac {16}{15}\right ) \log (8)\right ) \log ^2(2 x)-2 \log \left (\frac {16}{15}\right ) \log ^3(2 x)+2 \log (x) \log (2 x) \left (e^3 \left (1+\log \left (\frac {16}{15}\right )\right )-\log \left (\frac {16}{15}\right ) \log (8)+3 \log \left (\frac {16}{15}\right ) \log (2 x)\right )-\log ^2(x) \left (e^3 \left (1+\log \left (\frac {16}{15}\right )\right )-\log \left (\frac {16}{15}\right ) \log (8)+6 \log \left (\frac {16}{15}\right ) \log (2 x)\right )\right )}{x+\log (x)-\log (2 x)}}{(\log (x)-\log (2 x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^6*x + 2*E^3*x^2 + (-2*E^3*x^2 - 2*x^3)*Log[(-15*x)/16] + (4*E^3*x + (-2*E^3*x - 6*x^2)*Log[(-15
*x)/16])*Log[x] + (2*E^3 - 6*x*Log[(-15*x)/16])*Log[x]^2 - 2*Log[(-15*x)/16]*Log[x]^3 + (-4*E^3*x + (2*E^3*x +
6*x^2)*Log[(-15*x)/16] + (-4*E^3 + 12*x*Log[(-15*x)/16])*Log[x] + 6*Log[(-15*x)/16]*Log[x]^2)*Log[2*x] + (2*E
^3 - 6*x*Log[(-15*x)/16] - 6*Log[(-15*x)/16]*Log[x])*Log[2*x]^2 + 2*Log[(-15*x)/16]*Log[2*x]^3)/(-x^4 - 3*x^3*
Log[x] - 3*x^2*Log[x]^2 - x*Log[x]^3 + (3*x^3 + 6*x^2*Log[x] + 3*x*Log[x]^2)*Log[2*x] + (-3*x^2 - 3*x*Log[x])*
Log[2*x]^2 + x*Log[2*x]^3),x]

[Out]

-((Log[2]^3*Log[-x]^2 - 2*Log[x]*(Log[16/15]*Log[2]^3 + (-Log[4]^2 + Log[2]*Log[16])*Log[x]) + (E^6*Log[2]*(Lo
g[x] - Log[2*x])^2)/(x + Log[x] - Log[2*x])^2 + (2*Log[-x]*(E^3*Log[2]^2 + (-Log[4]^2 + Log[2]*Log[16])*Log[x]
)*(Log[x] - Log[2*x]))/(x + Log[x] - Log[2*x]) + (2*(Log[x] - Log[2*x])*(Log[2]^2*(E^3 - Log[16/15]*Log[2]) +
2*Log[16/15]*Log[x]^3 + (-(E^3*(1 + Log[16/15])) + Log[16/15]*Log[8])*Log[2*x]^2 - 2*Log[16/15]*Log[2*x]^3 + 2
*Log[x]*Log[2*x]*(E^3*(1 + Log[16/15]) - Log[16/15]*Log[8] + 3*Log[16/15]*Log[2*x]) - Log[x]^2*(E^3*(1 + Log[1
6/15]) - Log[16/15]*Log[8] + 6*Log[16/15]*Log[2*x])))/(x + Log[x] - Log[2*x]))/(Log[x] - Log[2*x])^3)

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fricas [B]  time = 0.87, size = 85, normalized size = 3.15 \begin {gather*} \frac {{\left (x^{2} - 2 \, {\left (x + \log \left (\frac {16}{15}\right )\right )} \log \left (\frac {32}{15}\right ) + \log \left (\frac {32}{15}\right )^{2} + 2 \, x \log \left (\frac {16}{15}\right ) + \log \left (\frac {16}{15}\right )^{2}\right )} \log \left (-\frac {15}{16} \, x\right )^{2} - 2 \, {\left (x e^{3} - e^{3} \log \left (\frac {32}{15}\right ) + e^{3} \log \left (\frac {16}{15}\right )\right )} \log \left (-\frac {15}{16} \, x\right ) + e^{6}}{x^{2} - 2 \, {\left (x + \log \left (\frac {16}{15}\right )\right )} \log \left (\frac {32}{15}\right ) + \log \left (\frac {32}{15}\right )^{2} + 2 \, x \log \left (\frac {16}{15}\right ) + \log \left (\frac {16}{15}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(-15/16*x)*log(2*x)^3+(-6*log(-15/16*x)*log(x)-6*x*log(-15/16*x)+2*exp(3))*log(2*x)^2+(6*log(-
15/16*x)*log(x)^2+(12*x*log(-15/16*x)-4*exp(3))*log(x)+(2*x*exp(3)+6*x^2)*log(-15/16*x)-4*x*exp(3))*log(2*x)-2
*log(-15/16*x)*log(x)^3+(-6*x*log(-15/16*x)+2*exp(3))*log(x)^2+((-2*x*exp(3)-6*x^2)*log(-15/16*x)+4*x*exp(3))*
log(x)+(-2*x^2*exp(3)-2*x^3)*log(-15/16*x)+2*x*exp(3)^2+2*x^2*exp(3))/(x*log(2*x)^3+(-3*x*log(x)-3*x^2)*log(2*
x)^2+(3*x*log(x)^2+6*x^2*log(x)+3*x^3)*log(2*x)-x*log(x)^3-3*x^2*log(x)^2-3*x^3*log(x)-x^4),x, algorithm="fric
as")

[Out]

((x^2 - 2*(x + log(16/15))*log(32/15) + log(32/15)^2 + 2*x*log(16/15) + log(16/15)^2)*log(-15/16*x)^2 - 2*(x*e
^3 - e^3*log(32/15) + e^3*log(16/15))*log(-15/16*x) + e^6)/(x^2 - 2*(x + log(16/15))*log(32/15) + log(32/15)^2
+ 2*x*log(16/15) + log(16/15)^2)

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giac [B]  time = 0.24, size = 85, normalized size = 3.15 \begin {gather*} \frac {x^{2} \log \left (-\frac {15}{16} \, x\right )^{2} - 2 \, x \log \relax (2) \log \left (-\frac {15}{16} \, x\right )^{2} + \log \relax (2)^{2} \log \left (-\frac {15}{16} \, x\right )^{2} + 8 \, x e^{3} \log \relax (2) - 8 \, e^{3} \log \relax (2)^{2} - 2 \, x e^{3} \log \left (-15 \, x\right ) + 2 \, e^{3} \log \relax (2) \log \left (-15 \, x\right ) + e^{6}}{x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(-15/16*x)*log(2*x)^3+(-6*log(-15/16*x)*log(x)-6*x*log(-15/16*x)+2*exp(3))*log(2*x)^2+(6*log(-
15/16*x)*log(x)^2+(12*x*log(-15/16*x)-4*exp(3))*log(x)+(2*x*exp(3)+6*x^2)*log(-15/16*x)-4*x*exp(3))*log(2*x)-2
*log(-15/16*x)*log(x)^3+(-6*x*log(-15/16*x)+2*exp(3))*log(x)^2+((-2*x*exp(3)-6*x^2)*log(-15/16*x)+4*x*exp(3))*
log(x)+(-2*x^2*exp(3)-2*x^3)*log(-15/16*x)+2*x*exp(3)^2+2*x^2*exp(3))/(x*log(2*x)^3+(-3*x*log(x)-3*x^2)*log(2*
x)^2+(3*x*log(x)^2+6*x^2*log(x)+3*x^3)*log(2*x)-x*log(x)^3-3*x^2*log(x)^2-3*x^3*log(x)-x^4),x, algorithm="giac
")

[Out]

(x^2*log(-15/16*x)^2 - 2*x*log(2)*log(-15/16*x)^2 + log(2)^2*log(-15/16*x)^2 + 8*x*e^3*log(2) - 8*e^3*log(2)^2
- 2*x*e^3*log(-15*x) + 2*e^3*log(2)*log(-15*x) + e^6)/(x^2 - 2*x*log(2) + log(2)^2)

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maple [B]  time = 0.57, size = 120, normalized size = 4.44

method	result	size
default	\(\frac {{\mathrm e}^{6}}{\left (x -\ln \relax (2)\right )^{2}}-\frac {2 \,{\mathrm e}^{3} \ln \left (x -\ln \relax (2)\right )}{\ln \relax (2)}+\frac {8 \,{\mathrm e}^{3} \ln \relax (2)}{x -\ln \relax (2)}-8 \ln \relax (2) \ln \relax (x )+\frac {2 \ln \relax (x ) {\mathrm e}^{3}}{\ln \relax (2)}-\frac {2 \ln \left (15\right ) {\mathrm e}^{3}}{x -\ln \relax (2)}+2 \ln \relax (x ) \ln \left (15\right )+\frac {2 \,{\mathrm e}^{3} \ln \left (\ln \relax (2)-x \right )}{\ln \relax (2)}+\frac {2 \,{\mathrm e}^{3} \ln \left (-x \right ) x}{\ln \relax (2) \left (\ln \relax (2)-x \right )}+\ln \left (-x \right )^{2}\)	\(120\)
risch	\(\ln \relax (x )^{2}+\frac {4 i {\mathrm e}^{3} \ln \relax (x )}{2 i \ln \relax (2)-2 i x}+\frac {-8 x \,{\mathrm e}^{3} \ln \relax (5)-8 x \,{\mathrm e}^{3} \ln \relax (3)+4 \,{\mathrm e}^{6}+8 i \pi \,{\mathrm e}^{3} \ln \relax (2)-8 i \pi \,{\mathrm e}^{3} x +8 i \pi \ln \relax (2)^{2} \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right )+8 i \pi \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x^{2}-16 \ln \relax (3) \ln \relax (2) \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x -16 \ln \relax (2) \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) \ln \relax (5) x +32 x \,{\mathrm e}^{3} \ln \relax (2)-32 \,{\mathrm e}^{3} \ln \relax (2)^{2}+8 \ln \relax (3) {\mathrm e}^{3} \ln \relax (2)+8 \,{\mathrm e}^{3} \ln \relax (2) \ln \relax (5)+8 \ln \relax (3) \ln \relax (2)^{2} \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right )+8 \ln \relax (3) \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x^{2}+8 \ln \relax (2)^{2} \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) \ln \relax (5)+64 \ln \relax (2)^{2} \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x -32 \ln \relax (2) \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x^{2}+8 \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) \ln \relax (5) x^{2}-16 i \pi \ln \relax (2) \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x \mathrm {csgn}\left (i x \right )^{3}+16 i \pi \ln \relax (2) \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x \mathrm {csgn}\left (i x \right )^{2}-32 \ln \relax (2)^{3} \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right )+8 i \pi \,{\mathrm e}^{3} x \mathrm {csgn}\left (i x \right )^{2}-8 i \pi \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x^{2} \mathrm {csgn}\left (i x \right )^{2}-8 i \pi \,{\mathrm e}^{3} \ln \relax (2) \mathrm {csgn}\left (i x \right )^{2}-8 i \pi \,{\mathrm e}^{3} x \mathrm {csgn}\left (i x \right )^{3}+8 i \pi \,{\mathrm e}^{3} \ln \relax (2) \mathrm {csgn}\left (i x \right )^{3}+8 i \pi \ln \relax (2)^{2} \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) \mathrm {csgn}\left (i x \right )^{3}-8 i \pi \ln \relax (2)^{2} \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) \mathrm {csgn}\left (i x \right )^{2}+8 i \pi \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x^{2} \mathrm {csgn}\left (i x \right )^{3}-16 i \pi \ln \relax (2) \ln \left (\left (-2 \pi \mathrm {csgn}\left (i x \right )^{2}+2 \pi \mathrm {csgn}\left (i x \right )^{3}+2 \pi -2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (5)\right ) x \right ) x}{4 \ln \relax (2)^{2}-8 x \ln \relax (2)+4 x^{2}}\)	\(1096\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(-15/16*x)*ln(2*x)^3+(-6*ln(-15/16*x)*ln(x)-6*x*ln(-15/16*x)+2*exp(3))*ln(2*x)^2+(6*ln(-15/16*x)*ln(x
)^2+(12*x*ln(-15/16*x)-4*exp(3))*ln(x)+(2*x*exp(3)+6*x^2)*ln(-15/16*x)-4*x*exp(3))*ln(2*x)-2*ln(-15/16*x)*ln(x
)^3+(-6*x*ln(-15/16*x)+2*exp(3))*ln(x)^2+((-2*x*exp(3)-6*x^2)*ln(-15/16*x)+4*x*exp(3))*ln(x)+(-2*x^2*exp(3)-2*
x^3)*ln(-15/16*x)+2*x*exp(3)^2+2*x^2*exp(3))/(x*ln(2*x)^3+(-3*x*ln(x)-3*x^2)*ln(2*x)^2+(3*x*ln(x)^2+6*x^2*ln(x
)+3*x^3)*ln(2*x)-x*ln(x)^3-3*x^2*ln(x)^2-3*x^3*ln(x)-x^4),x,method=_RETURNVERBOSE)

[Out]

exp(6)/(x-ln(2))^2-2*exp(3)/ln(2)*ln(x-ln(2))+8*exp(3)*ln(2)/(x-ln(2))-8*ln(2)*ln(x)+2/ln(2)*ln(x)*exp(3)-2*ln
(15)*exp(3)/(x-ln(2))+2*ln(x)*ln(15)+2*exp(3)/ln(2)*ln(ln(2)-x)+2*exp(3)*ln(-x)*x/ln(2)/(ln(2)-x)+ln(-x)^2

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maxima [C]  time = 0.55, size = 315, normalized size = 11.67 \begin {gather*} {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)} - \frac {\log \relax (x)}{\log \relax (2)} - \frac {1}{x - \log \relax (2)}\right )} e^{3} - \frac {{\left (2 \, x - \log \relax (2)\right )} e^{3} \log \left (-\frac {15}{16} \, x\right )}{x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2}} + \frac {{\left (2 \, x - \log \relax (2)\right )} e^{3}}{x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2}} - \frac {e^{3} \log \left (x - \log \relax (2)\right )}{\log \relax (2)} - \frac {x e^{3} \log \relax (2) - {\left (x^{2} \log \relax (2) - 2 \, x \log \relax (2)^{2} + \log \relax (2)^{3}\right )} \log \relax (x)^{2} - {\left ({\left (i \, \pi + \log \relax (5)\right )} \log \relax (2)^{2} + \log \relax (3) \log \relax (2)^{2} - 4 \, \log \relax (2)^{3}\right )} e^{3} + {\left (2 \, {\left (-i \, \pi - \log \relax (5)\right )} \log \relax (2)^{3} - 2 \, \log \relax (3) \log \relax (2)^{3} + 8 \, \log \relax (2)^{4} + {\left (2 \, {\left (-i \, \pi - \log \relax (5)\right )} \log \relax (2) - 2 \, \log \relax (3) \log \relax (2) + 8 \, \log \relax (2)^{2} - e^{3}\right )} x^{2} - 2 \, e^{3} \log \relax (2)^{2} + 2 \, {\left (2 \, {\left (i \, \pi + \log \relax (5)\right )} \log \relax (2)^{2} + 2 \, \log \relax (3) \log \relax (2)^{2} - 8 \, \log \relax (2)^{3} + e^{3} \log \relax (2)\right )} x\right )} \log \relax (x)}{x^{2} \log \relax (2) - 2 \, x \log \relax (2)^{2} + \log \relax (2)^{3}} + \frac {e^{6}}{x^{2} - 2 \, x \log \relax (2) + \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(-15/16*x)*log(2*x)^3+(-6*log(-15/16*x)*log(x)-6*x*log(-15/16*x)+2*exp(3))*log(2*x)^2+(6*log(-
15/16*x)*log(x)^2+(12*x*log(-15/16*x)-4*exp(3))*log(x)+(2*x*exp(3)+6*x^2)*log(-15/16*x)-4*x*exp(3))*log(2*x)-2
*log(-15/16*x)*log(x)^3+(-6*x*log(-15/16*x)+2*exp(3))*log(x)^2+((-2*x*exp(3)-6*x^2)*log(-15/16*x)+4*x*exp(3))*
log(x)+(-2*x^2*exp(3)-2*x^3)*log(-15/16*x)+2*x*exp(3)^2+2*x^2*exp(3))/(x*log(2*x)^3+(-3*x*log(x)-3*x^2)*log(2*
x)^2+(3*x*log(x)^2+6*x^2*log(x)+3*x^3)*log(2*x)-x*log(x)^3-3*x^2*log(x)^2-3*x^3*log(x)-x^4),x, algorithm="maxi
ma")

[Out]

(log(x - log(2))/log(2) - log(x)/log(2) - 1/(x - log(2)))*e^3 - (2*x - log(2))*e^3*log(-15/16*x)/(x^2 - 2*x*lo
g(2) + log(2)^2) + (2*x - log(2))*e^3/(x^2 - 2*x*log(2) + log(2)^2) - e^3*log(x - log(2))/log(2) - (x*e^3*log(
2) - (x^2*log(2) - 2*x*log(2)^2 + log(2)^3)*log(x)^2 - ((I*pi + log(5))*log(2)^2 + log(3)*log(2)^2 - 4*log(2)^
3)*e^3 + (2*(-I*pi - log(5))*log(2)^3 - 2*log(3)*log(2)^3 + 8*log(2)^4 + (2*(-I*pi - log(5))*log(2) - 2*log(3)
*log(2) + 8*log(2)^2 - e^3)*x^2 - 2*e^3*log(2)^2 + 2*(2*(I*pi + log(5))*log(2)^2 + 2*log(3)*log(2)^2 - 8*log(2
)^3 + e^3*log(2))*x)*log(x))/(x^2*log(2) - 2*x*log(2)^2 + log(2)^3) + e^6/(x^2 - 2*x*log(2) + log(2)^2)

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mupad [B]  time = 3.56, size = 105, normalized size = 3.89 \begin {gather*} 2\,\ln \relax (x)\,\left (\ln \left (-\frac {15\,x}{16}\right )-\ln \relax (x)\right )+{\ln \relax (x)}^2+\frac {{\mathrm {e}}^6-2\,x\,{\mathrm {e}}^3\,\left (\ln \left (-\frac {15\,x}{16}\right )-\ln \relax (x)\right )+2\,{\mathrm {e}}^3\,\left (\ln \left (2\,x\right )-\ln \relax (x)\right )\,\left (\ln \left (-\frac {15\,x}{16}\right )-\ln \relax (x)\right )}{{\left (\ln \left (2\,x\right )-\ln \relax (x)\right )}^2-2\,x\,\left (\ln \left (2\,x\right )-\ln \relax (x)\right )+x^2}-\frac {2\,{\mathrm {e}}^3\,\ln \relax (x)}{x-\ln \left (2\,x\right )+\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*exp(6) - log(2*x)*(4*x*exp(3) - log(-(15*x)/16)*(2*x*exp(3) + 6*x^2) + log(x)*(4*exp(3) - 12*x*log(-
(15*x)/16)) - 6*log(-(15*x)/16)*log(x)^2) + 2*log(2*x)^3*log(-(15*x)/16) + log(x)*(4*x*exp(3) - log(-(15*x)/16
)*(2*x*exp(3) + 6*x^2)) + 2*x^2*exp(3) - log(2*x)^2*(6*x*log(-(15*x)/16) - 2*exp(3) + 6*log(-(15*x)/16)*log(x)
) - log(-(15*x)/16)*(2*x^2*exp(3) + 2*x^3) + log(x)^2*(2*exp(3) - 6*x*log(-(15*x)/16)) - 2*log(-(15*x)/16)*log
(x)^3)/(x*log(x)^3 + 3*x^3*log(x) + log(2*x)^2*(3*x*log(x) + 3*x^2) - log(2*x)*(3*x*log(x)^2 + 6*x^2*log(x) +
3*x^3) - x*log(2*x)^3 + 3*x^2*log(x)^2 + x^4),x)

[Out]

2*log(x)*(log(-(15*x)/16) - log(x)) + log(x)^2 + (exp(6) - 2*x*exp(3)*(log(-(15*x)/16) - log(x)) + 2*exp(3)*(l
og(2*x) - log(x))*(log(-(15*x)/16) - log(x)))/((log(2*x) - log(x))^2 - 2*x*(log(2*x) - log(x)) + x^2) - (2*exp
(3)*log(x))/(x - log(2*x) + log(x))

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sympy [C]  time = 10.79, size = 114, normalized size = 4.22 \begin {gather*} \log {\relax (x )}^{2} + 2 \left (- 4 \log {\relax (2 )} + \log {\left (15 \right )} + i \pi \right ) \log {\relax (x )} + \frac {x \left (- 8 e^{3} \log {\relax (2 )} + 2 e^{3} \log {\left (15 \right )} + 2 i \pi e^{3}\right ) - e^{6} - 2 e^{3} \log {\relax (2 )} \log {\left (15 \right )} + 8 e^{3} \log {\relax (2 )}^{2} - 2 i \pi e^{3} \log {\relax (2 )}}{- x^{2} + 2 x \log {\relax (2 )} - \log {\relax (2 )}^{2}} - \frac {2 e^{3} \log {\relax (x )}}{x - \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(-15/16*x)*ln(2*x)**3+(-6*ln(-15/16*x)*ln(x)-6*x*ln(-15/16*x)+2*exp(3))*ln(2*x)**2+(6*ln(-15/16
*x)*ln(x)**2+(12*x*ln(-15/16*x)-4*exp(3))*ln(x)+(2*x*exp(3)+6*x**2)*ln(-15/16*x)-4*x*exp(3))*ln(2*x)-2*ln(-15/
16*x)*ln(x)**3+(-6*x*ln(-15/16*x)+2*exp(3))*ln(x)**2+((-2*x*exp(3)-6*x**2)*ln(-15/16*x)+4*x*exp(3))*ln(x)+(-2*
x**2*exp(3)-2*x**3)*ln(-15/16*x)+2*x*exp(3)**2+2*x**2*exp(3))/(x*ln(2*x)**3+(-3*x*ln(x)-3*x**2)*ln(2*x)**2+(3*
x*ln(x)**2+6*x**2*ln(x)+3*x**3)*ln(2*x)-x*ln(x)**3-3*x**2*ln(x)**2-3*x**3*ln(x)-x**4),x)

[Out]

log(x)**2 + 2*(-4*log(2) + log(15) + I*pi)*log(x) + (x*(-8*exp(3)*log(2) + 2*exp(3)*log(15) + 2*I*pi*exp(3)) -
exp(6) - 2*exp(3)*log(2)*log(15) + 8*exp(3)*log(2)**2 - 2*I*pi*exp(3)*log(2))/(-x**2 + 2*x*log(2) - log(2)**2
) - 2*exp(3)*log(x)/(x - log(2))

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