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

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

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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 = {} $$\begin{array}{c}\int \frac{2e^{6}x+2e^3{x}^2+(-2e^3{x}^2-2x^3)\log (-\frac{15x}{16})+(4e^{3}x+(-2e^{3}x-6x^2)\log (-\frac{15x}{16}))\log (x)+(2e^3-6x\log (-\frac{15x}{16})){\log }^2(x)-2\log (-\frac{15x}{16}){\log }^3(x)+(-4e^{3}x+(2e^{3}x+6x^2)\log (-\frac{15x}{16})+(-4e^3+12x\log (-\frac{15x}{16}))\log (x)+6\log (-\frac{15x}{16}){\log }^2(x))\log (2x)+(2e^3-6x\log (-\frac{15x}{16})-6\log (-\frac{15x}{16})\log (x)){\log }^2(2x)+2\log (-\frac{15x}{16}){\log }^3(2x)}{-x^4-3x^3\log (x)-3x^2{\log }^2(x)-x{\log }^3(x)+(3x^3+6x^2\log (x)+3x{\log }^2(x))\log (2x)+(-3x^2-3x\log (x)){\log }^2(2x)+x{\log }^3(2x)}dx\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2(x^2+{\log }^2(2)+x(e^3-\log (4)))(-e^3-\log \left(\frac{16}{15}\right)(x-\log (2))-(-x+\log (2))\log (-x))}{x(x+\log (x)-\log (2x){)}^3}dx\\ & =2\int \frac{(x^2+{\log }^2(2)+x(e^3-\log (4)))(-e^3-\log \left(\frac{16}{15}\right)(x-\log (2))-(-x+\log (2))\log (-x))}{x(x+\log (x)-\log (2x){)}^3}dx\\ & =2\int (\frac{x(-x\log \left(\frac{16}{15}\right)-e^3(1-\frac{\log \left(\frac{16}{15}\right)\log (2)}{e^3})+x\log (-x)-\log (2)\log (-x))}{(x+\log (x)-\log (2x){)}^3}+\frac{{\log }^2(2)(-x\log \left(\frac{16}{15}\right)-e^3(1-\frac{\log \left(\frac{16}{15}\right)\log (2)}{e^3})+x\log (-x)-\log (2)\log (-x))}{x(x+\log (x)-\log (2x){)}^3}+\frac{e^3(1-\frac{\log (4)}{e^3})(-x\log \left(\frac{16}{15}\right)-e^3(1-\frac{\log \left(\frac{16}{15}\right)\log (2)}{e^3})+x\log (-x)-\log (2)\log (-x))}{(x+\log (x)-\log (2x){)}^3})dx\\ & =2\int \frac{x(-x\log \left(\frac{16}{15}\right)-e^3(1-\frac{\log \left(\frac{16}{15}\right)\log (2)}{e^3})+x\log (-x)-\log (2)\log (-x))}{(x+\log (x)-\log (2x){)}^3}dx+(2{\log }^2(2))\int \frac{-x\log \left(\frac{16}{15}\right)-e^3(1-\frac{\log \left(\frac{16}{15}\right)\log (2)}{e^3})+x\log (-x)-\log (2)\log (-x)}{x(x+\log (x)-\log (2x){)}^3}dx+\left(2(e^3-\log (4))\right)\int \frac{-x\log \left(\frac{16}{15}\right)-e^3(1-\frac{\log \left(\frac{16}{15}\right)\log (2)}{e^3})+x\log (-x)-\log (2)\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx\\ & =2\int (-\frac{x^2\log \left(\frac{16}{15}\right)}{(x+\log (x)-\log (2x){)}^3}-\frac{x(e^3-\log \left(\frac{16}{15}\right)\log (2))}{(x+\log (x)-\log (2x){)}^3}+\frac{x^2\log (-x)}{(x+\log (x)-\log (2x){)}^3}-\frac{x\log (2)\log (-x)}{(x+\log (x)-\log (2x){)}^3})dx+(2{\log }^2(2))\int \frac{-e^3+\log \left(\frac{16}{15}\right)(-x+\log (2))+(x-\log (2))\log (-x)}{x(x+\log (x)-\log (2x){)}^3}dx+\left(2(e^3-\log (4))\right)\int \frac{-e^3+\log \left(\frac{16}{15}\right)(-x+\log (2))+(x-\log (2))\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx\\ & =2\int \frac{x^2\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx-(2\log \left(\frac{16}{15}\right))\int \frac{x^2}{(x+\log (x)-\log (2x){)}^3}dx-(2\log (2))\int \frac{x\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx+(2{\log }^2(2))\int (-\frac{\log \left(\frac{16}{15}\right)}{(x+\log (x)-\log (2x){)}^3}-\frac{e^3(1-\frac{\log \left(\frac{16}{15}\right)\log (2)}{e^3})}{x(x+\log (x)-\log (2x){)}^3}+\frac{\log (-x)}{(x+\log (x)-\log (2x){)}^3}-\frac{\log (2)\log (-x)}{x(x+\log (x)-\log (2x){)}^3})dx-\left(2(e^3-\log \left(\frac{16}{15}\right)\log (2))\right)\int \frac{x}{(x+\log (x)-\log (2x){)}^3}dx+\left(2(e^3-\log (4))\right)\int (-\frac{x\log \left(\frac{16}{15}\right)}{(x+\log (x)-\log (2x){)}^3}-\frac{e^3(1-\frac{\log \left(\frac{16}{15}\right)\log (2)}{e^3})}{(x+\log (x)-\log (2x){)}^3}+\frac{x\log (-x)}{(x+\log (x)-\log (2x){)}^3}-\frac{\log (2)\log (-x)}{(x+\log (x)-\log (2x){)}^3})dx\\ & =2\int \frac{x^2\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx-(2\log \left(\frac{16}{15}\right))\int \frac{x^2}{(x+\log (x)-\log (2x){)}^3}dx-(2\log (2))\int \frac{x\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx+(2{\log }^2(2))\int \frac{\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx-(2\log \left(\frac{16}{15}\right){\log }^2(2))\int \frac{1}{(x+\log (x)-\log (2x){)}^3}dx-(2{\log }^3(2))\int \frac{\log (-x)}{x(x+\log (x)-\log (2x){)}^3}dx-\left(2(e^3-\log \left(\frac{16}{15}\right)\log (2))\right)\int \frac{x}{(x+\log (x)-\log (2x){)}^3}dx-(2{\log }^2(2)(e^3-\log \left(\frac{16}{15}\right)\log (2)))\int \frac{1}{x(x+\log (x)-\log (2x){)}^3}dx+\left(2(e^3-\log (4))\right)\int \frac{x\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx-(2\log \left(\frac{16}{15}\right)(e^3-\log (4)))\int \frac{x}{(x+\log (x)-\log (2x){)}^3}dx-(2\log (2)(e^3-\log (4)))\int \frac{\log (-x)}{(x+\log (x)-\log (2x){)}^3}dx-\left(2(e^3-\log \left(\frac{16}{15}\right)\log (2))(e^3-\log (4))\right)\int \frac{1}{(x+\log (x)-\log (2x){)}^3}dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{(x^2-2(x+\log \left(\frac{16}{15}\right))\log \left(\frac{32}{15}\right)+\log {\left(\frac{32}{15}\right)}^2+2x\log \left(\frac{16}{15}\right)+\log {\left(\frac{16}{15}\right)}^2)\log {(-\frac{15}{16}x)}^2-2(x{e}^3-e^3\log \left(\frac{32}{15}\right)+e^3\log \left(\frac{16}{15}\right))\log (-\frac{15}{16}x)+e^6}{x^2-2(x+\log \left(\frac{16}{15}\right))\log \left(\frac{32}{15}\right)+\log {\left(\frac{32}{15}\right)}^2+2x\log \left(\frac{16}{15}\right)+\log {\left(\frac{16}{15}\right)}^2}\end{array}$$

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 $$\begin{array}{c}\frac{x^2\log {(-\frac{15}{16}x)}^2-2x\log (2)\log {(-\frac{15}{16}x)}^2+\log (2{)}^2\log {(-\frac{15}{16}x)}^2+8x{e}^3\log (2)-8e^3\log (2{)}^2-2x{e}^3\log (-15x)+2e^3\log (2)\log (-15x)+e^6}{x^2-2x\log (2)+\log (2{)}^2}\end{array}$$

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	$\frac{{\mathrm{e}}^6}{{(x-\ln (2))}^2}-\frac{2{\mathrm{e}}^3\ln (x-\ln (2))}{\ln (2)}+\frac{8{\mathrm{e}}^3\ln (2)}{x-\ln (2)}-8\ln (2)\ln (x)+\frac{2\ln (x){\mathrm{e}}^3}{\ln (2)}-\frac{2\ln \left(15\right){\mathrm{e}}^3}{x-\ln (2)}+2\ln (x)\ln \left(15\right)+\frac{2{\mathrm{e}}^3\ln (\ln (2)-x)}{\ln (2)}+\frac{2{\mathrm{e}}^3\ln (-x)x}{\ln (2)(\ln (2)-x)}+\ln {(-x)}^2$	$120$
risch	$\ln (x{)}^2+\frac{4i{\mathrm{e}}^3\ln (x)}{2i\ln (2)-2ix}+\frac{-8x{\mathrm{e}}^3\ln (5)-8x{\mathrm{e}}^3\ln (3)+4{\mathrm{e}}^6+8i\pi {\mathrm{e}}^3\ln (2)-8i\pi {\mathrm{e}}^{3}x+8i\pi \ln (2{)}^2\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)+8i\pi \ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x^2-16\ln (3)\ln (2)\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x-16\ln (2)\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)\ln (5)x+32x{\mathrm{e}}^3\ln (2)-32{\mathrm{e}}^3\ln (2{)}^2+8\ln (3){\mathrm{e}}^3\ln (2)+8{\mathrm{e}}^3\ln (2)\ln (5)+8\ln (3)\ln (2{)}^2\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)+8\ln (3)\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x^2+8\ln (2{)}^2\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)\ln (5)+64\ln (2{)}^2\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x-32\ln (2)\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x^2+8\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)\ln (5)x^2-16i\pi \ln (2)\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x\mathrm{csgn}{\left(ix\right)}^3+16i\pi \ln (2)\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x\mathrm{csgn}{\left(ix\right)}^2-32\ln (2{)}^3\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)+8i\pi {\mathrm{e}}^{3}x\mathrm{csgn}{\left(ix\right)}^2-8i\pi \ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x^2\mathrm{csgn}{\left(ix\right)}^2-8i\pi {\mathrm{e}}^3\ln (2)\mathrm{csgn}{\left(ix\right)}^2-8i\pi {\mathrm{e}}^{3}x\mathrm{csgn}{\left(ix\right)}^3+8i\pi {\mathrm{e}}^3\ln (2)\mathrm{csgn}{\left(ix\right)}^3+8i\pi \ln (2{)}^2\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)\mathrm{csgn}{\left(ix\right)}^3-8i\pi \ln (2{)}^2\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)\mathrm{csgn}{\left(ix\right)}^2+8i\pi \ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x^2\mathrm{csgn}{\left(ix\right)}^3-16i\pi \ln (2)\ln \left((-2\pi \mathrm{csgn}{\left(ix\right)}^2+2\pi \mathrm{csgn}{\left(ix\right)}^3+2\pi -2i\ln (3)+8i\ln (2)-2i\ln (5))x\right)x}{4\ln (2{)}^2-8x\ln (2)+4x^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 $$\begin{array}{c}(\frac{\log (x-\log (2))}{\log (2)}-\frac{\log (x)}{\log (2)}-\frac{1}{x-\log (2)})e^3-\frac{(2x-\log (2))e^3\log (-\frac{15}{16}x)}{x^2-2x\log (2)+\log (2{)}^2}+\frac{(2x-\log (2))e^3}{x^2-2x\log (2)+\log (2{)}^2}-\frac{e^3\log (x-\log (2))}{\log (2)}-\frac{x{e}^3\log (2)-(x^2\log (2)-2x\log (2{)}^2+\log (2{)}^3)\log (x{)}^2-((i\pi +\log (5))\log (2{)}^2+\log (3)\log (2{)}^2-4\log (2{)}^3)e^3+(2(-i\pi -\log (5))\log (2{)}^3-2\log (3)\log (2{)}^3+8\log (2{)}^4+(2(-i\pi -\log (5))\log (2)-2\log (3)\log (2)+8\log (2{)}^2-e^3)x^2-2e^3\log (2{)}^2+2(2(i\pi +\log (5))\log (2{)}^2+2\log (3)\log (2{)}^2-8\log (2{)}^3+e^3\log (2))x)\log (x)}{x^2\log (2)-2x\log (2{)}^2+\log (2{)}^3}+\frac{e^6}{x^2-2x\log (2)+\log (2{)}^2}\end{array}$$

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 $$\begin{array}{c}2\ln (x)(\ln (-\frac{15x}{16})-\ln (x))+{\ln (x)}^2+\frac{{\mathrm{e}}^6-2x{\mathrm{e}}^3(\ln (-\frac{15x}{16})-\ln (x))+2{\mathrm{e}}^3(\ln \left(2x\right)-\ln (x))(\ln (-\frac{15x}{16})-\ln (x))}{{(\ln \left(2x\right)-\ln (x))}^2-2x(\ln \left(2x\right)-\ln (x))+x^2}-\frac{2{\mathrm{e}}^3\ln (x)}{x-\ln \left(2x\right)+\ln (x)}\end{array}$$

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 $$\begin{array}{c}\log {(x)}^2+2(-4\log (2)+\log \left(15\right)+i\pi )\log (x)+\frac{x(-8e^3\log (2)+2e^3\log \left(15\right)+2i\pi e^3)-e^6-2e^3\log (2)\log \left(15\right)+8e^3\log {(2)}^2-2i\pi e^3\log (2)}{-x^2+2x\log (2)-\log {(2)}^2}-\frac{2e^3\log (x)}{x-\log (2)}\end{array}$$

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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