3.33.58 $\int \frac{(5-2x)\log (x)+(-5+x)\log (-5x+x^2)+(5x-x^2+(-5x+x^2)\log (x)){\log }^2(-5x+x^2)+(-5x+x^2+(-5x+x^2)\log (x))\log (x^2){\log }^2(-5x+x^2)}{(-5x+x^2){\log }^2(-5x+x^2)}dx$

Optimal. Leaf size=21 $$\log (x)(-x+\frac{1}{\log ((-5+x)x)}+x\log \left(x^2\right))$$

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Rubi [F]  time = 1.92, 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 = {} $$\begin{array}{c}\int \frac{(5-2x)\log (x)+(-5+x)\log (-5x+x^2)+(5x-x^2+(-5x+x^2)\log (x)){\log }^2(-5x+x^2)+(-5x+x^2+(-5x+x^2)\log (x))\log \left(x^2\right){\log }^2(-5x+x^2)}{(-5x+x^2){\log }^2(-5x+x^2)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[((5 - 2*x)*Log[x] + (-5 + x)*Log[-5*x + x^2] + (5*x - x^2 + (-5*x + x^2)*Log[x])*Log[-5*x + x^2]^2 + (-5*x
+ x^2 + (-5*x + x^2)*Log[x])*Log[x^2]*Log[-5*x + x^2]^2)/((-5*x + x^2)*Log[-5*x + x^2]^2),x]

[Out]

-(x*Log[x]) + x*Log[x]*Log[x^2] - Defer[Int][Log[x]/((-5 + x)*Log[(-5 + x)*x]^2), x] - Defer[Int][Log[x]/(x*Lo
g[(-5 + x)*x]^2), x] + Defer[Int][1/(x*Log[(-5 + x)*x]), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{(5-2x)\log (x)+(-5+x)\log (-5x+x^2)+(5x-x^2+(-5x+x^2)\log (x)){\log }^2(-5x+x^2)+(-5x+x^2+(-5x+x^2)\log (x))\log \left(x^2\right){\log }^2(-5x+x^2)}{(-5+x)x{\log }^2(-5x+x^2)}dx\\ & =\int \frac{-((-5+x)\log ((-5+x)x)(1+x\log ((-5+x)x)(-1+\log \left(x^2\right))))-\log (x)(5-2x+(-5+x)x{\log }^2((-5+x)x)(1+\log \left(x^2\right)))}{(5-x)x{\log }^2((-5+x)x)}dx\\ & =\int (\frac{5\log (x)-2x\log (x)-5\log ((-5+x)x)+x\log ((-5+x)x)+5x{\log }^2((-5+x)x)-x^2{\log }^2((-5+x)x)-5x\log (x){\log }^2((-5+x)x)+x^2\log (x){\log }^2((-5+x)x)}{(-5+x)x{\log }^2((-5+x)x)}+(1+\log (x))\log \left(x^2\right))dx\\ & =\int \frac{5\log (x)-2x\log (x)-5\log ((-5+x)x)+x\log ((-5+x)x)+5x{\log }^2((-5+x)x)-x^2{\log }^2((-5+x)x)-5x\log (x){\log }^2((-5+x)x)+x^2\log (x){\log }^2((-5+x)x)}{(-5+x)x{\log }^2((-5+x)x)}dx+\int (1+\log (x))\log \left(x^2\right)dx\\ & =x\log (x)\log \left(x^2\right)-2\int \log (x)dx+\int \frac{(-5+x)\log ((-5+x)x)(-1+x\log ((-5+x)x))-\log (x)(5-2x+(-5+x)x{\log }^2((-5+x)x))}{(5-x)x{\log }^2((-5+x)x)}dx\\ & =2x-2x\log (x)+x\log (x)\log \left(x^2\right)+\int (-1+\log (x)-\frac{(-5+2x)\log (x)}{(-5+x)x{\log }^2((-5+x)x)}+\frac{1}{x\log ((-5+x)x)})dx\\ & =x-2x\log (x)+x\log (x)\log \left(x^2\right)+\int \log (x)dx-\int \frac{(-5+2x)\log (x)}{(-5+x)x{\log }^2((-5+x)x)}dx+\int \frac{1}{x\log ((-5+x)x)}dx\\ & =-x\log (x)+x\log (x)\log \left(x^2\right)-\int (\frac{\log (x)}{(-5+x){\log }^2((-5+x)x)}+\frac{\log (x)}{x{\log }^2((-5+x)x)})dx+\int \frac{1}{x\log ((-5+x)x)}dx\\ & =-x\log (x)+x\log (x)\log \left(x^2\right)-\int \frac{\log (x)}{(-5+x){\log }^2((-5+x)x)}dx-\int \frac{\log (x)}{x{\log }^2((-5+x)x)}dx+\int \frac{1}{x\log ((-5+x)x)}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.17, size = 20, normalized size = 0.95 $$\begin{array}{c}\log (x)(\frac{1}{\log ((-5+x)x)}+x(-1+\log \left(x^2\right)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[((5 - 2*x)*Log[x] + (-5 + x)*Log[-5*x + x^2] + (5*x - x^2 + (-5*x + x^2)*Log[x])*Log[-5*x + x^2]^2 +
(-5*x + x^2 + (-5*x + x^2)*Log[x])*Log[x^2]*Log[-5*x + x^2]^2)/((-5*x + x^2)*Log[-5*x + x^2]^2),x]

[Out]

Log[x]*(Log[(-5 + x)*x]^(-1) + x*(-1 + Log[x^2]))

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fricas [A]  time = 0.56, size = 36, normalized size = 1.71 $$\begin{array}{c}\frac{(2x\log (x{)}^2-x\log (x))\log (x^2-5x)+\log (x)}{\log (x^2-5x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-5*x)*log(x)+x^2-5*x)*log(x^2-5*x)^2*log(x^2)+((x^2-5*x)*log(x)-x^2+5*x)*log(x^2-5*x)^2+(x-5)*
log(x^2-5*x)+(-2*x+5)*log(x))/(x^2-5*x)/log(x^2-5*x)^2,x, algorithm="fricas")

[Out]

((2*x*log(x)^2 - x*log(x))*log(x^2 - 5*x) + log(x))/log(x^2 - 5*x)

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giac [A]  time = 0.35, size = 25, normalized size = 1.19 $$\begin{array}{c}2x\log (x{)}^2-x\log (x)+\frac{\log (x)}{\log (x-5)+\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-5*x)*log(x)+x^2-5*x)*log(x^2-5*x)^2*log(x^2)+((x^2-5*x)*log(x)-x^2+5*x)*log(x^2-5*x)^2+(x-5)*
log(x^2-5*x)+(-2*x+5)*log(x))/(x^2-5*x)/log(x^2-5*x)^2,x, algorithm="giac")

[Out]

2*x*log(x)^2 - x*log(x) + log(x)/(log(x - 5) + log(x))

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maple [C]  time = 0.72, size = 167, normalized size = 7.95

method	result	size
risch	$2x\ln (x{)}^2-\frac{i\pi x\mathrm{csgn}{\left(ix\right)}^2\mathrm{csgn}\left(i{x}^2\right)\ln (x)}{2}+i\pi x\mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^2\right)}^2\ln (x)-\frac{i\pi x\mathrm{csgn}{\left(i{x}^2\right)}^3\ln (x)}{2}-x\ln (x)+\frac{2\ln (x)}{2\ln (x)+2\ln (x-5)-i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i(x-5)\right)\mathrm{csgn}\left(ix(x-5)\right)+i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(ix(x-5)\right)}^2+i\pi \mathrm{csgn}\left(i(x-5)\right)\mathrm{csgn}{\left(ix(x-5)\right)}^2-i\pi \mathrm{csgn}{\left(ix(x-5)\right)}^3}$	$167$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-5*x)*ln(x)+x^2-5*x)*ln(x^2-5*x)^2*ln(x^2)+((x^2-5*x)*ln(x)-x^2+5*x)*ln(x^2-5*x)^2+(x-5)*ln(x^2-5*x)
+(-2*x+5)*ln(x))/(x^2-5*x)/ln(x^2-5*x)^2,x,method=_RETURNVERBOSE)

[Out]

2*x*ln(x)^2-1/2*I*Pi*x*csgn(I*x)^2*csgn(I*x^2)*ln(x)+I*Pi*x*csgn(I*x)*csgn(I*x^2)^2*ln(x)-1/2*I*Pi*x*csgn(I*x^
2)^3*ln(x)-x*ln(x)+2*ln(x)/(2*ln(x)+2*ln(x-5)-I*Pi*csgn(I*x)*csgn(I*(x-5))*csgn(I*x*(x-5))+I*Pi*csgn(I*x)*csgn
(I*x*(x-5))^2+I*Pi*csgn(I*(x-5))*csgn(I*x*(x-5))^2-I*Pi*csgn(I*x*(x-5))^3)

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maxima [B]  time = 0.47, size = 45, normalized size = 2.14 $$\begin{array}{c}\frac{2x\log (x{)}^3-x\log (x{)}^2+(2x\log (x{)}^2-x\log (x))\log (x-5)+\log (x)}{\log (x-5)+\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-5*x)*log(x)+x^2-5*x)*log(x^2-5*x)^2*log(x^2)+((x^2-5*x)*log(x)-x^2+5*x)*log(x^2-5*x)^2+(x-5)*
log(x^2-5*x)+(-2*x+5)*log(x))/(x^2-5*x)/log(x^2-5*x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.18, size = 55, normalized size = 2.62 $$\begin{array}{c}\frac{\ln (x)}{\ln (x^2-5x)}-\frac{5}{4(x-\frac{5}{2})}-\frac{x}{2x-5}+\frac{5}{2x-5}-x\ln (x)+x\ln \left(x^2\right)\ln (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(2*x - 5) - log(x^2 - 5*x)*(x - 5) + log(x^2 - 5*x)^2*(log(x)*(5*x - x^2) - 5*x + x^2) + log(x^2)*
log(x^2 - 5*x)^2*(5*x + log(x)*(5*x - x^2) - x^2))/(log(x^2 - 5*x)^2*(5*x - x^2)),x)

[Out]

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

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sympy [A]  time = 0.33, size = 24, normalized size = 1.14 $$\begin{array}{c}2x\log {(x)}^2-x\log (x)+\frac{\log (x)}{\log (x^2-5x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-5*x)*ln(x)+x**2-5*x)*ln(x**2-5*x)**2*ln(x**2)+((x**2-5*x)*ln(x)-x**2+5*x)*ln(x**2-5*x)**2+(x
-5)*ln(x**2-5*x)+(-2*x+5)*ln(x))/(x**2-5*x)/ln(x**2-5*x)**2,x)

[Out]

2*x*log(x)**2 - x*log(x) + log(x)/log(x**2 - 5*x)

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