3.17.50 $\int \frac{-2\log (x)-\log (\frac{5x}{4})+(-x+x\log (x)){\log }^3(\frac{5x}{4})}{x{\log }^2(x){\log }^3(\frac{5x}{4})}dx$

Optimal. Leaf size=17 $$2+\frac{x+\frac{1}{{\log }^2\left(\frac{5x}{4}\right)}}{\log (x)}$$

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Rubi [F]  time = 0.42, 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 = {} $$\begin{array}{c}\int \frac{-2\log (x)-\log \left(\frac{5x}{4}\right)+(-x+x\log (x)){\log }^3\left(\frac{5x}{4}\right)}{x{\log }^2(x){\log }^3\left(\frac{5x}{4}\right)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-2*Log[x] - Log[(5*x)/4] + (-x + x*Log[x])*Log[(5*x)/4]^3)/(x*Log[x]^2*Log[(5*x)/4]^3),x]

[Out]

x/Log[x] - 2*Defer[Int][1/(x*Log[x]*Log[(5*x)/4]^3), x] - Defer[Int][1/(x*Log[x]^2*Log[(5*x)/4]^2), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{-1+\log (x)}{{\log }^2(x)}-\frac{2}{x\log (x){\log }^3\left(\frac{5x}{4}\right)}-\frac{1}{x{\log }^2(x){\log }^2\left(\frac{5x}{4}\right)})dx\\ & =-(2\int \frac{1}{x\log (x){\log }^3\left(\frac{5x}{4}\right)}dx)+\int \frac{-1+\log (x)}{{\log }^2(x)}dx-\int \frac{1}{x{\log }^2(x){\log }^2\left(\frac{5x}{4}\right)}dx\\ & =-(2\int \frac{1}{x\log (x){\log }^3\left(\frac{5x}{4}\right)}dx)+\int (-\frac{1}{{\log }^2(x)}+\frac{1}{\log (x)})dx-\int \frac{1}{x{\log }^2(x){\log }^2\left(\frac{5x}{4}\right)}dx\\ & =-(2\int \frac{1}{x\log (x){\log }^3\left(\frac{5x}{4}\right)}dx)-\int \frac{1}{{\log }^2(x)}dx+\int \frac{1}{\log (x)}dx-\int \frac{1}{x{\log }^2(x){\log }^2\left(\frac{5x}{4}\right)}dx\\ & =\frac{x}{\log (x)}+\text{li}(x)-2\int \frac{1}{x\log (x){\log }^3\left(\frac{5x}{4}\right)}dx-\int \frac{1}{\log (x)}dx-\int \frac{1}{x{\log }^2(x){\log }^2\left(\frac{5x}{4}\right)}dx\\ & =\frac{x}{\log (x)}-2\int \frac{1}{x\log (x){\log }^3\left(\frac{5x}{4}\right)}dx-\int \frac{1}{x{\log }^2(x){\log }^2\left(\frac{5x}{4}\right)}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.05, size = 15, normalized size = 0.88 $$\begin{array}{c}\frac{x+\frac{1}{{\log }^2\left(\frac{5x}{4}\right)}}{\log (x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-2*Log[x] - Log[(5*x)/4] + (-x + x*Log[x])*Log[(5*x)/4]^3)/(x*Log[x]^2*Log[(5*x)/4]^3),x]

[Out]

(x + Log[(5*x)/4]^(-2))/Log[x]

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fricas [B]  time = 0.87, size = 44, normalized size = 2.59 $$\begin{array}{c}\frac{x\log {\left(\frac{5}{4}\right)}^2+2x\log \left(\frac{5}{4}\right)\log (x)+x\log (x{)}^2+1}{\log {\left(\frac{5}{4}\right)}^2\log (x)+2\log \left(\frac{5}{4}\right)\log (x{)}^2+\log (x{)}^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x)*log(5/4*x)^3-log(5/4*x)-2*log(x))/x/log(x)^2/log(5/4*x)^3,x, algorithm="fricas")

[Out]

(x*log(5/4)^2 + 2*x*log(5/4)*log(x) + x*log(x)^2 + 1)/(log(5/4)^2*log(x) + 2*log(5/4)*log(x)^2 + log(x)^3)

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giac [B]  time = 0.26, size = 167, normalized size = 9.82 $$\begin{array}{c}\frac{x\log (5{)}^2-4x\log (5)\log (2)+4x\log (2{)}^2+1}{\log (5{)}^2\log (x)-4\log (5)\log (2)\log (x)+4\log (2{)}^2\log (x)}-\frac{2\log (5)-4\log (2)+\log (x)}{\log (5{)}^4-8\log (5{)}^3\log (2)+24\log (5{)}^2\log (2{)}^2-32\log (5)\log (2{)}^3+16\log (2{)}^4+2\log (5{)}^3\log (x)-12\log (5{)}^2\log (2)\log (x)+24\log (5)\log (2{)}^2\log (x)-16\log (2{)}^3\log (x)+\log (5{)}^2\log (x{)}^2-4\log (5)\log (2)\log (x{)}^2+4\log (2{)}^2\log (x{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x)*log(5/4*x)^3-log(5/4*x)-2*log(x))/x/log(x)^2/log(5/4*x)^3,x, algorithm="giac")

[Out]

(x*log(5)^2 - 4*x*log(5)*log(2) + 4*x*log(2)^2 + 1)/(log(5)^2*log(x) - 4*log(5)*log(2)*log(x) + 4*log(2)^2*log
(x)) - (2*log(5) - 4*log(2) + log(x))/(log(5)^4 - 8*log(5)^3*log(2) + 24*log(5)^2*log(2)^2 - 32*log(5)*log(2)^
3 + 16*log(2)^4 + 2*log(5)^3*log(x) - 12*log(5)^2*log(2)*log(x) + 24*log(5)*log(2)^2*log(x) - 16*log(2)^3*log(
x) + log(5)^2*log(x)^2 - 4*log(5)*log(2)*log(x)^2 + 4*log(2)^2*log(x)^2)

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maple [B]  time = 0.22, size = 65, normalized size = 3.82

method	result	size
risch	$\frac{4-16x\ln (2)\ln (x)-16x\ln (2)\ln (5)+16x\ln (2{)}^2+4x\ln (x{)}^2+4x\ln (5{)}^2+8x\ln (5)\ln (x)}{{(2\ln (5)-4\ln (2)+2\ln (x))}^2\ln (x)}$	$65$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)-x)*ln(5/4*x)^3-ln(5/4*x)-2*ln(x))/x/ln(x)^2/ln(5/4*x)^3,x,method=_RETURNVERBOSE)

[Out]

(4-16*x*ln(2)*ln(x)-16*x*ln(2)*ln(5)+16*x*ln(2)^2+4*x*ln(x)^2+4*x*ln(5)^2+8*x*ln(5)*ln(x))/(2*ln(5)-4*ln(2)+2*
ln(x))^2/ln(x)

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maxima [C]  time = 0.64, size = 186, normalized size = 10.94 $$\begin{array}{c}-\frac{3\log (5)-6\log (2)+2\log (x)}{\log (5{)}^4-8\log (5{)}^3\log (2)+24\log (5{)}^2\log (2{)}^2-32\log (5)\log (2{)}^3+16\log (2{)}^4+(\log (5{)}^2-4\log (5)\log (2)+4\log (2{)}^2)\log (x{)}^2+2(\log (5{)}^3-6\log (5{)}^2\log (2)+12\log (5)\log (2{)}^2-8\log (2{)}^3)\log (x)}+\frac{\log (5)-2\log (2)+2\log (x)}{(\log (5{)}^2-4\log (5)\log (2)+4\log (2{)}^2)\log (x{)}^2+(\log (5{)}^3-6\log (5{)}^2\log (2)+12\log (5)\log (2{)}^2-8\log (2{)}^3)\log (x)}+\mathrm{E}\mathrm{i}(\log (x))-\mathrm{\Gamma }(-1,-\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x)*log(5/4*x)^3-log(5/4*x)-2*log(x))/x/log(x)^2/log(5/4*x)^3,x, algorithm="maxima")

[Out]

-(3*log(5) - 6*log(2) + 2*log(x))/(log(5)^4 - 8*log(5)^3*log(2) + 24*log(5)^2*log(2)^2 - 32*log(5)*log(2)^3 +
16*log(2)^4 + (log(5)^2 - 4*log(5)*log(2) + 4*log(2)^2)*log(x)^2 + 2*(log(5)^3 - 6*log(5)^2*log(2) + 12*log(5)
*log(2)^2 - 8*log(2)^3)*log(x)) + (log(5) - 2*log(2) + 2*log(x))/((log(5)^2 - 4*log(5)*log(2) + 4*log(2)^2)*lo
g(x)^2 + (log(5)^3 - 6*log(5)^2*log(2) + 12*log(5)*log(2)^2 - 8*log(2)^3)*log(x)) + Ei(log(x)) - gamma(-1, -lo
g(x))

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mupad [B]  time = 1.18, size = 21, normalized size = 1.24 $$\begin{array}{c}\frac{x{\ln \left(\frac{5x}{4}\right)}^2+1}{{\ln \left(\frac{5x}{4}\right)}^2\ln (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((5*x)/4) + 2*log(x) + log((5*x)/4)^3*(x - x*log(x)))/(x*log((5*x)/4)^3*log(x)^2),x)

[Out]

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

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sympy [B]  time = 0.43, size = 94, normalized size = 5.53 $$\begin{array}{c}\frac{x\log {(x)}^2-4x\log (2)\log (5)+4x\log {(2)}^2+x\log {(5)}^2+(-4x\log (2)+2x\log (5))\log (x)+1}{\log {(x)}^3+(-4\log (2)+2\log (5))\log {(x)}^2+(-4\log (2)\log (5)+4\log {(2)}^2+\log {(5)}^2)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)-x)*ln(5/4*x)**3-ln(5/4*x)-2*ln(x))/x/ln(x)**2/ln(5/4*x)**3,x)

[Out]

(x*log(x)**2 - 4*x*log(2)*log(5) + 4*x*log(2)**2 + x*log(5)**2 + (-4*x*log(2) + 2*x*log(5))*log(x) + 1)/(log(x
)**3 + (-4*log(2) + 2*log(5))*log(x)**2 + (-4*log(2)*log(5) + 4*log(2)**2 + log(5)**2)*log(x))

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