3.89.59 $\int \frac{2-2x+(1-x)\log (x)+(25+x+x\log (x))\log (25+x+x\log (x))}{25-49x+23x^2+x^3+(x-2x^2+x^3)\log (x)}dx$

Optimal. Leaf size=18 $$-5+\frac{\log (25+x+x\log (x))}{1-x}$$

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Rubi [F]  time = 0.99, 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 = {} $$\begin{array}{c}\int \frac{2-2x+(1-x)\log (x)+(25+x+x\log (x))\log (25+x+x\log (x))}{25-49x+23x^2+x^3+(x-2x^2+x^3)\log (x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(2 - 2*x + (1 - x)*Log[x] + (25 + x + x*Log[x])*Log[25 + x + x*Log[x]])/(25 - 49*x + 23*x^2 + x^3 + (x - 2
*x^2 + x^3)*Log[x]),x]

[Out]

-Log[1 - x] + Log[x] + 24*Defer[Int][1/((-1 + x)*(25 + x + x*Log[x])), x] - 25*Defer[Int][1/(x*(25 + x + x*Log
[x])), x] + Defer[Int][Log[25 + x + x*Log[x]]/(-1 + x)^2, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2-2x+(1-x)\log (x)+(25+x+x\log (x))\log (25+x+x\log (x))}{(1-x{)}^2(25+x+x\log (x))}dx\\ & =\int (\frac{2}{(-1+x{)}^2(25+x+x\log (x))}-\frac{2x}{(-1+x{)}^2(25+x+x\log (x))}-\frac{\log (x)}{(-1+x)(25+x+x\log (x))}+\frac{\log (25+x+x\log (x))}{(-1+x{)}^2})dx\\ & =2\int \frac{1}{(-1+x{)}^2(25+x+x\log (x))}dx-2\int \frac{x}{(-1+x{)}^2(25+x+x\log (x))}dx-\int \frac{\log (x)}{(-1+x)(25+x+x\log (x))}dx+\int \frac{\log (25+x+x\log (x))}{(-1+x{)}^2}dx\\ & =2\int \frac{1}{(-1+x{)}^2(25+x+x\log (x))}dx-2\int (\frac{1}{(-1+x{)}^2(25+x+x\log (x))}+\frac{1}{(-1+x)(25+x+x\log (x))})dx-\int (\frac{1}{(-1+x)x}+\frac{-25-x}{(-1+x)x(25+x+x\log (x))})dx+\int \frac{\log (25+x+x\log (x))}{(-1+x{)}^2}dx\\ & =-(2\int \frac{1}{(-1+x)(25+x+x\log (x))}dx)-\int \frac{1}{(-1+x)x}dx-\int \frac{-25-x}{(-1+x)x(25+x+x\log (x))}dx+\int \frac{\log (25+x+x\log (x))}{(-1+x{)}^2}dx\\ & =-(2\int \frac{1}{(-1+x)(25+x+x\log (x))}dx)-\int \frac{1}{-1+x}dx+\int \frac{1}{x}dx-\int (-\frac{26}{(-1+x)(25+x+x\log (x))}+\frac{25}{x(25+x+x\log (x))})dx+\int \frac{\log (25+x+x\log (x))}{(-1+x{)}^2}dx\\ & =-\log (1-x)+\log (x)-2\int \frac{1}{(-1+x)(25+x+x\log (x))}dx-25\int \frac{1}{x(25+x+x\log (x))}dx+26\int \frac{1}{(-1+x)(25+x+x\log (x))}dx+\int \frac{\log (25+x+x\log (x))}{(-1+x{)}^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.18, size = 15, normalized size = 0.83 $$\begin{array}{c}-\frac{\log (25+x+x\log (x))}{-1+x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(2 - 2*x + (1 - x)*Log[x] + (25 + x + x*Log[x])*Log[25 + x + x*Log[x]])/(25 - 49*x + 23*x^2 + x^3 +
(x - 2*x^2 + x^3)*Log[x]),x]

[Out]

-(Log[25 + x + x*Log[x]]/(-1 + x))

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fricas [A]  time = 0.65, size = 15, normalized size = 0.83 $$\begin{array}{c}-\frac{\log (x\log (x)+x+25)}{x-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)+x+25)*log(x*log(x)+x+25)+(-x+1)*log(x)-2*x+2)/((x^3-2*x^2+x)*log(x)+x^3+23*x^2-49*x+25),x
, algorithm="fricas")

[Out]

-log(x*log(x) + x + 25)/(x - 1)

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giac [A]  time = 0.19, size = 15, normalized size = 0.83 $$\begin{array}{c}-\frac{\log (x\log (x)+x+25)}{x-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)+x+25)*log(x*log(x)+x+25)+(-x+1)*log(x)-2*x+2)/((x^3-2*x^2+x)*log(x)+x^3+23*x^2-49*x+25),x
, algorithm="giac")

[Out]

-log(x*log(x) + x + 25)/(x - 1)

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maple [A]  time = 0.07, size = 16, normalized size = 0.89

method	result	size
risch	$-\frac{\ln (x\ln (x)+x+25)}{x-1}$	$16$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)+x+25)*ln(x*ln(x)+x+25)+(1-x)*ln(x)-2*x+2)/((x^3-2*x^2+x)*ln(x)+x^3+23*x^2-49*x+25),x,method=_RET
URNVERBOSE)

[Out]

-1/(x-1)*ln(x*ln(x)+x+25)

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maxima [A]  time = 0.41, size = 15, normalized size = 0.83 $$\begin{array}{c}-\frac{\log (x\log (x)+x+25)}{x-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)+x+25)*log(x*log(x)+x+25)+(-x+1)*log(x)-2*x+2)/((x^3-2*x^2+x)*log(x)+x^3+23*x^2-49*x+25),x
, algorithm="maxima")

[Out]

-log(x*log(x) + x + 25)/(x - 1)

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mupad [B]  time = 5.39, size = 15, normalized size = 0.83 $$\begin{array}{c}-\frac{\ln (x+x\ln (x)+25)}{x-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + log(x)*(x - 1) - log(x + x*log(x) + 25)*(x + x*log(x) + 25) - 2)/(log(x)*(x - 2*x^2 + x^3) - 49*x
+ 23*x^2 + x^3 + 25),x)

[Out]

-log(x + x*log(x) + 25)/(x - 1)

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sympy [A]  time = 0.37, size = 14, normalized size = 0.78 $$\begin{array}{c}-\frac{\log (x\log (x)+x+25)}{x-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)+x+25)*ln(x*ln(x)+x+25)+(-x+1)*ln(x)-2*x+2)/((x**3-2*x**2+x)*ln(x)+x**3+23*x**2-49*x+25),x)

[Out]

-log(x*log(x) + x + 25)/(x - 1)

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