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

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 {gather*} \int \frac {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+\left (x-2 x^2+x^3\right ) \log (x)} \, dx \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {2-2 x+(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 \left (\frac {2}{(-1+x)^2 (25+x+x \log (x))}-\frac {2 x}{(-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}\right ) \, 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 \left (\frac {1}{(-1+x)^2 (25+x+x \log (x))}+\frac {1}{(-1+x) (25+x+x \log (x))}\right ) \, dx-\int \left (\frac {1}{(-1+x) x}+\frac {-25-x}{(-1+x) x (25+x+x \log (x))}\right ) \, dx+\int \frac {\log (25+x+x \log (x))}{(-1+x)^2} \, dx\\ &=-\left (2 \int \frac {1}{(-1+x) (25+x+x \log (x))} \, dx\right )-\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\\ &=-\left (2 \int \frac {1}{(-1+x) (25+x+x \log (x))} \, dx\right )-\int \frac {1}{-1+x} \, dx+\int \frac {1}{x} \, dx-\int \left (-\frac {26}{(-1+x) (25+x+x \log (x))}+\frac {25}{x (25+x+x \log (x))}\right ) \, 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 {aligned} \end {gather*}

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

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 {gather*} -\frac {\log \left (x \log \relax (x) + x + 25\right )}{x - 1} \end {gather*}

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 {gather*} -\frac {\log \left (x \log \relax (x) + x + 25\right )}{x - 1} \end {gather*}

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	\(-\frac {\ln \left (x \ln \relax (x )+x +25\right )}{x -1}\)	\(16\)

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 {gather*} -\frac {\log \left (x \log \relax (x) + x + 25\right )}{x - 1} \end {gather*}

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 {gather*} -\frac {\ln \left (x+x\,\ln \relax (x)+25\right )}{x-1} \end {gather*}

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 {gather*} - \frac {\log {\left (x \log {\relax (x )} + x + 25 \right )}}{x - 1} \end {gather*}

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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