3.93.11 \(\int \frac {-96 x-x^2+x^3+(5-6 x+11 x^2-2 x^3) \log (-5+x)+x \log (x)}{(-5 x+x^2) \log ^2(-5+x)} \, dx\)

Optimal. Leaf size=19 \[ \frac {96+x-x^2-\log (x)}{\log (-5+x)} \]

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Rubi [F]  time = 0.90, 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 {-96 x-x^2+x^3+\left (5-6 x+11 x^2-2 x^3\right ) \log (-5+x)+x \log (x)}{\left (-5 x+x^2\right ) \log ^2(-5+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-96*x - x^2 + x^3 + (5 - 6*x + 11*x^2 - 2*x^3)*Log[-5 + x] + x*Log[x])/((-5*x + x^2)*Log[-5 + x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-96 x-x^2+x^3+\left (5-6 x+11 x^2-2 x^3\right ) \log (-5+x)+x \log (x)}{(-5+x) x \log ^2(-5+x)} \, dx\\ &=\int \left (\frac {-96 x-x^2+x^3+5 \log (-5+x)-6 x \log (-5+x)+11 x^2 \log (-5+x)-2 x^3 \log (-5+x)}{(-5+x) x \log ^2(-5+x)}+\frac {\log (x)}{(-5+x) \log ^2(-5+x)}\right ) \, dx\\ &=\int \frac {-96 x-x^2+x^3+5 \log (-5+x)-6 x \log (-5+x)+11 x^2 \log (-5+x)-2 x^3 \log (-5+x)}{(-5+x) x \log ^2(-5+x)} \, dx+\int \frac {\log (x)}{(-5+x) \log ^2(-5+x)} \, dx\\ &=\int \left (\frac {-96-x+x^2}{(-5+x) \log ^2(-5+x)}+\frac {-1+x-2 x^2}{x \log (-5+x)}\right ) \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=\int \frac {-96-x+x^2}{(-5+x) \log ^2(-5+x)} \, dx+\int \frac {-1+x-2 x^2}{x \log (-5+x)} \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=\int \left (\frac {4}{\log ^2(-5+x)}-\frac {76}{(-5+x) \log ^2(-5+x)}+\frac {x}{\log ^2(-5+x)}\right ) \, dx+\int \left (\frac {1}{\log (-5+x)}-\frac {1}{x \log (-5+x)}-\frac {2 x}{\log (-5+x)}\right ) \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=-\left (2 \int \frac {x}{\log (-5+x)} \, dx\right )+4 \int \frac {1}{\log ^2(-5+x)} \, dx-76 \int \frac {1}{(-5+x) \log ^2(-5+x)} \, dx+\int \frac {x}{\log ^2(-5+x)} \, dx+\int \frac {1}{\log (-5+x)} \, dx-\int \frac {1}{x \log (-5+x)} \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=\frac {(5-x) x}{\log (-5+x)}-2 \int \left (\frac {5}{\log (-5+x)}+\frac {-5+x}{\log (-5+x)}\right ) \, dx+2 \int \frac {x}{\log (-5+x)} \, dx+4 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-5+x\right )-5 \int \frac {1}{\log (-5+x)} \, dx-76 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-5+x\right )-\int \frac {1}{x \log (-5+x)} \, dx+\operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-5+x\right )+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=\frac {4 (5-x)}{\log (-5+x)}+\frac {(5-x) x}{\log (-5+x)}+\text {li}(-5+x)+2 \int \left (\frac {5}{\log (-5+x)}+\frac {-5+x}{\log (-5+x)}\right ) \, dx-2 \int \frac {-5+x}{\log (-5+x)} \, dx+4 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-5+x\right )-5 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-5+x\right )-10 \int \frac {1}{\log (-5+x)} \, dx-76 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-5+x)\right )-\int \frac {1}{x \log (-5+x)} \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=\frac {76}{\log (-5+x)}+\frac {4 (5-x)}{\log (-5+x)}+\frac {(5-x) x}{\log (-5+x)}+2 \int \frac {-5+x}{\log (-5+x)} \, dx-2 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-5+x\right )+10 \int \frac {1}{\log (-5+x)} \, dx-10 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-5+x\right )-\int \frac {1}{x \log (-5+x)} \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=\frac {76}{\log (-5+x)}+\frac {4 (5-x)}{\log (-5+x)}+\frac {(5-x) x}{\log (-5+x)}-10 \text {li}(-5+x)-2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-5+x)\right )+2 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-5+x\right )+10 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-5+x\right )-\int \frac {1}{x \log (-5+x)} \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=-2 \text {Ei}(2 \log (-5+x))+\frac {76}{\log (-5+x)}+\frac {4 (5-x)}{\log (-5+x)}+\frac {(5-x) x}{\log (-5+x)}+2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-5+x)\right )-\int \frac {1}{x \log (-5+x)} \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=\frac {76}{\log (-5+x)}+\frac {4 (5-x)}{\log (-5+x)}+\frac {(5-x) x}{\log (-5+x)}-\int \frac {1}{x \log (-5+x)} \, dx+\operatorname {Subst}\left (\int \frac {\log (5+x)}{x \log ^2(x)} \, dx,x,-5+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 19, normalized size = 1.00 \begin {gather*} \frac {96+x-x^2-\log (x)}{\log (-5+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-96*x - x^2 + x^3 + (5 - 6*x + 11*x^2 - 2*x^3)*Log[-5 + x] + x*Log[x])/((-5*x + x^2)*Log[-5 + x]^2)
,x]

[Out]

(96 + x - x^2 - Log[x])/Log[-5 + x]

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fricas [A]  time = 0.51, size = 18, normalized size = 0.95 \begin {gather*} -\frac {x^{2} - x + \log \relax (x) - 96}{\log \left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+(-2*x^3+11*x^2-6*x+5)*log(x-5)+x^3-x^2-96*x)/(x^2-5*x)/log(x-5)^2,x, algorithm="fricas")

[Out]

-(x^2 - x + log(x) - 96)/log(x - 5)

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giac [A]  time = 0.22, size = 18, normalized size = 0.95 \begin {gather*} -\frac {x^{2} - x + \log \relax (x) - 96}{\log \left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+(-2*x^3+11*x^2-6*x+5)*log(x-5)+x^3-x^2-96*x)/(x^2-5*x)/log(x-5)^2,x, algorithm="giac")

[Out]

-(x^2 - x + log(x) - 96)/log(x - 5)

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maple [A]  time = 0.53, size = 19, normalized size = 1.00

method	result	size
risch	\(-\frac {x^{2}-x +\ln \relax (x )-96}{\ln \left (x -5\right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(x)+(-2*x^3+11*x^2-6*x+5)*ln(x-5)+x^3-x^2-96*x)/(x^2-5*x)/ln(x-5)^2,x,method=_RETURNVERBOSE)

[Out]

-(x^2-x+ln(x)-96)/ln(x-5)

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maxima [A]  time = 0.40, size = 18, normalized size = 0.95 \begin {gather*} -\frac {x^{2} - x + \log \relax (x) - 96}{\log \left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+(-2*x^3+11*x^2-6*x+5)*log(x-5)+x^3-x^2-96*x)/(x^2-5*x)/log(x-5)^2,x, algorithm="maxima")

[Out]

-(x^2 - x + log(x) - 96)/log(x - 5)

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mupad [B]  time = 7.41, size = 38, normalized size = 2.00 \begin {gather*} \frac {x}{\ln \left (x-5\right )}+\frac {96}{\ln \left (x-5\right )}-\frac {x^2}{\ln \left (x-5\right )}-\frac {\ln \relax (x)}{\ln \left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((96*x + log(x - 5)*(6*x - 11*x^2 + 2*x^3 - 5) - x*log(x) + x^2 - x^3)/(log(x - 5)^2*(5*x - x^2)),x)

[Out]

x/log(x - 5) + 96/log(x - 5) - x^2/log(x - 5) - log(x)/log(x - 5)

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sympy [A]  time = 0.34, size = 14, normalized size = 0.74 \begin {gather*} \frac {- x^{2} + x - \log {\relax (x )} + 96}{\log {\left (x - 5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(x)+(-2*x**3+11*x**2-6*x+5)*ln(x-5)+x**3-x**2-96*x)/(x**2-5*x)/ln(x-5)**2,x)

[Out]

(-x**2 + x - log(x) + 96)/log(x - 5)

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