3.64.84 \(\int \frac {-10 x^2+(-5 x-10 x^2) \log (5+10 x)+x^{\frac {1}{x}} (5+10 x+(-5-10 x) \log (x)) \log (5+10 x)}{(x^2+2 x^3) \log (5+10 x)} \, dx\)

Optimal. Leaf size=26 \[ 5 \left (x^{\frac {1}{x}}-\log \left (\frac {1}{2} x \log (7-2 (1-5 x))\right )\right ) \]

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Rubi [F]  time = 0.83, 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 {-10 x^2+\left (-5 x-10 x^2\right ) \log (5+10 x)+x^{\frac {1}{x}} (5+10 x+(-5-10 x) \log (x)) \log (5+10 x)}{\left (x^2+2 x^3\right ) \log (5+10 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*x^2 + (-5*x - 10*x^2)*Log[5 + 10*x] + x^x^(-1)*(5 + 10*x + (-5 - 10*x)*Log[x])*Log[5 + 10*x])/((x^2 +
2*x^3)*Log[5 + 10*x]),x]

[Out]

-5*Log[x] - 5*Log[Log[5 + 10*x]] + 5*Defer[Int][x^(-2 + x^(-1)), x] - 5*Log[x]*Defer[Int][x^(-2 + x^(-1)), x]
+ 5*Defer[Int][Defer[Int][x^(-2 + x^(-1)), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 x^2+\left (-5 x-10 x^2\right ) \log (5+10 x)+x^{\frac {1}{x}} (5+10 x+(-5-10 x) \log (x)) \log (5+10 x)}{x^2 (1+2 x) \log (5+10 x)} \, dx\\ &=\int \left (-\frac {5 \left (x-x^{\frac {1}{x}}+x^{\frac {1}{x}} \log (x)\right )}{x^2}-\frac {10}{(1+2 x) \log (5+10 x)}\right ) \, dx\\ &=-\left (5 \int \frac {x-x^{\frac {1}{x}}+x^{\frac {1}{x}} \log (x)}{x^2} \, dx\right )-10 \int \frac {1}{(1+2 x) \log (5+10 x)} \, dx\\ &=-\left (5 \int \left (\frac {1}{x}+x^{-2+\frac {1}{x}} (-1+\log (x))\right ) \, dx\right )-\operatorname {Subst}\left (\int \frac {5}{x \log (x)} \, dx,x,5+10 x\right )\\ &=-5 \log (x)-5 \int x^{-2+\frac {1}{x}} (-1+\log (x)) \, dx-5 \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,5+10 x\right )\\ &=-5 \log (x)-5 \int \left (-x^{-2+\frac {1}{x}}+x^{-2+\frac {1}{x}} \log (x)\right ) \, dx-5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (5+10 x)\right )\\ &=-5 \log (x)-5 \log (\log (5+10 x))+5 \int x^{-2+\frac {1}{x}} \, dx-5 \int x^{-2+\frac {1}{x}} \log (x) \, dx\\ &=-5 \log (x)-5 \log (\log (5+10 x))+5 \int x^{-2+\frac {1}{x}} \, dx+5 \int \frac {\int x^{-2+\frac {1}{x}} \, dx}{x} \, dx-(5 \log (x)) \int x^{-2+\frac {1}{x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 21, normalized size = 0.81 \begin {gather*} 5 x^{\frac {1}{x}}-5 \log (x)-5 \log (\log (5+10 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x^2 + (-5*x - 10*x^2)*Log[5 + 10*x] + x^x^(-1)*(5 + 10*x + (-5 - 10*x)*Log[x])*Log[5 + 10*x])/(
(x^2 + 2*x^3)*Log[5 + 10*x]),x]

[Out]

5*x^x^(-1) - 5*Log[x] - 5*Log[Log[5 + 10*x]]

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fricas [A]  time = 0.85, size = 21, normalized size = 0.81 \begin {gather*} 5 \, x^{\left (\frac {1}{x}\right )} - 5 \, \log \relax (x) - 5 \, \log \left (\log \left (10 \, x + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-5)*log(x)+10*x+5)*log(10*x+5)*exp(log(x)/x)+(-10*x^2-5*x)*log(10*x+5)-10*x^2)/(2*x^3+x^2)/l
og(10*x+5),x, algorithm="fricas")

[Out]

5*x^(1/x) - 5*log(x) - 5*log(log(10*x + 5))

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giac [A]  time = 0.39, size = 24, normalized size = 0.92 \begin {gather*} 5 \, x^{\left (\frac {1}{x}\right )} - 5 \, \log \relax (x) - 5 \, \log \left (\log \relax (5) + \log \left (2 \, x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-5)*log(x)+10*x+5)*log(10*x+5)*exp(log(x)/x)+(-10*x^2-5*x)*log(10*x+5)-10*x^2)/(2*x^3+x^2)/l
og(10*x+5),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.38, size = 22, normalized size = 0.85

method	result	size
risch	\(5 x^{\frac {1}{x}}-5 \ln \relax (x )-5 \ln \left (\ln \left (10 x +5\right )\right )\)	\(22\)
default	\(5 \,{\mathrm e}^{\frac {\ln \relax (x )}{x}}-5 \ln \relax (x )-5 \ln \left (\ln \left (10 x +5\right )\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-10*x-5)*ln(x)+10*x+5)*ln(10*x+5)*exp(ln(x)/x)+(-10*x^2-5*x)*ln(10*x+5)-10*x^2)/(2*x^3+x^2)/ln(10*x+5),
x,method=_RETURNVERBOSE)

[Out]

5*x^(1/x)-5*ln(x)-5*ln(ln(10*x+5))

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maxima [B]  time = 0.60, size = 63, normalized size = 2.42 \begin {gather*} 5 \, {\left (\log \relax (5) + \log \left (2 \, x + 1\right )\right )} \log \left (\log \relax (5) + \log \left (2 \, x + 1\right )\right ) - 5 \, \log \left (10 \, x + 5\right ) \log \left (\log \relax (5) + \log \left (2 \, x + 1\right )\right ) + 5 \, x^{\left (\frac {1}{x}\right )} - 5 \, \log \relax (x) - 5 \, \log \left (\log \relax (5) + \log \left (2 \, x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-5)*log(x)+10*x+5)*log(10*x+5)*exp(log(x)/x)+(-10*x^2-5*x)*log(10*x+5)-10*x^2)/(2*x^3+x^2)/l
og(10*x+5),x, algorithm="maxima")

[Out]

5*(log(5) + log(2*x + 1))*log(log(5) + log(2*x + 1)) - 5*log(10*x + 5)*log(log(5) + log(2*x + 1)) + 5*x^(1/x)
- 5*log(x) - 5*log(log(5) + log(2*x + 1))

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mupad [B]  time = 4.54, size = 21, normalized size = 0.81 \begin {gather*} 5\,x^{1/x}-5\,\ln \relax (x)-5\,\ln \left (\ln \left (10\,x+5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(10*x + 5)*(5*x + 10*x^2) + 10*x^2 - exp(log(x)/x)*log(10*x + 5)*(10*x - log(x)*(10*x + 5) + 5))/(log
(10*x + 5)*(x^2 + 2*x^3)),x)

[Out]

5*x^(1/x) - 5*log(x) - 5*log(log(10*x + 5))

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sympy [A]  time = 0.52, size = 22, normalized size = 0.85 \begin {gather*} 5 e^{\frac {\log {\relax (x )}}{x}} - 5 \log {\relax (x )} - 5 \log {\left (\log {\left (10 x + 5 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-5)*ln(x)+10*x+5)*ln(10*x+5)*exp(ln(x)/x)+(-10*x**2-5*x)*ln(10*x+5)-10*x**2)/(2*x**3+x**2)/l
n(10*x+5),x)

[Out]

5*exp(log(x)/x) - 5*log(x) - 5*log(log(10*x + 5))

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