3.67.54 \(\int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+(-10-10 x+(2+2 x) \log ^2(x)) \log (\frac {5 e^{25}-e^{25} \log ^2(x)}{x})}{-5+\log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ -5+x (2+x) \log \left (\frac {e^{25} \left (5-\log ^2(x)\right )}{x}\right ) \]

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Rubi [F]  time = 0.63, 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+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10 + 5*x + (4 + 2*x)*Log[x] + (-2 - x)*Log[x]^2 + (-10 - 10*x + (2 + 2*x)*Log[x]^2)*Log[(5*E^25 - E^25*Lo
g[x]^2)/x])/(-5 + Log[x]^2),x]

[Out]

-2*x - x^2/2 + 25*(1 + x)^2 + 2*E^Sqrt[5]*ExpIntegralEi[-Sqrt[5] + Log[x]] + (2*ExpIntegralEi[Sqrt[5] + Log[x]
])/E^Sqrt[5] + 2*Defer[Int][(x*Log[x])/(-5 + Log[x]^2), x] + 2*Defer[Int][Log[5/x - Log[x]^2/x], x] + 2*Defer[
Int][x*Log[5/x - Log[x]^2/x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {(2+x) \left (-5-2 \log (x)+\log ^2(x)\right )}{-5+\log ^2(x)}+2 (1+x) \left (25+\log \left (-\frac {-5+\log ^2(x)}{x}\right )\right )\right ) \, dx\\ &=2 \int (1+x) \left (25+\log \left (-\frac {-5+\log ^2(x)}{x}\right )\right ) \, dx-\int \frac {(2+x) \left (-5-2 \log (x)+\log ^2(x)\right )}{-5+\log ^2(x)} \, dx\\ &=2 \int \left (25 (1+x)+(1+x) \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right )\right ) \, dx-\int \left (2+x-\frac {2 (2+x) \log (x)}{-5+\log ^2(x)}\right ) \, dx\\ &=-2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \frac {(2+x) \log (x)}{-5+\log ^2(x)} \, dx+2 \int (1+x) \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx\\ &=-2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \left (\frac {2 \log (x)}{-5+\log ^2(x)}+\frac {x \log (x)}{-5+\log ^2(x)}\right ) \, dx+2 \int \left (\log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right )+x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right )\right ) \, dx\\ &=-2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+4 \int \frac {\log (x)}{-5+\log ^2(x)} \, dx\\ &=-2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+4 \int \left (-\frac {1}{2 \left (\sqrt {5}-\log (x)\right )}+\frac {1}{2 \left (\sqrt {5}+\log (x)\right )}\right ) \, dx\\ &=-2 x-\frac {x^2}{2}+25 (1+x)^2-2 \int \frac {1}{\sqrt {5}-\log (x)} \, dx+2 \int \frac {1}{\sqrt {5}+\log (x)} \, dx+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx\\ &=-2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx-2 \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {5}-x} \, dx,x,\log (x)\right )+2 \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {5}+x} \, dx,x,\log (x)\right )\\ &=-2 x-\frac {x^2}{2}+25 (1+x)^2+2 e^{\sqrt {5}} \text {Ei}\left (-\sqrt {5}+\log (x)\right )+2 e^{-\sqrt {5}} \text {Ei}\left (\sqrt {5}+\log (x)\right )+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 19, normalized size = 0.83 \begin {gather*} x (2+x) \left (25+\log \left (-\frac {-5+\log ^2(x)}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + 5*x + (4 + 2*x)*Log[x] + (-2 - x)*Log[x]^2 + (-10 - 10*x + (2 + 2*x)*Log[x]^2)*Log[(5*E^25 - E
^25*Log[x]^2)/x])/(-5 + Log[x]^2),x]

[Out]

x*(2 + x)*(25 + Log[-((-5 + Log[x]^2)/x)])

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fricas [A]  time = 0.56, size = 26, normalized size = 1.13 \begin {gather*} {\left (x^{2} + 2 \, x\right )} \log \left (-\frac {e^{25} \log \relax (x)^{2} - 5 \, e^{25}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+2)*log(x)^2-10*x-10)*log((-exp(25)*log(x)^2+5*exp(25))/x)+(-x-2)*log(x)^2+(2*x+4)*log(x)+5*x+
10)/(log(x)^2-5),x, algorithm="fricas")

[Out]

(x^2 + 2*x)*log(-(e^25*log(x)^2 - 5*e^25)/x)

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giac [A]  time = 0.31, size = 37, normalized size = 1.61 \begin {gather*} 25 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (-\log \relax (x)^{2} + 5\right ) - {\left (x^{2} + 2 \, x\right )} \log \relax (x) + 50 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+2)*log(x)^2-10*x-10)*log((-exp(25)*log(x)^2+5*exp(25))/x)+(-x-2)*log(x)^2+(2*x+4)*log(x)+5*x+
10)/(log(x)^2-5),x, algorithm="giac")

[Out]

25*x^2 + (x^2 + 2*x)*log(-log(x)^2 + 5) - (x^2 + 2*x)*log(x) + 50*x

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maple [A]  time = 0.20, size = 45, normalized size = 1.96

method	result	size
norman	\(x^{2} \ln \left (\frac {-{\mathrm e}^{25} \ln \relax (x )^{2}+5 \,{\mathrm e}^{25}}{x}\right )+2 x \ln \left (\frac {-{\mathrm e}^{25} \ln \relax (x )^{2}+5 \,{\mathrm e}^{25}}{x}\right )\)	\(45\)
risch	\(\left (x^{2}+2 x \right ) \ln \left (\ln \relax (x )^{2}-5\right )-x^{2} \ln \relax (x )-2 x \ln \relax (x )-i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )^{2}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )+i \pi \,x^{2}+2 i \pi x +i \pi x \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )^{3}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )^{3}}{2}-i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )^{2}+i \pi x \,\mathrm {csgn}\left (i \left (\ln \relax (x )^{2}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )^{2}-2 i \pi x \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )^{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \left (\ln \relax (x )^{2}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )^{2}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )^{2}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )}{2}+i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )^{2}-5\right )}{x}\right )^{2}+25 x^{2}+50 x\)	\(323\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x+2)*ln(x)^2-10*x-10)*ln((-exp(25)*ln(x)^2+5*exp(25))/x)+(-x-2)*ln(x)^2+(2*x+4)*ln(x)+5*x+10)/(ln(x)^
2-5),x,method=_RETURNVERBOSE)

[Out]

x^2*ln((-exp(25)*ln(x)^2+5*exp(25))/x)+2*x*ln((-exp(25)*ln(x)^2+5*exp(25))/x)

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maxima [A]  time = 0.42, size = 37, normalized size = 1.61 \begin {gather*} 25 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (-\log \relax (x)^{2} + 5\right ) - {\left (x^{2} + 2 \, x\right )} \log \relax (x) + 50 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+2)*log(x)^2-10*x-10)*log((-exp(25)*log(x)^2+5*exp(25))/x)+(-x-2)*log(x)^2+(2*x+4)*log(x)+5*x+
10)/(log(x)^2-5),x, algorithm="maxima")

[Out]

25*x^2 + (x^2 + 2*x)*log(-log(x)^2 + 5) - (x^2 + 2*x)*log(x) + 50*x

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mupad [B]  time = 4.50, size = 23, normalized size = 1.00 \begin {gather*} x\,\ln \left (\frac {5\,{\mathrm {e}}^{25}-{\mathrm {e}}^{25}\,{\ln \relax (x)}^2}{x}\right )\,\left (x+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x - log((5*exp(25) - exp(25)*log(x)^2)/x)*(10*x - log(x)^2*(2*x + 2) + 10) + log(x)*(2*x + 4) - log(x)^
2*(x + 2) + 10)/(log(x)^2 - 5),x)

[Out]

x*log((5*exp(25) - exp(25)*log(x)^2)/x)*(x + 2)

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sympy [A]  time = 0.42, size = 22, normalized size = 0.96 \begin {gather*} \left (x^{2} + 2 x\right ) \log {\left (\frac {- e^{25} \log {\relax (x )}^{2} + 5 e^{25}}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+2)*ln(x)**2-10*x-10)*ln((-exp(25)*ln(x)**2+5*exp(25))/x)+(-x-2)*ln(x)**2+(2*x+4)*ln(x)+5*x+10
)/(ln(x)**2-5),x)

[Out]

(x**2 + 2*x)*log((-exp(25)*log(x)**2 + 5*exp(25))/x)

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