3.63.29 \(\int \frac {188-62 \log (x^2)+(-3+\log (x^2)) \log (3-\log (x^2))}{-45+15 \log (x^2)} \, dx\)

Optimal. Leaf size=21 \[ -1-4 x+\frac {1}{15} x \left (-2+\log \left (3-\log \left (x^2\right )\right )\right ) \]

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Rubi [F]  time = 0.14, 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 {188-62 \log \left (x^2\right )+\left (-3+\log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )}{-45+15 \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(188 - 62*Log[x^2] + (-3 + Log[x^2])*Log[3 - Log[x^2]])/(-45 + 15*Log[x^2]),x]

[Out]

(-62*x)/15 + (E^(3/2)*x*ExpIntegralEi[(-3 + Log[x^2])/2])/(15*Sqrt[x^2]) + Defer[Int][Log[3 - Log[x^2]], x]/15

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-188+62 \log \left (x^2\right )-\left (-3+\log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )}{15 \left (3-\log \left (x^2\right )\right )} \, dx\\ &=\frac {1}{15} \int \frac {-188+62 \log \left (x^2\right )-\left (-3+\log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )}{3-\log \left (x^2\right )} \, dx\\ &=\frac {1}{15} \int \left (-\frac {2 \left (-94+31 \log \left (x^2\right )\right )}{-3+\log \left (x^2\right )}+\log \left (3-\log \left (x^2\right )\right )\right ) \, dx\\ &=\frac {1}{15} \int \log \left (3-\log \left (x^2\right )\right ) \, dx-\frac {2}{15} \int \frac {-94+31 \log \left (x^2\right )}{-3+\log \left (x^2\right )} \, dx\\ &=\frac {1}{15} \int \log \left (3-\log \left (x^2\right )\right ) \, dx-\frac {2}{15} \int \left (31-\frac {1}{-3+\log \left (x^2\right )}\right ) \, dx\\ &=-\frac {62 x}{15}+\frac {1}{15} \int \log \left (3-\log \left (x^2\right )\right ) \, dx+\frac {2}{15} \int \frac {1}{-3+\log \left (x^2\right )} \, dx\\ &=-\frac {62 x}{15}+\frac {1}{15} \int \log \left (3-\log \left (x^2\right )\right ) \, dx+\frac {x \operatorname {Subst}\left (\int \frac {e^{x/2}}{-3+x} \, dx,x,\log \left (x^2\right )\right )}{15 \sqrt {x^2}}\\ &=-\frac {62 x}{15}+\frac {e^{3/2} x \text {Ei}\left (\frac {1}{2} \left (-3+\log \left (x^2\right )\right )\right )}{15 \sqrt {x^2}}+\frac {1}{15} \int \log \left (3-\log \left (x^2\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 19, normalized size = 0.90 \begin {gather*} \frac {1}{15} \left (-62 x+x \log \left (3-\log \left (x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(188 - 62*Log[x^2] + (-3 + Log[x^2])*Log[3 - Log[x^2]])/(-45 + 15*Log[x^2]),x]

[Out]

(-62*x + x*Log[3 - Log[x^2]])/15

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fricas [A]  time = 0.81, size = 16, normalized size = 0.76 \begin {gather*} \frac {1}{15} \, x \log \left (-\log \left (x^{2}\right ) + 3\right ) - \frac {62}{15} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-3)*log(3-log(x^2))-62*log(x^2)+188)/(15*log(x^2)-45),x, algorithm="fricas")

[Out]

1/15*x*log(-log(x^2) + 3) - 62/15*x

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giac [A]  time = 0.19, size = 16, normalized size = 0.76 \begin {gather*} \frac {1}{15} \, x \log \left (-\log \left (x^{2}\right ) + 3\right ) - \frac {62}{15} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-3)*log(3-log(x^2))-62*log(x^2)+188)/(15*log(x^2)-45),x, algorithm="giac")

[Out]

1/15*x*log(-log(x^2) + 3) - 62/15*x

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maple [A]  time = 0.03, size = 17, normalized size = 0.81

method	result	size
norman	\(-\frac {62 x}{15}+\frac {x \ln \left (3-\ln \left (x^{2}\right )\right )}{15}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x^2)-3)*ln(3-ln(x^2))-62*ln(x^2)+188)/(15*ln(x^2)-45),x,method=_RETURNVERBOSE)

[Out]

-62/15*x+1/15*x*ln(3-ln(x^2))

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maxima [A]  time = 0.37, size = 14, normalized size = 0.67 \begin {gather*} \frac {1}{15} \, x \log \left (-2 \, \log \relax (x) + 3\right ) - \frac {62}{15} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-3)*log(3-log(x^2))-62*log(x^2)+188)/(15*log(x^2)-45),x, algorithm="maxima")

[Out]

1/15*x*log(-2*log(x) + 3) - 62/15*x

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mupad [B]  time = 4.23, size = 14, normalized size = 0.67 \begin {gather*} \frac {x\,\left (\ln \left (3-\ln \left (x^2\right )\right )-62\right )}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3 - log(x^2))*(log(x^2) - 3) - 62*log(x^2) + 188)/(15*log(x^2) - 45),x)

[Out]

(x*(log(3 - log(x^2)) - 62))/15

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sympy [A]  time = 0.32, size = 15, normalized size = 0.71 \begin {gather*} \frac {x \log {\left (3 - \log {\left (x^{2} \right )} \right )}}{15} - \frac {62 x}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x**2)-3)*ln(3-ln(x**2))-62*ln(x**2)+188)/(15*ln(x**2)-45),x)

[Out]

x*log(3 - log(x**2))/15 - 62*x/15

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