3.19.84 \(\int \frac {-30-62 x^2+56 x^3+4 x^4+e^{\frac {\log ^2(x)}{5}} (30 x^3-6 x \log (x))}{-30 x+28 x^2+15 e^{\frac {\log ^2(x)}{5}} x^2+2 x^3} \, dx\)

Optimal. Leaf size=37 \[ x^2-\log \left (e^{\frac {\log ^2(x)}{5}}+\frac {\left (3+\frac {x}{5}\right ) (-2+2 x)}{3 x}\right ) \]

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Rubi [F]  time = 1.87, 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 {-30-62 x^2+56 x^3+4 x^4+e^{\frac {\log ^2(x)}{5}} \left (30 x^3-6 x \log (x)\right )}{-30 x+28 x^2+15 e^{\frac {\log ^2(x)}{5}} x^2+2 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-30 - 62*x^2 + 56*x^3 + 4*x^4 + E^(Log[x]^2/5)*(30*x^3 - 6*x*Log[x]))/(-30*x + 28*x^2 + 15*E^(Log[x]^2/5)
*x^2 + 2*x^3),x]

[Out]

x^2 - Log[x]^2/5 - 30*Defer[Int][1/(x*(-30 + 28*x + 15*E^(Log[x]^2/5)*x + 2*x^2)), x] - 2*Defer[Int][x/(-30 +
28*x + 15*E^(Log[x]^2/5)*x + 2*x^2), x] + (56*Defer[Int][Log[x]/(-30 + 28*x + 15*E^(Log[x]^2/5)*x + 2*x^2), x]
)/5 - 12*Defer[Int][Log[x]/(x*(-30 + 28*x + 15*E^(Log[x]^2/5)*x + 2*x^2)), x] + (4*Defer[Int][(x*Log[x])/(-30
+ 28*x + 15*E^(Log[x]^2/5)*x + 2*x^2), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (5 x^2-\log (x)\right )}{5 x}+\frac {2 \left (-75-5 x^2-30 \log (x)+28 x \log (x)+2 x^2 \log (x)\right )}{5 x \left (-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2\right )}\right ) \, dx\\ &=\frac {2}{5} \int \frac {5 x^2-\log (x)}{x} \, dx+\frac {2}{5} \int \frac {-75-5 x^2-30 \log (x)+28 x \log (x)+2 x^2 \log (x)}{x \left (-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2\right )} \, dx\\ &=\frac {2}{5} \int \left (5 x-\frac {\log (x)}{x}\right ) \, dx+\frac {2}{5} \int \left (-\frac {75}{x \left (-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2\right )}-\frac {5 x}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2}+\frac {28 \log (x)}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2}-\frac {30 \log (x)}{x \left (-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2\right )}+\frac {2 x \log (x)}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2}\right ) \, dx\\ &=x^2-\frac {2}{5} \int \frac {\log (x)}{x} \, dx+\frac {4}{5} \int \frac {x \log (x)}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2} \, dx-2 \int \frac {x}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2} \, dx+\frac {56}{5} \int \frac {\log (x)}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2} \, dx-12 \int \frac {\log (x)}{x \left (-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2\right )} \, dx-30 \int \frac {1}{x \left (-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2\right )} \, dx\\ &=x^2-\frac {\log ^2(x)}{5}+\frac {4}{5} \int \frac {x \log (x)}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2} \, dx-2 \int \frac {x}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2} \, dx+\frac {56}{5} \int \frac {\log (x)}{-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2} \, dx-12 \int \frac {\log (x)}{x \left (-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2\right )} \, dx-30 \int \frac {1}{x \left (-30+28 x+15 e^{\frac {\log ^2(x)}{5}} x+2 x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 32, normalized size = 0.86 \begin {gather*} x^2+\log (x)-\log \left (30-28 x-15 e^{\frac {\log ^2(x)}{5}} x-2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30 - 62*x^2 + 56*x^3 + 4*x^4 + E^(Log[x]^2/5)*(30*x^3 - 6*x*Log[x]))/(-30*x + 28*x^2 + 15*E^(Log[x
]^2/5)*x^2 + 2*x^3),x]

[Out]

x^2 + Log[x] - Log[30 - 28*x - 15*E^(Log[x]^2/5)*x - 2*x^2]

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fricas [A]  time = 0.79, size = 31, normalized size = 0.84 \begin {gather*} x^{2} - \log \left (\frac {2 \, x^{2} + 15 \, x e^{\left (\frac {1}{5} \, \log \relax (x)^{2}\right )} + 28 \, x - 30}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x*log(x)+30*x^3)*exp(1/5*log(x)^2)+4*x^4+56*x^3-62*x^2-30)/(15*x^2*exp(1/5*log(x)^2)+2*x^3+28*x
^2-30*x),x, algorithm="fricas")

[Out]

x^2 - log((2*x^2 + 15*x*e^(1/5*log(x)^2) + 28*x - 30)/x)

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giac [A]  time = 0.23, size = 29, normalized size = 0.78 \begin {gather*} x^{2} - \log \left (2 \, x^{2} + 15 \, x e^{\left (\frac {1}{5} \, \log \relax (x)^{2}\right )} + 28 \, x - 30\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x*log(x)+30*x^3)*exp(1/5*log(x)^2)+4*x^4+56*x^3-62*x^2-30)/(15*x^2*exp(1/5*log(x)^2)+2*x^3+28*x
^2-30*x),x, algorithm="giac")

[Out]

x^2 - log(2*x^2 + 15*x*e^(1/5*log(x)^2) + 28*x - 30) + log(x)

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maple [A]  time = 0.04, size = 29, normalized size = 0.78

method	result	size
risch	\(x^{2}-\ln \left ({\mathrm e}^{\frac {\ln \relax (x )^{2}}{5}}+\frac {\frac {2}{15} x^{2}+\frac {28}{15} x -2}{x}\right )\)	\(29\)
norman	\(x^{2}+\ln \relax (x )-\ln \left (15 x \,{\mathrm e}^{\frac {\ln \relax (x )^{2}}{5}}+2 x^{2}+28 x -30\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x*ln(x)+30*x^3)*exp(1/5*ln(x)^2)+4*x^4+56*x^3-62*x^2-30)/(15*x^2*exp(1/5*ln(x)^2)+2*x^3+28*x^2-30*x),
x,method=_RETURNVERBOSE)

[Out]

x^2-ln(exp(1/5*ln(x)^2)+2/15*(x^2+14*x-15)/x)

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maxima [A]  time = 0.58, size = 32, normalized size = 0.86 \begin {gather*} x^{2} - \log \left (\frac {2 \, x^{2} + 15 \, x e^{\left (\frac {1}{5} \, \log \relax (x)^{2}\right )} + 28 \, x - 30}{15 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x*log(x)+30*x^3)*exp(1/5*log(x)^2)+4*x^4+56*x^3-62*x^2-30)/(15*x^2*exp(1/5*log(x)^2)+2*x^3+28*x
^2-30*x),x, algorithm="maxima")

[Out]

x^2 - log(1/15*(2*x^2 + 15*x*e^(1/5*log(x)^2) + 28*x - 30)/x)

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mupad [B]  time = 1.43, size = 31, normalized size = 0.84 \begin {gather*} x^2-\ln \left (\frac {28\,x+2\,x^2+15\,x\,{\mathrm {e}}^{\frac {{\ln \relax (x)}^2}{5}}-30}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(x)^2/5)*(6*x*log(x) - 30*x^3) + 62*x^2 - 56*x^3 - 4*x^4 + 30)/(15*x^2*exp(log(x)^2/5) - 30*x + 2
8*x^2 + 2*x^3),x)

[Out]

x^2 - log((28*x + 2*x^2 + 15*x*exp(log(x)^2/5) - 30)/x)

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sympy [A]  time = 0.39, size = 26, normalized size = 0.70 \begin {gather*} x^{2} - \log {\left (e^{\frac {\log {\relax (x )}^{2}}{5}} + \frac {2 x^{2} + 28 x - 30}{15 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x*ln(x)+30*x**3)*exp(1/5*ln(x)**2)+4*x**4+56*x**3-62*x**2-30)/(15*x**2*exp(1/5*ln(x)**2)+2*x**3
+28*x**2-30*x),x)

[Out]

x**2 - log(exp(log(x)**2/5) + (2*x**2 + 28*x - 30)/(15*x))

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