3.51.4 \(\int \frac {e^x (-x^2+x^3)+e^{\frac {3 x^2+30 \log (x)}{x}} (-120-12 x^2+4 x^3+120 \log (x))}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx\)

Optimal. Leaf size=30 \[ \frac {x^2}{2}-\log \left (e^x+4 e^{3 x+\frac {30 \log (x)}{x}}\right ) \]

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Rubi [F]  time = 1.25, 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 {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-x^2 + x^3) + E^((3*x^2 + 30*Log[x])/x)*(-120 - 12*x^2 + 4*x^3 + 120*Log[x]))/(E^x*x^2 + 4*E^((3*x^2
+ 30*Log[x])/x)*x^2),x]

[Out]

-3*x + x^2/2 - (30*Log[x])/x + 2*Defer[Int][(1 + 4*E^(2*x)*x^(30/x))^(-1), x] + 30*Defer[Int][1/(x^2*(1 + 4*E^
(2*x)*x^(30/x))), x] - 30*Log[x]*Defer[Int][1/(x^2*(1 + 4*E^(2*x)*x^(30/x))), x] + 30*Defer[Int][Defer[Int][(x
^2 + 4*E^(2*x)*x^(2 + 30/x))^(-1), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x^2+x^3+4 e^{2 x} x^{30/x} \left (-30-3 x^2+x^3+30 \log (x)\right )}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx\\ &=\int \left (\frac {2 \left (15+x^2-15 \log (x)\right )}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )}+\frac {-30-3 x^2+x^3+30 \log (x)}{x^2}\right ) \, dx\\ &=2 \int \frac {15+x^2-15 \log (x)}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+\int \frac {-30-3 x^2+x^3+30 \log (x)}{x^2} \, dx\\ &=2 \int \left (\frac {1}{1+4 e^{2 x} x^{30/x}}+\frac {15}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )}-\frac {15 \log (x)}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )}\right ) \, dx+\int \left (\frac {-30-3 x^2+x^3}{x^2}+\frac {30 \log (x)}{x^2}\right ) \, dx\\ &=2 \int \frac {1}{1+4 e^{2 x} x^{30/x}} \, dx+30 \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+30 \int \frac {\log (x)}{x^2} \, dx-30 \int \frac {\log (x)}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+\int \frac {-30-3 x^2+x^3}{x^2} \, dx\\ &=-\frac {30}{x}-\frac {30 \log (x)}{x}+2 \int \frac {1}{1+4 e^{2 x} x^{30/x}} \, dx+30 \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+30 \int \frac {\int \frac {1}{x^2+4 e^{2 x} x^{2+\frac {30}{x}}} \, dx}{x} \, dx-(30 \log (x)) \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+\int \left (-3-\frac {30}{x^2}+x\right ) \, dx\\ &=-3 x+\frac {x^2}{2}-\frac {30 \log (x)}{x}+2 \int \frac {1}{1+4 e^{2 x} x^{30/x}} \, dx+30 \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+30 \int \frac {\int \frac {1}{x^2+4 e^{2 x} x^{2+\frac {30}{x}}} \, dx}{x} \, dx-(30 \log (x)) \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 30, normalized size = 1.00 \begin {gather*} -x+\frac {x^2}{2}-\log \left (1+4 e^{2 x} x^{30/x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-x^2 + x^3) + E^((3*x^2 + 30*Log[x])/x)*(-120 - 12*x^2 + 4*x^3 + 120*Log[x]))/(E^x*x^2 + 4*E^(
(3*x^2 + 30*Log[x])/x)*x^2),x]

[Out]

-x + x^2/2 - Log[1 + 4*E^(2*x)*x^(30/x)]

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fricas [A]  time = 0.97, size = 28, normalized size = 0.93 \begin {gather*} \frac {1}{2} \, x^{2} - \log \left (e^{x} + 4 \, e^{\left (\frac {3 \, {\left (x^{2} + 10 \, \log \relax (x)\right )}}{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2)*exp(x))/(4*x^2*exp((30*log(x)+3*x^
2)/x)+exp(x)*x^2),x, algorithm="fricas")

[Out]

1/2*x^2 - log(e^x + 4*e^(3*(x^2 + 10*log(x))/x))

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giac [A]  time = 0.18, size = 28, normalized size = 0.93 \begin {gather*} \frac {1}{2} \, x^{2} - \log \left (e^{x} + 4 \, e^{\left (\frac {3 \, {\left (x^{2} + 10 \, \log \relax (x)\right )}}{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2)*exp(x))/(4*x^2*exp((30*log(x)+3*x^
2)/x)+exp(x)*x^2),x, algorithm="giac")

[Out]

1/2*x^2 - log(e^x + 4*e^(3*(x^2 + 10*log(x))/x))

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maple [A]  time = 0.04, size = 51, normalized size = 1.70

method	result	size
risch	\(-\frac {30 \ln \relax (x )}{x}+\frac {x^{2}}{2}-3 x +\frac {30 \ln \relax (x )+3 x^{2}}{x}-\ln \left (\frac {{\mathrm e}^{x}}{4}+x^{\frac {30}{x}} {\mathrm e}^{3 x}\right )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((120*ln(x)+4*x^3-12*x^2-120)*exp((30*ln(x)+3*x^2)/x)+(x^3-x^2)*exp(x))/(4*x^2*exp((30*ln(x)+3*x^2)/x)+exp
(x)*x^2),x,method=_RETURNVERBOSE)

[Out]

-30*ln(x)/x+1/2*x^2-3*x+(30*ln(x)+3*x^2)/x-ln(1/4*exp(x)+x^(30/x)*exp(3*x))

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maxima [A]  time = 0.40, size = 34, normalized size = 1.13 \begin {gather*} \frac {1}{2} \, x^{2} - 3 \, x - \log \left (\frac {1}{4} \, {\left (4 \, e^{\left (2 \, x + \frac {30 \, \log \relax (x)}{x}\right )} + 1\right )} e^{\left (-2 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2)*exp(x))/(4*x^2*exp((30*log(x)+3*x^
2)/x)+exp(x)*x^2),x, algorithm="maxima")

[Out]

1/2*x^2 - 3*x - log(1/4*(4*e^(2*x + 30*log(x)/x) + 1)*e^(-2*x))

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mupad [B]  time = 3.35, size = 26, normalized size = 0.87 \begin {gather*} \frac {x^2}{2}-\ln \left (\frac {{\mathrm {e}}^x}{4}+x^{30/x}\,{\mathrm {e}}^{3\,x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x^2 - x^3) - exp((30*log(x) + 3*x^2)/x)*(120*log(x) - 12*x^2 + 4*x^3 - 120))/(x^2*exp(x) + 4*x^2
*exp((30*log(x) + 3*x^2)/x)),x)

[Out]

x^2/2 - log(exp(x)/4 + x^(30/x)*exp(3*x))

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sympy [A]  time = 0.39, size = 24, normalized size = 0.80 \begin {gather*} \frac {x^{2}}{2} - \log {\left (\frac {e^{x}}{4} + e^{\frac {3 x^{2} + 30 \log {\relax (x )}}{x}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((120*ln(x)+4*x**3-12*x**2-120)*exp((30*ln(x)+3*x**2)/x)+(x**3-x**2)*exp(x))/(4*x**2*exp((30*ln(x)+3
*x**2)/x)+exp(x)*x**2),x)

[Out]

x**2/2 - log(exp(x)/4 + exp((3*x**2 + 30*log(x))/x))

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