3.94.84 \(\int \frac {-1+7 x^2-2 x^4-x \log (2 e^{\frac {1+x^2}{x}})}{3 x^3} \, dx\)

Optimal. Leaf size=32 \[ -2+\frac {1}{3} \left (-x^2+\frac {x+\log \left (2 e^{\frac {1}{x}+x}\right )}{x}\right )+\log \left (x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 29, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14, 2551} \begin {gather*} -\frac {x^2}{3}+2 \log (x)+\frac {\log \left (2 e^{x+\frac {1}{x}}\right )}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 7*x^2 - 2*x^4 - x*Log[2*E^((1 + x^2)/x)])/(3*x^3),x]

[Out]

-1/3*x^2 + Log[2*E^(x^(-1) + x)]/(3*x) + 2*Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{x^3} \, dx\\ &=\frac {1}{3} \int \left (\frac {-1+7 x^2-2 x^4}{x^3}-\frac {\log \left (2 e^{\frac {1}{x}+x}\right )}{x^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {-1+7 x^2-2 x^4}{x^3} \, dx-\frac {1}{3} \int \frac {\log \left (2 e^{\frac {1}{x}+x}\right )}{x^2} \, dx\\ &=\frac {\log \left (2 e^{\frac {1}{x}+x}\right )}{3 x}+\frac {1}{3} \int \left (-\frac {1}{x^3}+\frac {7}{x}-2 x\right ) \, dx-\frac {1}{3} \int \frac {-1+x^2}{x^3} \, dx\\ &=\frac {1}{6 x^2}-\frac {x^2}{3}+\frac {\log \left (2 e^{\frac {1}{x}+x}\right )}{3 x}+\frac {7 \log (x)}{3}-\frac {1}{3} \int \left (-\frac {1}{x^3}+\frac {1}{x}\right ) \, dx\\ &=-\frac {x^2}{3}+\frac {\log \left (2 e^{\frac {1}{x}+x}\right )}{3 x}+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.91 \begin {gather*} -\frac {x^2}{3}+\frac {\log \left (2 e^{\frac {1}{x}+x}\right )}{3 x}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 7*x^2 - 2*x^4 - x*Log[2*E^((1 + x^2)/x)])/(3*x^3),x]

[Out]

-1/3*x^2 + Log[2*E^(x^(-1) + x)]/(3*x) + 2*Log[x]

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fricas [A]  time = 0.61, size = 22, normalized size = 0.69 \begin {gather*} -\frac {x^{4} - 6 \, x^{2} \log \relax (x) - x \log \relax (2) - 1}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-x*log(2*exp((x^2+1)/x))-2*x^4+7*x^2-1)/x^3,x, algorithm="fricas")

[Out]

-1/3*(x^4 - 6*x^2*log(x) - x*log(2) - 1)/x^2

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giac [A]  time = 0.13, size = 22, normalized size = 0.69 \begin {gather*} -\frac {1}{3} \, x^{2} + \frac {x \log \relax (2) + 1}{3 \, x^{2}} + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-x*log(2*exp((x^2+1)/x))-2*x^4+7*x^2-1)/x^3,x, algorithm="giac")

[Out]

-1/3*x^2 + 1/3*(x*log(2) + 1)/x^2 + 2*log(abs(x))

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maple [A]  time = 0.11, size = 29, normalized size = 0.91

method	result	size
default	\(\frac {\ln \left (2 \,{\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3 x}+2 \ln \relax (x )-\frac {x^{2}}{3}\)	\(29\)
norman	\(\frac {-\frac {x^{4}}{3}+\frac {x \ln \left (2 \,{\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3}}{x^{2}}+2 \ln \relax (x )\)	\(32\)
risch	\(\frac {\ln \left ({\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3 x}+\frac {-2 x^{3}+12 x \ln \relax (x )+2 \ln \relax (2)}{6 x}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-x*ln(2*exp((x^2+1)/x))-2*x^4+7*x^2-1)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 30, normalized size = 0.94 \begin {gather*} -\frac {1}{3} \, x^{2} + \frac {\log \left (2 \, e^{\left (x + \frac {1}{x}\right )}\right )}{3 \, x} - \frac {1}{6} \, \log \left (x^{2}\right ) + \frac {7}{3} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-x*log(2*exp((x^2+1)/x))-2*x^4+7*x^2-1)/x^3,x, algorithm="maxima")

[Out]

-1/3*x^2 + 1/3*log(2*e^(x + 1/x))/x - 1/6*log(x^2) + 7/3*log(x)

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mupad [B]  time = 7.96, size = 21, normalized size = 0.66 \begin {gather*} 2\,\ln \relax (x)-\frac {x^2}{3}+\frac {\frac {x\,\ln \relax (2)}{3}+\frac {1}{3}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x*log(2*exp((x^2 + 1)/x)))/3 - (7*x^2)/3 + (2*x^4)/3 + 1/3)/x^3,x)

[Out]

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

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sympy [A]  time = 0.30, size = 22, normalized size = 0.69 \begin {gather*} - \frac {x^{2}}{3} + 2 \log {\relax (x )} - \frac {- x \log {\relax (2 )} - 1}{3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-x*ln(2*exp((x**2+1)/x))-2*x**4+7*x**2-1)/x**3,x)

[Out]

-x**2/3 + 2*log(x) - (-x*log(2) - 1)/(3*x**2)

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