\(\int \frac {-1+7 x^2-2 x^4-x \log (2 e^{\frac {1+x^2}{x}})}{3 x^3} \, dx\) [9384]

Optimal result

Integrand size = 36, antiderivative size = 32 \[ \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{3 x^3} \, dx=-2+\frac {1}{3} \left (-x^2+\frac {x+\log \left (2 e^{\frac {1}{x}+x}\right )}{x}\right )+\log \left (x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14, 2631} \[ \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{3 x^3} \, dx=-\frac {x^2}{3}+2 \log (x)+\frac {\log \left (2 e^{x+\frac {1}{x}}\right )}{3 x} \]

[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 2631

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{3 x^3} \, dx=-\frac {x^2}{3}+\frac {\log \left (2 e^{\frac {1}{x}+x}\right )}{3 x}+2 \log (x) \]

[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]

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\ln \left (2 \,{\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3 x}+2 \ln \left (x \right )-\frac {x^{2}}{3}\)	\(29\)
parts	\(\frac {\ln \left (2 \,{\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3 x}+2 \ln \left (x \right )-\frac {x^{2}}{3}\)	\(29\)
parallelrisch	\(\frac {-x^{4}+6 x^{2} \ln \left (x \right )+x \ln \left (2 \,{\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3 x^{2}}\)	\(34\)
risch	\(\frac {\ln \left ({\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3 x}+\frac {-2 x^{3}+12 x \ln \left (x \right )+2 \ln \left (2\right )}{6 x}\)	\(38\)
norman	\(\frac {-\frac {2 x^{3} \ln \left (2 \,{\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3}-\frac {x \ln \left (2 \,{\mathrm e}^{\frac {x^{2}+1}{x}}\right )}{3}+\frac {x^{2} \ln \left (2 \,{\mathrm e}^{\frac {x^{2}+1}{x}}\right )^{2}}{3}+\frac {1}{3}}{x^{2}}+2 \ln \left (x \right )\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{3 x^3} \, dx=-\frac {x^{4} - 6 \, x^{2} \log \left (x\right ) - x \log \left (2\right ) - 1}{3 \, x^{2}} \]

[In]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{3 x^3} \, dx=- \frac {x^{2}}{3} + 2 \log {\left (x \right )} - \frac {- x \log {\left (2 \right )} - 1}{3 x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{3 x^3} \, dx=-\frac {1}{3} \, x^{2} + \frac {\log \left (2 \, e^{\left (x + \frac {1}{x}\right )}\right )}{3 \, x} - \frac {1}{6} \, \log \left (x^{2}\right ) + \frac {7}{3} \, \log \left (x\right ) \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{3 x^3} \, dx=-\frac {1}{3} \, x^{2} + \frac {x \log \left (2\right ) + 1}{3 \, x^{2}} + 2 \, \log \left ({\left | x \right |}\right ) \]

[In]

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

Mupad [B] (verification not implemented)

Time = 14.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {-1+7 x^2-2 x^4-x \log \left (2 e^{\frac {1+x^2}{x}}\right )}{3 x^3} \, dx=2\,\ln \left (x\right )-\frac {x^2}{3}+\frac {\frac {x\,\ln \left (2\right )}{3}+\frac {1}{3}}{x^2} \]

[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