3.42.35 \(\int \frac {8+x-x^3-10 e^{10} x^{12}-x \log (x)}{x^3} \, dx\)

Optimal. Leaf size=28 \[ 4-e^{5 \left (2+\log \left (x^2\right )\right )}-\frac {4}{x^2}-x+\frac {\log (x)}{x} \]

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Rubi [A]  time = 0.03, antiderivative size = 23, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {14, 2304} \begin {gather*} -e^{10} x^{10}-\frac {4}{x^2}-x+\frac {\log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 + x - x^3 - 10*E^10*x^12 - x*Log[x])/x^3,x]

[Out]

-4/x^2 - x - E^10*x^10 + Log[x]/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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {8+x-x^3-10 e^{10} x^{12}}{x^3}-\frac {\log (x)}{x^2}\right ) \, dx\\ &=\int \frac {8+x-x^3-10 e^{10} x^{12}}{x^3} \, dx-\int \frac {\log (x)}{x^2} \, dx\\ &=\frac {1}{x}+\frac {\log (x)}{x}+\int \left (-1+\frac {8}{x^3}+\frac {1}{x^2}-10 e^{10} x^9\right ) \, dx\\ &=-\frac {4}{x^2}-x-e^{10} x^{10}+\frac {\log (x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 0.82 \begin {gather*} -\frac {4}{x^2}-x-e^{10} x^{10}+\frac {\log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + x - x^3 - 10*E^10*x^12 - x*Log[x])/x^3,x]

[Out]

-4/x^2 - x - E^10*x^10 + Log[x]/x

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fricas [A]  time = 1.02, size = 21, normalized size = 0.75 \begin {gather*} -\frac {x^{12} e^{10} + x^{3} - x \log \relax (x) + 4}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^2*exp(5*log(x^2)+10)-x*log(x)-x^3+x+8)/x^3,x, algorithm="fricas")

[Out]

-(x^12*e^10 + x^3 - x*log(x) + 4)/x^2

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giac [A]  time = 0.14, size = 21, normalized size = 0.75 \begin {gather*} -\frac {x^{12} e^{10} + x^{3} - x \log \relax (x) + 4}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^2*exp(5*log(x^2)+10)-x*log(x)-x^3+x+8)/x^3,x, algorithm="giac")

[Out]

-(x^12*e^10 + x^3 - x*log(x) + 4)/x^2

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maple [A]  time = 0.03, size = 27, normalized size = 0.96

method	result	size
default	\(-{\mathrm e}^{5 \ln \left (x^{2}\right )+10}-x -\frac {4}{x^{2}}+\frac {\ln \relax (x )}{x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*x^2*exp(5*ln(x^2)+10)-x*ln(x)-x^3+x+8)/x^3,x,method=_RETURNVERBOSE)

[Out]

-exp(5*ln(x^2)+10)-x-4/x^2+ln(x)/x

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maxima [A]  time = 0.36, size = 22, normalized size = 0.79 \begin {gather*} -x^{10} e^{10} - x + \frac {\log \relax (x)}{x} - \frac {4}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^2*exp(5*log(x^2)+10)-x*log(x)-x^3+x+8)/x^3,x, algorithm="maxima")

[Out]

-x^10*e^10 - x + log(x)/x - 4/x^2

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mupad [B]  time = 3.36, size = 21, normalized size = 0.75 \begin {gather*} \frac {x\,\ln \relax (x)-4}{x^2}-x-x^{10}\,{\mathrm {e}}^{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x^2*exp(5*log(x^2) + 10) - x + x*log(x) + x^3 - 8)/x^3,x)

[Out]

(x*log(x) - 4)/x^2 - x - x^10*exp(10)

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sympy [A]  time = 0.12, size = 17, normalized size = 0.61 \begin {gather*} - x^{10} e^{10} - x + \frac {\log {\relax (x )}}{x} - \frac {4}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x**2*exp(5*ln(x**2)+10)-x*ln(x)-x**3+x+8)/x**3,x)

[Out]

-x**10*exp(10) - x + log(x)/x - 4/x**2

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