3.62.48 \(\int \frac {-2 x+15 x^3+30 \log (4)}{15 x^3} \, dx\)

Optimal. Leaf size=23 \[ x-\frac {-\frac {2}{15}+\frac {-4 x^2+\log (4)}{x}}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.70, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 14} \begin {gather*} -\frac {\log (4)}{x^2}+x+\frac {2}{15 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x + 15*x^3 + 30*Log[4])/(15*x^3),x]

[Out]

2/(15*x) + x - Log[4]/x^2

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]]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 18, normalized size = 0.78 \begin {gather*} \frac {2}{15 x}+x-\frac {\log (16)}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + 15*x^3 + 30*Log[4])/(15*x^3),x]

[Out]

2/(15*x) + x - Log[16]/(2*x^2)

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fricas [A]  time = 0.61, size = 18, normalized size = 0.78 \begin {gather*} \frac {15 \, x^{3} + 2 \, x - 30 \, \log \relax (2)}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(60*log(2)+15*x^3-2*x)/x^3,x, algorithm="fricas")

[Out]

1/15*(15*x^3 + 2*x - 30*log(2))/x^2

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giac [A]  time = 0.14, size = 13, normalized size = 0.57 \begin {gather*} x + \frac {2 \, {\left (x - 15 \, \log \relax (2)\right )}}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(60*log(2)+15*x^3-2*x)/x^3,x, algorithm="giac")

[Out]

x + 2/15*(x - 15*log(2))/x^2

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maple [A]  time = 0.06, size = 15, normalized size = 0.65




method result size



default \(x +\frac {2}{15 x}-\frac {2 \ln \relax (2)}{x^{2}}\) \(15\)
norman \(\frac {x^{3}+\frac {2 x}{15}-2 \ln \relax (2)}{x^{2}}\) \(16\)
risch \(x +\frac {-30 \ln \relax (2)+2 x}{15 x^{2}}\) \(16\)
gosper \(-\frac {-15 x^{3}+30 \ln \relax (2)-2 x}{15 x^{2}}\) \(19\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*(60*ln(2)+15*x^3-2*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

x+2/15/x-2*ln(2)/x^2

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maxima [A]  time = 0.35, size = 13, normalized size = 0.57 \begin {gather*} x + \frac {2 \, {\left (x - 15 \, \log \relax (2)\right )}}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(60*log(2)+15*x^3-2*x)/x^3,x, algorithm="maxima")

[Out]

x + 2/15*(x - 15*log(2))/x^2

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mupad [B]  time = 0.04, size = 15, normalized size = 0.65 \begin {gather*} x+\frac {2\,x-30\,\ln \relax (2)}{15\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(2) - (2*x)/15 + x^3)/x^3,x)

[Out]

x + (2*x - 30*log(2))/(15*x^2)

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sympy [A]  time = 0.09, size = 14, normalized size = 0.61 \begin {gather*} x + \frac {2 x - 30 \log {\relax (2 )}}{15 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(60*ln(2)+15*x**3-2*x)/x**3,x)

[Out]

x + (2*x - 30*log(2))/(15*x**2)

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