3.26.5 \(\int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx\)

Optimal. Leaf size=21 \[ 3-\frac {1}{x}+x^2-2 \left (4+\frac {5 \log (3)}{x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.71, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14} \begin {gather*} x^2-\frac {1+10 \log (3)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^3 + 10*Log[3])/x^2,x]

[Out]

x^2 - (1 + 10*Log[3])/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} &=\int \left (2 x+\frac {1+10 \log (3)}{x^2}\right ) \, dx\\ &=x^2-\frac {1+10 \log (3)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.67 \begin {gather*} x^2+\frac {-1-10 \log (3)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^3 + 10*Log[3])/x^2,x]

[Out]

x^2 + (-1 - 10*Log[3])/x

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fricas [A]  time = 1.02, size = 13, normalized size = 0.62 \begin {gather*} \frac {x^{3} - 10 \, \log \relax (3) - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(3)+2*x^3+1)/x^2,x, algorithm="fricas")

[Out]

(x^3 - 10*log(3) - 1)/x

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giac [A]  time = 0.20, size = 15, normalized size = 0.71 \begin {gather*} x^{2} - \frac {10 \, \log \relax (3) + 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(3)+2*x^3+1)/x^2,x, algorithm="giac")

[Out]

x^2 - (10*log(3) + 1)/x

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maple [A]  time = 0.02, size = 14, normalized size = 0.67

method	result	size
norman	\(\frac {x^{3}-1-10 \ln \relax (3)}{x}\)	\(14\)
default	\(x^{2}-\frac {10 \ln \relax (3)+1}{x}\)	\(16\)
gosper	\(-\frac {-x^{3}+10 \ln \relax (3)+1}{x}\)	\(17\)
risch	\(x^{2}-\frac {10 \ln \relax (3)}{x}-\frac {1}{x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*ln(3)+2*x^3+1)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/x*(x^3-1-10*ln(3))

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maxima [A]  time = 0.38, size = 15, normalized size = 0.71 \begin {gather*} x^{2} - \frac {10 \, \log \relax (3) + 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(3)+2*x^3+1)/x^2,x, algorithm="maxima")

[Out]

x^2 - (10*log(3) + 1)/x

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mupad [B]  time = 1.33, size = 15, normalized size = 0.71 \begin {gather*} x^2-\frac {10\,\ln \relax (3)+1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*log(3) + 2*x^3 + 1)/x^2,x)

[Out]

x^2 - (10*log(3) + 1)/x

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sympy [A]  time = 0.08, size = 12, normalized size = 0.57 \begin {gather*} x^{2} + \frac {- 10 \log {\relax (3 )} - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*ln(3)+2*x**3+1)/x**2,x)

[Out]

x**2 + (-10*log(3) - 1)/x

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