3.17.100 \(\int \frac {5 x^2+30 \log (2)}{x^2 \log (2)} \, dx\)

Optimal. Leaf size=18 \[ \frac {5 \left (-6-2 x+\frac {x^2}{\log (2)}\right )}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 14} \begin {gather*} \frac {5 x}{\log (2)}-\frac {30}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x^2 + 30*Log[2])/(x^2*Log[2]),x]

[Out]

-30/x + (5*x)/Log[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 {\int \frac {5 x^2+30 \log (2)}{x^2} \, dx}{\log (2)}\\ &=\frac {\int \left (5+\frac {30 \log (2)}{x^2}\right ) \, dx}{\log (2)}\\ &=-\frac {30}{x}+\frac {5 x}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.83 \begin {gather*} \frac {5 \left (x-\frac {\log (64)}{x}\right )}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^2 + 30*Log[2])/(x^2*Log[2]),x]

[Out]

(5*(x - Log[64]/x))/Log[2]

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fricas [A]  time = 0.61, size = 17, normalized size = 0.94 \begin {gather*} \frac {5 \, {\left (x^{2} - 6 \, \log \relax (2)\right )}}{x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((30*log(2)+5*x^2)/x^2/log(2),x, algorithm="fricas")

[Out]

5*(x^2 - 6*log(2))/(x*log(2))

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giac [A]  time = 0.37, size = 15, normalized size = 0.83 \begin {gather*} \frac {5 \, {\left (x - \frac {6 \, \log \relax (2)}{x}\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((30*log(2)+5*x^2)/x^2/log(2),x, algorithm="giac")

[Out]

5*(x - 6*log(2)/x)/log(2)

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maple [A]  time = 0.03, size = 14, normalized size = 0.78

method	result	size
risch	\(\frac {5 x}{\ln \relax (2)}-\frac {30}{x}\)	\(14\)
default	\(\frac {5 x -\frac {30 \ln \relax (2)}{x}}{\ln \relax (2)}\)	\(16\)
norman	\(\frac {-30+\frac {5 x^{2}}{\ln \relax (2)}}{x}\)	\(16\)
gosper	\(-\frac {5 \left (-x^{2}+6 \ln \relax (2)\right )}{x \ln \relax (2)}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((30*ln(2)+5*x^2)/x^2/ln(2),x,method=_RETURNVERBOSE)

[Out]

5*x/ln(2)-30/x

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maxima [A]  time = 0.39, size = 15, normalized size = 0.83 \begin {gather*} \frac {5 \, {\left (x - \frac {6 \, \log \relax (2)}{x}\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((30*log(2)+5*x^2)/x^2/log(2),x, algorithm="maxima")

[Out]

5*(x - 6*log(2)/x)/log(2)

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mupad [B]  time = 1.02, size = 13, normalized size = 0.72 \begin {gather*} \frac {5\,x}{\ln \relax (2)}-\frac {30}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((30*log(2) + 5*x^2)/(x^2*log(2)),x)

[Out]

(5*x)/log(2) - 30/x

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sympy [A]  time = 0.09, size = 12, normalized size = 0.67 \begin {gather*} \frac {5 x - \frac {30 \log {\relax (2 )}}{x}}{\log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((30*ln(2)+5*x**2)/x**2/ln(2),x)

[Out]

(5*x - 30*log(2)/x)/log(2)

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