3.93.25 \(\int \frac {45+x^2+45 \log (-\frac {1}{3 x})}{x^2} \, dx\)

Optimal. Leaf size=16 \[ 4+x-\frac {45 \log \left (-\frac {1}{3 x}\right )}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2304} \begin {gather*} x-\frac {45 \log \left (-\frac {1}{3 x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(45 + x^2 + 45*Log[-1/3*1/x])/x^2,x]

[Out]

x - (45*Log[-1/3*1/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 {45+x^2}{x^2}+\frac {45 \log \left (-\frac {1}{3 x}\right )}{x^2}\right ) \, dx\\ &=45 \int \frac {\log \left (-\frac {1}{3 x}\right )}{x^2} \, dx+\int \frac {45+x^2}{x^2} \, dx\\ &=\frac {45}{x}-\frac {45 \log \left (-\frac {1}{3 x}\right )}{x}+\int \left (1+\frac {45}{x^2}\right ) \, dx\\ &=x-\frac {45 \log \left (-\frac {1}{3 x}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.94 \begin {gather*} x-\frac {45 \log \left (-\frac {1}{3 x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(45 + x^2 + 45*Log[-1/3*1/x])/x^2,x]

[Out]

x - (45*Log[-1/3*1/x])/x

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fricas [A]  time = 0.87, size = 16, normalized size = 1.00 \begin {gather*} \frac {x^{2} - 45 \, \log \left (-\frac {1}{3 \, x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(-1/3/x)+x^2+45)/x^2,x, algorithm="fricas")

[Out]

(x^2 - 45*log(-1/3/x))/x

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giac [A]  time = 0.24, size = 13, normalized size = 0.81 \begin {gather*} x - \frac {45 \, \log \left (-\frac {1}{3 \, x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(-1/3/x)+x^2+45)/x^2,x, algorithm="giac")

[Out]

x - 45*log(-1/3/x)/x

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maple [A]  time = 0.06, size = 14, normalized size = 0.88

method	result	size
derivativedivides	\(x -\frac {45 \ln \left (-\frac {1}{3 x}\right )}{x}\)	\(14\)
default	\(x -\frac {45 \ln \left (-\frac {1}{3 x}\right )}{x}\)	\(14\)
risch	\(x -\frac {45 \ln \left (-\frac {1}{3 x}\right )}{x}\)	\(14\)
norman	\(\frac {x^{2}-45 \ln \left (-\frac {1}{3 x}\right )}{x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*ln(-1/3/x)+x^2+45)/x^2,x,method=_RETURNVERBOSE)

[Out]

x-45*ln(-1/3/x)/x

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maxima [A]  time = 0.36, size = 13, normalized size = 0.81 \begin {gather*} x - \frac {45 \, \log \left (-\frac {1}{3 \, x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(-1/3/x)+x^2+45)/x^2,x, algorithm="maxima")

[Out]

x - 45*log(-1/3/x)/x

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mupad [B]  time = 7.25, size = 13, normalized size = 0.81 \begin {gather*} x-\frac {45\,\ln \left (-\frac {1}{3\,x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*log(-1/(3*x)) + x^2 + 45)/x^2,x)

[Out]

x - (45*log(-1/(3*x)))/x

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sympy [A]  time = 0.10, size = 12, normalized size = 0.75 \begin {gather*} x - \frac {45 \log {\left (- \frac {1}{3 x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*ln(-1/3/x)+x**2+45)/x**2,x)

[Out]

x - 45*log(-1/(3*x))/x

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