\(\int \frac {45+x^2+45 \log (-\frac {1}{3 x})}{x^2} \, dx\) [9225]

Optimal result

Integrand size = 19, antiderivative size = 16 \[ \int \frac {45+x^2+45 \log \left (-\frac {1}{3 x}\right )}{x^2} \, dx=4+x-\frac {45 \log \left (-\frac {1}{3 x}\right )}{x} \]

[Out]

-45*ln(-1/3/x)/x+4+x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2341} \[ \int \frac {45+x^2+45 \log \left (-\frac {1}{3 x}\right )}{x^2} \, dx=x-\frac {45 \log \left (-\frac {1}{3 x}\right )}{x} \]

[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 2341

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {45+x^2+45 \log \left (-\frac {1}{3 x}\right )}{x^2} \, dx=x-\frac {45 \log \left (-\frac {1}{3 x}\right )}{x} \]

[In]

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

[Out]

x - (45*Log[-1/3*1/x])/x

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 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	\(x -\frac {45 \ln \left (-\frac {1}{3 x}\right )}{x}\)	\(14\)
parts	\(x -\frac {45 \ln \left (-\frac {1}{3 x}\right )}{x}\)	\(14\)
norman	\(\frac {x^{2}-45 \ln \left (-\frac {1}{3 x}\right )}{x}\)	\(17\)
parallelrisch	\(\frac {x^{2}-45 \ln \left (-\frac {1}{3 x}\right )}{x}\)	\(17\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {45+x^2+45 \log \left (-\frac {1}{3 x}\right )}{x^2} \, dx=\frac {x^{2} - 45 \, \log \left (-\frac {1}{3 \, x}\right )}{x} \]

[In]

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

[Out]

(x^2 - 45*log(-1/3/x))/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {45+x^2+45 \log \left (-\frac {1}{3 x}\right )}{x^2} \, dx=x - \frac {45 \log {\left (- \frac {1}{3 x} \right )}}{x} \]

[In]

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

[Out]

x - 45*log(-1/(3*x))/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {45+x^2+45 \log \left (-\frac {1}{3 x}\right )}{x^2} \, dx=x - \frac {45 \, \log \left (-\frac {1}{3 \, x}\right )}{x} \]

[In]

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

[Out]

x - 45*log(-1/3/x)/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {45+x^2+45 \log \left (-\frac {1}{3 x}\right )}{x^2} \, dx=x - \frac {45 \, \log \left (-\frac {1}{3 \, x}\right )}{x} \]

[In]

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

[Out]

x - 45*log(-1/3/x)/x

Mupad [B] (verification not implemented)

Time = 14.43 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {45+x^2+45 \log \left (-\frac {1}{3 x}\right )}{x^2} \, dx=x-\frac {45\,\ln \left (-\frac {1}{3\,x}\right )}{x} \]

[In]

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

[Out]

x - (45*log(-1/(3*x)))/x