\(\int \frac {-300-750 x-900 \log (\frac {x^2}{3})}{x^3} \, dx\) [6454]

Optimal result

Integrand size = 19, antiderivative size = 24 \[ \int \frac {-300-750 x-900 \log \left (\frac {x^2}{3}\right )}{x^3} \, dx=\frac {150 \left (4+5 x+3 \left (x^2+\log \left (\frac {x^2}{3}\right )\right )\right )}{x^2} \]

[Out]

150*(5*x+3*x^2+3*ln(1/3*x^2)+4)/x^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {14, 37, 2341} \[ \int \frac {-300-750 x-900 \log \left (\frac {x^2}{3}\right )}{x^3} \, dx=\frac {75 (5 x+2)^2}{2 x^2}+\frac {450}{x^2}+\frac {450 \log \left (\frac {x^2}{3}\right )}{x^2} \]

[In]

Int[(-300 - 750*x - 900*Log[x^2/3])/x^3,x]

[Out]

450/x^2 + (75*(2 + 5*x)^2)/(2*x^2) + (450*Log[x^2/3])/x^2

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 {150 (2+5 x)}{x^3}-\frac {900 \log \left (\frac {x^2}{3}\right )}{x^3}\right ) \, dx \\ & = -\left (150 \int \frac {2+5 x}{x^3} \, dx\right )-900 \int \frac {\log \left (\frac {x^2}{3}\right )}{x^3} \, dx \\ & = \frac {450}{x^2}+\frac {75 (2+5 x)^2}{2 x^2}+\frac {450 \log \left (\frac {x^2}{3}\right )}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-300-750 x-900 \log \left (\frac {x^2}{3}\right )}{x^3} \, dx=\frac {600}{x^2}+\frac {750}{x}+\frac {450 \log \left (\frac {x^2}{3}\right )}{x^2} \]

[In]

Integrate[(-300 - 750*x - 900*Log[x^2/3])/x^3,x]

[Out]

600/x^2 + 750/x + (450*Log[x^2/3])/x^2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method	result	size
norman	\(\frac {600+750 x +450 \ln \left (\frac {x^{2}}{3}\right )}{x^{2}}\)	\(18\)
parallelrisch	\(-\frac {-600-750 x -450 \ln \left (\frac {x^{2}}{3}\right )}{x^{2}}\)	\(19\)
risch	\(\frac {450 \ln \left (\frac {x^{2}}{3}\right )}{x^{2}}+\frac {750 x +600}{x^{2}}\)	\(23\)
default	\(-\frac {450 \ln \left (3\right )}{x^{2}}+\frac {450 \ln \left (x^{2}\right )}{x^{2}}+\frac {600}{x^{2}}+\frac {750}{x}\)	\(28\)
parts	\(-\frac {450 \ln \left (3\right )}{x^{2}}+\frac {450 \ln \left (x^{2}\right )}{x^{2}}+\frac {600}{x^{2}}+\frac {750}{x}\)	\(28\)

[In]

int((-900*ln(1/3*x^2)-750*x-300)/x^3,x,method=_RETURNVERBOSE)

[Out]

(600+750*x+450*ln(1/3*x^2))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-300-750 x-900 \log \left (\frac {x^2}{3}\right )}{x^3} \, dx=\frac {150 \, {\left (5 \, x + 3 \, \log \left (\frac {1}{3} \, x^{2}\right ) + 4\right )}}{x^{2}} \]

[In]

integrate((-900*log(1/3*x^2)-750*x-300)/x^3,x, algorithm="fricas")

[Out]

150*(5*x + 3*log(1/3*x^2) + 4)/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-300-750 x-900 \log \left (\frac {x^2}{3}\right )}{x^3} \, dx=- \frac {- 750 x - 600}{x^{2}} + \frac {450 \log {\left (\frac {x^{2}}{3} \right )}}{x^{2}} \]

[In]

integrate((-900*ln(1/3*x**2)-750*x-300)/x**3,x)

[Out]

-(-750*x - 600)/x**2 + 450*log(x**2/3)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-300-750 x-900 \log \left (\frac {x^2}{3}\right )}{x^3} \, dx=\frac {750}{x} + \frac {450 \, \log \left (\frac {1}{3} \, x^{2}\right )}{x^{2}} + \frac {600}{x^{2}} \]

[In]

integrate((-900*log(1/3*x^2)-750*x-300)/x^3,x, algorithm="maxima")

[Out]

750/x + 450*log(1/3*x^2)/x^2 + 600/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-300-750 x-900 \log \left (\frac {x^2}{3}\right )}{x^3} \, dx=\frac {150 \, {\left (5 \, x + 4\right )}}{x^{2}} + \frac {450 \, \log \left (\frac {1}{3} \, x^{2}\right )}{x^{2}} \]

[In]

integrate((-900*log(1/3*x^2)-750*x-300)/x^3,x, algorithm="giac")

[Out]

150*(5*x + 4)/x^2 + 450*log(1/3*x^2)/x^2

Mupad [B] (verification not implemented)

Time = 11.55 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-300-750 x-900 \log \left (\frac {x^2}{3}\right )}{x^3} \, dx=\frac {150\,\left (5\,x+\ln \left (\frac {x^6}{27}\right )+4\right )}{x^2} \]

[In]

int(-(750*x + 900*log(x^2/3) + 300)/x^3,x)

[Out]

(150*(5*x + log(x^6/27) + 4))/x^2