\(\int \frac {1}{10} (-22 x-5 \log (x^2)) \, dx\) [4836]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 14, antiderivative size = 27 \[ \int \frac {1}{10} \left (-22 x-5 \log \left (x^2\right )\right ) \, dx=x \left (-\frac {8 x}{5}+\frac {1+x}{x}+\frac {1}{2} \left (x-\log \left (x^2\right )\right )\right ) \]

[Out]

x*((1+x)/x-11/10*x-1/2*ln(x^2))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2332} \[ \int \frac {1}{10} \left (-22 x-5 \log \left (x^2\right )\right ) \, dx=-\frac {11 x^2}{10}-\frac {1}{2} x \log \left (x^2\right )+x \]

[In]

Int[(-22*x - 5*Log[x^2])/10,x]

[Out]

x - (11*x^2)/10 - (x*Log[x^2])/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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{10} \int \left (-22 x-5 \log \left (x^2\right )\right ) \, dx \\ & = -\frac {11 x^2}{10}-\frac {1}{2} \int \log \left (x^2\right ) \, dx \\ & = x-\frac {11 x^2}{10}-\frac {1}{2} x \log \left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {1}{10} \left (-22 x-5 \log \left (x^2\right )\right ) \, dx=x-\frac {11 x^2}{10}-\frac {1}{2} x \log \left (x^2\right ) \]

[In]

Integrate[(-22*x - 5*Log[x^2])/10,x]

[Out]

x - (11*x^2)/10 - (x*Log[x^2])/2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56

method result size
default \(-\frac {11 x^{2}}{10}-\frac {x \ln \left (x^{2}\right )}{2}+x\) \(15\)
norman \(-\frac {11 x^{2}}{10}-\frac {x \ln \left (x^{2}\right )}{2}+x\) \(15\)
risch \(-\frac {11 x^{2}}{10}-\frac {x \ln \left (x^{2}\right )}{2}+x\) \(15\)
parallelrisch \(-\frac {11 x^{2}}{10}-\frac {x \ln \left (x^{2}\right )}{2}+x\) \(15\)
parts \(-\frac {11 x^{2}}{10}-\frac {x \ln \left (x^{2}\right )}{2}+x\) \(15\)

[In]

int(-1/2*ln(x^2)-11/5*x,x,method=_RETURNVERBOSE)

[Out]

-11/10*x^2-1/2*x*ln(x^2)+x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52 \[ \int \frac {1}{10} \left (-22 x-5 \log \left (x^2\right )\right ) \, dx=-\frac {11}{10} \, x^{2} - \frac {1}{2} \, x \log \left (x^{2}\right ) + x \]

[In]

integrate(-1/2*log(x^2)-11/5*x,x, algorithm="fricas")

[Out]

-11/10*x^2 - 1/2*x*log(x^2) + x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {1}{10} \left (-22 x-5 \log \left (x^2\right )\right ) \, dx=- \frac {11 x^{2}}{10} - \frac {x \log {\left (x^{2} \right )}}{2} + x \]

[In]

integrate(-1/2*ln(x**2)-11/5*x,x)

[Out]

-11*x**2/10 - x*log(x**2)/2 + x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52 \[ \int \frac {1}{10} \left (-22 x-5 \log \left (x^2\right )\right ) \, dx=-\frac {11}{10} \, x^{2} - \frac {1}{2} \, x \log \left (x^{2}\right ) + x \]

[In]

integrate(-1/2*log(x^2)-11/5*x,x, algorithm="maxima")

[Out]

-11/10*x^2 - 1/2*x*log(x^2) + x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52 \[ \int \frac {1}{10} \left (-22 x-5 \log \left (x^2\right )\right ) \, dx=-\frac {11}{10} \, x^{2} - \frac {1}{2} \, x \log \left (x^{2}\right ) + x \]

[In]

integrate(-1/2*log(x^2)-11/5*x,x, algorithm="giac")

[Out]

-11/10*x^2 - 1/2*x*log(x^2) + x

Mupad [B] (verification not implemented)

Time = 10.68 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.44 \[ \int \frac {1}{10} \left (-22 x-5 \log \left (x^2\right )\right ) \, dx=-\frac {x\,\left (11\,x+\ln \left (x^{10}\right )-10\right )}{10} \]

[In]

int(- (11*x)/5 - log(x^2)/2,x)

[Out]

-(x*(11*x + log(x^10) - 10))/10