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

Optimal. Leaf size=27 \[ x \left (-\frac {8 x}{5}+\frac {1+x}{x}+\frac {1}{2} \left (x-\log \left (x^2\right )\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 0.67, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2295} \begin {gather*} -\frac {11 x^2}{10}-\frac {1}{2} x \log \left (x^2\right )+x \end {gather*}

Antiderivative was successfully verified.

[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 2295

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 0.67 \begin {gather*} x-\frac {11 x^2}{10}-\frac {1}{2} x \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-22*x - 5*Log[x^2])/10,x]

[Out]

x - (11*x^2)/10 - (x*Log[x^2])/2

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fricas [A]  time = 0.77, size = 14, normalized size = 0.52 \begin {gather*} -\frac {11}{10} \, x^{2} - \frac {1}{2} \, x \log \left (x^{2}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.21, size = 14, normalized size = 0.52 \begin {gather*} -\frac {11}{10} \, x^{2} - \frac {1}{2} \, x \log \left (x^{2}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.01, size = 15, normalized size = 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	\(-\frac {11 x^{2}}{10}-\frac {x \ln \left (x^{2}\right )}{2}+x\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.38, size = 14, normalized size = 0.52 \begin {gather*} -\frac {11}{10} \, x^{2} - \frac {1}{2} \, x \log \left (x^{2}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 3.26, size = 12, normalized size = 0.44 \begin {gather*} -\frac {x\,\left (11\,x+\ln \left (x^{10}\right )-10\right )}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- (11*x)/5 - log(x^2)/2,x)

[Out]

-(x*(11*x + log(x^10) - 10))/10

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sympy [A]  time = 0.09, size = 15, normalized size = 0.56 \begin {gather*} - \frac {11 x^{2}}{10} - \frac {x \log {\left (x^{2} \right )}}{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11*x**2/10 - x*log(x**2)/2 + x

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