3.24.56 \(\int \frac {3 x+8 \log (x^2)}{2 x} \, dx\)

Optimal. Leaf size=13 \[ 1+\frac {3 x}{2}+\log ^2\left (x^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 14, 2301} \begin {gather*} \log ^2\left (x^2\right )+\frac {3 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*x + 8*Log[x^2])/(2*x),x]

[Out]

(3*x)/2 + 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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {3 x+8 \log \left (x^2\right )}{x} \, dx\\ &=\frac {1}{2} \int \left (3+\frac {8 \log \left (x^2\right )}{x}\right ) \, dx\\ &=\frac {3 x}{2}+4 \int \frac {\log \left (x^2\right )}{x} \, dx\\ &=\frac {3 x}{2}+\log ^2\left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.92 \begin {gather*} \frac {3 x}{2}+\log ^2\left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*x + 8*Log[x^2])/(2*x),x]

[Out]

(3*x)/2 + Log[x^2]^2

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fricas [A]  time = 0.49, size = 10, normalized size = 0.77 \begin {gather*} \log \left (x^{2}\right )^{2} + \frac {3}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(8*log(x^2)+3*x)/x,x, algorithm="fricas")

[Out]

log(x^2)^2 + 3/2*x

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giac [A]  time = 0.21, size = 10, normalized size = 0.77 \begin {gather*} \log \left (x^{2}\right )^{2} + \frac {3}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(8*log(x^2)+3*x)/x,x, algorithm="giac")

[Out]

log(x^2)^2 + 3/2*x

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maple [A]  time = 0.03, size = 11, normalized size = 0.85

method	result	size
default	\(\frac {3 x}{2}+\ln \left (x^{2}\right )^{2}\)	\(11\)
norman	\(\frac {3 x}{2}+\ln \left (x^{2}\right )^{2}\)	\(11\)
risch	\(\frac {3 x}{2}+\ln \left (x^{2}\right )^{2}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(8*ln(x^2)+3*x)/x,x,method=_RETURNVERBOSE)

[Out]

3/2*x+ln(x^2)^2

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maxima [A]  time = 0.35, size = 10, normalized size = 0.77 \begin {gather*} \log \left (x^{2}\right )^{2} + \frac {3}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(8*log(x^2)+3*x)/x,x, algorithm="maxima")

[Out]

log(x^2)^2 + 3/2*x

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mupad [B]  time = 1.50, size = 10, normalized size = 0.77 \begin {gather*} {\ln \left (x^2\right )}^2+\frac {3\,x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x)/2 + 4*log(x^2))/x,x)

[Out]

(3*x)/2 + log(x^2)^2

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sympy [A]  time = 0.09, size = 10, normalized size = 0.77 \begin {gather*} \frac {3 x}{2} + \log {\left (x^{2} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(8*ln(x**2)+3*x)/x,x)

[Out]

3*x/2 + log(x**2)**2

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