3.97.56 \(\int \frac {-8+e^3 (4-8 x)+32 x^2+(4-8 x) \log (x^2)}{x} \, dx\)

Optimal. Leaf size=14 \[ \left (-2+e^3-4 x+\log \left (x^2\right )\right )^2 \]

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Rubi [B]  time = 0.05, antiderivative size = 39, normalized size of antiderivative = 2.79, number of steps used = 7, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {14, 2346, 2301, 2295} \begin {gather*} 16 x^2+\log ^2\left (x^2\right )-8 x \log \left (x^2\right )-8 e^3 x+16 x-4 \left (2-e^3\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 + E^3*(4 - 8*x) + 32*x^2 + (4 - 8*x)*Log[x^2])/x,x]

[Out]

16*x - 8*E^3*x + 16*x^2 - 4*(2 - E^3)*Log[x] - 8*x*Log[x^2] + Log[x^2]^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 2295

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

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]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4 \left (-2+e^3-2 e^3 x+8 x^2\right )}{x}-\frac {4 (-1+2 x) \log \left (x^2\right )}{x}\right ) \, dx\\ &=4 \int \frac {-2+e^3-2 e^3 x+8 x^2}{x} \, dx-4 \int \frac {(-1+2 x) \log \left (x^2\right )}{x} \, dx\\ &=4 \int \left (-2 e^3+\frac {-2+e^3}{x}+8 x\right ) \, dx+4 \int \frac {\log \left (x^2\right )}{x} \, dx-8 \int \log \left (x^2\right ) \, dx\\ &=16 x-8 e^3 x+16 x^2-4 \left (2-e^3\right ) \log (x)-8 x \log \left (x^2\right )+\log ^2\left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 39, normalized size = 2.79 \begin {gather*} 16 x-8 e^3 x+16 x^2-8 \log (x)+4 e^3 \log (x)-8 x \log \left (x^2\right )+\log ^2\left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + E^3*(4 - 8*x) + 32*x^2 + (4 - 8*x)*Log[x^2])/x,x]

[Out]

16*x - 8*E^3*x + 16*x^2 - 8*Log[x] + 4*E^3*Log[x] - 8*x*Log[x^2] + Log[x^2]^2

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fricas [B]  time = 0.53, size = 35, normalized size = 2.50 \begin {gather*} 16 \, x^{2} - 8 \, x e^{3} - 2 \, {\left (4 \, x - e^{3} + 2\right )} \log \left (x^{2}\right ) + \log \left (x^{2}\right )^{2} + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+4)*log(x^2)+(-8*x+4)*exp(3)+32*x^2-8)/x,x, algorithm="fricas")

[Out]

16*x^2 - 8*x*e^3 - 2*(4*x - e^3 + 2)*log(x^2) + log(x^2)^2 + 16*x

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giac [B]  time = 0.22, size = 37, normalized size = 2.64 \begin {gather*} 16 \, x^{2} - 8 \, x e^{3} - 8 \, x \log \left (x^{2}\right ) + \log \left (x^{2}\right )^{2} + 4 \, e^{3} \log \relax (x) + 16 \, x - 8 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+4)*log(x^2)+(-8*x+4)*exp(3)+32*x^2-8)/x,x, algorithm="giac")

[Out]

16*x^2 - 8*x*e^3 - 8*x*log(x^2) + log(x^2)^2 + 4*e^3*log(x) + 16*x - 8*log(x)

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maple [B]  time = 0.04, size = 39, normalized size = 2.79

method	result	size
norman	\(\ln \left (x^{2}\right )^{2}+\left (16-8 \,{\mathrm e}^{3}\right ) x +\left (2 \,{\mathrm e}^{3}-4\right ) \ln \left (x^{2}\right )+16 x^{2}-8 x \ln \left (x^{2}\right )\)	\(39\)
default	\(16 x^{2}-8 x \,{\mathrm e}^{3}+4 \ln \relax (x ) {\mathrm e}^{3}-8 \ln \relax (x )-8 x \ln \left (x^{2}\right )+16 x +4 \ln \relax (x ) \ln \left (x^{2}\right )-4 \ln \relax (x )^{2}\)	\(46\)
risch	\(16 x^{2}-8 x \,{\mathrm e}^{3}+4 \ln \relax (x ) {\mathrm e}^{3}-8 \ln \relax (x )-8 x \ln \left (x^{2}\right )+16 x +4 \ln \relax (x ) \ln \left (x^{2}\right )-4 \ln \relax (x )^{2}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x+4)*ln(x^2)+(-8*x+4)*exp(3)+32*x^2-8)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x^2)^2+(16-8*exp(3))*x+(2*exp(3)-4)*ln(x^2)+16*x^2-8*x*ln(x^2)

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maxima [B]  time = 0.36, size = 37, normalized size = 2.64 \begin {gather*} 16 \, x^{2} - 8 \, x e^{3} - 8 \, x \log \left (x^{2}\right ) + \log \left (x^{2}\right )^{2} + 4 \, e^{3} \log \relax (x) + 16 \, x - 8 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+4)*log(x^2)+(-8*x+4)*exp(3)+32*x^2-8)/x,x, algorithm="maxima")

[Out]

16*x^2 - 8*x*e^3 - 8*x*log(x^2) + log(x^2)^2 + 4*e^3*log(x) + 16*x - 8*log(x)

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mupad [B]  time = 7.54, size = 26, normalized size = 1.86 \begin {gather*} \left (4\,x-\ln \left (x^2\right )\right )\,\left (4\,x-\ln \left (x^2\right )-2\,{\mathrm {e}}^3+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2)*(8*x - 4) - 32*x^2 + exp(3)*(8*x - 4) + 8)/x,x)

[Out]

(4*x - log(x^2))*(4*x - log(x^2) - 2*exp(3) + 4)

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sympy [B]  time = 0.16, size = 37, normalized size = 2.64 \begin {gather*} 16 x^{2} - 8 x \log {\left (x^{2} \right )} + x \left (16 - 8 e^{3}\right ) + 4 \left (-2 + e^{3}\right ) \log {\relax (x )} + \log {\left (x^{2} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+4)*ln(x**2)+(-8*x+4)*exp(3)+32*x**2-8)/x,x)

[Out]

16*x**2 - 8*x*log(x**2) + x*(16 - 8*exp(3)) + 4*(-2 + exp(3))*log(x) + log(x**2)**2

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