3.40.70 \(\int (5-16 x+3 x^2+(-6+4 x) \log (x^2)+\log ^2(x^2)) \, dx\)

Optimal. Leaf size=17 \[ x \left (-\frac {4}{x}+\left (-5+x+\log \left (x^2\right )\right )^2\right ) \]

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Rubi [B]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 3.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2313, 9, 2296, 2295} \begin {gather*} x^3-8 x^2+x \log ^2\left (x^2\right )-4 x \log \left (x^2\right )-2 \left (3 x-x^2\right ) \log \left (x^2\right )+13 x-2 (3-x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[5 - 16*x + 3*x^2 + (-6 + 4*x)*Log[x^2] + Log[x^2]^2,x]

[Out]

-2*(3 - x)^2 + 13*x - 8*x^2 + x^3 - 4*x*Log[x^2] - 2*(3*x - x^2)*Log[x^2] + x*Log[x^2]^2

Rule 9

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

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=5 x-8 x^2+x^3+\int (-6+4 x) \log \left (x^2\right ) \, dx+\int \log ^2\left (x^2\right ) \, dx\\ &=5 x-8 x^2+x^3-2 \left (3 x-x^2\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )-2 \int 2 (-3+x) \, dx-4 \int \log \left (x^2\right ) \, dx\\ &=-2 (3-x)^2+13 x-8 x^2+x^3-4 x \log \left (x^2\right )-2 \left (3 x-x^2\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.00, size = 36, normalized size = 2.12 \begin {gather*} 25 x-10 x^2+x^3-10 x \log \left (x^2\right )+2 x^2 \log \left (x^2\right )+x \log ^2\left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[5 - 16*x + 3*x^2 + (-6 + 4*x)*Log[x^2] + Log[x^2]^2,x]

[Out]

25*x - 10*x^2 + x^3 - 10*x*Log[x^2] + 2*x^2*Log[x^2] + x*Log[x^2]^2

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fricas [A]  time = 0.79, size = 33, normalized size = 1.94 \begin {gather*} x^{3} + x \log \left (x^{2}\right )^{2} - 10 \, x^{2} + 2 \, {\left (x^{2} - 5 \, x\right )} \log \left (x^{2}\right ) + 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x^2)^2+(4*x-6)*log(x^2)+3*x^2-16*x+5,x, algorithm="fricas")

[Out]

x^3 + x*log(x^2)^2 - 10*x^2 + 2*(x^2 - 5*x)*log(x^2) + 25*x

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giac [B]  time = 0.13, size = 40, normalized size = 2.35 \begin {gather*} x^{3} + x \log \left (x^{2}\right )^{2} - 10 \, x^{2} + 2 \, {\left (x^{2} - 3 \, x\right )} \log \left (x^{2}\right ) - 4 \, x \log \left (x^{2}\right ) + 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x^2)^2+(4*x-6)*log(x^2)+3*x^2-16*x+5,x, algorithm="giac")

[Out]

x^3 + x*log(x^2)^2 - 10*x^2 + 2*(x^2 - 3*x)*log(x^2) - 4*x*log(x^2) + 25*x

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maple [B]  time = 0.02, size = 37, normalized size = 2.18

method	result	size
default	\(25 x +x \ln \left (x^{2}\right )^{2}-10 x \ln \left (x^{2}\right )+2 x^{2} \ln \left (x^{2}\right )-10 x^{2}+x^{3}\)	\(37\)
norman	\(25 x +x \ln \left (x^{2}\right )^{2}-10 x \ln \left (x^{2}\right )+2 x^{2} \ln \left (x^{2}\right )-10 x^{2}+x^{3}\)	\(37\)
risch	\(25 x +x \ln \left (x^{2}\right )^{2}-10 x \ln \left (x^{2}\right )+2 x^{2} \ln \left (x^{2}\right )-10 x^{2}+x^{3}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x^2)^2+(4*x-6)*ln(x^2)+3*x^2-16*x+5,x,method=_RETURNVERBOSE)

[Out]

25*x+x*ln(x^2)^2-10*x*ln(x^2)+2*x^2*ln(x^2)-10*x^2+x^3

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maxima [B]  time = 0.43, size = 40, normalized size = 2.35 \begin {gather*} x^{3} + x \log \left (x^{2}\right )^{2} - 10 \, x^{2} + 2 \, {\left (x^{2} - 3 \, x\right )} \log \left (x^{2}\right ) - 4 \, x \log \left (x^{2}\right ) + 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x^2)^2+(4*x-6)*log(x^2)+3*x^2-16*x+5,x, algorithm="maxima")

[Out]

x^3 + x*log(x^2)^2 - 10*x^2 + 2*(x^2 - 3*x)*log(x^2) - 4*x*log(x^2) + 25*x

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mupad [B]  time = 2.58, size = 11, normalized size = 0.65 \begin {gather*} x\,{\left (x+\ln \left (x^2\right )-5\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x^2)^2 - 16*x + 3*x^2 + log(x^2)*(4*x - 6) + 5,x)

[Out]

x*(x + log(x^2) - 5)^2

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sympy [B]  time = 0.11, size = 32, normalized size = 1.88 \begin {gather*} x^{3} - 10 x^{2} + x \log {\left (x^{2} \right )}^{2} + 25 x + \left (2 x^{2} - 10 x\right ) \log {\left (x^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x**2)**2+(4*x-6)*ln(x**2)+3*x**2-16*x+5,x)

[Out]

x**3 - 10*x**2 + x*log(x**2)**2 + 25*x + (2*x**2 - 10*x)*log(x**2)

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