3.18.1 \(\int \frac {2 x-2 x^2+\frac {1}{27} e^{2 \log (3) \log (x^2)} (x+(-16+4 x) \log (3))}{x} \, dx\)

Optimal. Leaf size=24 \[ 2-(4-x) \left (-2+3^{-3+2 \log \left (x^2\right )}-x\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {14, 2274, 12, 15, 43} \begin {gather*} \frac {1}{27} x \left (x^2\right )^{\log (9)}-\frac {4}{27} \left (x^2\right )^{\log (9)}-(1-x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x - 2*x^2 + (E^(2*Log[3]*Log[x^2])*(x + (-16 + 4*x)*Log[3]))/27)/x,x]

[Out]

-(1 - x)^2 - (4*(x^2)^Log[9])/27 + (x*(x^2)^Log[9])/27

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 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2 (-1+x)+\frac {3^{-3+2 \log \left (x^2\right )} (-16 \log (3)+x (1+\log (81)))}{x}\right ) \, dx\\ &=-(1-x)^2+\int \frac {3^{-3+2 \log \left (x^2\right )} (-16 \log (3)+x (1+\log (81)))}{x} \, dx\\ &=-(1-x)^2+\int \frac {\left (x^2\right )^{2 \log (3)} (-16 \log (3)+x (1+\log (81)))}{27 x} \, dx\\ &=-(1-x)^2+\frac {1}{27} \int \frac {\left (x^2\right )^{2 \log (3)} (-16 \log (3)+x (1+\log (81)))}{x} \, dx\\ &=-(1-x)^2+\frac {1}{27} \left (x^{-4 \log (3)} \left (x^2\right )^{2 \log (3)}\right ) \int x^{-1+4 \log (3)} (-16 \log (3)+x (1+\log (81))) \, dx\\ &=-(1-x)^2+\frac {1}{27} \left (x^{-4 \log (3)} \left (x^2\right )^{2 \log (3)}\right ) \int \left (-16 x^{-1+4 \log (3)} \log (3)+x^{\log (81)} (1+\log (81))\right ) \, dx\\ &=-(1-x)^2-\frac {4}{27} \left (x^2\right )^{\log (9)}+\frac {1}{27} x \left (x^2\right )^{\log (9)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 23, normalized size = 0.96 \begin {gather*} 3^{-3+2 \log \left (x^2\right )} (-4+x)+2 x-x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - 2*x^2 + (E^(2*Log[3]*Log[x^2])*(x + (-16 + 4*x)*Log[3]))/27)/x,x]

[Out]

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

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fricas [A]  time = 0.70, size = 27, normalized size = 1.12 \begin {gather*} -x^{2} + {\left (x - 4\right )} e^{\left (2 \, \log \relax (3) \log \left (x^{2}\right ) - 3 \, \log \relax (3)\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.41, size = 32, normalized size = 1.33 \begin {gather*} -x^{2} + \frac {1}{27} \, x e^{\left (2 \, \log \relax (3) \log \left (x^{2}\right )\right )} + 2 \, x - \frac {4}{27} \, e^{\left (2 \, \log \relax (3) \log \left (x^{2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 + 1/27*x*e^(2*log(3)*log(x^2)) + 2*x - 4/27*e^(2*log(3)*log(x^2))

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maple [A]  time = 0.11, size = 23, normalized size = 0.96

method	result	size
risch	\(-x^{2}+2 x +\frac {\left (x -4\right ) \left (x^{2}\right )^{2 \ln \relax (3)}}{27}\)	\(23\)
norman	\({\mathrm e}^{2 \ln \relax (3) \ln \left (x^{2}\right )-3 \ln \relax (3)} x +2 x -x^{2}-4 \,{\mathrm e}^{2 \ln \relax (3) \ln \left (x^{2}\right )-3 \ln \relax (3)}\)	\(42\)
default	\(2 x +\frac {3^{2 \ln \left (x^{2}\right )-4 \ln \relax (x )} x \,{\mathrm e}^{4 \ln \relax (3) \ln \relax (x )}}{108 \ln \relax (3)+27}-\frac {4 \,{\mathrm e}^{4 \ln \relax (3) \ln \relax (x )} 3^{2 \ln \left (x^{2}\right )-4 \ln \relax (x )}}{27}+\frac {4 \,3^{2 \ln \left (x^{2}\right )-4 \ln \relax (x )} \ln \relax (3) x \,{\mathrm e}^{4 \ln \relax (3) \ln \relax (x )}}{27 \left (4 \ln \relax (3)+1\right )}-x^{2}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2+2*x+1/27*(x-4)*(x^2)^(2*ln(3))

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maxima [B]  time = 0.62, size = 58, normalized size = 2.42 \begin {gather*} -x^{2} + \frac {4 \, x e^{\left (4 \, \log \relax (3) \log \relax (x)\right )} \log \relax (3)}{27 \, {\left (4 \, \log \relax (3) + 1\right )}} + 2 \, x + \frac {x e^{\left (4 \, \log \relax (3) \log \relax (x)\right )}}{27 \, {\left (4 \, \log \relax (3) + 1\right )}} - \frac {4}{27} \, e^{\left (2 \, \log \relax (3) \log \left (x^{2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 + 4/27*x*e^(4*log(3)*log(x))*log(3)/(4*log(3) + 1) + 2*x + 1/27*x*e^(4*log(3)*log(x))/(4*log(3) + 1) - 4/
27*e^(2*log(3)*log(x^2))

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mupad [B]  time = 1.09, size = 30, normalized size = 1.25 \begin {gather*} 2\,x-\frac {4\,{\left (x^2\right )}^{2\,\ln \relax (3)}}{27}-x^2+\frac {x\,{\left (x^2\right )}^{2\,\ln \relax (3)}}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - 2*x^2 + exp(2*log(x^2)*log(3) - 3*log(3))*(x + log(3)*(4*x - 16)))/x,x)

[Out]

2*x - (4*(x^2)^(2*log(3)))/27 - x^2 + (x*(x^2)^(2*log(3)))/27

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sympy [A]  time = 0.81, size = 32, normalized size = 1.33 \begin {gather*} - x^{2} + \frac {x e^{4 \log {\relax (3 )} \log {\relax (x )}}}{27} + 2 x - \frac {4 e^{4 \log {\relax (3 )} \log {\relax (x )}}}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-16)*ln(3)+x)*exp(2*ln(3)*ln(x**2)-3*ln(3))-2*x**2+2*x)/x,x)

[Out]

-x**2 + x*exp(4*log(3)*log(x))/27 + 2*x - 4*exp(4*log(3)*log(x))/27

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