3.57.34 \(\int \frac {(-8-24 x^2+e^3 (4+12 x^2)) \log ^2(e^{-2+3 x^2} x^2)^{-2+e^3}}{x \log (e^{-2+3 x^2} x^2)} \, dx\)

Optimal. Leaf size=22 \[ \log ^2\left (e^{-2+3 x^2} x^2\right )^{-2+e^3} \]

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Rubi [A]  time = 0.42, antiderivative size = 37, normalized size of antiderivative = 1.68, number of steps used = 3, number of rules used = 3, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6697, 15, 30} \begin {gather*} \frac {\log ^2\left (e^{3 x^2-2} x^2\right )^{e^3}}{\log ^4\left (e^{3 x^2-2} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-8 - 24*x^2 + E^3*(4 + 12*x^2))*(Log[E^(-2 + 3*x^2)*x^2]^2)^(-2 + E^3))/(x*Log[E^(-2 + 3*x^2)*x^2]),x]

[Out]

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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6697

Int[(u_.)*(v_)^(m_.)*((a_.) + (b_.)*(y_)^(n_))^(p_.), x_Symbol] :> Module[{q, r}, Dist[q*r, Subst[Int[x^m*(a +
b*x^n)^p, x], x, y], x] /;  !FalseQ[r = Divides[y^m, v^m, x]] &&  !FalseQ[q = DerivativeDivides[y, u, x]]] /;
FreeQ[{a, b, m, n, p}, x]

Rubi steps

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

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Mathematica [B]  time = 0.03, size = 52, normalized size = 2.36 \begin {gather*} \frac {2 \left (-2+e^3\right ) \log ^2\left (e^{-2+3 x^2} x^2\right )^{e^3}}{\left (-4+2 e^3\right ) \log ^4\left (e^{-2+3 x^2} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-8 - 24*x^2 + E^3*(4 + 12*x^2))*(Log[E^(-2 + 3*x^2)*x^2]^2)^(-2 + E^3))/(x*Log[E^(-2 + 3*x^2)*x^2]
),x]

[Out]

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

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fricas [A]  time = 0.56, size = 20, normalized size = 0.91 \begin {gather*} {\left (\log \left (x^{2} e^{\left (3 \, x^{2} - 2\right )}\right )^{2}\right )}^{e^{3} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2+4)*exp(3)-24*x^2-8)*exp((exp(3)-2)*log(log(x^2*exp(3*x^2-2))^2))/x/log(x^2*exp(3*x^2-2)),x,
algorithm="fricas")

[Out]

(log(x^2*e^(3*x^2 - 2))^2)^(e^3 - 2)

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giac [A]  time = 0.23, size = 35, normalized size = 1.59 \begin {gather*} \left (\log \left (x^{2} e^{\left (3 \, x^{2} - 2\right )}\right ) \mathrm {sgn}\left (\log \left (x^{2} e^{\left (3 \, x^{2} - 2\right )}\right )\right )\right )^{2 \, e^{3} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2+4)*exp(3)-24*x^2-8)*exp((exp(3)-2)*log(log(x^2*exp(3*x^2-2))^2))/x/log(x^2*exp(3*x^2-2)),x,
algorithm="giac")

[Out]

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

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (12 x^{2}+4\right ) {\mathrm e}^{3}-24 x^{2}-8\right ) {\mathrm e}^{\left ({\mathrm e}^{3}-2\right ) \ln \left (\ln \left (x^{2} {\mathrm e}^{3 x^{2}-2}\right )^{2}\right )}}{x \ln \left (x^{2} {\mathrm e}^{3 x^{2}-2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x^2+4)*exp(3)-24*x^2-8)*exp((exp(3)-2)*ln(ln(x^2*exp(3*x^2-2))^2))/x/ln(x^2*exp(3*x^2-2)),x)

[Out]

int(((12*x^2+4)*exp(3)-24*x^2-8)*exp((exp(3)-2)*ln(ln(x^2*exp(3*x^2-2))^2))/x/ln(x^2*exp(3*x^2-2)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -4 \, \int \frac {{\left (6 \, x^{2} - {\left (3 \, x^{2} + 1\right )} e^{3} + 2\right )} {\left (\log \left (x^{2} e^{\left (3 \, x^{2} - 2\right )}\right )^{2}\right )}^{e^{3} - 2}}{x \log \left (x^{2} e^{\left (3 \, x^{2} - 2\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2+4)*exp(3)-24*x^2-8)*exp((exp(3)-2)*log(log(x^2*exp(3*x^2-2))^2))/x/log(x^2*exp(3*x^2-2)),x,
algorithm="maxima")

[Out]

-4*integrate((6*x^2 - (3*x^2 + 1)*e^3 + 2)*(log(x^2*e^(3*x^2 - 2))^2)^(e^3 - 2)/(x*log(x^2*e^(3*x^2 - 2))), x)

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mupad [B]  time = 3.81, size = 18, normalized size = 0.82 \begin {gather*} {\left ({\left (\ln \left (x^2\right )+3\,x^2-2\right )}^2\right )}^{{\mathrm {e}}^3-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((log(x^2*exp(3*x^2 - 2))^2)^(exp(3) - 2)*(24*x^2 - exp(3)*(12*x^2 + 4) + 8))/(x*log(x^2*exp(3*x^2 - 2)))
,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x**2+4)*exp(3)-24*x**2-8)*exp((exp(3)-2)*ln(ln(x**2*exp(3*x**2-2))**2))/x/ln(x**2*exp(3*x**2-2)
),x)

[Out]

Timed out

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