3.37.42 \(\int \frac {74 e^{e^3} x-56 e^{e^3} x \log (3 x^2)+10 e^{e^3} x \log ^2(3 x^2)}{9-6 \log (3 x^2)+\log ^2(3 x^2)} \, dx\)

Optimal. Leaf size=25 \[ e^{e^3} x^2 \left (5-\frac {2}{3-\log \left (3 x^2\right )}\right ) \]

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Rubi [A]  time = 0.31, antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 11, number of rules used = 7, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.117, Rules used = {6688, 12, 6715, 6742, 2297, 2299, 2178} \begin {gather*} 5 e^{e^3} x^2-\frac {2 e^{e^3} x^2}{3-\log \left (3 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(74*E^E^3*x - 56*E^E^3*x*Log[3*x^2] + 10*E^E^3*x*Log[3*x^2]^2)/(9 - 6*Log[3*x^2] + Log[3*x^2]^2),x]

[Out]

5*E^E^3*x^2 - (2*E^E^3*x^2)/(3 - Log[3*x^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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

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

Rule 2299

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{e^3} x \left (37-28 \log \left (3 x^2\right )+5 \log ^2\left (3 x^2\right )\right )}{\left (3-\log \left (3 x^2\right )\right )^2} \, dx\\ &=\left (2 e^{e^3}\right ) \int \frac {x \left (37-28 \log \left (3 x^2\right )+5 \log ^2\left (3 x^2\right )\right )}{\left (3-\log \left (3 x^2\right )\right )^2} \, dx\\ &=e^{e^3} \operatorname {Subst}\left (\int \frac {37-28 \log (3 x)+5 \log ^2(3 x)}{(3-\log (3 x))^2} \, dx,x,x^2\right )\\ &=\frac {1}{3} e^{e^3} \operatorname {Subst}\left (\int \frac {37-28 \log (x)+5 \log ^2(x)}{(3-\log (x))^2} \, dx,x,3 x^2\right )\\ &=\frac {1}{3} e^{e^3} \operatorname {Subst}\left (\int \left (5-\frac {2}{(-3+\log (x))^2}+\frac {2}{-3+\log (x)}\right ) \, dx,x,3 x^2\right )\\ &=5 e^{e^3} x^2-\frac {1}{3} \left (2 e^{e^3}\right ) \operatorname {Subst}\left (\int \frac {1}{(-3+\log (x))^2} \, dx,x,3 x^2\right )+\frac {1}{3} \left (2 e^{e^3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3+\log (x)} \, dx,x,3 x^2\right )\\ &=5 e^{e^3} x^2-\frac {2 e^{e^3} x^2}{3-\log \left (3 x^2\right )}+\frac {1}{3} \left (2 e^{e^3}\right ) \operatorname {Subst}\left (\int \frac {e^x}{-3+x} \, dx,x,\log \left (3 x^2\right )\right )-\frac {1}{3} \left (2 e^{e^3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3+\log (x)} \, dx,x,3 x^2\right )\\ &=5 e^{e^3} x^2+\frac {2}{3} e^{3+e^3} \text {Ei}\left (-3+\log \left (3 x^2\right )\right )-\frac {2 e^{e^3} x^2}{3-\log \left (3 x^2\right )}-\frac {1}{3} \left (2 e^{e^3}\right ) \operatorname {Subst}\left (\int \frac {e^x}{-3+x} \, dx,x,\log \left (3 x^2\right )\right )\\ &=5 e^{e^3} x^2-\frac {2 e^{e^3} x^2}{3-\log \left (3 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 29, normalized size = 1.16 \begin {gather*} 2 e^{e^3} \left (\frac {5 x^2}{2}+\frac {x^2}{-3+\log \left (3 x^2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(74*E^E^3*x - 56*E^E^3*x*Log[3*x^2] + 10*E^E^3*x*Log[3*x^2]^2)/(9 - 6*Log[3*x^2] + Log[3*x^2]^2),x]

[Out]

2*E^E^3*((5*x^2)/2 + x^2/(-3 + Log[3*x^2]))

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fricas [A]  time = 0.79, size = 34, normalized size = 1.36 \begin {gather*} \frac {5 \, x^{2} e^{\left (e^{3}\right )} \log \left (3 \, x^{2}\right ) - 13 \, x^{2} e^{\left (e^{3}\right )}}{\log \left (3 \, x^{2}\right ) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x*exp(exp(3))*log(3*x^2)^2-56*x*exp(exp(3))*log(3*x^2)+74*x*exp(exp(3)))/(log(3*x^2)^2-6*log(3*x
^2)+9),x, algorithm="fricas")

[Out]

(5*x^2*e^(e^3)*log(3*x^2) - 13*x^2*e^(e^3))/(log(3*x^2) - 3)

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giac [B]  time = 0.18, size = 43, normalized size = 1.72 \begin {gather*} \frac {5 \, x^{2} e^{\left (e^{3}\right )} \log \left (3 \, x^{2}\right )}{\log \left (3 \, x^{2}\right ) - 3} - \frac {13 \, x^{2} e^{\left (e^{3}\right )}}{\log \left (3 \, x^{2}\right ) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x*exp(exp(3))*log(3*x^2)^2-56*x*exp(exp(3))*log(3*x^2)+74*x*exp(exp(3)))/(log(3*x^2)^2-6*log(3*x
^2)+9),x, algorithm="giac")

[Out]

5*x^2*e^(e^3)*log(3*x^2)/(log(3*x^2) - 3) - 13*x^2*e^(e^3)/(log(3*x^2) - 3)

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maple [A]  time = 0.11, size = 28, normalized size = 1.12

method	result	size
risch	\(5 x^{2} {\mathrm e}^{{\mathrm e}^{3}}+\frac {2 x^{2} {\mathrm e}^{{\mathrm e}^{3}}}{\ln \left (3 x^{2}\right )-3}\)	\(28\)
default	\(\frac {{\mathrm e}^{{\mathrm e}^{3}} x^{2} \left (5 \ln \left (x^{2}\right )-13+5 \ln \relax (3)\right )}{\ln \relax (3)+\ln \left (x^{2}\right )-3}\)	\(30\)
norman	\(\frac {-13 x^{2} {\mathrm e}^{{\mathrm e}^{3}}+5 x^{2} {\mathrm e}^{{\mathrm e}^{3}} \ln \left (3 x^{2}\right )}{\ln \left (3 x^{2}\right )-3}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x*exp(exp(3))*ln(3*x^2)^2-56*x*exp(exp(3))*ln(3*x^2)+74*x*exp(exp(3)))/(ln(3*x^2)^2-6*ln(3*x^2)+9),x,m
ethod=_RETURNVERBOSE)

[Out]

5*x^2*exp(exp(3))+2*x^2*exp(exp(3))/(ln(3*x^2)-3)

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maxima [A]  time = 0.77, size = 35, normalized size = 1.40 \begin {gather*} \frac {x^{2} {\left (5 \, \log \relax (3) - 13\right )} e^{\left (e^{3}\right )} + 10 \, x^{2} e^{\left (e^{3}\right )} \log \relax (x)}{\log \relax (3) + 2 \, \log \relax (x) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x*exp(exp(3))*log(3*x^2)^2-56*x*exp(exp(3))*log(3*x^2)+74*x*exp(exp(3)))/(log(3*x^2)^2-6*log(3*x
^2)+9),x, algorithm="maxima")

[Out]

(x^2*(5*log(3) - 13)*e^(e^3) + 10*x^2*e^(e^3)*log(x))/(log(3) + 2*log(x) - 3)

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mupad [B]  time = 2.20, size = 27, normalized size = 1.08 \begin {gather*} 5\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^3}+\frac {2\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^3}}{\ln \left (3\,x^2\right )-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((74*x*exp(exp(3)) - 56*x*exp(exp(3))*log(3*x^2) + 10*x*exp(exp(3))*log(3*x^2)^2)/(log(3*x^2)^2 - 6*log(3*x
^2) + 9),x)

[Out]

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

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sympy [A]  time = 0.14, size = 27, normalized size = 1.08 \begin {gather*} 5 x^{2} e^{e^{3}} + \frac {2 x^{2} e^{e^{3}}}{\log {\left (3 x^{2} \right )} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x*exp(exp(3))*ln(3*x**2)**2-56*x*exp(exp(3))*ln(3*x**2)+74*x*exp(exp(3)))/(ln(3*x**2)**2-6*ln(3*
x**2)+9),x)

[Out]

5*x**2*exp(exp(3)) + 2*x**2*exp(exp(3))/(log(3*x**2) - 3)

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