3.70.91 \(\int \frac {90 x^2 \log ^2(3)+e^{x^2} (-300 x-300 x^3) \log ^2(3)}{225+60 x^3 \log ^2(3)+100 e^{2 x^2} x^4 \log ^4(3)+4 x^6 \log ^4(3)+e^{x^2} (-300 x^2 \log ^2(3)-40 x^5 \log ^4(3))} \, dx\)

Optimal. Leaf size=26 \[ \frac {3}{-3+2 \left (e^{x^2}-\frac {x}{5}\right ) x^2 \log ^2(3)} \]

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Rubi [A]  time = 0.30, antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, integrand size = 93, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {15}{2 x^3 \log ^2(3)-10 e^{x^2} x^2 \log ^2(3)+15} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(90*x^2*Log[3]^2 + E^x^2*(-300*x - 300*x^3)*Log[3]^2)/(225 + 60*x^3*Log[3]^2 + 100*E^(2*x^2)*x^4*Log[3]^4
+ 4*x^6*Log[3]^4 + E^x^2*(-300*x^2*Log[3]^2 - 40*x^5*Log[3]^4)),x]

[Out]

-15/(15 - 10*E^x^2*x^2*Log[3]^2 + 2*x^3*Log[3]^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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {30 x \left (3 x-10 e^{x^2} \left (1+x^2\right )\right ) \log ^2(3)}{\left (15-10 e^{x^2} x^2 \log ^2(3)+2 x^3 \log ^2(3)\right )^2} \, dx\\ &=\left (30 \log ^2(3)\right ) \int \frac {x \left (3 x-10 e^{x^2} \left (1+x^2\right )\right )}{\left (15-10 e^{x^2} x^2 \log ^2(3)+2 x^3 \log ^2(3)\right )^2} \, dx\\ &=-\frac {15}{15-10 e^{x^2} x^2 \log ^2(3)+2 x^3 \log ^2(3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 38, normalized size = 1.46 \begin {gather*} -\frac {30 \log ^2(3)}{30 \log ^2(3)-20 e^{x^2} x^2 \log ^4(3)+4 x^3 \log ^4(3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(90*x^2*Log[3]^2 + E^x^2*(-300*x - 300*x^3)*Log[3]^2)/(225 + 60*x^3*Log[3]^2 + 100*E^(2*x^2)*x^4*Log
[3]^4 + 4*x^6*Log[3]^4 + E^x^2*(-300*x^2*Log[3]^2 - 40*x^5*Log[3]^4)),x]

[Out]

(-30*Log[3]^2)/(30*Log[3]^2 - 20*E^x^2*x^2*Log[3]^4 + 4*x^3*Log[3]^4)

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fricas [A]  time = 1.09, size = 28, normalized size = 1.08 \begin {gather*} -\frac {15}{2 \, x^{3} \log \relax (3)^{2} - 10 \, x^{2} e^{\left (x^{2}\right )} \log \relax (3)^{2} + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-300*x^3-300*x)*log(3)^2*exp(x^2)+90*x^2*log(3)^2)/(100*x^4*log(3)^4*exp(x^2)^2+(-40*x^5*log(3)^4-
300*x^2*log(3)^2)*exp(x^2)+4*x^6*log(3)^4+60*x^3*log(3)^2+225),x, algorithm="fricas")

[Out]

-15/(2*x^3*log(3)^2 - 10*x^2*e^(x^2)*log(3)^2 + 15)

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giac [A]  time = 0.25, size = 28, normalized size = 1.08 \begin {gather*} -\frac {15}{2 \, x^{3} \log \relax (3)^{2} - 10 \, x^{2} e^{\left (x^{2}\right )} \log \relax (3)^{2} + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-300*x^3-300*x)*log(3)^2*exp(x^2)+90*x^2*log(3)^2)/(100*x^4*log(3)^4*exp(x^2)^2+(-40*x^5*log(3)^4-
300*x^2*log(3)^2)*exp(x^2)+4*x^6*log(3)^4+60*x^3*log(3)^2+225),x, algorithm="giac")

[Out]

-15/(2*x^3*log(3)^2 - 10*x^2*e^(x^2)*log(3)^2 + 15)

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maple [A]  time = 0.25, size = 29, normalized size = 1.12




method result size



risch \(-\frac {15}{2 x^{3} \ln \relax (3)^{2}-10 \ln \relax (3)^{2} {\mathrm e}^{x^{2}} x^{2}+15}\) \(29\)
norman \(\frac {2 x^{3} \ln \relax (3)^{2}-10 \ln \relax (3)^{2} {\mathrm e}^{x^{2}} x^{2}}{2 x^{3} \ln \relax (3)^{2}-10 \ln \relax (3)^{2} {\mathrm e}^{x^{2}} x^{2}+15}\) \(51\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-300*x^3-300*x)*ln(3)^2*exp(x^2)+90*x^2*ln(3)^2)/(100*x^4*ln(3)^4*exp(x^2)^2+(-40*x^5*ln(3)^4-300*x^2*ln
(3)^2)*exp(x^2)+4*x^6*ln(3)^4+60*x^3*ln(3)^2+225),x,method=_RETURNVERBOSE)

[Out]

-15/(2*x^3*ln(3)^2-10*ln(3)^2*exp(x^2)*x^2+15)

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maxima [A]  time = 0.52, size = 28, normalized size = 1.08 \begin {gather*} -\frac {15}{2 \, x^{3} \log \relax (3)^{2} - 10 \, x^{2} e^{\left (x^{2}\right )} \log \relax (3)^{2} + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-300*x^3-300*x)*log(3)^2*exp(x^2)+90*x^2*log(3)^2)/(100*x^4*log(3)^4*exp(x^2)^2+(-40*x^5*log(3)^4-
300*x^2*log(3)^2)*exp(x^2)+4*x^6*log(3)^4+60*x^3*log(3)^2+225),x, algorithm="maxima")

[Out]

-15/(2*x^3*log(3)^2 - 10*x^2*e^(x^2)*log(3)^2 + 15)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {90\,x^2\,{\ln \relax (3)}^2-{\mathrm {e}}^{x^2}\,{\ln \relax (3)}^2\,\left (300\,x^3+300\,x\right )}{60\,x^3\,{\ln \relax (3)}^2-{\mathrm {e}}^{x^2}\,\left (40\,{\ln \relax (3)}^4\,x^5+300\,{\ln \relax (3)}^2\,x^2\right )+4\,x^6\,{\ln \relax (3)}^4+100\,x^4\,{\mathrm {e}}^{2\,x^2}\,{\ln \relax (3)}^4+225} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*x^2*log(3)^2 - exp(x^2)*log(3)^2*(300*x + 300*x^3))/(60*x^3*log(3)^2 - exp(x^2)*(300*x^2*log(3)^2 + 40
*x^5*log(3)^4) + 4*x^6*log(3)^4 + 100*x^4*exp(2*x^2)*log(3)^4 + 225),x)

[Out]

int((90*x^2*log(3)^2 - exp(x^2)*log(3)^2*(300*x + 300*x^3))/(60*x^3*log(3)^2 - exp(x^2)*(300*x^2*log(3)^2 + 40
*x^5*log(3)^4) + 4*x^6*log(3)^4 + 100*x^4*exp(2*x^2)*log(3)^4 + 225), x)

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sympy [A]  time = 0.18, size = 27, normalized size = 1.04 \begin {gather*} \frac {15}{- 2 x^{3} \log {\relax (3 )}^{2} + 10 x^{2} e^{x^{2}} \log {\relax (3 )}^{2} - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-300*x**3-300*x)*ln(3)**2*exp(x**2)+90*x**2*ln(3)**2)/(100*x**4*ln(3)**4*exp(x**2)**2+(-40*x**5*ln
(3)**4-300*x**2*ln(3)**2)*exp(x**2)+4*x**6*ln(3)**4+60*x**3*ln(3)**2+225),x)

[Out]

15/(-2*x**3*log(3)**2 + 10*x**2*exp(x**2)*log(3)**2 - 15)

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