3.78.46 \(\int \frac {-13 e^{x/2}-52 e^{x^2} x+(-78+26 e^{x/2}+26 e^{x^2}) \log (-3+e^{x/2}+e^{x^2})}{(-36 x^2+12 e^{x/2} x^2+12 e^{x^2} x^2) \log (-3+e^{x/2}+e^{x^2})+(72 x-24 e^{x/2} x-24 e^{x^2} x) \log (-3+e^{x/2}+e^{x^2}) \log (\log (-3+e^{x/2}+e^{x^2}))+(-36+12 e^{x/2}+12 e^{x^2}) \log (-3+e^{x/2}+e^{x^2}) \log ^2(\log (-3+e^{x/2}+e^{x^2}))} \, dx\)

Optimal. Leaf size=26 \[ \frac {13}{6 \left (-x+\log \left (\log \left (-3+e^{x/2}+e^{x^2}\right )\right )\right )} \]

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Rubi [A]  time = 0.63, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 206, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {13}{6 \left (x-\log \left (\log \left (e^{x^2}+e^{x/2}-3\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-13*E^(x/2) - 52*E^x^2*x + (-78 + 26*E^(x/2) + 26*E^x^2)*Log[-3 + E^(x/2) + E^x^2])/((-36*x^2 + 12*E^(x/2
)*x^2 + 12*E^x^2*x^2)*Log[-3 + E^(x/2) + E^x^2] + (72*x - 24*E^(x/2)*x - 24*E^x^2*x)*Log[-3 + E^(x/2) + E^x^2]
*Log[Log[-3 + E^(x/2) + E^x^2]] + (-36 + 12*E^(x/2) + 12*E^x^2)*Log[-3 + E^(x/2) + E^x^2]*Log[Log[-3 + E^(x/2)
+ E^x^2]]^2),x]

[Out]

-13/(6*(x - Log[Log[-3 + E^(x/2) + E^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 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 {13 \left (e^{x/2}+4 e^{x^2} x-2 \left (-3+e^{x/2}+e^{x^2}\right ) \log \left (-3+e^{x/2}+e^{x^2}\right )\right )}{12 \left (3-e^{x/2}-e^{x^2}\right ) \log \left (-3+e^{x/2}+e^{x^2}\right ) \left (x-\log \left (\log \left (-3+e^{x/2}+e^{x^2}\right )\right )\right )^2} \, dx\\ &=\frac {13}{12} \int \frac {e^{x/2}+4 e^{x^2} x-2 \left (-3+e^{x/2}+e^{x^2}\right ) \log \left (-3+e^{x/2}+e^{x^2}\right )}{\left (3-e^{x/2}-e^{x^2}\right ) \log \left (-3+e^{x/2}+e^{x^2}\right ) \left (x-\log \left (\log \left (-3+e^{x/2}+e^{x^2}\right )\right )\right )^2} \, dx\\ &=-\frac {13}{6 \left (x-\log \left (\log \left (-3+e^{x/2}+e^{x^2}\right )\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 26, normalized size = 1.00 \begin {gather*} \frac {13}{6 \left (-x+\log \left (\log \left (-3+e^{x/2}+e^{x^2}\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-13*E^(x/2) - 52*E^x^2*x + (-78 + 26*E^(x/2) + 26*E^x^2)*Log[-3 + E^(x/2) + E^x^2])/((-36*x^2 + 12*
E^(x/2)*x^2 + 12*E^x^2*x^2)*Log[-3 + E^(x/2) + E^x^2] + (72*x - 24*E^(x/2)*x - 24*E^x^2*x)*Log[-3 + E^(x/2) +
E^x^2]*Log[Log[-3 + E^(x/2) + E^x^2]] + (-36 + 12*E^(x/2) + 12*E^x^2)*Log[-3 + E^(x/2) + E^x^2]*Log[Log[-3 + E
^(x/2) + E^x^2]]^2),x]

[Out]

13/(6*(-x + Log[Log[-3 + E^(x/2) + E^x^2]]))

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fricas [A]  time = 1.01, size = 20, normalized size = 0.77 \begin {gather*} -\frac {13}{6 \, {\left (x - \log \left (\log \left (e^{\left (x^{2}\right )} + e^{\left (\frac {1}{2} \, x\right )} - 3\right )\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((26*exp(x^2)+26*exp(1/2*x)-78)*log(exp(x^2)+exp(1/2*x)-3)-52*exp(x^2)*x-13*exp(1/2*x))/((12*exp(x^2
)+12*exp(1/2*x)-36)*log(exp(x^2)+exp(1/2*x)-3)*log(log(exp(x^2)+exp(1/2*x)-3))^2+(-24*exp(x^2)*x-24*x*exp(1/2*
x)+72*x)*log(exp(x^2)+exp(1/2*x)-3)*log(log(exp(x^2)+exp(1/2*x)-3))+(12*x^2*exp(x^2)+12*x^2*exp(1/2*x)-36*x^2)
*log(exp(x^2)+exp(1/2*x)-3)),x, algorithm="fricas")

[Out]

-13/6/(x - log(log(e^(x^2) + e^(1/2*x) - 3)))

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giac [A]  time = 1.97, size = 20, normalized size = 0.77 \begin {gather*} -\frac {13}{6 \, {\left (x - \log \left (\log \left (e^{\left (x^{2}\right )} + e^{\left (\frac {1}{2} \, x\right )} - 3\right )\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((26*exp(x^2)+26*exp(1/2*x)-78)*log(exp(x^2)+exp(1/2*x)-3)-52*exp(x^2)*x-13*exp(1/2*x))/((12*exp(x^2
)+12*exp(1/2*x)-36)*log(exp(x^2)+exp(1/2*x)-3)*log(log(exp(x^2)+exp(1/2*x)-3))^2+(-24*exp(x^2)*x-24*x*exp(1/2*
x)+72*x)*log(exp(x^2)+exp(1/2*x)-3)*log(log(exp(x^2)+exp(1/2*x)-3))+(12*x^2*exp(x^2)+12*x^2*exp(1/2*x)-36*x^2)
*log(exp(x^2)+exp(1/2*x)-3)),x, algorithm="giac")

[Out]

-13/6/(x - log(log(e^(x^2) + e^(1/2*x) - 3)))

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maple [A]  time = 0.04, size = 21, normalized size = 0.81

method	result	size
risch	\(-\frac {13}{6 \left (x -\ln \left (\ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{\frac {x}{2}}-3\right )\right )\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((26*exp(x^2)+26*exp(1/2*x)-78)*ln(exp(x^2)+exp(1/2*x)-3)-52*exp(x^2)*x-13*exp(1/2*x))/((12*exp(x^2)+12*ex
p(1/2*x)-36)*ln(exp(x^2)+exp(1/2*x)-3)*ln(ln(exp(x^2)+exp(1/2*x)-3))^2+(-24*exp(x^2)*x-24*x*exp(1/2*x)+72*x)*l
n(exp(x^2)+exp(1/2*x)-3)*ln(ln(exp(x^2)+exp(1/2*x)-3))+(12*x^2*exp(x^2)+12*x^2*exp(1/2*x)-36*x^2)*ln(exp(x^2)+
exp(1/2*x)-3)),x,method=_RETURNVERBOSE)

[Out]

-13/6/(x-ln(ln(exp(x^2)+exp(1/2*x)-3)))

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maxima [A]  time = 0.46, size = 20, normalized size = 0.77 \begin {gather*} -\frac {13}{6 \, {\left (x - \log \left (\log \left (e^{\left (x^{2}\right )} + e^{\left (\frac {1}{2} \, x\right )} - 3\right )\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((26*exp(x^2)+26*exp(1/2*x)-78)*log(exp(x^2)+exp(1/2*x)-3)-52*exp(x^2)*x-13*exp(1/2*x))/((12*exp(x^2
)+12*exp(1/2*x)-36)*log(exp(x^2)+exp(1/2*x)-3)*log(log(exp(x^2)+exp(1/2*x)-3))^2+(-24*exp(x^2)*x-24*x*exp(1/2*
x)+72*x)*log(exp(x^2)+exp(1/2*x)-3)*log(log(exp(x^2)+exp(1/2*x)-3))+(12*x^2*exp(x^2)+12*x^2*exp(1/2*x)-36*x^2)
*log(exp(x^2)+exp(1/2*x)-3)),x, algorithm="maxima")

[Out]

-13/6/(x - log(log(e^(x^2) + e^(1/2*x) - 3)))

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mupad [B]  time = 5.76, size = 22, normalized size = 0.85 \begin {gather*} -\frac {13}{6\,\left (x-\ln \left (\ln \left ({\mathrm {e}}^{x/2}+{\mathrm {e}}^{x^2}-3\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(13*exp(x/2) + 52*x*exp(x^2) - log(exp(x/2) + exp(x^2) - 3)*(26*exp(x/2) + 26*exp(x^2) - 78))/(log(exp(x/
2) + exp(x^2) - 3)*(12*x^2*exp(x/2) + 12*x^2*exp(x^2) - 36*x^2) + log(exp(x/2) + exp(x^2) - 3)*log(log(exp(x/2
) + exp(x^2) - 3))^2*(12*exp(x/2) + 12*exp(x^2) - 36) - log(exp(x/2) + exp(x^2) - 3)*log(log(exp(x/2) + exp(x^
2) - 3))*(24*x*exp(x/2) - 72*x + 24*x*exp(x^2))),x)

[Out]

-13/(6*(x - log(log(exp(x/2) + exp(x^2) - 3))))

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sympy [A]  time = 40.27, size = 20, normalized size = 0.77 \begin {gather*} \frac {13}{- 6 x + 6 \log {\left (\log {\left (e^{\frac {x}{2}} + e^{x^{2}} - 3 \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((26*exp(x**2)+26*exp(1/2*x)-78)*ln(exp(x**2)+exp(1/2*x)-3)-52*exp(x**2)*x-13*exp(1/2*x))/((12*exp(x
**2)+12*exp(1/2*x)-36)*ln(exp(x**2)+exp(1/2*x)-3)*ln(ln(exp(x**2)+exp(1/2*x)-3))**2+(-24*exp(x**2)*x-24*x*exp(
1/2*x)+72*x)*ln(exp(x**2)+exp(1/2*x)-3)*ln(ln(exp(x**2)+exp(1/2*x)-3))+(12*x**2*exp(x**2)+12*x**2*exp(1/2*x)-3
6*x**2)*ln(exp(x**2)+exp(1/2*x)-3)),x)

[Out]

13/(-6*x + 6*log(log(exp(x/2) + exp(x**2) - 3)))

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