3.97.95 \(\int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} (5 x^2+4 \log (x)-2 \log ^3(x))+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} (-20 x^2-16 \log (x)+8 \log ^3(x))}{4 x} \, dx\)

Optimal. Leaf size=32 \[ \frac {1}{16} \left (4-e^{5 \left (6+x^2\right )-\left (2-\log ^2(x)\right )^2}\right )^2 \]

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Rubi [A]  time = 0.60, antiderivative size = 53, normalized size of antiderivative = 1.66, number of steps used = 5, number of rules used = 3, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {12, 14, 6706} \begin {gather*} \frac {1}{16} e^{2 \left (5 x^2-\log ^4(x)+4 \log ^2(x)+26\right )}-\frac {1}{2} e^{5 x^2-\log ^4(x)+4 \log ^2(x)+26} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(52 + 10*x^2 + 8*Log[x]^2 - 2*Log[x]^4)*(5*x^2 + 4*Log[x] - 2*Log[x]^3) + E^(26 + 5*x^2 + 4*Log[x]^2 -
Log[x]^4)*(-20*x^2 - 16*Log[x] + 8*Log[x]^3))/(4*x),x]

[Out]

-1/2*E^(26 + 5*x^2 + 4*Log[x]^2 - Log[x]^4) + E^(2*(26 + 5*x^2 + 4*Log[x]^2 - Log[x]^4))/16

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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{x} \, dx\\ &=\frac {1}{4} \int \left (-\frac {4 e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )}{x}+\frac {e^{2 \left (26+5 x^2+4 \log ^2(x)-\log ^4(x)\right )} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )}{x}\right ) \, dx\\ &=\frac {1}{4} \int \frac {e^{2 \left (26+5 x^2+4 \log ^2(x)-\log ^4(x)\right )} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )}{x} \, dx-\int \frac {e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )}{x} \, dx\\ &=-\frac {1}{2} e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)}+\frac {1}{16} e^{2 \left (26+5 x^2+4 \log ^2(x)-\log ^4(x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.70, size = 49, normalized size = 1.53 \begin {gather*} \frac {1}{16} e^{-2 \log ^4(x)} \left (e^{52+10 x^2+8 \log ^2(x)}-8 e^{26+5 x^2+4 \log ^2(x)+\log ^4(x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(52 + 10*x^2 + 8*Log[x]^2 - 2*Log[x]^4)*(5*x^2 + 4*Log[x] - 2*Log[x]^3) + E^(26 + 5*x^2 + 4*Log[x
]^2 - Log[x]^4)*(-20*x^2 - 16*Log[x] + 8*Log[x]^3))/(4*x),x]

[Out]

(E^(52 + 10*x^2 + 8*Log[x]^2) - 8*E^(26 + 5*x^2 + 4*Log[x]^2 + Log[x]^4))/(16*E^(2*Log[x]^4))

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fricas [A]  time = 0.94, size = 45, normalized size = 1.41 \begin {gather*} -\frac {1}{2} \, e^{\left (-\log \relax (x)^{4} + 5 \, x^{2} + 4 \, \log \relax (x)^{2} + 26\right )} + \frac {1}{16} \, e^{\left (-2 \, \log \relax (x)^{4} + 10 \, x^{2} + 8 \, \log \relax (x)^{2} + 52\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*log(x)^3+4*log(x)+5*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2+26)^2+(8*log(x)^3-16*log(x)-20*x^2)
*exp(-log(x)^4+4*log(x)^2+5*x^2+26))/x,x, algorithm="fricas")

[Out]

-1/2*e^(-log(x)^4 + 5*x^2 + 4*log(x)^2 + 26) + 1/16*e^(-2*log(x)^4 + 10*x^2 + 8*log(x)^2 + 52)

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giac [A]  time = 0.22, size = 45, normalized size = 1.41 \begin {gather*} -\frac {1}{2} \, e^{\left (-\log \relax (x)^{4} + 5 \, x^{2} + 4 \, \log \relax (x)^{2} + 26\right )} + \frac {1}{16} \, e^{\left (-2 \, \log \relax (x)^{4} + 10 \, x^{2} + 8 \, \log \relax (x)^{2} + 52\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*log(x)^3+4*log(x)+5*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2+26)^2+(8*log(x)^3-16*log(x)-20*x^2)
*exp(-log(x)^4+4*log(x)^2+5*x^2+26))/x,x, algorithm="giac")

[Out]

-1/2*e^(-log(x)^4 + 5*x^2 + 4*log(x)^2 + 26) + 1/16*e^(-2*log(x)^4 + 10*x^2 + 8*log(x)^2 + 52)

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maple [A]  time = 0.05, size = 46, normalized size = 1.44

method	result	size
risch	\(\frac {{\mathrm e}^{-2 \ln \relax (x )^{4}+8 \ln \relax (x )^{2}+10 x^{2}+52}}{16}-\frac {{\mathrm e}^{-\ln \relax (x )^{4}+4 \ln \relax (x )^{2}+5 x^{2}+26}}{2}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-2*ln(x)^3+4*ln(x)+5*x^2)*exp(-ln(x)^4+4*ln(x)^2+5*x^2+26)^2+(8*ln(x)^3-16*ln(x)-20*x^2)*exp(-ln(x)^
4+4*ln(x)^2+5*x^2+26))/x,x,method=_RETURNVERBOSE)

[Out]

1/16*exp(-2*ln(x)^4+8*ln(x)^2+10*x^2+52)-1/2*exp(-ln(x)^4+4*ln(x)^2+5*x^2+26)

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maxima [A]  time = 0.53, size = 45, normalized size = 1.41 \begin {gather*} -\frac {1}{2} \, e^{\left (-\log \relax (x)^{4} + 5 \, x^{2} + 4 \, \log \relax (x)^{2} + 26\right )} + \frac {1}{16} \, e^{\left (-2 \, \log \relax (x)^{4} + 10 \, x^{2} + 8 \, \log \relax (x)^{2} + 52\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*log(x)^3+4*log(x)+5*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2+26)^2+(8*log(x)^3-16*log(x)-20*x^2)
*exp(-log(x)^4+4*log(x)^2+5*x^2+26))/x,x, algorithm="maxima")

[Out]

-1/2*e^(-log(x)^4 + 5*x^2 + 4*log(x)^2 + 26) + 1/16*e^(-2*log(x)^4 + 10*x^2 + 8*log(x)^2 + 52)

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mupad [B]  time = 6.12, size = 49, normalized size = 1.53 \begin {gather*} -\frac {{\mathrm {e}}^{4\,{\ln \relax (x)}^2}\,{\mathrm {e}}^{-2\,{\ln \relax (x)}^4}\,{\mathrm {e}}^{26}\,{\mathrm {e}}^{5\,x^2}\,\left (8\,{\mathrm {e}}^{{\ln \relax (x)}^4}-{\mathrm {e}}^{4\,{\ln \relax (x)}^2}\,{\mathrm {e}}^{26}\,{\mathrm {e}}^{5\,x^2}\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(4*log(x)^2 - log(x)^4 + 5*x^2 + 26)*(16*log(x) - 8*log(x)^3 + 20*x^2))/4 - (exp(8*log(x)^2 - 2*log(
x)^4 + 10*x^2 + 52)*(4*log(x) - 2*log(x)^3 + 5*x^2))/4)/x,x)

[Out]

-(exp(4*log(x)^2)*exp(-2*log(x)^4)*exp(26)*exp(5*x^2)*(8*exp(log(x)^4) - exp(4*log(x)^2)*exp(26)*exp(5*x^2)))/
16

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sympy [B]  time = 0.36, size = 44, normalized size = 1.38 \begin {gather*} - \frac {e^{5 x^{2} - \log {\relax (x )}^{4} + 4 \log {\relax (x )}^{2} + 26}}{2} + \frac {e^{10 x^{2} - 2 \log {\relax (x )}^{4} + 8 \log {\relax (x )}^{2} + 52}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*ln(x)**3+4*ln(x)+5*x**2)*exp(-ln(x)**4+4*ln(x)**2+5*x**2+26)**2+(8*ln(x)**3-16*ln(x)-20*x**
2)*exp(-ln(x)**4+4*ln(x)**2+5*x**2+26))/x,x)

[Out]

-exp(5*x**2 - log(x)**4 + 4*log(x)**2 + 26)/2 + exp(10*x**2 - 2*log(x)**4 + 8*log(x)**2 + 52)/16

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