3.95.96 \(\int \frac {e^{\frac {35+18 x-12 x^2-5 x^3}{\log (\log (5))}} (18-24 x-15 x^2)-\log (\log (5))}{\log (\log (5))} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {\left (5-x-x^2\right ) (-5+x+4 (3+x))}{\log (\log (5))}}-x \]

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Rubi [A]  time = 0.11, antiderivative size = 27, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 6706} \begin {gather*} e^{\frac {-5 x^3-12 x^2+18 x+35}{\log (\log (5))}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((35 + 18*x - 12*x^2 - 5*x^3)/Log[Log[5]])*(18 - 24*x - 15*x^2) - Log[Log[5]])/Log[Log[5]],x]

[Out]

E^((35 + 18*x - 12*x^2 - 5*x^3)/Log[Log[5]]) - x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {\int \left (e^{\frac {35+18 x-12 x^2-5 x^3}{\log (\log (5))}} \left (18-24 x-15 x^2\right )-\log (\log (5))\right ) \, dx}{\log (\log (5))}\\ &=-x+\frac {\int e^{\frac {35+18 x-12 x^2-5 x^3}{\log (\log (5))}} \left (18-24 x-15 x^2\right ) \, dx}{\log (\log (5))}\\ &=e^{\frac {35+18 x-12 x^2-5 x^3}{\log (\log (5))}}-x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 27, normalized size = 0.90 \begin {gather*} e^{\frac {35+18 x-12 x^2-5 x^3}{\log (\log (5))}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((35 + 18*x - 12*x^2 - 5*x^3)/Log[Log[5]])*(18 - 24*x - 15*x^2) - Log[Log[5]])/Log[Log[5]],x]

[Out]

E^((35 + 18*x - 12*x^2 - 5*x^3)/Log[Log[5]]) - x

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fricas [A]  time = 0.61, size = 27, normalized size = 0.90 \begin {gather*} -x + e^{\left (-\frac {5 \, x^{3} + 12 \, x^{2} - 18 \, x - 35}{\log \left (\log \relax (5)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^2-24*x+18)*exp((-5*x^3-12*x^2+18*x+35)/log(log(5)))-log(log(5)))/log(log(5)),x, algorithm="f
ricas")

[Out]

-x + e^(-(5*x^3 + 12*x^2 - 18*x - 35)/log(log(5)))

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giac [B]  time = 3.00, size = 55, normalized size = 1.83 \begin {gather*} -\frac {x \log \left (\log \relax (5)\right ) - e^{\left (-\frac {5 \, x^{3}}{\log \left (\log \relax (5)\right )} - \frac {12 \, x^{2}}{\log \left (\log \relax (5)\right )} + \frac {18 \, x}{\log \left (\log \relax (5)\right )} + \frac {35}{\log \left (\log \relax (5)\right )}\right )} \log \left (\log \relax (5)\right )}{\log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^2-24*x+18)*exp((-5*x^3-12*x^2+18*x+35)/log(log(5)))-log(log(5)))/log(log(5)),x, algorithm="g
iac")

[Out]

-(x*log(log(5)) - e^(-5*x^3/log(log(5)) - 12*x^2/log(log(5)) + 18*x/log(log(5)) + 35/log(log(5)))*log(log(5)))
/log(log(5))

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maple [A]  time = 0.09, size = 24, normalized size = 0.80

method	result	size
risch	\(-x +{\mathrm e}^{-\frac {\left (5 x +7\right ) \left (x^{2}+x -5\right )}{\ln \left (\ln \relax (5)\right )}}\)	\(24\)
norman	\(-x +{\mathrm e}^{\frac {-5 x^{3}-12 x^{2}+18 x +35}{\ln \left (\ln \relax (5)\right )}}\)	\(27\)
default	\(\frac {\ln \left (\ln \relax (5)\right ) {\mathrm e}^{\frac {-5 x^{3}-12 x^{2}+18 x +35}{\ln \left (\ln \relax (5)\right )}}-x \ln \left (\ln \relax (5)\right )}{\ln \left (\ln \relax (5)\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x^2-24*x+18)*exp((-5*x^3-12*x^2+18*x+35)/ln(ln(5)))-ln(ln(5)))/ln(ln(5)),x,method=_RETURNVERBOSE)

[Out]

-x+exp(-(5*x+7)*(x^2+x-5)/ln(ln(5)))

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maxima [B]  time = 0.51, size = 55, normalized size = 1.83 \begin {gather*} -\frac {x \log \left (\log \relax (5)\right ) - e^{\left (-\frac {5 \, x^{3}}{\log \left (\log \relax (5)\right )} - \frac {12 \, x^{2}}{\log \left (\log \relax (5)\right )} + \frac {18 \, x}{\log \left (\log \relax (5)\right )} + \frac {35}{\log \left (\log \relax (5)\right )}\right )} \log \left (\log \relax (5)\right )}{\log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^2-24*x+18)*exp((-5*x^3-12*x^2+18*x+35)/log(log(5)))-log(log(5)))/log(log(5)),x, algorithm="m
axima")

[Out]

-(x*log(log(5)) - e^(-5*x^3/log(log(5)) - 12*x^2/log(log(5)) + 18*x/log(log(5)) + 35/log(log(5)))*log(log(5)))
/log(log(5))

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mupad [B]  time = 0.19, size = 44, normalized size = 1.47 \begin {gather*} {\mathrm {e}}^{\frac {18\,x}{\ln \left (\ln \relax (5)\right )}}\,{\mathrm {e}}^{-\frac {5\,x^3}{\ln \left (\ln \relax (5)\right )}}\,{\mathrm {e}}^{-\frac {12\,x^2}{\ln \left (\ln \relax (5)\right )}}\,{\mathrm {e}}^{\frac {35}{\ln \left (\ln \relax (5)\right )}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(5)) + exp((18*x - 12*x^2 - 5*x^3 + 35)/log(log(5)))*(24*x + 15*x^2 - 18))/log(log(5)),x)

[Out]

exp((18*x)/log(log(5)))*exp(-(5*x^3)/log(log(5)))*exp(-(12*x^2)/log(log(5)))*exp(35/log(log(5))) - x

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sympy [A]  time = 0.14, size = 22, normalized size = 0.73 \begin {gather*} - x + e^{\frac {- 5 x^{3} - 12 x^{2} + 18 x + 35}{\log {\left (\log {\relax (5 )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x**2-24*x+18)*exp((-5*x**3-12*x**2+18*x+35)/ln(ln(5)))-ln(ln(5)))/ln(ln(5)),x)

[Out]

-x + exp((-5*x**3 - 12*x**2 + 18*x + 35)/log(log(5)))

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