3.21.24 \(\int \frac {e^{\frac {6 x+e^4 (6 x-24 x^2-x^4)+(x+e^4 (x-4 x^2)) \log (5)}{6+\log (5)}} (6+e^4 (6-48 x-4 x^3)+(1+e^4 (1-8 x)) \log (5))}{6+\log (5)} \, dx\)

Optimal. Leaf size=27 \[ e^{x+e^4 \left (x-x^2 \left (4+\frac {x^2}{6+\log (5)}\right )\right )} \]

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Rubi [A]  time = 0.44, antiderivative size = 54, normalized size of antiderivative = 2.00, number of steps used = 2, number of rules used = 2, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {12, 6706} \begin {gather*} 5^{\frac {e^4 \left (x-4 x^2\right )+x}{6+\log (5)}} \exp \left (\frac {e^4 \left (-x^4-24 x^2+6 x\right )+6 x}{6+\log (5)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((6*x + E^4*(6*x - 24*x^2 - x^4) + (x + E^4*(x - 4*x^2))*Log[5])/(6 + Log[5]))*(6 + E^4*(6 - 48*x - 4*x
^3) + (1 + E^4*(1 - 8*x))*Log[5]))/(6 + Log[5]),x]

[Out]

5^((x + E^4*(x - 4*x^2))/(6 + Log[5]))*E^((6*x + E^4*(6*x - 24*x^2 - x^4))/(6 + Log[5]))

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 \exp \left (\frac {6 x+e^4 \left (6 x-24 x^2-x^4\right )+\left (x+e^4 \left (x-4 x^2\right )\right ) \log (5)}{6+\log (5)}\right ) \left (6+e^4 \left (6-48 x-4 x^3\right )+\left (1+e^4 (1-8 x)\right ) \log (5)\right ) \, dx}{6+\log (5)}\\ &=5^{\frac {x+e^4 \left (x-4 x^2\right )}{6+\log (5)}} e^{\frac {6 x+e^4 \left (6 x-24 x^2-x^4\right )}{6+\log (5)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.13, size = 47, normalized size = 1.74 \begin {gather*} 5^{\frac {x+e^4 (1-4 x) x}{6+\log (5)}} e^{\frac {x \left (6-e^4 \left (-6+24 x+x^3\right )\right )}{6+\log (5)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((6*x + E^4*(6*x - 24*x^2 - x^4) + (x + E^4*(x - 4*x^2))*Log[5])/(6 + Log[5]))*(6 + E^4*(6 - 48*x
- 4*x^3) + (1 + E^4*(1 - 8*x))*Log[5]))/(6 + Log[5]),x]

[Out]

5^((x + E^4*(1 - 4*x)*x)/(6 + Log[5]))*E^((x*(6 - E^4*(-6 + 24*x + x^3)))/(6 + Log[5]))

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fricas [A]  time = 0.90, size = 47, normalized size = 1.74 \begin {gather*} e^{\left (-\frac {{\left (x^{4} + 24 \, x^{2} - 6 \, x\right )} e^{4} + {\left ({\left (4 \, x^{2} - x\right )} e^{4} - x\right )} \log \relax (5) - 6 \, x}{\log \relax (5) + 6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+1)*exp(4)+1)*log(5)+(-4*x^3-48*x+6)*exp(4)+6)*exp((((-4*x^2+x)*exp(4)+x)*log(5)+(-x^4-24*x^2
+6*x)*exp(4)+6*x)/(log(5)+6))/(log(5)+6),x, algorithm="fricas")

[Out]

e^(-((x^4 + 24*x^2 - 6*x)*e^4 + ((4*x^2 - x)*e^4 - x)*log(5) - 6*x)/(log(5) + 6))

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giac [B]  time = 0.21, size = 85, normalized size = 3.15 \begin {gather*} e^{\left (-\frac {x^{4} e^{4}}{\log \relax (5) + 6} - \frac {4 \, x^{2} e^{4} \log \relax (5)}{\log \relax (5) + 6} - \frac {24 \, x^{2} e^{4}}{\log \relax (5) + 6} + \frac {x e^{4} \log \relax (5)}{\log \relax (5) + 6} + \frac {6 \, x e^{4}}{\log \relax (5) + 6} + \frac {x \log \relax (5)}{\log \relax (5) + 6} + \frac {6 \, x}{\log \relax (5) + 6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+1)*exp(4)+1)*log(5)+(-4*x^3-48*x+6)*exp(4)+6)*exp((((-4*x^2+x)*exp(4)+x)*log(5)+(-x^4-24*x^2
+6*x)*exp(4)+6*x)/(log(5)+6))/(log(5)+6),x, algorithm="giac")

[Out]

e^(-x^4*e^4/(log(5) + 6) - 4*x^2*e^4*log(5)/(log(5) + 6) - 24*x^2*e^4/(log(5) + 6) + x*e^4*log(5)/(log(5) + 6)
+ 6*x*e^4/(log(5) + 6) + x*log(5)/(log(5) + 6) + 6*x/(log(5) + 6))

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maple [A]  time = 0.09, size = 45, normalized size = 1.67

method	result	size
gosper	\({\mathrm e}^{-\frac {x \left (x^{3} {\mathrm e}^{4}+4 x \,{\mathrm e}^{4} \ln \relax (5)-{\mathrm e}^{4} \ln \relax (5)+24 x \,{\mathrm e}^{4}-\ln \relax (5)-6 \,{\mathrm e}^{4}-6\right )}{\ln \relax (5)+6}}\)	\(45\)
derivativedivides	\({\mathrm e}^{\frac {\left (\left (-4 x^{2}+x \right ) {\mathrm e}^{4}+x \right ) \ln \relax (5)+\left (-x^{4}-24 x^{2}+6 x \right ) {\mathrm e}^{4}+6 x}{\ln \relax (5)+6}}\)	\(45\)
default	\({\mathrm e}^{\frac {\left (\left (-4 x^{2}+x \right ) {\mathrm e}^{4}+x \right ) \ln \relax (5)+\left (-x^{4}-24 x^{2}+6 x \right ) {\mathrm e}^{4}+6 x}{\ln \relax (5)+6}}\)	\(45\)
norman	\({\mathrm e}^{\frac {\left (\left (-4 x^{2}+x \right ) {\mathrm e}^{4}+x \right ) \ln \relax (5)+\left (-x^{4}-24 x^{2}+6 x \right ) {\mathrm e}^{4}+6 x}{\ln \relax (5)+6}}\)	\(45\)
risch	\({\mathrm e}^{-\frac {x \left (x^{3} {\mathrm e}^{4}+4 x \,{\mathrm e}^{4} \ln \relax (5)-{\mathrm e}^{4} \ln \relax (5)+24 x \,{\mathrm e}^{4}-\ln \relax (5)-6 \,{\mathrm e}^{4}-6\right )}{\ln \relax (5)+6}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x+1)*exp(4)+1)*ln(5)+(-4*x^3-48*x+6)*exp(4)+6)*exp((((-4*x^2+x)*exp(4)+x)*ln(5)+(-x^4-24*x^2+6*x)*ex
p(4)+6*x)/(ln(5)+6))/(ln(5)+6),x,method=_RETURNVERBOSE)

[Out]

exp(-x*(x^3*exp(4)+4*x*exp(4)*ln(5)-exp(4)*ln(5)+24*x*exp(4)-ln(5)-6*exp(4)-6)/(ln(5)+6))

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maxima [B]  time = 1.11, size = 85, normalized size = 3.15 \begin {gather*} e^{\left (-\frac {x^{4} e^{4}}{\log \relax (5) + 6} - \frac {4 \, x^{2} e^{4} \log \relax (5)}{\log \relax (5) + 6} - \frac {24 \, x^{2} e^{4}}{\log \relax (5) + 6} + \frac {x e^{4} \log \relax (5)}{\log \relax (5) + 6} + \frac {6 \, x e^{4}}{\log \relax (5) + 6} + \frac {x \log \relax (5)}{\log \relax (5) + 6} + \frac {6 \, x}{\log \relax (5) + 6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+1)*exp(4)+1)*log(5)+(-4*x^3-48*x+6)*exp(4)+6)*exp((((-4*x^2+x)*exp(4)+x)*log(5)+(-x^4-24*x^2
+6*x)*exp(4)+6*x)/(log(5)+6))/(log(5)+6),x, algorithm="maxima")

[Out]

e^(-x^4*e^4/(log(5) + 6) - 4*x^2*e^4*log(5)/(log(5) + 6) - 24*x^2*e^4/(log(5) + 6) + x*e^4*log(5)/(log(5) + 6)
+ 6*x*e^4/(log(5) + 6) + x*log(5)/(log(5) + 6) + 6*x/(log(5) + 6))

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mupad [B]  time = 0.42, size = 73, normalized size = 2.70 \begin {gather*} 5^{\frac {x+x\,{\mathrm {e}}^4-4\,x^2\,{\mathrm {e}}^4}{\ln \relax (5)+6}}\,{\mathrm {e}}^{-\frac {x^4\,{\mathrm {e}}^4}{\ln \relax (5)+6}}\,{\mathrm {e}}^{-\frac {24\,x^2\,{\mathrm {e}}^4}{\ln \relax (5)+6}}\,{\mathrm {e}}^{\frac {6\,x}{\ln \relax (5)+6}}\,{\mathrm {e}}^{\frac {6\,x\,{\mathrm {e}}^4}{\ln \relax (5)+6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((6*x - exp(4)*(24*x^2 - 6*x + x^4) + log(5)*(x + exp(4)*(x - 4*x^2)))/(log(5) + 6))*(log(5)*(exp(4)*
(8*x - 1) - 1) + exp(4)*(48*x + 4*x^3 - 6) - 6))/(log(5) + 6),x)

[Out]

5^((x + x*exp(4) - 4*x^2*exp(4))/(log(5) + 6))*exp(-(x^4*exp(4))/(log(5) + 6))*exp(-(24*x^2*exp(4))/(log(5) +
6))*exp((6*x)/(log(5) + 6))*exp((6*x*exp(4))/(log(5) + 6))

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sympy [A]  time = 0.25, size = 39, normalized size = 1.44 \begin {gather*} e^{\frac {6 x + \left (x + \left (- 4 x^{2} + x\right ) e^{4}\right ) \log {\relax (5 )} + \left (- x^{4} - 24 x^{2} + 6 x\right ) e^{4}}{\log {\relax (5 )} + 6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+1)*exp(4)+1)*ln(5)+(-4*x**3-48*x+6)*exp(4)+6)*exp((((-4*x**2+x)*exp(4)+x)*ln(5)+(-x**4-24*x*
*2+6*x)*exp(4)+6*x)/(ln(5)+6))/(ln(5)+6),x)

[Out]

exp((6*x + (x + (-4*x**2 + x)*exp(4))*log(5) + (-x**4 - 24*x**2 + 6*x)*exp(4))/(log(5) + 6))

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