3.65.11 \(\int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} (1-4 x^3-\log (x))}{5 x^2} \, dx\)

Optimal. Leaf size=28 \[ e^{\frac {1}{5} \left (e^5+\frac {2+e}{3}-2 x^2+\frac {\log (x)}{x}\right )} \]

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Rubi [A]  time = 0.45, antiderivative size = 37, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {12, 6706} \begin {gather*} e^{\frac {-6 x^3+3 e^5 x+e x+2 x}{15 x}} x^{\left .\frac {1}{5}\right /x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*x + E*x + 3*E^5*x - 6*x^3 + 3*Log[x])/(15*x))*(1 - 4*x^3 - Log[x]))/(5*x^2),x]

[Out]

E^((2*x + E*x + 3*E^5*x - 6*x^3)/(15*x))*x^(1/(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 {1}{5} \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{x^2} \, dx\\ &=e^{\frac {2 x+e x+3 e^5 x-6 x^3}{15 x}} x^{\left .\frac {1}{5}\right /x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.04 \begin {gather*} e^{\frac {1}{15} \left (2+e+3 e^5-6 x^2\right )} x^{\left .\frac {1}{5}\right /x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*x + E*x + 3*E^5*x - 6*x^3 + 3*Log[x])/(15*x))*(1 - 4*x^3 - Log[x]))/(5*x^2),x]

[Out]

E^((2 + E + 3*E^5 - 6*x^2)/15)*x^(1/(5*x))

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fricas [A]  time = 0.90, size = 29, normalized size = 1.04 \begin {gather*} e^{\left (-\frac {6 \, x^{3} - 3 \, x e^{5} - x e - 2 \, x - 3 \, \log \relax (x)}{15 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-log(x)-4*x^3+1)*exp(1/15*(3*log(x)+3*x*exp(5)+x*exp(1)-6*x^3+2*x)/x)/x^2,x, algorithm="fricas"
)

[Out]

e^(-1/15*(6*x^3 - 3*x*e^5 - x*e - 2*x - 3*log(x))/x)

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giac [A]  time = 0.15, size = 23, normalized size = 0.82 \begin {gather*} e^{\left (-\frac {2}{5} \, x^{2} + \frac {\log \relax (x)}{5 \, x} + \frac {1}{5} \, e^{5} + \frac {1}{15} \, e + \frac {2}{15}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-log(x)-4*x^3+1)*exp(1/15*(3*log(x)+3*x*exp(5)+x*exp(1)-6*x^3+2*x)/x)/x^2,x, algorithm="giac")

[Out]

e^(-2/5*x^2 + 1/5*log(x)/x + 1/5*e^5 + 1/15*e + 2/15)

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maple [A]  time = 0.04, size = 25, normalized size = 0.89

method	result	size
risch	\(x^{\frac {1}{5 x}} {\mathrm e}^{\frac {{\mathrm e}^{5}}{5}-\frac {2 x^{2}}{5}+\frac {{\mathrm e}}{15}+\frac {2}{15}}\)	\(25\)
norman	\({\mathrm e}^{\frac {3 \ln \relax (x )+3 x \,{\mathrm e}^{5}+x \,{\mathrm e}-6 x^{3}+2 x}{15 x}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-ln(x)-4*x^3+1)*exp(1/15*(3*ln(x)+3*x*exp(5)+x*exp(1)-6*x^3+2*x)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

x^(1/5/x)*exp(1/5*exp(5)-2/5*x^2+1/15*exp(1)+2/15)

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maxima [A]  time = 0.45, size = 23, normalized size = 0.82 \begin {gather*} e^{\left (-\frac {2}{5} \, x^{2} + \frac {\log \relax (x)}{5 \, x} + \frac {1}{5} \, e^{5} + \frac {1}{15} \, e + \frac {2}{15}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-log(x)-4*x^3+1)*exp(1/15*(3*log(x)+3*x*exp(5)+x*exp(1)-6*x^3+2*x)/x)/x^2,x, algorithm="maxima"
)

[Out]

e^(-2/5*x^2 + 1/5*log(x)/x + 1/5*e^5 + 1/15*e + 2/15)

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mupad [B]  time = 4.33, size = 26, normalized size = 0.93 \begin {gather*} x^{\frac {1}{5\,x}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^5}{5}}\,{\mathrm {e}}^{\frac {\mathrm {e}}{15}}\,{\mathrm {e}}^{2/15}\,{\mathrm {e}}^{-\frac {2\,x^2}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(((2*x)/15 + log(x)/5 + (x*exp(1))/15 + (x*exp(5))/5 - (2*x^3)/5)/x)*(log(x) + 4*x^3 - 1))/(5*x^2),x)

[Out]

x^(1/(5*x))*exp(exp(5)/5)*exp(exp(1)/15)*exp(2/15)*exp(-(2*x^2)/5)

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sympy [A]  time = 0.32, size = 32, normalized size = 1.14 \begin {gather*} e^{\frac {- \frac {2 x^{3}}{5} + \frac {2 x}{15} + \frac {e x}{15} + \frac {x e^{5}}{5} + \frac {\log {\relax (x )}}{5}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-ln(x)-4*x**3+1)*exp(1/15*(3*ln(x)+3*x*exp(5)+x*exp(1)-6*x**3+2*x)/x)/x**2,x)

[Out]

exp((-2*x**3/5 + 2*x/15 + E*x/15 + x*exp(5)/5 + log(x)/5)/x)

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