\(\int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} (1-4 x^3-\log (x))}{5 x^2} \, dx\) [6411]

Optimal result

Integrand size = 49, antiderivative size = 28 \[ \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{5 x^2} \, dx=e^{\frac {1}{5} \left (e^5+\frac {2+e}{3}-2 x^2+\frac {\log (x)}{x}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {12, 6838} \[ \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{5 x^2} \, dx=e^{\frac {-6 x^3+3 e^5 x+e x+2 x}{15 x}} x^{\left .\frac {1}{5}\right /x} \]

[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 6838

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{5 x^2} \, dx=e^{\frac {1}{15} \left (2+e+3 e^5-6 x^2\right )} x^{\left .\frac {1}{5}\right /x} \]

[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))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 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	\({\mathrm e}^{\frac {3 \ln \left (x \right )+3 x \,{\mathrm e}^{5}+x \,{\mathrm e}-6 x^{3}+2 x}{15 x}}\)	\(29\)
parallelrisch	\({\mathrm e}^{\frac {3 \ln \left (x \right )+3 x \,{\mathrm e}^{5}+x \,{\mathrm e}-6 x^{3}+2 x}{15 x}}\)	\(29\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{5 x^2} \, dx=e^{\left (-\frac {6 \, x^{3} - 3 \, x e^{5} - x e - 2 \, x - 3 \, \log \left (x\right )}{15 \, x}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{5 x^2} \, dx=e^{\frac {- \frac {2 x^{3}}{5} + \frac {2 x}{15} + \frac {e x}{15} + \frac {x e^{5}}{5} + \frac {\log {\left (x \right )}}{5}}{x}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{5 x^2} \, dx=e^{\left (-\frac {2}{5} \, x^{2} + \frac {\log \left (x\right )}{5 \, x} + \frac {1}{5} \, e^{5} + \frac {1}{15} \, e + \frac {2}{15}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{5 x^2} \, dx=e^{\left (-\frac {2}{5} \, x^{2} + \frac {\log \left (x\right )}{5 \, x} + \frac {1}{5} \, e^{5} + \frac {1}{15} \, e + \frac {2}{15}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 13.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {2 x+e x+3 e^5 x-6 x^3+3 \log (x)}{15 x}} \left (1-4 x^3-\log (x)\right )}{5 x^2} \, dx=x^{\frac {1}{5\,x}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^5}{5}}\,{\mathrm {e}}^{\frac {\mathrm {e}}{15}}\,{\mathrm {e}}^{2/15}\,{\mathrm {e}}^{-\frac {2\,x^2}{5}} \]

[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)