3.11.38 \(\int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} (24+26 x+5 x^2-4 x^3+5 x^4+(-6 x-5 x^2) \log (x)+10 x^2 \log (6+5 x)+(-6 x-5 x^2) \log ^2(6+5 x))}{6 x^3+5 x^4} \, dx\)

Optimal. Leaf size=32 \[ e^{\frac {-\frac {2}{x}+\log (x)+(-x+\log (-x+3 (2+2 x)))^2}{x}} \]

________________________________________________________________________________________

Rubi [A]  time = 3.12, antiderivative size = 38, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {1593, 6706} \begin {gather*} \frac {x^{\frac {1}{x}} e^{-\frac {-x^3-x \log ^2(5 x+6)+2}{x^2}}}{(5 x+6)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-2 + x^3 + x*Log[x] - 2*x^2*Log[6 + 5*x] + x*Log[6 + 5*x]^2)/x^2)*(24 + 26*x + 5*x^2 - 4*x^3 + 5*x^4
+ (-6*x - 5*x^2)*Log[x] + 10*x^2*Log[6 + 5*x] + (-6*x - 5*x^2)*Log[6 + 5*x]^2))/(6*x^3 + 5*x^4),x]

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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} &=\int \frac {\exp \left (\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}\right ) \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{x^3 (6+5 x)} \, dx\\ &=\frac {e^{-\frac {2-x^3-x \log ^2(6+5 x)}{x^2}} x^{\frac {1}{x}}}{(6+5 x)^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 34, normalized size = 1.06 \begin {gather*} \frac {e^{\frac {-2+x^3+x \log ^2(6+5 x)}{x^2}} x^{\frac {1}{x}}}{(6+5 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-2 + x^3 + x*Log[x] - 2*x^2*Log[6 + 5*x] + x*Log[6 + 5*x]^2)/x^2)*(24 + 26*x + 5*x^2 - 4*x^3 +
5*x^4 + (-6*x - 5*x^2)*Log[x] + 10*x^2*Log[6 + 5*x] + (-6*x - 5*x^2)*Log[6 + 5*x]^2))/(6*x^3 + 5*x^4),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 35, normalized size = 1.09 \begin {gather*} e^{\left (\frac {x^{3} - 2 \, x^{2} \log \left (5 \, x + 6\right ) + x \log \left (5 \, x + 6\right )^{2} + x \log \relax (x) - 2}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2-6*x)*log(5*x+6)^2+10*x^2*log(5*x+6)+(-5*x^2-6*x)*log(x)+5*x^4-4*x^3+5*x^2+26*x+24)*exp((x*l
og(5*x+6)^2-2*x^2*log(5*x+6)+x*log(x)+x^3-2)/x^2)/(5*x^4+6*x^3),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.59, size = 34, normalized size = 1.06 \begin {gather*} e^{\left (x + \frac {\log \left (5 \, x + 6\right )^{2}}{x} + \frac {\log \relax (x)}{x} - \frac {2}{x^{2}} - 2 \, \log \left (5 \, x + 6\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2-6*x)*log(5*x+6)^2+10*x^2*log(5*x+6)+(-5*x^2-6*x)*log(x)+5*x^4-4*x^3+5*x^2+26*x+24)*exp((x*l
og(5*x+6)^2-2*x^2*log(5*x+6)+x*log(x)+x^3-2)/x^2)/(5*x^4+6*x^3),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.17, size = 34, normalized size = 1.06

method	result	size
risch	\(\frac {x^{\frac {1}{x}} {\mathrm e}^{\frac {x \ln \left (5 x +6\right )^{2}+x^{3}-2}{x^{2}}}}{\left (5 x +6\right )^{2}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^2-6*x)*ln(5*x+6)^2+10*x^2*ln(5*x+6)+(-5*x^2-6*x)*ln(x)+5*x^4-4*x^3+5*x^2+26*x+24)*exp((x*ln(5*x+6)^
2-2*x^2*ln(5*x+6)+x*ln(x)+x^3-2)/x^2)/(5*x^4+6*x^3),x,method=_RETURNVERBOSE)

[Out]

1/(5*x+6)^2*x^(1/x)*exp((x*ln(5*x+6)^2+x^3-2)/x^2)

________________________________________________________________________________________

maxima [A]  time = 0.80, size = 39, normalized size = 1.22 \begin {gather*} \frac {e^{\left (x + \frac {\log \left (5 \, x + 6\right )^{2}}{x} + \frac {\log \relax (x)}{x} - \frac {2}{x^{2}}\right )}}{25 \, x^{2} + 60 \, x + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2-6*x)*log(5*x+6)^2+10*x^2*log(5*x+6)+(-5*x^2-6*x)*log(x)+5*x^4-4*x^3+5*x^2+26*x+24)*exp((x*l
og(5*x+6)^2-2*x^2*log(5*x+6)+x*log(x)+x^3-2)/x^2)/(5*x^4+6*x^3),x, algorithm="maxima")

[Out]

e^(x + log(5*x + 6)^2/x + log(x)/x - 2/x^2)/(25*x^2 + 60*x + 36)

________________________________________________________________________________________

mupad [B]  time = 0.96, size = 34, normalized size = 1.06 \begin {gather*} \frac {x^{1/x}\,{\mathrm {e}}^{-\frac {2}{x^2}}\,{\mathrm {e}}^{\frac {{\ln \left (5\,x+6\right )}^2}{x}}\,{\mathrm {e}}^x}{{\left (5\,x+6\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x*log(x) + x^3 + x*log(5*x + 6)^2 - 2*x^2*log(5*x + 6) - 2)/x^2)*(26*x - log(x)*(6*x + 5*x^2) - log(
5*x + 6)^2*(6*x + 5*x^2) + 5*x^2 - 4*x^3 + 5*x^4 + 10*x^2*log(5*x + 6) + 24))/(6*x^3 + 5*x^4),x)

[Out]

(x^(1/x)*exp(-2/x^2)*exp(log(5*x + 6)^2/x)*exp(x))/(5*x + 6)^2

________________________________________________________________________________________

sympy [A]  time = 0.86, size = 36, normalized size = 1.12 \begin {gather*} e^{\frac {x^{3} - 2 x^{2} \log {\left (5 x + 6 \right )} + x \log {\relax (x )} + x \log {\left (5 x + 6 \right )}^{2} - 2}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**2-6*x)*ln(5*x+6)**2+10*x**2*ln(5*x+6)+(-5*x**2-6*x)*ln(x)+5*x**4-4*x**3+5*x**2+26*x+24)*exp(
(x*ln(5*x+6)**2-2*x**2*ln(5*x+6)+x*ln(x)+x**3-2)/x**2)/(5*x**4+6*x**3),x)

[Out]

exp((x**3 - 2*x**2*log(5*x + 6) + x*log(x) + x*log(5*x + 6)**2 - 2)/x**2)

________________________________________________________________________________________