\(\int \frac {e^{-4 \log ^2(\frac {1}{3} (-15+x^2))} (15-x^2+e^{4 \log ^2(\frac {1}{3} (-15+x^2))} (30 x-2 x^3)+16 x^2 \log (\frac {5 x}{2}) \log (\frac {1}{3} (-15+x^2)))}{-15 x+x^3} \, dx\) [7875]

Optimal result

Integrand size = 80, antiderivative size = 29 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=8-2 x-e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \log \left (\frac {5 x}{2}\right ) \]

[Out]

8-2*x-ln(5/2*x)/exp(4*ln(1/3*x^2-5)^2)

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1607, 6857, 2326} \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-e^{-4 \log ^2\left (\frac {x^2}{3}-5\right )} \log \left (\frac {5 x}{2}\right )-2 x \]

[In]

Int[(15 - x^2 + E^(4*Log[(-15 + x^2)/3]^2)*(30*x - 2*x^3) + 16*x^2*Log[(5*x)/2]*Log[(-15 + x^2)/3])/(E^(4*Log[
(-15 + x^2)/3]^2)*(-15*x + x^3)),x]

[Out]

-2*x - Log[(5*x)/2]/E^(4*Log[-5 + x^2/3]^2)

Rule 1607

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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{x \left (-15+x^2\right )} \, dx \\ & = \int \left (-2+\frac {e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \left (15-x^2+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (-5+\frac {x^2}{3}\right )\right )}{x \left (-15+x^2\right )}\right ) \, dx \\ & = -2 x+\int \frac {e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \left (15-x^2+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (-5+\frac {x^2}{3}\right )\right )}{x \left (-15+x^2\right )} \, dx \\ & = -2 x-e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \log \left (\frac {5 x}{2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 1.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-2 x-e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \log \left (\frac {5 x}{2}\right ) \]

[In]

Integrate[(15 - x^2 + E^(4*Log[(-15 + x^2)/3]^2)*(30*x - 2*x^3) + 16*x^2*Log[(5*x)/2]*Log[(-15 + x^2)/3])/(E^(
4*Log[(-15 + x^2)/3]^2)*(-15*x + x^3)),x]

[Out]

-2*x - Log[(5*x)/2]/E^(4*Log[-5 + x^2/3]^2)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83

method	result	size
risch	\(-2 x -\ln \left (\frac {5 x}{2}\right ) {\mathrm e}^{-4 \ln \left (\frac {x^{2}}{3}-5\right )^{2}}\)	\(24\)
parallelrisch	\(-\frac {\left (900 x \,{\mathrm e}^{4 \ln \left (\frac {x^{2}}{3}-5\right )^{2}}+450 \ln \left (\frac {5 x}{2}\right )\right ) {\mathrm e}^{-4 \ln \left (\frac {x^{2}}{3}-5\right )^{2}}}{450}\)	\(41\)

[In]

int(((-2*x^3+30*x)*exp(4*ln(1/3*x^2-5)^2)+16*x^2*ln(5/2*x)*ln(1/3*x^2-5)-x^2+15)/(x^3-15*x)/exp(4*ln(1/3*x^2-5
)^2),x,method=_RETURNVERBOSE)

[Out]

-2*x-ln(5/2*x)*exp(-4*ln(1/3*x^2-5)^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-{\left (2 \, x e^{\left (4 \, \log \left (\frac {1}{3} \, x^{2} - 5\right )^{2}\right )} + \log \left (\frac {5}{2} \, x\right )\right )} e^{\left (-4 \, \log \left (\frac {1}{3} \, x^{2} - 5\right )^{2}\right )} \]

[In]

integrate(((-2*x^3+30*x)*exp(4*log(1/3*x^2-5)^2)+16*x^2*log(5/2*x)*log(1/3*x^2-5)-x^2+15)/(x^3-15*x)/exp(4*log
(1/3*x^2-5)^2),x, algorithm="fricas")

[Out]

-(2*x*e^(4*log(1/3*x^2 - 5)^2) + log(5/2*x))*e^(-4*log(1/3*x^2 - 5)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=\text {Timed out} \]

[In]

integrate(((-2*x**3+30*x)*exp(4*ln(1/3*x**2-5)**2)+16*x**2*ln(5/2*x)*ln(1/3*x**2-5)-x**2+15)/(x**3-15*x)/exp(4
*ln(1/3*x**2-5)**2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).

Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-{\left (2 \, x e^{\left (4 \, \log \left (3\right )^{2} + 4 \, \log \left (x^{2} - 15\right )^{2}\right )} + {\left (\log \left (5\right ) - \log \left (2\right ) + \log \left (x\right )\right )} e^{\left (8 \, \log \left (3\right ) \log \left (x^{2} - 15\right )\right )}\right )} e^{\left (-4 \, \log \left (3\right )^{2} - 4 \, \log \left (x^{2} - 15\right )^{2}\right )} \]

[In]

integrate(((-2*x^3+30*x)*exp(4*log(1/3*x^2-5)^2)+16*x^2*log(5/2*x)*log(1/3*x^2-5)-x^2+15)/(x^3-15*x)/exp(4*log
(1/3*x^2-5)^2),x, algorithm="maxima")

[Out]

-(2*x*e^(4*log(3)^2 + 4*log(x^2 - 15)^2) + (log(5) - log(2) + log(x))*e^(8*log(3)*log(x^2 - 15)))*e^(-4*log(3)
^2 - 4*log(x^2 - 15)^2)

Giac [F]

\[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=\int { \frac {{\left (16 \, x^{2} \log \left (\frac {1}{3} \, x^{2} - 5\right ) \log \left (\frac {5}{2} \, x\right ) - x^{2} - 2 \, {\left (x^{3} - 15 \, x\right )} e^{\left (4 \, \log \left (\frac {1}{3} \, x^{2} - 5\right )^{2}\right )} + 15\right )} e^{\left (-4 \, \log \left (\frac {1}{3} \, x^{2} - 5\right )^{2}\right )}}{x^{3} - 15 \, x} \,d x } \]

[In]

integrate(((-2*x^3+30*x)*exp(4*log(1/3*x^2-5)^2)+16*x^2*log(5/2*x)*log(1/3*x^2-5)-x^2+15)/(x^3-15*x)/exp(4*log
(1/3*x^2-5)^2),x, algorithm="giac")

[Out]

integrate((16*x^2*log(1/3*x^2 - 5)*log(5/2*x) - x^2 - 2*(x^3 - 15*x)*e^(4*log(1/3*x^2 - 5)^2) + 15)*e^(-4*log(
1/3*x^2 - 5)^2)/(x^3 - 15*x), x)

Mupad [B] (verification not implemented)

Time = 12.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-2\,x-\ln \left (\frac {5\,x}{2}\right )\,{\mathrm {e}}^{-4\,{\ln \left (\frac {x^2}{3}-5\right )}^2} \]

[In]

int(-(exp(-4*log(x^2/3 - 5)^2)*(exp(4*log(x^2/3 - 5)^2)*(30*x - 2*x^3) - x^2 + 16*x^2*log(x^2/3 - 5)*log((5*x)
/2) + 15))/(15*x - x^3),x)

[Out]

- 2*x - log((5*x)/2)*exp(-4*log(x^2/3 - 5)^2)