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

Optimal. Leaf size=29 \[ 8-2 x-e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \log \left (\frac {5 x}{2}\right ) \]

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Rubi [A]  time = 1.71, antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1593, 6725, 2288} \begin {gather*} -e^{-4 \log ^2\left (\frac {x^2}{3}-5\right )} \log \left (\frac {5 x}{2}\right )-2 x \end {gather*}

Antiderivative was successfully verified.

[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 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 2288

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 6725

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 28, normalized size = 0.97 \begin {gather*} -2 x-e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \log \left (\frac {5 x}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.77, size = 36, normalized size = 1.24 \begin {gather*} -{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.23, size = 24, normalized size = 0.83

method	result	size
risch	\(-2 x -\ln \left (\frac {5 x}{2}\right ) {\mathrm e}^{-4 \ln \left (\frac {x^{2}}{3}-5\right )^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [B]  time = 0.49, size = 63, normalized size = 2.17 \begin {gather*} -{\left (2 \, x e^{\left (4 \, \log \relax (3)^{2} + 4 \, \log \left (x^{2} - 15\right )^{2}\right )} + {\left (\log \relax (5) - \log \relax (2) + \log \relax (x)\right )} e^{\left (8 \, \log \relax (3) \log \left (x^{2} - 15\right )\right )}\right )} e^{\left (-4 \, \log \relax (3)^{2} - 4 \, \log \left (x^{2} - 15\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 6.31, size = 23, normalized size = 0.79 \begin {gather*} -2\,x-\ln \left (\frac {5\,x}{2}\right )\,{\mathrm {e}}^{-4\,{\ln \left (\frac {x^2}{3}-5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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