3.12.100 \(\int \frac {x (-3 x^2+48 x^3-144 x^4+(-1+144 x^2) \log (2))}{e^2 (-x^2-\log (2)) (36 x^5+36 x^3 \log (2))} \, dx\)

Optimal. Leaf size=31 \[ \frac {\left (4-\frac {1}{3 x}\right )^2}{4 e^2 \left (-x-\frac {\log (2)}{x}\right )} \]

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Rubi [A]  time = 0.09, antiderivative size = 47, normalized size of antiderivative = 1.52, number of steps used = 6, number of rules used = 5, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 1584, 21, 1805, 30} \begin {gather*} \frac {x (1-144 \log (2))+24 \log (2)}{36 e^2 \log (2) \left (x^2+\log (2)\right )}-\frac {1}{36 e^2 x \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(-3*x^2 + 48*x^3 - 144*x^4 + (-1 + 144*x^2)*Log[2]))/(E^2*(-x^2 - Log[2])*(36*x^5 + 36*x^3*Log[2])),x]

[Out]

-1/36*1/(E^2*x*Log[2]) + (x*(1 - 144*Log[2]) + 24*Log[2])/(36*E^2*Log[2]*(x^2 + Log[2]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 1584

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{\left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx}{e^2}\\ &=\frac {\int \frac {-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)}{x^2 \left (-x^2-\log (2)\right ) \left (36 x^2+36 \log (2)\right )} \, dx}{e^2}\\ &=-\frac {\int \frac {-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)}{x^2 \left (-x^2-\log (2)\right )^2} \, dx}{36 e^2}\\ &=\frac {x (1-144 \log (2))+24 \log (2)}{36 e^2 \log (2) \left (x^2+\log (2)\right )}-\frac {\int \frac {2 x^2+\log (4)}{x^2 \left (-x^2-\log (2)\right )} \, dx}{72 e^2 \log (2)}\\ &=\frac {x (1-144 \log (2))+24 \log (2)}{36 e^2 \log (2) \left (x^2+\log (2)\right )}+\frac {\int \frac {1}{x^2} \, dx}{36 e^2 \log (2)}\\ &=-\frac {1}{36 e^2 x \log (2)}+\frac {x (1-144 \log (2))+24 \log (2)}{36 e^2 \log (2) \left (x^2+\log (2)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 0.81 \begin {gather*} -\frac {(1-12 x)^2}{36 e^2 x \left (x^2+\log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(-3*x^2 + 48*x^3 - 144*x^4 + (-1 + 144*x^2)*Log[2]))/(E^2*(-x^2 - Log[2])*(36*x^5 + 36*x^3*Log[2]
)),x]

[Out]

-1/36*(1 - 12*x)^2/(E^2*x*(x^2 + Log[2]))

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fricas [A]  time = 0.50, size = 27, normalized size = 0.87 \begin {gather*} -\frac {144 \, x^{2} - 24 \, x + 1}{36 \, {\left (x^{3} e^{2} + x e^{2} \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((144*x^2-1)*log(2)-144*x^4+48*x^3-3*x^2)/(36*x^3*log(2)+36*x^5)/exp(log((-log(2)-x^2)/x)+2),x, algo
rithm="fricas")

[Out]

-1/36*(144*x^2 - 24*x + 1)/(x^3*e^2 + x*e^2*log(2))

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giac [A]  time = 0.28, size = 24, normalized size = 0.77 \begin {gather*} -\frac {{\left (144 \, x^{2} - 24 \, x + 1\right )} e^{\left (-2\right )}}{36 \, {\left (x^{3} + x \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((144*x^2-1)*log(2)-144*x^4+48*x^3-3*x^2)/(36*x^3*log(2)+36*x^5)/exp(log((-log(2)-x^2)/x)+2),x, algo
rithm="giac")

[Out]

-1/36*(144*x^2 - 24*x + 1)*e^(-2)/(x^3 + x*log(2))

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maple [A]  time = 0.25, size = 25, normalized size = 0.81

method	result	size
risch	\(\frac {{\mathrm e}^{-2} \left (-4 x^{2}+\frac {2}{3} x -\frac {1}{36}\right )}{x \left (\ln \relax (2)+x^{2}\right )}\)	\(25\)
gosper	\(-\frac {\left (12 x -1\right )^{2} {\mathrm e}^{-2}}{36 x \left (\ln \relax (2)+x^{2}\right )}\)	\(30\)
norman	\(\frac {\frac {2 x^{2} {\mathrm e}^{-2}}{3}-\frac {x \,{\mathrm e}^{-2}}{36}-4 \,{\mathrm e}^{-2} x^{3}}{x^{2} \left (\ln \relax (2)+x^{2}\right )}\)	\(39\)
default	\(\frac {{\mathrm e}^{-2} \left (-\frac {\left (144 \ln \relax (2)-1\right ) x -24 \ln \relax (2)}{\ln \relax (2) \left (\ln \relax (2)+x^{2}\right )}-\frac {1}{x \ln \relax (2)}\right )}{36}\)	\(42\)
meijerg	\(-\frac {{\mathrm e}^{-2} \left (-\ln \relax (2)-x^{2}\right )^{x -1} \left (-\ln \relax (2)\right )^{-x} \left (\frac {\Gamma \left (3+x \right ) x^{2} \hypergeom \left (\left [1, 1, 3+x \right ], \left [2, 3\right ], -\frac {x^{2}}{\ln \relax (2)}\right )}{2 \ln \relax (2)}-\left (\Psi \left (2+x \right )+\gamma -1+2 \ln \relax (x )-\ln \left (\ln \relax (2)\right )\right ) \Gamma \left (2+x \right )-\frac {\Gamma \left (x +1\right ) \ln \relax (2)}{x^{2}}\right )}{72 \ln \relax (2) \Gamma \relax (x )}-\frac {{\mathrm e}^{-2} \left (-\ln \relax (2)-x^{2}\right )^{x -1} \left (-\ln \relax (2)\right )^{-x} \left (-\frac {\Gamma \left (2+x \right ) x^{2} \hypergeom \left (\left [1, 1, 2+x \right ], \left [2, 2\right ], -\frac {x^{2}}{\ln \relax (2)}\right )}{\ln \relax (2)}+\left (\Psi \left (x +1\right )+\gamma +2 \ln \relax (x )-\ln \left (\ln \relax (2)\right )\right ) \Gamma \left (x +1\right )\right )}{24 \ln \relax (2) \Gamma \relax (x )}+\frac {2 \,{\mathrm e}^{-2} \left (-\ln \relax (2)-x^{2}\right )^{x -1} \left (-\ln \relax (2)\right )^{-x} \left (-\frac {\Gamma \left (2+x \right ) x^{2} \hypergeom \left (\left [1, 1, 2+x \right ], \left [2, 2\right ], -\frac {x^{2}}{\ln \relax (2)}\right )}{\ln \relax (2)}+\left (\Psi \left (x +1\right )+\gamma +2 \ln \relax (x )-\ln \left (\ln \relax (2)\right )\right ) \Gamma \left (x +1\right )\right )}{\Gamma \relax (x )}-\frac {2 \,{\mathrm e}^{-2} \left (-\ln \relax (2)-x^{2}\right )^{x -1} \left (-\ln \relax (2)\right )^{-x} x^{3} \hypergeom \left (\left [1, x +1\right ], \relax [2], -\frac {x^{2}}{\ln \relax (2)}\right )}{\ln \relax (2)}+\frac {4 \,{\mathrm e}^{-2} \left (-\ln \relax (2)-x^{2}\right )^{x -1} \left (-\ln \relax (2)\right )^{-x} x^{2} \hypergeom \left (\left [\frac {1}{2}, x +1\right ], \left [\frac {3}{2}\right ], -\frac {x^{2}}{\ln \relax (2)}\right )}{3 \ln \relax (2)}\)	\(372\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((144*x^2-1)*ln(2)-144*x^4+48*x^3-3*x^2)/(36*x^3*ln(2)+36*x^5)/exp(ln((-ln(2)-x^2)/x)+2),x,method=_RETURNV
ERBOSE)

[Out]

exp(-2)*(-4*x^2+2/3*x-1/36)/x/(ln(2)+x^2)

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maxima [A]  time = 0.47, size = 24, normalized size = 0.77 \begin {gather*} -\frac {{\left (144 \, x^{2} - 24 \, x + 1\right )} e^{\left (-2\right )}}{36 \, {\left (x^{3} + x \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((144*x^2-1)*log(2)-144*x^4+48*x^3-3*x^2)/(36*x^3*log(2)+36*x^5)/exp(log((-log(2)-x^2)/x)+2),x, algo
rithm="maxima")

[Out]

-1/36*(144*x^2 - 24*x + 1)*e^(-2)/(x^3 + x*log(2))

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mupad [B]  time = 0.14, size = 34, normalized size = 1.10 \begin {gather*} -\frac {{\mathrm {e}}^{-2}\,\left (24\,x^3+144\,\ln \relax (2)\,x^2+\ln \relax (2)\right )}{36\,x\,\ln \relax (2)\,\left (x^2+\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(- log(-(log(2) + x^2)/x) - 2)*(log(2)*(144*x^2 - 1) - 3*x^2 + 48*x^3 - 144*x^4))/(36*x^3*log(2) + 36*
x^5),x)

[Out]

-(exp(-2)*(log(2) + 144*x^2*log(2) + 24*x^3))/(36*x*log(2)*(log(2) + x^2))

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sympy [A]  time = 0.88, size = 27, normalized size = 0.87 \begin {gather*} \frac {- 144 x^{2} + 24 x - 1}{36 x^{3} e^{2} + 36 x e^{2} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((144*x**2-1)*ln(2)-144*x**4+48*x**3-3*x**2)/(36*x**3*ln(2)+36*x**5)/exp(ln((-ln(2)-x**2)/x)+2),x)

[Out]

(-144*x**2 + 24*x - 1)/(36*x**3*exp(2) + 36*x*exp(2)*log(2))

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