3.7.39 \(\int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2(\frac {16}{x^2})} x^2 \log (\frac {16}{x^2})+e^{\log ^2(\frac {16}{x^2})} (12 x+(48 x-16 x^2) \log (\frac {16}{x^2}))}{x^3} \, dx\)

Optimal. Leaf size=26 \[ 2 \left (-1+\left (1+e^{\log ^2\left (\frac {16}{x^2}\right )}-\frac {3}{x}\right )^2-x\right ) \]

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Rubi [B]  time = 0.21, antiderivative size = 69, normalized size of antiderivative = 2.65, number of steps used = 7, number of rules used = 5, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {14, 2276, 2204, 2209, 2288} \begin {gather*} \frac {18}{x^2}+2 e^{2 \log ^2\left (\frac {16}{x^2}\right )}-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )}-2 x-\frac {12}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-36 + 12*x - 2*x^3 - 16*E^(2*Log[16/x^2]^2)*x^2*Log[16/x^2] + E^Log[16/x^2]^2*(12*x + (48*x - 16*x^2)*Log
[16/x^2]))/x^3,x]

[Out]

2*E^(2*Log[16/x^2]^2) + 18/x^2 - 12/x - 2*x - (4*E^Log[16/x^2]^2*(3*Log[16/x^2] - x*Log[16/x^2]))/(x*Log[16/x^
2])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2276

Int[(F_)^(((a_.) + Log[(c_.)*(x_)^(n_.)]^2*(b_.))*(d_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)/(
e*n*(c*x^n)^((m + 1)/n)), Subst[Int[E^(a*d*Log[F] + ((m + 1)*x)/n + b*d*Log[F]*x^2), x], x, Log[c*x^n]], x] /;
FreeQ[{F, a, b, c, d, e, m, n}, x]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 \left (18-6 x+x^3\right )}{x^3}-\frac {16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} \log \left (\frac {16}{x^2}\right )}{x}-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (-3-12 \log \left (\frac {16}{x^2}\right )+4 x \log \left (\frac {16}{x^2}\right )\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {18-6 x+x^3}{x^3} \, dx\right )-4 \int \frac {e^{\log ^2\left (\frac {16}{x^2}\right )} \left (-3-12 \log \left (\frac {16}{x^2}\right )+4 x \log \left (\frac {16}{x^2}\right )\right )}{x^2} \, dx-16 \int \frac {e^{2 \log ^2\left (\frac {16}{x^2}\right )} \log \left (\frac {16}{x^2}\right )}{x} \, dx\\ &=-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )}-2 \int \left (1+\frac {18}{x^3}-\frac {6}{x^2}\right ) \, dx+8 \operatorname {Subst}\left (\int e^{2 x^2} x \, dx,x,\log \left (\frac {16}{x^2}\right )\right )\\ &=2 e^{2 \log ^2\left (\frac {16}{x^2}\right )}+\frac {18}{x^2}-\frac {12}{x}-2 x-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 46, normalized size = 1.77 \begin {gather*} -2 \left (-e^{2 \log ^2\left (\frac {16}{x^2}\right )}-\frac {9}{x^2}+\frac {6}{x}-\frac {2 e^{\log ^2\left (\frac {16}{x^2}\right )} (-3+x)}{x}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-36 + 12*x - 2*x^3 - 16*E^(2*Log[16/x^2]^2)*x^2*Log[16/x^2] + E^Log[16/x^2]^2*(12*x + (48*x - 16*x^
2)*Log[16/x^2]))/x^3,x]

[Out]

-2*(-E^(2*Log[16/x^2]^2) - 9/x^2 + 6/x - (2*E^Log[16/x^2]^2*(-3 + x))/x + x)

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fricas [A]  time = 0.76, size = 47, normalized size = 1.81 \begin {gather*} -\frac {2 \, {\left (x^{3} - x^{2} e^{\left (2 \, \log \left (\frac {16}{x^{2}}\right )^{2}\right )} - 2 \, {\left (x^{2} - 3 \, x\right )} e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x - 9\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^2*log(16/x^2)*exp(log(16/x^2)^2)^2+((-16*x^2+48*x)*log(16/x^2)+12*x)*exp(log(16/x^2)^2)-2*x^3
+12*x-36)/x^3,x, algorithm="fricas")

[Out]

-2*(x^3 - x^2*e^(2*log(16/x^2)^2) - 2*(x^2 - 3*x)*e^(log(16/x^2)^2) + 6*x - 9)/x^2

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giac [B]  time = 0.72, size = 55, normalized size = 2.12 \begin {gather*} -\frac {2 \, {\left (x^{3} - x^{2} e^{\left (2 \, \log \left (\frac {16}{x^{2}}\right )^{2}\right )} - 2 \, x^{2} e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x - 9\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^2*log(16/x^2)*exp(log(16/x^2)^2)^2+((-16*x^2+48*x)*log(16/x^2)+12*x)*exp(log(16/x^2)^2)-2*x^3
+12*x-36)/x^3,x, algorithm="giac")

[Out]

-2*(x^3 - x^2*e^(2*log(16/x^2)^2) - 2*x^2*e^(log(16/x^2)^2) + 6*x*e^(log(16/x^2)^2) + 6*x - 9)/x^2

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maple [A]  time = 0.08, size = 44, normalized size = 1.69

method	result	size
risch	\(-2 x +\frac {-12 x +18}{x^{2}}+2 \,{\mathrm e}^{2 \ln \left (\frac {16}{x^{2}}\right )^{2}}+\frac {4 \left (x -3\right ) {\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}}}{x}\)	\(44\)
default	\(\frac {4 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}} x -12 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}}}{x}-2 x +\frac {18}{x^{2}}-\frac {12}{x}+2 \,{\mathrm e}^{2 \ln \left (\frac {16}{x^{2}}\right )^{2}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*x^2*ln(16/x^2)*exp(ln(16/x^2)^2)^2+((-16*x^2+48*x)*ln(16/x^2)+12*x)*exp(ln(16/x^2)^2)-2*x^3+12*x-36)/
x^3,x,method=_RETURNVERBOSE)

[Out]

-2*x+(-12*x+18)/x^2+2*exp(2*ln(16/x^2)^2)+4*(x-3)/x*exp(ln(16/x^2)^2)

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maxima [B]  time = 0.63, size = 77, normalized size = 2.96 \begin {gather*} -2 \, x + \frac {2 \, {\left (x e^{\left (32 \, \log \relax (2)^{2} - 32 \, \log \relax (2) \log \relax (x) + 8 \, \log \relax (x)^{2}\right )} + 2 \, {\left (x e^{\left (16 \, \log \relax (2)^{2}\right )} - 3 \, e^{\left (16 \, \log \relax (2)^{2}\right )}\right )} e^{\left (-16 \, \log \relax (2) \log \relax (x) + 4 \, \log \relax (x)^{2}\right )}\right )}}{x} - \frac {12}{x} + \frac {18}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^2*log(16/x^2)*exp(log(16/x^2)^2)^2+((-16*x^2+48*x)*log(16/x^2)+12*x)*exp(log(16/x^2)^2)-2*x^3
+12*x-36)/x^3,x, algorithm="maxima")

[Out]

-2*x + 2*(x*e^(32*log(2)^2 - 32*log(2)*log(x) + 8*log(x)^2) + 2*(x*e^(16*log(2)^2) - 3*e^(16*log(2)^2))*e^(-16
*log(2)*log(x) + 4*log(x)^2))/x - 12/x + 18/x^2

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mupad [B]  time = 0.64, size = 93, normalized size = 3.58 \begin {gather*} 2\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x^2}\right )}^2+32\,{\ln \relax (2)}^2}\,{\left (\frac {1}{x^2}\right )}^{16\,\ln \relax (2)}-2\,x+4\,{\mathrm {e}}^{{\ln \left (\frac {1}{x^2}\right )}^2+16\,{\ln \relax (2)}^2}\,{\left (\frac {1}{x^2}\right )}^{8\,\ln \relax (2)}+\frac {18\,x-x^2\,\left (12\,{\mathrm {e}}^{{\ln \left (\frac {1}{x^2}\right )}^2+16\,{\ln \relax (2)}^2}\,{\left (\frac {1}{x^2}\right )}^{8\,\ln \relax (2)}+12\right )}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^3 - exp(log(16/x^2)^2)*(12*x + log(16/x^2)*(48*x - 16*x^2)) - 12*x + 16*x^2*exp(2*log(16/x^2)^2)*log
(16/x^2) + 36)/x^3,x)

[Out]

2*exp(2*log(1/x^2)^2 + 32*log(2)^2)*(1/x^2)^(16*log(2)) - 2*x + 4*exp(log(1/x^2)^2 + 16*log(2)^2)*(1/x^2)^(8*l
og(2)) + (18*x - x^2*(12*exp(log(1/x^2)^2 + 16*log(2)^2)*(1/x^2)^(8*log(2)) + 12))/x^3

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sympy [A]  time = 0.34, size = 42, normalized size = 1.62 \begin {gather*} - 2 x + \frac {2 x e^{2 \log {\left (\frac {16}{x^{2}} \right )}^{2}} + \left (4 x - 12\right ) e^{\log {\left (\frac {16}{x^{2}} \right )}^{2}}}{x} - \frac {12 x - 18}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x**2*ln(16/x**2)*exp(ln(16/x**2)**2)**2+((-16*x**2+48*x)*ln(16/x**2)+12*x)*exp(ln(16/x**2)**2)-
2*x**3+12*x-36)/x**3,x)

[Out]

-2*x + (2*x*exp(2*log(16/x**2)**2) + (4*x - 12)*exp(log(16/x**2)**2))/x - (12*x - 18)/x**2

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