3.3.98 \(\int \frac {-64+2^{-x^2} (-16-288 x^2 \log (2))+2^{-2 x^2} (-1-36 x^2 \log (2))+(-2^{2-2 x^2} x^2 \log (2)-2^{5-x^2} x^2 \log (2)) \log (x)}{81 x+18 x \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {\left (8+2^{-x^2}\right )^2}{9+\log (x)} \]

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Rubi [B]  time = 2.18, antiderivative size = 87, normalized size of antiderivative = 4.83, number of steps used = 7, number of rules used = 5, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6688, 6742, 2302, 30, 2288} \begin {gather*} \frac {2^{-2 x^2-1} \left (x^2 \log (16) \log (x)+36 x^2 \log (2)\right )}{x^2 \log (4) (\log (x)+9)^2}+\frac {2^{2-x^2} \left (x^2 \log (16) \log (x)+36 x^2 \log (2)\right )}{x^2 \log (2) (\log (x)+9)^2}+\frac {64}{\log (x)+9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64 + (-16 - 288*x^2*Log[2])/2^x^2 + (-1 - 36*x^2*Log[2])/2^(2*x^2) + (-(2^(2 - 2*x^2)*x^2*Log[2]) - 2^(5
- x^2)*x^2*Log[2])*Log[x])/(81*x + 18*x*Log[x] + x*Log[x]^2),x]

[Out]

64/(9 + Log[x]) + (2^(2 - x^2)*(36*x^2*Log[2] + x^2*Log[16]*Log[x]))/(x^2*Log[2]*(9 + Log[x])^2) + (2^(-1 - 2*
x^2)*(36*x^2*Log[2] + x^2*Log[16]*Log[x]))/(x^2*Log[4]*(9 + Log[x])^2)

Rule 30

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

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4^{-x^2} \left (1+2^{3+x^2}\right ) \left (-1-2^{3+x^2}-36 x^2 \log (2)-x^2 \log (16) \log (x)\right )}{x (9+\log (x))^2} \, dx\\ &=\int \left (-\frac {64}{x (9+\log (x))^2}+\frac {4^{-x^2} \left (-1-36 x^2 \log (2)-x^2 \log (16) \log (x)\right )}{x (9+\log (x))^2}-\frac {2^{3-x^2} \left (2+36 x^2 \log (2)+x^2 \log (16) \log (x)\right )}{x (9+\log (x))^2}\right ) \, dx\\ &=-\left (64 \int \frac {1}{x (9+\log (x))^2} \, dx\right )+\int \frac {4^{-x^2} \left (-1-36 x^2 \log (2)-x^2 \log (16) \log (x)\right )}{x (9+\log (x))^2} \, dx-\int \frac {2^{3-x^2} \left (2+36 x^2 \log (2)+x^2 \log (16) \log (x)\right )}{x (9+\log (x))^2} \, dx\\ &=\frac {2^{2-x^2} \left (36 x^2 \log (2)+x^2 \log (16) \log (x)\right )}{x^2 \log (2) (9+\log (x))^2}+\frac {2^{-1-2 x^2} \left (36 x^2 \log (2)+x^2 \log (16) \log (x)\right )}{x^2 \log (4) (9+\log (x))^2}-64 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,9+\log (x)\right )\\ &=\frac {64}{9+\log (x)}+\frac {2^{2-x^2} \left (36 x^2 \log (2)+x^2 \log (16) \log (x)\right )}{x^2 \log (2) (9+\log (x))^2}+\frac {2^{-1-2 x^2} \left (36 x^2 \log (2)+x^2 \log (16) \log (x)\right )}{x^2 \log (4) (9+\log (x))^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.68, size = 87, normalized size = 4.83 \begin {gather*} -\frac {2^{-1-2 x^2} \left (-36 \log (2) \left (\log (2)+2^{3+x^2} \left (1+2^{2+x^2}\right ) \log (4)\right )-\left (2^{3+x^2} \log (4) \log (16)+\log (2) \left (2^{7+2 x^2} \log (4)+\log (16)\right )\right ) \log (x)\right )}{\log (2) \log (4) (9+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64 + (-16 - 288*x^2*Log[2])/2^x^2 + (-1 - 36*x^2*Log[2])/2^(2*x^2) + (-(2^(2 - 2*x^2)*x^2*Log[2])
- 2^(5 - x^2)*x^2*Log[2])*Log[x])/(81*x + 18*x*Log[x] + x*Log[x]^2),x]

[Out]

-((2^(-1 - 2*x^2)*(-36*Log[2]*(Log[2] + 2^(3 + x^2)*(1 + 2^(2 + x^2))*Log[4]) - (2^(3 + x^2)*Log[4]*Log[16] +
Log[2]*(2^(7 + 2*x^2)*Log[4] + Log[16]))*Log[x]))/(Log[2]*Log[4]*(9 + Log[x])^2))

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fricas [B]  time = 0.76, size = 41, normalized size = 2.28 \begin {gather*} \frac {64 \cdot 2^{2 \, x^{2}} + 16 \cdot 2^{\left (x^{2}\right )} + 1}{2^{2 \, x^{2}} \log \relax (x) + 9 \cdot 2^{2 \, x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*log(2)*exp(-x^2*log(2))^2-32*x^2*log(2)*exp(-x^2*log(2)))*log(x)+(-36*x^2*log(2)-1)*exp(-x^
2*log(2))^2+(-288*x^2*log(2)-16)*exp(-x^2*log(2))-64)/(x*log(x)^2+18*x*log(x)+81*x),x, algorithm="fricas")

[Out]

(64*2^(2*x^2) + 16*2^(x^2) + 1)/(2^(2*x^2)*log(x) + 9*2^(2*x^2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, {\left (\frac {8 \, x^{2} \log \relax (2)}{2^{\left (x^{2}\right )}} + \frac {x^{2} \log \relax (2)}{2^{2 \, x^{2}}}\right )} \log \relax (x) + \frac {16 \, {\left (18 \, x^{2} \log \relax (2) + 1\right )}}{2^{\left (x^{2}\right )}} + \frac {36 \, x^{2} \log \relax (2) + 1}{2^{2 \, x^{2}}} + 64}{x \log \relax (x)^{2} + 18 \, x \log \relax (x) + 81 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*log(2)*exp(-x^2*log(2))^2-32*x^2*log(2)*exp(-x^2*log(2)))*log(x)+(-36*x^2*log(2)-1)*exp(-x^
2*log(2))^2+(-288*x^2*log(2)-16)*exp(-x^2*log(2))-64)/(x*log(x)^2+18*x*log(x)+81*x),x, algorithm="giac")

[Out]

integrate(-(4*(8*x^2*log(2)/2^(x^2) + x^2*log(2)/2^(2*x^2))*log(x) + 16*(18*x^2*log(2) + 1)/2^(x^2) + (36*x^2*
log(2) + 1)/2^(2*x^2) + 64)/(x*log(x)^2 + 18*x*log(x) + 81*x), x)

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maple [A]  time = 0.41, size = 24, normalized size = 1.33

method	result	size
risch	\(\frac {2^{-2 x^{2}}+16 \left (\frac {1}{2}\right )^{x^{2}}+64}{9+\ln \relax (x )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2*ln(2)*exp(-x^2*ln(2))^2-32*x^2*ln(2)*exp(-x^2*ln(2)))*ln(x)+(-36*x^2*ln(2)-1)*exp(-x^2*ln(2))^2+(
-288*x^2*ln(2)-16)*exp(-x^2*ln(2))-64)/(x*ln(x)^2+18*x*ln(x)+81*x),x,method=_RETURNVERBOSE)

[Out]

(((1/2)^(x^2))^2+16*(1/2)^(x^2)+64)/(9+ln(x))

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maxima [A]  time = 1.32, size = 35, normalized size = 1.94 \begin {gather*} \frac {\frac {1}{2^{2 \, x^{2}}} + \frac {16}{2^{\left (x^{2}\right )}}}{\log \relax (x) + 9} + \frac {64}{\log \relax (x) + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*log(2)*exp(-x^2*log(2))^2-32*x^2*log(2)*exp(-x^2*log(2)))*log(x)+(-36*x^2*log(2)-1)*exp(-x^
2*log(2))^2+(-288*x^2*log(2)-16)*exp(-x^2*log(2))-64)/(x*log(x)^2+18*x*log(x)+81*x),x, algorithm="maxima")

[Out]

(1/2^(2*x^2) + 16/2^(x^2))/(log(x) + 9) + 64/(log(x) + 9)

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mupad [B]  time = 0.67, size = 27, normalized size = 1.50 \begin {gather*} \frac {{\left (8\,2^{x^2}+1\right )}^2}{2^{2\,x^2}\,\left (\ln \relax (x)+9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x^2*log(2))*(36*x^2*log(2) + 1) + exp(-x^2*log(2))*(288*x^2*log(2) + 16) + log(x)*(32*x^2*exp(-x^
2*log(2))*log(2) + 4*x^2*exp(-2*x^2*log(2))*log(2)) + 64)/(81*x + x*log(x)^2 + 18*x*log(x)),x)

[Out]

(8*2^(x^2) + 1)^2/(2^(2*x^2)*(log(x) + 9))

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sympy [B]  time = 0.46, size = 48, normalized size = 2.67 \begin {gather*} \frac {\left (\log {\relax (x )} + 9\right ) e^{- 2 x^{2} \log {\relax (2 )}} + \left (16 \log {\relax (x )} + 144\right ) e^{- x^{2} \log {\relax (2 )}}}{\log {\relax (x )}^{2} + 18 \log {\relax (x )} + 81} + \frac {64}{\log {\relax (x )} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2*ln(2)*exp(-x**2*ln(2))**2-32*x**2*ln(2)*exp(-x**2*ln(2)))*ln(x)+(-36*x**2*ln(2)-1)*exp(-x*
*2*ln(2))**2+(-288*x**2*ln(2)-16)*exp(-x**2*ln(2))-64)/(x*ln(x)**2+18*x*ln(x)+81*x),x)

[Out]

((log(x) + 9)*exp(-2*x**2*log(2)) + (16*log(x) + 144)*exp(-x**2*log(2)))/(log(x)**2 + 18*log(x) + 81) + 64/(lo
g(x) + 9)

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