3.39.78 \(\int \frac {e^{\frac {1}{2} (-10+\log (15-12 x \log (2+\log ^2(4+x^2))))} (8 x^2 \log (4+x^2)+(16+4 x^2+(8+2 x^2) \log ^2(4+x^2)) \log (2+\log ^2(4+x^2)))}{-40-10 x^2+(-20-5 x^2) \log ^2(4+x^2)+(32 x+8 x^3+(16 x+4 x^3) \log ^2(4+x^2)) \log (2+\log ^2(4+x^2))} \, dx\)

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

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Rubi [A]  time = 0.89, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 137, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2274, 12, 6741, 6686} \begin {gather*} \frac {\sqrt {3} \sqrt {5-4 x \log \left (\log ^2\left (x^2+4\right )+2\right )}}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-10 + Log[15 - 12*x*Log[2 + Log[4 + x^2]^2]])/2)*(8*x^2*Log[4 + x^2] + (16 + 4*x^2 + (8 + 2*x^2)*Log[
4 + x^2]^2)*Log[2 + Log[4 + x^2]^2]))/(-40 - 10*x^2 + (-20 - 5*x^2)*Log[4 + x^2]^2 + (32*x + 8*x^3 + (16*x + 4
*x^3)*Log[4 + x^2]^2)*Log[2 + Log[4 + x^2]^2]),x]

[Out]

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

Rule 12

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

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\sqrt {15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{e^5 \left (-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )} \, dx\\ &=\frac {\int \frac {\sqrt {15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx}{e^5}\\ &=\frac {\int \frac {\sqrt {3} \left (-8 x^2 \log \left (4+x^2\right )-\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{\left (4+x^2\right ) \left (2+\log ^2\left (4+x^2\right )\right ) \sqrt {5-4 x \log \left (2+\log ^2\left (4+x^2\right )\right )}} \, dx}{e^5}\\ &=\frac {\sqrt {3} \int \frac {-8 x^2 \log \left (4+x^2\right )-\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )}{\left (4+x^2\right ) \left (2+\log ^2\left (4+x^2\right )\right ) \sqrt {5-4 x \log \left (2+\log ^2\left (4+x^2\right )\right )}} \, dx}{e^5}\\ &=\frac {\sqrt {3} \sqrt {5-4 x \log \left (2+\log ^2\left (4+x^2\right )\right )}}{e^5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 29, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \sqrt {5-4 x \log \left (2+\log ^2\left (4+x^2\right )\right )}}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-10 + Log[15 - 12*x*Log[2 + Log[4 + x^2]^2]])/2)*(8*x^2*Log[4 + x^2] + (16 + 4*x^2 + (8 + 2*x^2
)*Log[4 + x^2]^2)*Log[2 + Log[4 + x^2]^2]))/(-40 - 10*x^2 + (-20 - 5*x^2)*Log[4 + x^2]^2 + (32*x + 8*x^3 + (16
*x + 4*x^3)*Log[4 + x^2]^2)*Log[2 + Log[4 + x^2]^2]),x]

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+8)*log(x^2+4)^2+4*x^2+16)*log(log(x^2+4)^2+2)+8*x^2*log(x^2+4))*exp(1/2*log(-12*x*log(log(x
^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*log(x^2+4)^2+8*x^3+32*x)*log(log(x^2+4)^2+2)+(-5*x^2-20)*log(x^2+4)^2-10*x^2-
40),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [A]  time = 0.58, size = 22, normalized size = 0.76 \begin {gather*} e^{\left (\frac {1}{2} \, \log \left (-12 \, x \log \left (\log \left (x^{2} + 4\right )^{2} + 2\right ) + 15\right ) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+8)*log(x^2+4)^2+4*x^2+16)*log(log(x^2+4)^2+2)+8*x^2*log(x^2+4))*exp(1/2*log(-12*x*log(log(x
^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*log(x^2+4)^2+8*x^3+32*x)*log(log(x^2+4)^2+2)+(-5*x^2-20)*log(x^2+4)^2-10*x^2-
40),x, algorithm="giac")

[Out]

e^(1/2*log(-12*x*log(log(x^2 + 4)^2 + 2) + 15) - 5)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (2 x^{2}+8\right ) \ln \left (x^{2}+4\right )^{2}+4 x^{2}+16\right ) \ln \left (\ln \left (x^{2}+4\right )^{2}+2\right )+8 x^{2} \ln \left (x^{2}+4\right )\right ) {\mathrm e}^{\frac {\ln \left (-12 x \ln \left (\ln \left (x^{2}+4\right )^{2}+2\right )+15\right )}{2}-5}}{\left (\left (4 x^{3}+16 x \right ) \ln \left (x^{2}+4\right )^{2}+8 x^{3}+32 x \right ) \ln \left (\ln \left (x^{2}+4\right )^{2}+2\right )+\left (-5 x^{2}-20\right ) \ln \left (x^{2}+4\right )^{2}-10 x^{2}-40}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2+8)*ln(x^2+4)^2+4*x^2+16)*ln(ln(x^2+4)^2+2)+8*x^2*ln(x^2+4))*exp(1/2*ln(-12*x*ln(ln(x^2+4)^2+2)+15
)-5)/(((4*x^3+16*x)*ln(x^2+4)^2+8*x^3+32*x)*ln(ln(x^2+4)^2+2)+(-5*x^2-20)*ln(x^2+4)^2-10*x^2-40),x)

[Out]

int((((2*x^2+8)*ln(x^2+4)^2+4*x^2+16)*ln(ln(x^2+4)^2+2)+8*x^2*ln(x^2+4))*exp(1/2*ln(-12*x*ln(ln(x^2+4)^2+2)+15
)-5)/(((4*x^3+16*x)*ln(x^2+4)^2+8*x^3+32*x)*ln(ln(x^2+4)^2+2)+(-5*x^2-20)*ln(x^2+4)^2-10*x^2-40),x)

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maxima [C]  time = 0.67, size = 25, normalized size = 0.86 \begin {gather*} i \, \sqrt {3} \sqrt {4 \, x \log \left (\log \left (x^{2} + 4\right )^{2} + 2\right ) - 5} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+8)*log(x^2+4)^2+4*x^2+16)*log(log(x^2+4)^2+2)+8*x^2*log(x^2+4))*exp(1/2*log(-12*x*log(log(x
^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*log(x^2+4)^2+8*x^3+32*x)*log(log(x^2+4)^2+2)+(-5*x^2-20)*log(x^2+4)^2-10*x^2-
40),x, algorithm="maxima")

[Out]

I*sqrt(3)*sqrt(4*x*log(log(x^2 + 4)^2 + 2) - 5)*e^(-5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {\ln \left (15-12\,x\,\ln \left ({\ln \left (x^2+4\right )}^2+2\right )\right )}{2}-5}\,\left (\ln \left ({\ln \left (x^2+4\right )}^2+2\right )\,\left (4\,x^2+{\ln \left (x^2+4\right )}^2\,\left (2\,x^2+8\right )+16\right )+8\,x^2\,\ln \left (x^2+4\right )\right )}{10\,x^2-\ln \left ({\ln \left (x^2+4\right )}^2+2\right )\,\left (32\,x+{\ln \left (x^2+4\right )}^2\,\left (4\,x^3+16\,x\right )+8\,x^3\right )+{\ln \left (x^2+4\right )}^2\,\left (5\,x^2+20\right )+40} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(15 - 12*x*log(log(x^2 + 4)^2 + 2))/2 - 5)*(log(log(x^2 + 4)^2 + 2)*(4*x^2 + log(x^2 + 4)^2*(2*x^
2 + 8) + 16) + 8*x^2*log(x^2 + 4)))/(10*x^2 - log(log(x^2 + 4)^2 + 2)*(32*x + log(x^2 + 4)^2*(16*x + 4*x^3) +
8*x^3) + log(x^2 + 4)^2*(5*x^2 + 20) + 40),x)

[Out]

int(-(exp(log(15 - 12*x*log(log(x^2 + 4)^2 + 2))/2 - 5)*(log(log(x^2 + 4)^2 + 2)*(4*x^2 + log(x^2 + 4)^2*(2*x^
2 + 8) + 16) + 8*x^2*log(x^2 + 4)))/(10*x^2 - log(log(x^2 + 4)^2 + 2)*(32*x + log(x^2 + 4)^2*(16*x + 4*x^3) +
8*x^3) + log(x^2 + 4)^2*(5*x^2 + 20) + 40), x)

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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**2+8)*ln(x**2+4)**2+4*x**2+16)*ln(ln(x**2+4)**2+2)+8*x**2*ln(x**2+4))*exp(1/2*ln(-12*x*ln(ln(
x**2+4)**2+2)+15)-5)/(((4*x**3+16*x)*ln(x**2+4)**2+8*x**3+32*x)*ln(ln(x**2+4)**2+2)+(-5*x**2-20)*ln(x**2+4)**2
-10*x**2-40),x)

[Out]

Timed out

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