3.42.13 \(\int \frac {e^{\frac {1}{4} (2-\log (4))} ((-9-x^2) \log (4)+e^{\frac {1}{4} (-2+\log (4))} (18 x+2 x^3+4 x \log (4)))}{(9+x^2) \log (4)} \, dx\)

Optimal. Leaf size=32 \[ -e^{\frac {1}{4} (2-\log (4))} x+\frac {x^2}{\log (4)}+2 \log \left (9+x^2\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 29, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {12, 1810, 260} \begin {gather*} \frac {x^2}{\log (4)}+2 \log \left (x^2+9\right )-\sqrt {\frac {e}{2}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2 - Log[4])/4)*((-9 - x^2)*Log[4] + E^((-2 + Log[4])/4)*(18*x + 2*x^3 + 4*x*Log[4])))/((9 + x^2)*Log[
4]),x]

[Out]

-(Sqrt[E/2]*x) + x^2/Log[4] + 2*Log[9 + x^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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\sqrt {\frac {e}{2}} \int \frac {\left (-9-x^2\right ) \log (4)+e^{\frac {1}{4} (-2+\log (4))} \left (18 x+2 x^3+4 x \log (4)\right )}{9+x^2} \, dx}{\log (4)}\\ &=\frac {\sqrt {\frac {e}{2}} \int \left (2 \sqrt {\frac {2}{e}} x-\log (4)+\frac {4 \sqrt {\frac {2}{e}} x \log (4)}{9+x^2}\right ) \, dx}{\log (4)}\\ &=-\sqrt {\frac {e}{2}} x+\frac {x^2}{\log (4)}+4 \int \frac {x}{9+x^2} \, dx\\ &=-\sqrt {\frac {e}{2}} x+\frac {x^2}{\log (4)}+2 \log \left (9+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 33, normalized size = 1.03 \begin {gather*} \frac {18+2 x^2-\sqrt {2 e} x \log (4)+\log (256) \log \left (9+x^2\right )}{\log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2 - Log[4])/4)*((-9 - x^2)*Log[4] + E^((-2 + Log[4])/4)*(18*x + 2*x^3 + 4*x*Log[4])))/((9 + x^2
)*Log[4]),x]

[Out]

(18 + 2*x^2 - Sqrt[2*E]*x*Log[4] + Log[256]*Log[9 + x^2])/Log[16]

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fricas [A]  time = 0.59, size = 47, normalized size = 1.47 \begin {gather*} \frac {{\left (x^{2} e^{\left (\frac {1}{2} \, \log \relax (2) - \frac {1}{2}\right )} + 4 \, e^{\left (\frac {1}{2} \, \log \relax (2) - \frac {1}{2}\right )} \log \relax (2) \log \left (x^{2} + 9\right ) - 2 \, x \log \relax (2)\right )} e^{\left (-\frac {1}{2} \, \log \relax (2) + \frac {1}{2}\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x*log(2)+2*x^3+18*x)*exp(1/2*log(2)-1/2)+2*(-x^2-9)*log(2))/(x^2+9)/log(2)/exp(1/2*log(2)-1/
2),x, algorithm="fricas")

[Out]

1/2*(x^2*e^(1/2*log(2) - 1/2) + 4*e^(1/2*log(2) - 1/2)*log(2)*log(x^2 + 9) - 2*x*log(2))*e^(-1/2*log(2) + 1/2)
/log(2)

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giac [A]  time = 0.98, size = 47, normalized size = 1.47 \begin {gather*} \frac {{\left (x^{2} e^{\left (\frac {1}{2} \, \log \relax (2) - \frac {1}{2}\right )} + 4 \, e^{\left (\frac {1}{2} \, \log \relax (2) - \frac {1}{2}\right )} \log \relax (2) \log \left (x^{2} + 9\right ) - 2 \, x \log \relax (2)\right )} e^{\left (-\frac {1}{2} \, \log \relax (2) + \frac {1}{2}\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x*log(2)+2*x^3+18*x)*exp(1/2*log(2)-1/2)+2*(-x^2-9)*log(2))/(x^2+9)/log(2)/exp(1/2*log(2)-1/
2),x, algorithm="giac")

[Out]

1/2*(x^2*e^(1/2*log(2) - 1/2) + 4*e^(1/2*log(2) - 1/2)*log(2)*log(x^2 + 9) - 2*x*log(2))*e^(-1/2*log(2) + 1/2)
/log(2)

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maple [A]  time = 0.12, size = 27, normalized size = 0.84

method	result	size
risch	\(\frac {x^{2}}{2 \ln \relax (2)}-\frac {\sqrt {2}\, {\mathrm e}^{\frac {1}{2}} x}{2}+2 \ln \left (x^{2}+9\right )\)	\(27\)
norman	\(\frac {x^{2}}{2 \ln \relax (2)}-\frac {\sqrt {2}\, {\mathrm e}^{\frac {1}{2}} x}{2}+2 \ln \left (x^{2}+9\right )\)	\(29\)
default	\(\frac {{\mathrm e}^{\frac {1}{2}-\frac {\ln \relax (2)}{2}} \left ({\mathrm e}^{\frac {\ln \relax (2)}{2}-\frac {1}{2}} x^{2}-2 x \ln \relax (2)+4 \ln \relax (2) {\mathrm e}^{\frac {\ln \relax (2)}{2}-\frac {1}{2}} \ln \left (x^{2}+9\right )\right )}{2 \ln \relax (2)}\)	\(50\)
meijerg	\(-3 \,{\mathrm e}^{\frac {1}{2}-\frac {\ln \relax (2)}{2}} \arctan \left (\frac {x}{3}\right )-\frac {3 \,{\mathrm e}^{\frac {1}{2}-\frac {\ln \relax (2)}{2}} \left (\frac {2 x}{3}-2 \arctan \left (\frac {x}{3}\right )\right )}{2}+\frac {\left (4 \ln \relax (2)+9\right ) \ln \left (1+\frac {x^{2}}{9}\right )}{2 \ln \relax (2)}+\frac {\frac {x^{2}}{2}-\frac {9 \ln \left (1+\frac {x^{2}}{9}\right )}{2}}{\ln \relax (2)}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((8*x*ln(2)+2*x^3+18*x)*exp(1/2*ln(2)-1/2)+2*(-x^2-9)*ln(2))/(x^2+9)/ln(2)/exp(1/2*ln(2)-1/2),x,method
=_RETURNVERBOSE)

[Out]

1/2*x^2/ln(2)-1/2*2^(1/2)*exp(1/2)*x+2*ln(x^2+9)

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maxima [A]  time = 0.36, size = 51, normalized size = 1.59 \begin {gather*} \frac {{\left (4 \, e^{\left (\frac {1}{2} \, \log \relax (2) - \frac {1}{2}\right )} \log \relax (2) \log \left (x^{2} + 9\right ) - {\left (2 \, x e^{\frac {1}{2}} \log \relax (2) - \sqrt {2} x^{2}\right )} e^{\left (-\frac {1}{2}\right )}\right )} e^{\left (-\frac {1}{2} \, \log \relax (2) + \frac {1}{2}\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x*log(2)+2*x^3+18*x)*exp(1/2*log(2)-1/2)+2*(-x^2-9)*log(2))/(x^2+9)/log(2)/exp(1/2*log(2)-1/
2),x, algorithm="maxima")

[Out]

1/2*(4*e^(1/2*log(2) - 1/2)*log(2)*log(x^2 + 9) - (2*x*e^(1/2)*log(2) - sqrt(2)*x^2)*e^(-1/2))*e^(-1/2*log(2)
+ 1/2)/log(2)

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mupad [B]  time = 0.15, size = 26, normalized size = 0.81 \begin {gather*} 2\,\ln \left (x^2+9\right )+\frac {x^2}{2\,\ln \relax (2)}-\frac {\sqrt {2}\,x\,\sqrt {\mathrm {e}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1/2 - log(2)/2)*(log(2)*(x^2 + 9) - (exp(log(2)/2 - 1/2)*(18*x + 8*x*log(2) + 2*x^3))/2))/(log(2)*(x
^2 + 9)),x)

[Out]

2*log(x^2 + 9) + x^2/(2*log(2)) - (2^(1/2)*x*exp(1/2))/2

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sympy [A]  time = 0.20, size = 29, normalized size = 0.91 \begin {gather*} \frac {x^{2}}{2 \log {\relax (2 )}} - \frac {\sqrt {2} x e^{\frac {1}{2}}}{2} + 2 \log {\left (x^{2} + 9 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x*ln(2)+2*x**3+18*x)*exp(1/2*ln(2)-1/2)+2*(-x**2-9)*ln(2))/(x**2+9)/ln(2)/exp(1/2*ln(2)-1/2)
,x)

[Out]

x**2/(2*log(2)) - sqrt(2)*x*exp(1/2)/2 + 2*log(x**2 + 9)

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