3.51.3 \(\int \frac {12 x^2+4 x^3+(8+4 x) \log ^2(5 e^3)+(2 x^2+2 \log ^2(5 e^3)) \log (x^2+\log ^2(5 e^3))}{x^2+\log ^2(5 e^3)} \, dx\)

Optimal. Leaf size=29 \[ 3 (3+2 x)+2 \left (x+x \left (x+\log \left (x^2+\log ^2\left (5 e^3\right )\right )\right )\right ) \]

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Rubi [A]  time = 0.20, antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 8, number of rules used = 5, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6725, 1810, 203, 2448, 321} \begin {gather*} 2 x^2+2 x \log \left (x^2+(3+\log (5))^2\right )+8 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12*x^2 + 4*x^3 + (8 + 4*x)*Log[5*E^3]^2 + (2*x^2 + 2*Log[5*E^3]^2)*Log[x^2 + Log[5*E^3]^2])/(x^2 + Log[5*
E^3]^2),x]

[Out]

8*x + 2*x^2 + 2*x*Log[x^2 + (3 + Log[5])^2]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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]

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4 \left (3 x^2+x^3+2 (3+\log (5))^2+x (3+\log (5))^2\right )}{9+x^2+6 \log (5)+\log ^2(5)}+2 \log \left (x^2+(3+\log (5))^2\right )\right ) \, dx\\ &=2 \int \log \left (x^2+(3+\log (5))^2\right ) \, dx+4 \int \frac {3 x^2+x^3+2 (3+\log (5))^2+x (3+\log (5))^2}{9+x^2+6 \log (5)+\log ^2(5)} \, dx\\ &=2 x \log \left (x^2+(3+\log (5))^2\right )-4 \int \frac {x^2}{x^2+(3+\log (5))^2} \, dx+4 \int \left (3+x+\frac {-9-6 \log (5)-\log ^2(5)}{x^2+(3+\log (5))^2}\right ) \, dx\\ &=8 x+2 x^2+2 x \log \left (x^2+(3+\log (5))^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 48, normalized size = 1.66 \begin {gather*} 4 \tan ^{-1}\left (\frac {x}{3+\log (5)}\right ) (3+\log (5))+4 \tan ^{-1}\left (\frac {3+\log (5)}{x}\right ) (3+\log (5))+2 x \left (4+x+\log \left (x^2+(3+\log (5))^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*x^2 + 4*x^3 + (8 + 4*x)*Log[5*E^3]^2 + (2*x^2 + 2*Log[5*E^3]^2)*Log[x^2 + Log[5*E^3]^2])/(x^2 +
Log[5*E^3]^2),x]

[Out]

4*ArcTan[x/(3 + Log[5])]*(3 + Log[5]) + 4*ArcTan[(3 + Log[5])/x]*(3 + Log[5]) + 2*x*(4 + x + Log[x^2 + (3 + Lo
g[5])^2])

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fricas [A]  time = 0.94, size = 26, normalized size = 0.90 \begin {gather*} 2 \, x^{2} + 2 \, x \log \left (x^{2} + \log \relax (5)^{2} + 6 \, \log \relax (5) + 9\right ) + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5*exp(3))^2+2*x^2)*log(log(5*exp(3))^2+x^2)+(4*x+8)*log(5*exp(3))^2+4*x^3+12*x^2)/(log(5*exp
(3))^2+x^2),x, algorithm="fricas")

[Out]

2*x^2 + 2*x*log(x^2 + log(5)^2 + 6*log(5) + 9) + 8*x

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giac [A]  time = 0.14, size = 26, normalized size = 0.90 \begin {gather*} 2 \, x^{2} + 2 \, x \log \left (x^{2} + \log \relax (5)^{2} + 6 \, \log \relax (5) + 9\right ) + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5*exp(3))^2+2*x^2)*log(log(5*exp(3))^2+x^2)+(4*x+8)*log(5*exp(3))^2+4*x^3+12*x^2)/(log(5*exp
(3))^2+x^2),x, algorithm="giac")

[Out]

2*x^2 + 2*x*log(x^2 + log(5)^2 + 6*log(5) + 9) + 8*x

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maple [A]  time = 0.13, size = 24, normalized size = 0.83

method	result	size
risch	\(2 x^{2}+2 x \ln \left (\left (\ln \relax (5)+3\right )^{2}+x^{2}\right )+8 x\)	\(24\)
norman	\(8 x +2 x^{2}+2 x \ln \left (\ln \left (5 \,{\mathrm e}^{3}\right )^{2}+x^{2}\right )\)	\(25\)
default	\(8 x +2 x^{2}-\frac {4 \arctan \left (\frac {x}{\sqrt {\ln \relax (5)^{2}+6 \ln \relax (5)+9}}\right ) \ln \relax (5)^{2}}{\sqrt {\ln \relax (5)^{2}+6 \ln \relax (5)+9}}-\frac {24 \arctan \left (\frac {x}{\sqrt {\ln \relax (5)^{2}+6 \ln \relax (5)+9}}\right ) \ln \relax (5)}{\sqrt {\ln \relax (5)^{2}+6 \ln \relax (5)+9}}-\frac {36 \arctan \left (\frac {x}{\sqrt {\ln \relax (5)^{2}+6 \ln \relax (5)+9}}\right )}{\sqrt {\ln \relax (5)^{2}+6 \ln \relax (5)+9}}+2 x \ln \left (\ln \left (5 \,{\mathrm e}^{3}\right )^{2}+x^{2}\right )+4 \ln \left (5 \,{\mathrm e}^{3}\right ) \arctan \left (\frac {x}{\ln \left (5 \,{\mathrm e}^{3}\right )}\right )\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*ln(5*exp(3))^2+2*x^2)*ln(ln(5*exp(3))^2+x^2)+(4*x+8)*ln(5*exp(3))^2+4*x^3+12*x^2)/(ln(5*exp(3))^2+x^2)
,x,method=_RETURNVERBOSE)

[Out]

2*x^2+2*x*ln((ln(5)+3)^2+x^2)+8*x

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maxima [B]  time = 0.61, size = 70, normalized size = 2.41 \begin {gather*} 2 \, x^{2} + 2 \, x \log \left (x^{2} + \log \relax (5)^{2} + 6 \, \log \relax (5) + 9\right ) - 4 \, \arctan \left (\frac {x}{\log \left (5 \, e^{3}\right )}\right ) \log \left (5 \, e^{3}\right ) + 8 \, x + \frac {4 \, {\left (\log \relax (5)^{2} + 6 \, \log \relax (5) + 9\right )} \arctan \left (\frac {x}{\log \relax (5) + 3}\right )}{\log \relax (5) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5*exp(3))^2+2*x^2)*log(log(5*exp(3))^2+x^2)+(4*x+8)*log(5*exp(3))^2+4*x^3+12*x^2)/(log(5*exp
(3))^2+x^2),x, algorithm="maxima")

[Out]

2*x^2 + 2*x*log(x^2 + log(5)^2 + 6*log(5) + 9) - 4*arctan(x/log(5*e^3))*log(5*e^3) + 8*x + 4*(log(5)^2 + 6*log
(5) + 9)*arctan(x/(log(5) + 3))/(log(5) + 3)

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mupad [B]  time = 0.40, size = 18, normalized size = 0.62 \begin {gather*} 2\,x\,\left (x+\ln \left (x^2+{\ln \left (5\,{\mathrm {e}}^3\right )}^2\right )+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(5*exp(3))^2 + x^2)*(2*log(5*exp(3))^2 + 2*x^2) + log(5*exp(3))^2*(4*x + 8) + 12*x^2 + 4*x^3)/(log
(5*exp(3))^2 + x^2),x)

[Out]

2*x*(x + log(log(5*exp(3))^2 + x^2) + 4)

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sympy [A]  time = 0.17, size = 24, normalized size = 0.83 \begin {gather*} 2 x^{2} + 2 x \log {\left (x^{2} + \log {\left (5 e^{3} \right )}^{2} \right )} + 8 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*ln(5*exp(3))**2+2*x**2)*ln(ln(5*exp(3))**2+x**2)+(4*x+8)*ln(5*exp(3))**2+4*x**3+12*x**2)/(ln(5*e
xp(3))**2+x**2),x)

[Out]

2*x**2 + 2*x*log(x**2 + log(5*exp(3))**2) + 8*x

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