3.11 \(\int \frac{x \log (1+x^2) \log (x+\sqrt{1+x^2})}{\sqrt{1+x^2}} \, dx\)

Optimal. Leaf size=68 \[ x \left (-\log \left (x^2+1\right )\right )+\sqrt{x^2+1} \log \left (x^2+1\right ) \log \left (\sqrt{x^2+1}+x\right )-2 \sqrt{x^2+1} \log \left (\sqrt{x^2+1}+x\right )+4 x-2 \tan ^{-1}(x) \]

[Out]

4*x - 2*ArcTan[x] - x*Log[1 + x^2] - 2*Sqrt[1 + x^2]*Log[x + Sqrt[1 + x^2]] + Sqrt[1 + x^2]*Log[1 + x^2]*Log[x
+ Sqrt[1 + x^2]]

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Rubi [A]  time = 0.146424, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {261, 2554, 8, 2557, 12, 2448, 321, 203} \[ x \left (-\log \left (x^2+1\right )\right )+\sqrt{x^2+1} \log \left (x^2+1\right ) \log \left (\sqrt{x^2+1}+x\right )-2 \sqrt{x^2+1} \log \left (\sqrt{x^2+1}+x\right )+4 x-2 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*Log[1 + x^2]*Log[x + Sqrt[1 + x^2]])/Sqrt[1 + x^2],x]

[Out]

4*x - 2*ArcTan[x] - x*Log[1 + x^2] - 2*Sqrt[1 + x^2]*Log[x + Sqrt[1 + x^2]] + Sqrt[1 + x^2]*Log[1 + x^2]*Log[x
+ Sqrt[1 + x^2]]

Rule 261

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

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2557

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[Log[v]*Log[w], z, x] + (-Int[SimplifyInt
egrand[(z*Log[w]*D[v, x])/v, x], x] - Int[SimplifyIntegrand[(z*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFr
eeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]

Rule 12

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

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 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 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])

Rubi steps

\begin{align*} \int \frac{x \log \left (1+x^2\right ) \log \left (x+\sqrt{1+x^2}\right )}{\sqrt{1+x^2}} \, dx &=\sqrt{1+x^2} \log \left (1+x^2\right ) \log \left (x+\sqrt{1+x^2}\right )-\int \log \left (1+x^2\right ) \, dx-\int \frac{2 x \log \left (x+\sqrt{1+x^2}\right )}{\sqrt{1+x^2}} \, dx\\ &=-x \log \left (1+x^2\right )+\sqrt{1+x^2} \log \left (1+x^2\right ) \log \left (x+\sqrt{1+x^2}\right )+2 \int \frac{x^2}{1+x^2} \, dx-2 \int \frac{x \log \left (x+\sqrt{1+x^2}\right )}{\sqrt{1+x^2}} \, dx\\ &=2 x-x \log \left (1+x^2\right )-2 \sqrt{1+x^2} \log \left (x+\sqrt{1+x^2}\right )+\sqrt{1+x^2} \log \left (1+x^2\right ) \log \left (x+\sqrt{1+x^2}\right )+2 \int 1 \, dx-2 \int \frac{1}{1+x^2} \, dx\\ &=4 x-2 \tan ^{-1}(x)-x \log \left (1+x^2\right )-2 \sqrt{1+x^2} \log \left (x+\sqrt{1+x^2}\right )+\sqrt{1+x^2} \log \left (1+x^2\right ) \log \left (x+\sqrt{1+x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0492567, size = 64, normalized size = 0.94 \[ -2 \sqrt{x^2+1} \log \left (\sqrt{x^2+1}+x\right )+\log \left (x^2+1\right ) \left (\sqrt{x^2+1} \log \left (\sqrt{x^2+1}+x\right )-x\right )+4 x-2 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Log[1 + x^2]*Log[x + Sqrt[1 + x^2]])/Sqrt[1 + x^2],x]

[Out]

4*x - 2*ArcTan[x] - 2*Sqrt[1 + x^2]*Log[x + Sqrt[1 + x^2]] + Log[1 + x^2]*(-x + Sqrt[1 + x^2]*Log[x + Sqrt[1 +
x^2]])

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Maple [F]  time = 0.014, size = 0, normalized size = 0. \begin{align*} \int{x\ln \left ({x}^{2}+1 \right ) \ln \left ( x+\sqrt{{x}^{2}+1} \right ){\frac{1}{\sqrt{{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(x^2+1)*ln(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x)

[Out]

int(x*ln(x^2+1)*ln(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (2 \, x^{2} -{\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 2\right )} \log \left (x + \sqrt{x^{2} + 1}\right )}{\sqrt{x^{2} + 1}} + \int \frac{\log \left (x^{2} + 1\right ) - 2}{x^{2} + \sqrt{x^{2} + 1} x}\,{d x} - \int -\frac{2 \, x^{2} -{\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 2}{\sqrt{x^{2} + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x^2+1)*log(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-(2*x^2 - (x^2 + 1)*log(x^2 + 1) + 2)*log(x + sqrt(x^2 + 1))/sqrt(x^2 + 1) + integrate((log(x^2 + 1) - 2)/(x^2
+ sqrt(x^2 + 1)*x), x) - integrate(-(2*x^2 - (x^2 + 1)*log(x^2 + 1) + 2)/(sqrt(x^2 + 1)*x), x)

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Fricas [A]  time = 2.43639, size = 127, normalized size = 1.87 \begin{align*} \sqrt{x^{2} + 1}{\left (\log \left (x^{2} + 1\right ) - 2\right )} \log \left (x + \sqrt{x^{2} + 1}\right ) - x \log \left (x^{2} + 1\right ) + 4 \, x - 2 \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x^2+1)*log(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x, algorithm="fricas")

[Out]

sqrt(x^2 + 1)*(log(x^2 + 1) - 2)*log(x + sqrt(x^2 + 1)) - x*log(x^2 + 1) + 4*x - 2*arctan(x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(x**2+1)*ln(x+(x**2+1)**(1/2))/(x**2+1)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \log \left (x^{2} + 1\right ) \log \left (x + \sqrt{x^{2} + 1}\right )}{\sqrt{x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x^2+1)*log(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x*log(x^2 + 1)*log(x + sqrt(x^2 + 1))/sqrt(x^2 + 1), x)