3.1.17 \(\int \frac {\log (x+\sqrt {-1+x^2})}{(1+x^2)^{3/2}} \, dx\) [17]

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

[Out]

-1/2*arccosh(x^2)+x*ln(x+(x^2-1)^(1/2))/(x^2+1)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {197, 2634, 282, 54} \begin {gather*} \frac {x \log \left (\sqrt {x^2-1}+x\right )}{\sqrt {x^2+1}}-\frac {1}{2} \cosh ^{-1}\left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[x + Sqrt[-1 + x^2]]/(1 + x^2)^(3/2),x]

[Out]

-1/2*ArcCosh[x^2] + (x*Log[x + Sqrt[-1 + x^2]])/Sqrt[1 + x^2]

Rule 54

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[b*(x/a)]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 282

Int[(x_)^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m +
1, 2*n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a1 + b1*x^(n/k))^p*(a2 + b2*x^(n/k))^p, x], x, x^k], x] /; k
!= 1] /; FreeQ[{a1, b1, a2, b2, p}, x] && EqQ[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && IntegerQ[m]

Rule 2634

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]

Rubi steps

\begin {align*} \int \frac {\log \left (x+\sqrt {-1+x^2}\right )}{\left (1+x^2\right )^{3/2}} \, dx &=\frac {x \log \left (x+\sqrt {-1+x^2}\right )}{\sqrt {1+x^2}}-\int \frac {x}{\sqrt {-1+x^2} \sqrt {1+x^2}} \, dx\\ &=\frac {x \log \left (x+\sqrt {-1+x^2}\right )}{\sqrt {1+x^2}}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x} \sqrt {1+x}} \, dx,x,x^2\right )\\ &=-\frac {1}{2} \cosh ^{-1}\left (x^2\right )+\frac {x \log \left (x+\sqrt {-1+x^2}\right )}{\sqrt {1+x^2}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(32)=64\).
time = 0.07, size = 89, normalized size = 2.78 \begin {gather*} \frac {4 x \log \left (x+\sqrt {-1+x^2}\right )+\frac {\sqrt {-1+x^2} \left (1+x^2\right ) \left (\log \left (1-\frac {x^2}{\sqrt {-1+x^4}}\right )-\log \left (1+\frac {x^2}{\sqrt {-1+x^4}}\right )\right )}{\sqrt {-1+x^4}}}{4 \sqrt {1+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[x + Sqrt[-1 + x^2]]/(1 + x^2)^(3/2),x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\ln \left (x +\sqrt {x^{2}-1}\right )}{\left (x^{2}+1\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x+(x^2-1)^(1/2))/(x^2+1)^(3/2),x)

[Out]

int(ln(x+(x^2-1)^(1/2))/(x^2+1)^(3/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x+(x^2-1)^(1/2))/(x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(log(x + sqrt(x^2 - 1))/(x^2 + 1)^(3/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).
time = 0.51, size = 58, normalized size = 1.81 \begin {gather*} \frac {2 \, \sqrt {x^{2} + 1} x \log \left (x + \sqrt {x^{2} - 1}\right ) + {\left (x^{2} + 1\right )} \log \left (-x^{2} + \sqrt {x^{2} + 1} \sqrt {x^{2} - 1}\right )}{2 \, {\left (x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x+(x^2-1)^(1/2))/(x^2+1)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\log {\left (x + \sqrt {x^{2} - 1} \right )}}{\left (x^{2} + 1\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x+(x**2-1)**(1/2))/(x**2+1)**(3/2),x)

[Out]

Integral(log(x + sqrt(x**2 - 1))/(x**2 + 1)**(3/2), x)

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Giac [A]
time = 0.51, size = 36, normalized size = 1.12 \begin {gather*} \frac {x \log \left (x + \sqrt {x^{2} - 1}\right )}{\sqrt {x^{2} + 1}} + \frac {1}{2} \, \log \left (x^{2} - \sqrt {x^{4} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x+(x^2-1)^(1/2))/(x^2+1)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\ln \left (x+\sqrt {x^2-1}\right )}{{\left (x^2+1\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x + (x^2 - 1)^(1/2))/(x^2 + 1)^(3/2),x)

[Out]

int(log(x + (x^2 - 1)^(1/2))/(x^2 + 1)^(3/2), x)

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