3.19.97 \(\int \frac {-1+x}{(1+x) \sqrt {x+\sqrt {1+x^2}}} \, dx\) [1897]

Optimal. Leaf size=131 \[ \frac {2 (-6+3 x) \sqrt {1+x^2}+2 \left (1-6 x+3 x^2\right )}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}-4 \sqrt {-1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {1+\sqrt {2}}}\right )+4 \sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {-1+\sqrt {2}}}\right ) \]

[Out]

1/3*(2*(-6+3*x)*(x^2+1)^(1/2)+6*x^2-12*x+2)/(x+(x^2+1)^(1/2))^(3/2)-4*(2^(1/2)-1)^(1/2)*arctan((x+(x^2+1)^(1/2
))^(1/2)/(1+2^(1/2))^(1/2))+4*(1+2^(1/2))^(1/2)*arctanh((x+(x^2+1)^(1/2))^(1/2)/(2^(1/2)-1)^(1/2))

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Rubi [A]
time = 0.26, antiderivative size = 134, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6874, 2142, 14, 2144, 1642, 842, 840, 1180, 213, 209} \begin {gather*} -\frac {4 \text {ArcTan}\left (\sqrt {\sqrt {2}-1} \sqrt {\sqrt {x^2+1}+x}\right )}{\sqrt {1+\sqrt {2}}}+\sqrt {\sqrt {x^2+1}+x}-\frac {4}{\sqrt {\sqrt {x^2+1}+x}}-\frac {1}{3 \left (\sqrt {x^2+1}+x\right )^{3/2}}+\frac {4 \tanh ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {x^2+1}+x}\right )}{\sqrt {\sqrt {2}-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*1/(x + Sqrt[1 + x^2])^(3/2) - 4/Sqrt[x + Sqrt[1 + x^2]] + Sqrt[x + Sqrt[1 + x^2]] - (4*ArcTan[Sqrt[-1 + S
qrt[2]]*Sqrt[x + Sqrt[1 + x^2]]])/Sqrt[1 + Sqrt[2]] + (4*ArcTanh[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]])/S
qrt[-1 + Sqrt[2]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 209

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

Rule 213

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

Rule 840

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 842

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e
*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(d +
e*x)^(m + 1)*(Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c,
d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && FractionQ[m] && LtQ[m, -1]

Rule 1180

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2142

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[(g + h*x^n)^p*((d^2 + a*f^2 - 2*d*x + x^2)/(d - x)^2), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 2144

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(
m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*((-a)*f^2*h + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt
[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

Rule 6874

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

Rubi steps

\begin {align*} \int \frac {-1+x}{(1+x) \sqrt {x+\sqrt {1+x^2}}} \, dx &=\int \left (\frac {1}{\sqrt {x+\sqrt {1+x^2}}}-\frac {2}{(1+x) \sqrt {x+\sqrt {1+x^2}}}\right ) \, dx\\ &=-\left (2 \int \frac {1}{(1+x) \sqrt {x+\sqrt {1+x^2}}} \, dx\right )+\int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {1+x^2}{x^{5/2}} \, dx,x,x+\sqrt {1+x^2}\right )-2 \text {Subst}\left (\int \frac {1+x^2}{x^{3/2} \left (-1+2 x+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x^{5/2}}+\frac {1}{\sqrt {x}}\right ) \, dx,x,x+\sqrt {1+x^2}\right )-2 \text {Subst}\left (\int \left (\frac {1}{x^{3/2}}+\frac {2 (1-x)}{x^{3/2} \left (-1+2 x+x^2\right )}\right ) \, dx,x,x+\sqrt {1+x^2}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\frac {4}{\sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}-4 \text {Subst}\left (\int \frac {1-x}{x^{3/2} \left (-1+2 x+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {4}{\sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}+4 \text {Subst}\left (\int \frac {-1-x}{\sqrt {x} \left (-1+2 x+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {4}{\sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}+8 \text {Subst}\left (\int \frac {-1-x^2}{-1+2 x^2+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {4}{\sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}-4 \text {Subst}\left (\int \frac {1}{1-\sqrt {2}+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )-4 \text {Subst}\left (\int \frac {1}{1+\sqrt {2}+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {4}{\sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}-\frac {4 \tan ^{-1}\left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}+\frac {4 \tanh ^{-1}\left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {-1+\sqrt {2}}}\right )}{\sqrt {-1+\sqrt {2}}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 126, normalized size = 0.96 \begin {gather*} \frac {2-12 x+6 x^2+6 (-2+x) \sqrt {1+x^2}}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}-4 \sqrt {-1+\sqrt {2}} \text {ArcTan}\left (\sqrt {-1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right )+4 \sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2 - 12*x + 6*x^2 + 6*(-2 + x)*Sqrt[1 + x^2])/(3*(x + Sqrt[1 + x^2])^(3/2)) - 4*Sqrt[-1 + Sqrt[2]]*ArcTan[Sqrt
[-1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]] + 4*Sqrt[1 + Sqrt[2]]*ArcTanh[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]
]]

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {-1+x}{\left (1+x \right ) \sqrt {x +\sqrt {x^{2}+1}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((-1+x)/(1+x)/(x+(x^2+1)^(1/2))^(1/2),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((-1+x)/(1+x)/(x+(x^2+1)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 164, normalized size = 1.25 \begin {gather*} -\frac {2}{3} \, {\left (x^{2} - \sqrt {x^{2} + 1} {\left (x - 6\right )} - 6 \, x - 1\right )} \sqrt {x + \sqrt {x^{2} + 1}} + 8 \, \sqrt {\sqrt {2} - 1} \arctan \left (\sqrt {x + \sqrt {2} + \sqrt {x^{2} + 1} + 1} \sqrt {\sqrt {2} - 1} - \sqrt {x + \sqrt {x^{2} + 1}} \sqrt {\sqrt {2} - 1}\right ) + 2 \, \sqrt {\sqrt {2} + 1} \log \left (4 \, \sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )} + 4 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) - 2 \, \sqrt {\sqrt {2} + 1} \log \left (-4 \, \sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )} + 4 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(x^2 - sqrt(x^2 + 1)*(x - 6) - 6*x - 1)*sqrt(x + sqrt(x^2 + 1)) + 8*sqrt(sqrt(2) - 1)*arctan(sqrt(x + sqr
t(2) + sqrt(x^2 + 1) + 1)*sqrt(sqrt(2) - 1) - sqrt(x + sqrt(x^2 + 1))*sqrt(sqrt(2) - 1)) + 2*sqrt(sqrt(2) + 1)
*log(4*sqrt(sqrt(2) + 1)*(sqrt(2) - 1) + 4*sqrt(x + sqrt(x^2 + 1))) - 2*sqrt(sqrt(2) + 1)*log(-4*sqrt(sqrt(2)
+ 1)*(sqrt(2) - 1) + 4*sqrt(x + sqrt(x^2 + 1)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x-1}{\sqrt {x+\sqrt {x^2+1}}\,\left (x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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