3.19.54 \(\int \frac {(-1+x) \sqrt {x+\sqrt {1+x^2}}}{1+x} \, dx\)

Optimal. Leaf size=127 \[ \frac {2 \sqrt {x^2+1} (x-6)+2 \left (x^2-6 x-1\right )}{3 \sqrt {\sqrt {x^2+1}+x}}+4 \sqrt {1+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {\sqrt {x^2+1}+x}}{\sqrt {1+\sqrt {2}}}\right )+4 \sqrt {\sqrt {2}-1} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {x^2+1}+x}}{\sqrt {\sqrt {2}-1}}\right ) \]

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Rubi [A]  time = 0.32, antiderivative size = 136, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6742, 2117, 14, 2119, 1628, 826, 1166, 207, 203} \begin {gather*} \frac {1}{3} \left (\sqrt {x^2+1}+x\right )^{3/2}-4 \sqrt {\sqrt {x^2+1}+x}-\frac {1}{\sqrt {\sqrt {x^2+1}+x}}+4 \sqrt {1+\sqrt {2}} \tan ^{-1}\left (\sqrt {\sqrt {2}-1} \sqrt {\sqrt {x^2+1}+x}\right )+4 \sqrt {\sqrt {2}-1} \tanh ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {x^2+1}+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/Sqrt[x + Sqrt[1 + x^2]]) - 4*Sqrt[x + Sqrt[1 + x^2]] + (x + Sqrt[1 + x^2])^(3/2)/3 + 4*Sqrt[1 + Sqrt[2]]*A
rcTan[Sqrt[-1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]] + 4*Sqrt[-1 + Sqrt[2]]*ArcTanh[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sq
rt[1 + x^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 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 207

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

Rule 826

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 1166

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 1628

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 2117

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 2119

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 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 150, normalized size = 1.18 \begin {gather*} \frac {1}{3} \left (\sqrt {x^2+1}+x\right )^{3/2}-4 \sqrt {\sqrt {x^2+1}+x}-\frac {1}{\sqrt {\sqrt {x^2+1}+x}}+4 \sqrt {\sqrt {2}-1} \left (1+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {\sqrt {x^2+1}+x}}{\sqrt {1+\sqrt {2}}}\right )+4 \left (\sqrt {2}-1\right ) \sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {x^2+1}+x}}{\sqrt {\sqrt {2}-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.30, size = 127, normalized size = 1.00 \begin {gather*} \frac {2 (-6+x) \sqrt {1+x^2}+2 \left (-1-6 x+x^2\right )}{3 \sqrt {x+\sqrt {1+x^2}}}+4 \sqrt {1+\sqrt {2}} \tan ^{-1}\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]

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

[Out]

(2*(-6 + x)*Sqrt[1 + x^2] + 2*(-1 - 6*x + x^2))/(3*Sqrt[x + Sqrt[1 + x^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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fricas [A]  time = 0.46, size = 158, normalized size = 1.24 \begin {gather*} \frac {2}{3} \, {\left (2 \, x - \sqrt {x^{2} + 1} - 6\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} {\left (\sqrt {2} - 1\right )} - \sqrt {x + \sqrt {x^{2} + 1}} \sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )}\right ) + 2 \, \sqrt {\sqrt {2} - 1} \log \left (4 \, \sqrt {x + \sqrt {x^{2} + 1}} + 4 \, \sqrt {\sqrt {2} - 1}\right ) - 2 \, \sqrt {\sqrt {2} - 1} \log \left (4 \, \sqrt {x + \sqrt {x^{2} + 1}} - 4 \, \sqrt {\sqrt {2} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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