3.899 \(\int \frac{1}{(1+\frac{2 x}{1+x^2})^{3/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac{3 (x+2)}{2 \sqrt{\frac{2 x}{x^2+1}+1}}-\frac{x^2+1}{2 (x+1) \sqrt{\frac{2 x}{x^2+1}+1}}-\frac{3 (x+1) \sinh ^{-1}(x)}{\sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1}}-\frac{9 (x+1) \tanh ^{-1}\left (\frac{1-x}{\sqrt{2} \sqrt{x^2+1}}\right )}{2 \sqrt{2} \sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1}} \]

[Out]

(3*(2 + x))/(2*Sqrt[1 + (2*x)/(1 + x^2)]) - (1 + x^2)/(2*(1 + x)*Sqrt[1 + (2*x)/(1 + x^2)]) - (3*(1 + x)*ArcSi
nh[x])/(Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)]) - (9*(1 + x)*ArcTanh[(1 - x)/(Sqrt[2]*Sqrt[1 + x^2])])/(2*Sqr
t[2]*Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)])

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Rubi [A]  time = 0.0817175, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6723, 970, 733, 813, 844, 215, 725, 206} \[ \frac{3 (x+2)}{2 \sqrt{\frac{2 x}{x^2+1}+1}}-\frac{x^2+1}{2 (x+1) \sqrt{\frac{2 x}{x^2+1}+1}}-\frac{3 (x+1) \sinh ^{-1}(x)}{\sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1}}-\frac{9 (x+1) \tanh ^{-1}\left (\frac{1-x}{\sqrt{2} \sqrt{x^2+1}}\right )}{2 \sqrt{2} \sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + (2*x)/(1 + x^2))^(-3/2),x]

[Out]

(3*(2 + x))/(2*Sqrt[1 + (2*x)/(1 + x^2)]) - (1 + x^2)/(2*(1 + x)*Sqrt[1 + (2*x)/(1 + x^2)]) - (3*(1 + x)*ArcSi
nh[x])/(Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)]) - (9*(1 + x)*ArcTanh[(1 - x)/(Sqrt[2]*Sqrt[1 + x^2])])/(2*Sqr
t[2]*Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)])

Rule 6723

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_)*(x_)^(m_.))^(p_), x_Symbol] :> Dist[(a + b*x^m*v^n)^FracPart[p]/(v^(n*FracP
art[p])*(b*x^m + a/v^n)^FracPart[p]), Int[u*v^(n*p)*(b*x^m + a/v^n)^p, x], x] /; FreeQ[{a, b, m, p}, x] &&  !I
ntegerQ[p] && ILtQ[n, 0] && BinomialQ[v, x]

Rule 970

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[(a + b*x + c*x^2)^F
racPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(b + 2*c*x)^(2*p)*(d + f*x^2)^q, x], x] /; Free
Q[{a, b, c, d, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 733

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 1)), x] - Dist[(2*c*p)/(e*(m + 1)), Int[x*(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c,
d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

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

Rule 215

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{\left (1+\frac{2 x}{1+x^2}\right )^{3/2}} \, dx &=\frac{\sqrt{1+2 x+x^2} \int \frac{\left (1+x^2\right )^{3/2}}{\left (1+2 x+x^2\right )^{3/2}} \, dx}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{(4 (2+2 x)) \int \frac{\left (1+x^2\right )^{3/2}}{(2+2 x)^3} \, dx}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=-\frac{1+x^2}{2 (1+x) \sqrt{1+\frac{2 x}{1+x^2}}}+\frac{(3 (2+2 x)) \int \frac{x \sqrt{1+x^2}}{(2+2 x)^2} \, dx}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{3 (2+x)}{2 \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{1+x^2}{2 (1+x) \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{(3 (2+2 x)) \int \frac{-4+8 x}{(2+2 x) \sqrt{1+x^2}} \, dx}{8 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{3 (2+x)}{2 \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{1+x^2}{2 (1+x) \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{(3 (2+2 x)) \int \frac{1}{\sqrt{1+x^2}} \, dx}{2 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}+\frac{(9 (2+2 x)) \int \frac{1}{(2+2 x) \sqrt{1+x^2}} \, dx}{2 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{3 (2+x)}{2 \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{1+x^2}{2 (1+x) \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{3 (1+x) \sinh ^{-1}(x)}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{(9 (2+2 x)) \operatorname{Subst}\left (\int \frac{1}{8-x^2} \, dx,x,\frac{2-2 x}{\sqrt{1+x^2}}\right )}{2 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{3 (2+x)}{2 \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{1+x^2}{2 (1+x) \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{3 (1+x) \sinh ^{-1}(x)}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{9 (1+x) \tanh ^{-1}\left (\frac{1-x}{\sqrt{2} \sqrt{1+x^2}}\right )}{2 \sqrt{2} \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ \end{align*}

Mathematica [A]  time = 0.0901986, size = 95, normalized size = 0.66 \[ \frac{(x+1) \left (2 \sqrt{x^2+1} \left (2 x^2+9 x+5\right )+9 \sqrt{2} (x+1)^2 \tanh ^{-1}\left (\frac{x-1}{\sqrt{2} \sqrt{x^2+1}}\right )-12 (x+1)^2 \sinh ^{-1}(x)\right )}{4 \left (\frac{(x+1)^2}{x^2+1}\right )^{3/2} \left (x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + x)*(2*Sqrt[1 + x^2]*(5 + 9*x + 2*x^2) - 12*(1 + x)^2*ArcSinh[x] + 9*Sqrt[2]*(1 + x)^2*ArcTanh[(-1 + x)/(
Sqrt[2]*Sqrt[1 + x^2])]))/(4*((1 + x)^2/(1 + x^2))^(3/2)*(1 + x^2)^(3/2))

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Maple [A]  time = 0.01, size = 218, normalized size = 1.5 \begin{align*}{\frac{1+x}{8} \left ( \left ({x}^{2}+1 \right ) ^{{\frac{5}{2}}}x- \left ({x}^{2}+1 \right ) ^{{\frac{3}{2}}}{x}^{3}- \left ({x}^{2}+1 \right ) ^{{\frac{5}{2}}}+ \left ({x}^{2}+1 \right ) ^{{\frac{3}{2}}}{x}^{2}+18\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{ \left ( x-1 \right ) \sqrt{2}}{\sqrt{{x}^{2}+1}}} \right ){x}^{2}+5\,x \left ({x}^{2}+1 \right ) ^{3/2}-6\,\sqrt{{x}^{2}+1}{x}^{3}+36\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{ \left ( x-1 \right ) \sqrt{2}}{\sqrt{{x}^{2}+1}}} \right ) x-24\,{\it Arcsinh} \left ( x \right ){x}^{2}+3\, \left ({x}^{2}+1 \right ) ^{3/2}+6\,\sqrt{{x}^{2}+1}{x}^{2}+18\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{ \left ( x-1 \right ) \sqrt{2}}{\sqrt{{x}^{2}+1}}} \right ) -48\,{\it Arcsinh} \left ( x \right ) x+30\,x\sqrt{{x}^{2}+1}-24\,{\it Arcsinh} \left ( x \right ) +18\,\sqrt{{x}^{2}+1} \right ) \left ({\frac{{x}^{2}+2\,x+1}{{x}^{2}+1}} \right ) ^{-{\frac{3}{2}}} \left ({x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+2*x/(x^2+1))^(3/2),x)

[Out]

1/8/((x^2+2*x+1)/(x^2+1))^(3/2)*(1+x)*((x^2+1)^(5/2)*x-(x^2+1)^(3/2)*x^3-(x^2+1)^(5/2)+(x^2+1)^(3/2)*x^2+18*2^
(1/2)*arctanh(1/2*(x-1)*2^(1/2)/(x^2+1)^(1/2))*x^2+5*x*(x^2+1)^(3/2)-6*(x^2+1)^(1/2)*x^3+36*2^(1/2)*arctanh(1/
2*(x-1)*2^(1/2)/(x^2+1)^(1/2))*x-24*arcsinh(x)*x^2+3*(x^2+1)^(3/2)+6*(x^2+1)^(1/2)*x^2+18*2^(1/2)*arctanh(1/2*
(x-1)*2^(1/2)/(x^2+1)^(1/2))-48*arcsinh(x)*x+30*x*(x^2+1)^(1/2)-24*arcsinh(x)+18*(x^2+1)^(1/2))/(x^2+1)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (\frac{2 \, x}{x^{2} + 1} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.53229, size = 513, normalized size = 3.56 \begin{align*} \frac{10 \, x^{3} + 9 \, \sqrt{2}{\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} \log \left (-\frac{x^{2} + \sqrt{2}{\left (x^{2} - 1\right )} +{\left (2 \, x^{2} + \sqrt{2}{\left (x^{2} + 1\right )} + 2\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}} - 1}{x^{2} + 2 \, x + 1}\right ) + 30 \, x^{2} + 12 \,{\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} \log \left (-\frac{x^{2} -{\left (x^{2} + 1\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}} + x}{x + 1}\right ) + 2 \,{\left (2 \, x^{4} + 9 \, x^{3} + 7 \, x^{2} + 9 \, x + 5\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}} + 30 \, x + 10}{4 \,{\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(10*x^3 + 9*sqrt(2)*(x^3 + 3*x^2 + 3*x + 1)*log(-(x^2 + sqrt(2)*(x^2 - 1) + (2*x^2 + sqrt(2)*(x^2 + 1) + 2
)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) - 1)/(x^2 + 2*x + 1)) + 30*x^2 + 12*(x^3 + 3*x^2 + 3*x + 1)*log(-(x^2 - (x^2
+ 1)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) + x)/(x + 1)) + 2*(2*x^4 + 9*x^3 + 7*x^2 + 9*x + 5)*sqrt((x^2 + 2*x + 1)
/(x^2 + 1)) + 30*x + 10)/(x^3 + 3*x^2 + 3*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (\frac{2 \, x}{x^{2} + 1} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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