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

Optimal. Leaf size=90 \[ -(1-x) \sqrt{\frac{2 x}{x^2+1}+1} (x+1)-\frac{x \left (x^2+1\right ) \sqrt{\frac{2 x}{x^2+1}+1}}{x+1}+\frac{3 \sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1} \sinh ^{-1}(x)}{x+1} \]

[Out]

-((1 - x)*(1 + x)*Sqrt[1 + (2*x)/(1 + x^2)]) - (x*(1 + x^2)*Sqrt[1 + (2*x)/(1 + x^2)])/(1 + x) + (3*Sqrt[1 + x
^2]*Sqrt[1 + (2*x)/(1 + x^2)]*ArcSinh[x])/(1 + x)

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Rubi [A]  time = 0.0478881, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6723, 970, 739, 517, 388, 215} \[ -(1-x) \sqrt{\frac{2 x}{x^2+1}+1} (x+1)-\frac{x \left (x^2+1\right ) \sqrt{\frac{2 x}{x^2+1}+1}}{x+1}+\frac{3 \sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1} \sinh ^{-1}(x)}{x+1} \]

Antiderivative was successfully verified.

[In]

Int[(1 + (2*x)/(1 + x^2))^(3/2),x]

[Out]

-((1 - x)*(1 + x)*Sqrt[1 + (2*x)/(1 + x^2)]) - (x*(1 + x^2)*Sqrt[1 + (2*x)/(1 + x^2)])/(1 + x) + (3*Sqrt[1 + x
^2]*Sqrt[1 + (2*x)/(1 + x^2)]*ArcSinh[x])/(1 + x)

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 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 517

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^
(p_.), x_Symbol] :> Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x]
&& EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 388

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

Rubi steps

\begin{align*} \int \left (1+\frac{2 x}{1+x^2}\right )^{3/2} \, dx &=\frac{\left (\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}\right ) \int \frac{\left (1+2 x+x^2\right )^{3/2}}{\left (1+x^2\right )^{3/2}} \, dx}{\sqrt{1+2 x+x^2}}\\ &=\frac{\left (\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}\right ) \int \frac{(2+2 x)^3}{\left (1+x^2\right )^{3/2}} \, dx}{4 (2+2 x)}\\ &=-(1-x) (1+x) \sqrt{1+\frac{2 x}{1+x^2}}+\frac{\left (\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}\right ) \int \frac{(8-8 x) (2+2 x)}{\sqrt{1+x^2}} \, dx}{4 (2+2 x)}\\ &=-(1-x) (1+x) \sqrt{1+\frac{2 x}{1+x^2}}+\frac{\left (\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}\right ) \int \frac{16-16 x^2}{\sqrt{1+x^2}} \, dx}{4 (2+2 x)}\\ &=-(1-x) (1+x) \sqrt{1+\frac{2 x}{1+x^2}}-\frac{x \left (1+x^2\right ) \sqrt{1+\frac{2 x}{1+x^2}}}{1+x}+\frac{\left (6 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}\right ) \int \frac{1}{\sqrt{1+x^2}} \, dx}{2+2 x}\\ &=-(1-x) (1+x) \sqrt{1+\frac{2 x}{1+x^2}}-\frac{x \left (1+x^2\right ) \sqrt{1+\frac{2 x}{1+x^2}}}{1+x}+\frac{3 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}} \sinh ^{-1}(x)}{1+x}\\ \end{align*}

Mathematica [A]  time = 0.0340606, size = 44, normalized size = 0.49 \[ \frac{\sqrt{\frac{(x+1)^2}{x^2+1}} \left (x^2+3 \sqrt{x^2+1} \sinh ^{-1}(x)-2 x-1\right )}{x+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(1 + x)^2/(1 + x^2)]*(-1 - 2*x + x^2 + 3*Sqrt[1 + x^2]*ArcSinh[x]))/(1 + x)

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Maple [A]  time = 0.01, size = 49, normalized size = 0.5 \begin{align*}{\frac{{x}^{2}+1}{ \left ( 1+x \right ) ^{3}} \left ({\frac{{x}^{2}+2\,x+1}{{x}^{2}+1}} \right ) ^{{\frac{3}{2}}} \left ( 3\,{\it Arcsinh} \left ( x \right ) \sqrt{{x}^{2}+1}+{x}^{2}-2\,x-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^2+2*x+1)/(x^2+1))^(3/2)/(1+x)^3*(x^2+1)*(3*arcsinh(x)*(x^2+1)^(1/2)+x^2-2*x-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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+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 = 1.48585, size = 203, normalized size = 2.26 \begin{align*} -\frac{3 \,{\left (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 ) -{\left (x^{2} - 2 \, x - 1\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}} + 2 \, x + 2}{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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+2*x/(x**2+1))**(3/2),x)

[Out]

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

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Giac [A]  time = 1.19456, size = 90, normalized size = 1. \begin{align*} -{\left (\sqrt{2} - 3 \, \log \left (\sqrt{2} + 1\right )\right )} \mathrm{sgn}\left (x + 1\right ) - 3 \, \log \left (-x + \sqrt{x^{2} + 1}\right ) \mathrm{sgn}\left (x + 1\right ) + \frac{{\left (x \mathrm{sgn}\left (x + 1\right ) - 2 \, \mathrm{sgn}\left (x + 1\right )\right )} x - \mathrm{sgn}\left (x + 1\right )}{\sqrt{x^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2) - 3*log(sqrt(2) + 1))*sgn(x + 1) - 3*log(-x + sqrt(x^2 + 1))*sgn(x + 1) + ((x*sgn(x + 1) - 2*sgn(x +
1))*x - sgn(x + 1))/sqrt(x^2 + 1)