3.895 \(\int (1+\frac{2 x}{1+x^2})^{5/2} \, dx\)

Optimal. Leaf size=133 \[ -\frac{(1-x) \sqrt{\frac{2 x}{x^2+1}+1} (x+1)^3}{3 \left (x^2+1\right )}-\frac{4}{3} (1-2 x) \sqrt{\frac{2 x}{x^2+1}+1} (x+1)-\frac{(3 x+4) \left (x^2+1\right ) \sqrt{\frac{2 x}{x^2+1}+1}}{x+1}+\frac{5 \sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1} \sinh ^{-1}(x)}{x+1} \]

[Out]

(-4*(1 - 2*x)*(1 + x)*Sqrt[1 + (2*x)/(1 + x^2)])/3 - ((1 - x)*(1 + x)^3*Sqrt[1 + (2*x)/(1 + x^2)])/(3*(1 + x^2
)) - ((4 + 3*x)*(1 + x^2)*Sqrt[1 + (2*x)/(1 + x^2)])/(1 + x) + (5*Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)]*ArcS
inh[x])/(1 + x)

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Rubi [A]  time = 0.0737738, antiderivative size = 133, 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, 819, 780, 215} \[ -\frac{(1-x) \sqrt{\frac{2 x}{x^2+1}+1} (x+1)^3}{3 \left (x^2+1\right )}-\frac{4}{3} (1-2 x) \sqrt{\frac{2 x}{x^2+1}+1} (x+1)-\frac{(3 x+4) \left (x^2+1\right ) \sqrt{\frac{2 x}{x^2+1}+1}}{x+1}+\frac{5 \sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1} \sinh ^{-1}(x)}{x+1} \]

Antiderivative was successfully verified.

[In]

Int[(1 + (2*x)/(1 + x^2))^(5/2),x]

[Out]

(-4*(1 - 2*x)*(1 + x)*Sqrt[1 + (2*x)/(1 + x^2)])/3 - ((1 - x)*(1 + x)^3*Sqrt[1 + (2*x)/(1 + x^2)])/(3*(1 + x^2
)) - ((4 + 3*x)*(1 + x^2)*Sqrt[1 + (2*x)/(1 + x^2)])/(1 + x) + (5*Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)]*ArcS
inh[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 819

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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 )^{5/2} \, 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 )^{5/2}}{\left (1+x^2\right )^{5/2}} \, 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)^5}{\left (1+x^2\right )^{5/2}} \, dx}{16 (2+2 x)}\\ &=-\frac{(1-x) (1+x)^3 \sqrt{1+\frac{2 x}{1+x^2}}}{3 \left (1+x^2\right )}+\frac{\left (\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}\right ) \int \frac{(24-8 x) (2+2 x)^3}{\left (1+x^2\right )^{3/2}} \, dx}{48 (2+2 x)}\\ &=-\frac{4}{3} (1-2 x) (1+x) \sqrt{1+\frac{2 x}{1+x^2}}-\frac{(1-x) (1+x)^3 \sqrt{1+\frac{2 x}{1+x^2}}}{3 \left (1+x^2\right )}+\frac{\left (\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}\right ) \int \frac{(96-288 x) (2+2 x)}{\sqrt{1+x^2}} \, dx}{48 (2+2 x)}\\ &=-\frac{4}{3} (1-2 x) (1+x) \sqrt{1+\frac{2 x}{1+x^2}}-\frac{(1-x) (1+x)^3 \sqrt{1+\frac{2 x}{1+x^2}}}{3 \left (1+x^2\right )}-\frac{(4+3 x) \left (1+x^2\right ) \sqrt{1+\frac{2 x}{1+x^2}}}{1+x}+\frac{\left (10 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}\right ) \int \frac{1}{\sqrt{1+x^2}} \, dx}{2+2 x}\\ &=-\frac{4}{3} (1-2 x) (1+x) \sqrt{1+\frac{2 x}{1+x^2}}-\frac{(1-x) (1+x)^3 \sqrt{1+\frac{2 x}{1+x^2}}}{3 \left (1+x^2\right )}-\frac{(4+3 x) \left (1+x^2\right ) \sqrt{1+\frac{2 x}{1+x^2}}}{1+x}+\frac{5 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}} \sinh ^{-1}(x)}{1+x}\\ \end{align*}

Mathematica [A]  time = 0.0728741, size = 64, normalized size = 0.48 \[ \frac{(x+1) \left (3 x^4-8 x^3-18 x^2+15 \left (x^2+1\right )^{3/2} \sinh ^{-1}(x)-12 x-17\right )}{3 \sqrt{\frac{(x+1)^2}{x^2+1}} \left (x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + (2*x)/(1 + x^2))^(5/2),x]

[Out]

((1 + x)*(-17 - 12*x - 18*x^2 - 8*x^3 + 3*x^4 + 15*(1 + x^2)^(3/2)*ArcSinh[x]))/(3*Sqrt[(1 + x)^2/(1 + x^2)]*(
1 + x^2)^2)

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Maple [A]  time = 0.015, size = 62, normalized size = 0.5 \begin{align*}{\frac{{x}^{2}+1}{3\, \left ( 1+x \right ) ^{5}} \left ({\frac{{x}^{2}+2\,x+1}{{x}^{2}+1}} \right ) ^{{\frac{5}{2}}} \left ( 15\,{\it Arcsinh} \left ( x \right ) \left ({x}^{2}+1 \right ) ^{3/2}+3\,{x}^{4}-8\,{x}^{3}-18\,{x}^{2}-12\,x-17 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*x/(x^2+1))^(5/2),x)

[Out]

1/3*((x^2+2*x+1)/(x^2+1))^(5/2)/(1+x)^5*(x^2+1)*(15*arcsinh(x)*(x^2+1)^(3/2)+3*x^4-8*x^3-18*x^2-12*x-17)

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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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x/(x^2+1))^(5/2),x, algorithm="maxima")

[Out]

integrate((2*x/(x^2 + 1) + 1)^(5/2), x)

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Fricas [A]  time = 1.43611, size = 292, normalized size = 2.2 \begin{align*} -\frac{8 \, x^{3} + 8 \, x^{2} + 15 \,{\left (x^{3} + x^{2} + 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 (3 \, x^{4} - 8 \, x^{3} - 18 \, x^{2} - 12 \, x - 17\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}} + 8 \, x + 8}{3 \,{\left (x^{3} + x^{2} + x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x/(x^2+1))^(5/2),x, algorithm="fricas")

[Out]

-1/3*(8*x^3 + 8*x^2 + 15*(x^3 + x^2 + x + 1)*log(-(x^2 - (x^2 + 1)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) + x)/(x + 1
)) - (3*x^4 - 8*x^3 - 18*x^2 - 12*x - 17)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) + 8*x + 8)/(x^3 + 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{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x/(x**2+1))**(5/2),x)

[Out]

Integral((2*x/(x**2 + 1) + 1)**(5/2), x)

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Giac [A]  time = 1.13822, size = 116, normalized size = 0.87 \begin{align*}{\left (\sqrt{2} + 5 \, \log \left (\sqrt{2} + 1\right )\right )} \mathrm{sgn}\left (x + 1\right ) - 5 \, \log \left (-x + \sqrt{x^{2} + 1}\right ) \mathrm{sgn}\left (x + 1\right ) + \frac{{\left ({\left ({\left (3 \, x \mathrm{sgn}\left (x + 1\right ) - 8 \, \mathrm{sgn}\left (x + 1\right )\right )} x - 18 \, \mathrm{sgn}\left (x + 1\right )\right )} x - 12 \, \mathrm{sgn}\left (x + 1\right )\right )} x - 17 \, \mathrm{sgn}\left (x + 1\right )}{3 \,{\left (x^{2} + 1\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x/(x^2+1))^(5/2),x, algorithm="giac")

[Out]

(sqrt(2) + 5*log(sqrt(2) + 1))*sgn(x + 1) - 5*log(-x + sqrt(x^2 + 1))*sgn(x + 1) + 1/3*((((3*x*sgn(x + 1) - 8*
sgn(x + 1))*x - 18*sgn(x + 1))*x - 12*sgn(x + 1))*x - 17*sgn(x + 1))/(x^2 + 1)^(3/2)