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

Optimal. Leaf size=113 \[ -\frac {\sqrt {x}}{2 \left (x^2+1\right )}-\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{8 \sqrt {2}}+\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{8 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )}{4 \sqrt {2}} \]

[Out]

1/8*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)+1/8*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)-1/16*ln(1+x-2^(1/2)*x^(1/2))*2^(1
/2)+1/16*ln(1+x+2^(1/2)*x^(1/2))*2^(1/2)-1/2*x^(1/2)/(x^2+1)

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Rubi [A]  time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\sqrt {x}}{2 \left (x^2+1\right )}-\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{8 \sqrt {2}}+\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{8 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(1 + x^2)^2,x]

[Out]

-Sqrt[x]/(2*(1 + x^2)) - ArcTan[1 - Sqrt[2]*Sqrt[x]]/(4*Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[x]]/(4*Sqrt[2]) - L
og[1 - Sqrt[2]*Sqrt[x] + x]/(8*Sqrt[2]) + Log[1 + Sqrt[2]*Sqrt[x] + x]/(8*Sqrt[2])

Rule 204

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 106, normalized size = 0.94 \[ \frac {1}{16} \left (-\frac {8 \sqrt {x}}{x^2+1}-\sqrt {2} \log \left (x-\sqrt {2} \sqrt {x}+1\right )+\sqrt {2} \log \left (x+\sqrt {2} \sqrt {x}+1\right )-2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )+2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(1 + x^2)^2,x]

[Out]

((-8*Sqrt[x])/(1 + x^2) - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[x]] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[x]] - Sqrt
[2]*Log[1 - Sqrt[2]*Sqrt[x] + x] + Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[x] + x])/16

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fricas [A]  time = 1.10, size = 140, normalized size = 1.24 \[ -\frac {4 \, \sqrt {2} {\left (x^{2} + 1\right )} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} \sqrt {x} + x + 1} - \sqrt {2} \sqrt {x} - 1\right ) + 4 \, \sqrt {2} {\left (x^{2} + 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4} - \sqrt {2} \sqrt {x} + 1\right ) - \sqrt {2} {\left (x^{2} + 1\right )} \log \left (4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4\right ) + \sqrt {2} {\left (x^{2} + 1\right )} \log \left (-4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4\right ) + 8 \, \sqrt {x}}{16 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*sqrt(2)*(x^2 + 1)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) + 4*sqrt(2)*(x^
2 + 1)*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) - sqrt(2)*(x^2 + 1)*log(4*
sqrt(2)*sqrt(x) + 4*x + 4) + sqrt(2)*(x^2 + 1)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4) + 8*sqrt(x))/(x^2 + 1)

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giac [A]  time = 0.63, size = 86, normalized size = 0.76 \[ \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {\sqrt {x}}{2 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) + 1/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x)))
+ 1/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) - 1/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) - 1/2*sqrt(x)/(x^2 +
1)

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maple [A]  time = 0.01, size = 74, normalized size = 0.65 \[ \frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, \sqrt {x}-1\right )}{8}+\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, \sqrt {x}+1\right )}{8}+\frac {\sqrt {2}\, \ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )}{16}-\frac {\sqrt {x}}{2 \left (x^{2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(x^2+1)*x^(1/2)+1/8*2^(1/2)*arctan(2^(1/2)*x^(1/2)+1)+1/8*2^(1/2)*arctan(2^(1/2)*x^(1/2)-1)+1/16*2^(1/2)*
ln((x+2^(1/2)*x^(1/2)+1)/(x-2^(1/2)*x^(1/2)+1))

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maxima [A]  time = 2.96, size = 86, normalized size = 0.76 \[ \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {\sqrt {x}}{2 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) + 1/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x)))
+ 1/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) - 1/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) - 1/2*sqrt(x)/(x^2 +
1)

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mupad [B]  time = 4.68, size = 51, normalized size = 0.45 \[ -\frac {\sqrt {x}}{2\,\left (x^2+1\right )}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(x^2 + 1)^2,x)

[Out]

2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 - 1i/2))*(1/8 + 1i/8) + 2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 + 1i/2))*(1/8 - 1i
/8) - x^(1/2)/(2*(x^2 + 1))

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sympy [B]  time = 3.39, size = 257, normalized size = 2.27 \[ - \frac {8 \sqrt {x}}{16 x^{2} + 16} - \frac {\sqrt {2} x^{2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {\sqrt {2} x^{2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {2 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{16 x^{2} + 16} + \frac {2 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{16 x^{2} + 16} - \frac {\sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {\sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {2 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{16 x^{2} + 16} + \frac {2 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{16 x^{2} + 16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*sqrt(x)/(16*x**2 + 16) - sqrt(2)*x**2*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + sqrt(2)*x**2*log(4
*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + 2*sqrt(2)*x**2*atan(sqrt(2)*sqrt(x) - 1)/(16*x**2 + 16) + 2*sqrt(
2)*x**2*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16) - sqrt(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) +
sqrt(2)*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + 2*sqrt(2)*atan(sqrt(2)*sqrt(x) - 1)/(16*x**2 + 16) +
2*sqrt(2)*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16)

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