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

Optimal. Leaf size=122 \[ \frac{1}{2 \sqrt{x} \left (x^2+1\right )}-\frac{5}{2 \sqrt{x}}-\frac{5 \log \left (x-\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}+\frac{5 \log \left (x+\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}+\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}-\frac{5 \tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{4 \sqrt{2}} \]

[Out]

-5/(2*Sqrt[x]) + 1/(2*Sqrt[x]*(1 + x^2)) + (5*ArcTan[1 - Sqrt[2]*Sqrt[x]])/(4*Sqrt[2]) - (5*ArcTan[1 + Sqrt[2]
*Sqrt[x]])/(4*Sqrt[2]) - (5*Log[1 - Sqrt[2]*Sqrt[x] + x])/(8*Sqrt[2]) + (5*Log[1 + Sqrt[2]*Sqrt[x] + x])/(8*Sq
rt[2])

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Rubi [A]  time = 0.0648586, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {290, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{1}{2 \sqrt{x} \left (x^2+1\right )}-\frac{5}{2 \sqrt{x}}-\frac{5 \log \left (x-\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}+\frac{5 \log \left (x+\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}+\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}-\frac{5 \tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(1 + x^2)^2),x]

[Out]

-5/(2*Sqrt[x]) + 1/(2*Sqrt[x]*(1 + x^2)) + (5*ArcTan[1 - Sqrt[2]*Sqrt[x]])/(4*Sqrt[2]) - (5*ArcTan[1 + Sqrt[2]
*Sqrt[x]])/(4*Sqrt[2]) - (5*Log[1 - Sqrt[2]*Sqrt[x] + x])/(8*Sqrt[2]) + (5*Log[1 + Sqrt[2]*Sqrt[x] + x])/(8*Sq
rt[2])

Rule 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 297

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

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 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 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 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]

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]

Rubi steps

\begin{align*} \int \frac{1}{x^{3/2} \left (1+x^2\right )^2} \, dx &=\frac{1}{2 \sqrt{x} \left (1+x^2\right )}+\frac{5}{4} \int \frac{1}{x^{3/2} \left (1+x^2\right )} \, dx\\ &=-\frac{5}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \left (1+x^2\right )}-\frac{5}{4} \int \frac{\sqrt{x}}{1+x^2} \, dx\\ &=-\frac{5}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \left (1+x^2\right )}-\frac{5}{2} \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=-\frac{5}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \left (1+x^2\right )}+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{x}\right )-\frac{5}{4} \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=-\frac{5}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \left (1+x^2\right )}-\frac{5}{8} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{x}\right )-\frac{5}{8} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{x}\right )-\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2}}-\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2}}\\ &=-\frac{5}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \left (1+x^2\right )}-\frac{5 \log \left (1-\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}+\frac{5 \log \left (1+\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}\\ &=-\frac{5}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \left (1+x^2\right )}+\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}-\frac{5 \tan ^{-1}\left (1+\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}-\frac{5 \log \left (1-\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}+\frac{5 \log \left (1+\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.004854, size = 20, normalized size = 0.16 \[ -\frac{2 \, _2F_1\left (-\frac{1}{4},2;\frac{3}{4};-x^2\right )}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(1 + x^2)^2),x]

[Out]

(-2*Hypergeometric2F1[-1/4, 2, 3/4, -x^2])/Sqrt[x]

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Maple [A]  time = 0.011, size = 79, normalized size = 0.7 \begin{align*} -2\,{\frac{1}{\sqrt{x}}}-{\frac{1}{2\,{x}^{2}+2}{x}^{{\frac{3}{2}}}}-{\frac{5\,\sqrt{2}}{8}\arctan \left ( 1+\sqrt{2}\sqrt{x} \right ) }-{\frac{5\,\sqrt{2}}{8}\arctan \left ( -1+\sqrt{2}\sqrt{x} \right ) }-{\frac{5\,\sqrt{2}}{16}\ln \left ({ \left ( 1+x-\sqrt{2}\sqrt{x} \right ) \left ( 1+x+\sqrt{2}\sqrt{x} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/x^(1/2)-1/2*x^(3/2)/(x^2+1)-5/8*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)-5/8*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)-5/
16*2^(1/2)*ln((1+x-2^(1/2)*x^(1/2))/(1+x+2^(1/2)*x^(1/2)))

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Maxima [A]  time = 3.72872, size = 124, normalized size = 1.02 \begin{align*} -\frac{5}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{x}\right )}\right ) - \frac{5}{8} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{x}\right )}\right ) + \frac{5}{16} \, \sqrt{2} \log \left (\sqrt{2} \sqrt{x} + x + 1\right ) - \frac{5}{16} \, \sqrt{2} \log \left (-\sqrt{2} \sqrt{x} + x + 1\right ) - \frac{5 \, x^{2} + 4}{2 \,{\left (x^{\frac{5}{2}} + \sqrt{x}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 5/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x))
) + 5/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) - 5/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) - 1/2*(5*x^2 + 4)/(
x^(5/2) + sqrt(x))

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Fricas [A]  time = 1.34484, size = 462, normalized size = 3.79 \begin{align*} \frac{20 \, \sqrt{2}{\left (x^{3} + x\right )} \arctan \left (\sqrt{2} \sqrt{\sqrt{2} \sqrt{x} + x + 1} - \sqrt{2} \sqrt{x} - 1\right ) + 20 \, \sqrt{2}{\left (x^{3} + x\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4} - \sqrt{2} \sqrt{x} + 1\right ) + 5 \, \sqrt{2}{\left (x^{3} + x\right )} \log \left (4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) - 5 \, \sqrt{2}{\left (x^{3} + x\right )} \log \left (-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) - 8 \,{\left (5 \, x^{2} + 4\right )} \sqrt{x}}{16 \,{\left (x^{3} + x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(20*sqrt(2)*(x^3 + x)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) + 20*sqrt(2)*(x
^3 + x)*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) + 5*sqrt(2)*(x^3 + x)*log
(4*sqrt(2)*sqrt(x) + 4*x + 4) - 5*sqrt(2)*(x^3 + x)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4) - 8*(5*x^2 + 4)*sqrt(x))
/(x^3 + x)

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Sympy [B]  time = 6.46214, size = 366, normalized size = 3. \begin{align*} - \frac{5 \sqrt{2} x^{\frac{5}{2}} \log{\left (- 4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} + \frac{5 \sqrt{2} x^{\frac{5}{2}} \log{\left (4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} - \frac{10 \sqrt{2} x^{\frac{5}{2}} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} - 1 \right )}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} - \frac{10 \sqrt{2} x^{\frac{5}{2}} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} + 1 \right )}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} - \frac{5 \sqrt{2} \sqrt{x} \log{\left (- 4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} + \frac{5 \sqrt{2} \sqrt{x} \log{\left (4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} - \frac{10 \sqrt{2} \sqrt{x} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} - 1 \right )}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} - \frac{10 \sqrt{2} \sqrt{x} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} + 1 \right )}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} - \frac{40 x^{2}}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} - \frac{32}{16 x^{\frac{5}{2}} + 16 \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*sqrt(2)*x**(5/2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**(5/2) + 16*sqrt(x)) + 5*sqrt(2)*x**(5/2)*log(4*sq
rt(2)*sqrt(x) + 4*x + 4)/(16*x**(5/2) + 16*sqrt(x)) - 10*sqrt(2)*x**(5/2)*atan(sqrt(2)*sqrt(x) - 1)/(16*x**(5/
2) + 16*sqrt(x)) - 10*sqrt(2)*x**(5/2)*atan(sqrt(2)*sqrt(x) + 1)/(16*x**(5/2) + 16*sqrt(x)) - 5*sqrt(2)*sqrt(x
)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**(5/2) + 16*sqrt(x)) + 5*sqrt(2)*sqrt(x)*log(4*sqrt(2)*sqrt(x) + 4*x
+ 4)/(16*x**(5/2) + 16*sqrt(x)) - 10*sqrt(2)*sqrt(x)*atan(sqrt(2)*sqrt(x) - 1)/(16*x**(5/2) + 16*sqrt(x)) - 1
0*sqrt(2)*sqrt(x)*atan(sqrt(2)*sqrt(x) + 1)/(16*x**(5/2) + 16*sqrt(x)) - 40*x**2/(16*x**(5/2) + 16*sqrt(x)) -
32/(16*x**(5/2) + 16*sqrt(x))

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Giac [A]  time = 1.98657, size = 124, normalized size = 1.02 \begin{align*} -\frac{5}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{x}\right )}\right ) - \frac{5}{8} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{x}\right )}\right ) + \frac{5}{16} \, \sqrt{2} \log \left (\sqrt{2} \sqrt{x} + x + 1\right ) - \frac{5}{16} \, \sqrt{2} \log \left (-\sqrt{2} \sqrt{x} + x + 1\right ) - \frac{5 \, x^{2} + 4}{2 \,{\left (x^{\frac{5}{2}} + \sqrt{x}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 5/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x))
) + 5/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) - 5/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) - 1/2*(5*x^2 + 4)/(
x^(5/2) + sqrt(x))