3.4.25 \(\int \frac {1}{\sqrt {x} (1+x^2)^2} \, dx\) [325]

Optimal. Leaf size=113 \[ \frac {\sqrt {x}}{2 \left (1+x^2\right )}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}+\frac {3 \tan ^{-1}\left (1+\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}-\frac {3 \log \left (1-\sqrt {2} \sqrt {x}+x\right )}{8 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} \sqrt {x}+x\right )}{8 \sqrt {2}} \]

[Out]

3/8*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)+3/8*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)-3/16*ln(1+x-2^(1/2)*x^(1/2))*2^(1
/2)+3/16*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 = 0.04, 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 = {296, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {3 \text {ArcTan}\left (1-\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}+\frac {3 \text {ArcTan}\left (\sqrt {2} \sqrt {x}+1\right )}{4 \sqrt {2}}+\frac {\sqrt {x}}{2 \left (x^2+1\right )}-\frac {3 \log \left (x-\sqrt {2} \sqrt {x}+1\right )}{8 \sqrt {2}}+\frac {3 \log \left (x+\sqrt {2} \sqrt {x}+1\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(1 + x^2)^2),x]

[Out]

Sqrt[x]/(2*(1 + x^2)) - (3*ArcTan[1 - Sqrt[2]*Sqrt[x]])/(4*Sqrt[2]) + (3*ArcTan[1 + Sqrt[2]*Sqrt[x]])/(4*Sqrt[
2]) - (3*Log[1 - Sqrt[2]*Sqrt[x] + x])/(8*Sqrt[2]) + (3*Log[1 + Sqrt[2]*Sqrt[x] + x])/(8*Sqrt[2])

Rule 210

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

Rule 217

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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

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 631

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 642

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 1176

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 1179

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

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Mathematica [A]
time = 0.11, size = 65, normalized size = 0.58 \begin {gather*} \frac {1}{8} \left (\frac {4 \sqrt {x}}{1+x^2}+3 \sqrt {2} \tan ^{-1}\left (\frac {-1+x}{\sqrt {2} \sqrt {x}}\right )+3 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(1 + x^2)^2),x]

[Out]

((4*Sqrt[x])/(1 + x^2) + 3*Sqrt[2]*ArcTan[(-1 + x)/(Sqrt[2]*Sqrt[x])] + 3*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[x])/(1
+ x)])/8

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Maple [A]
time = 0.61, size = 69, normalized size = 0.61

method	result	size
derivativedivides	\(\frac {\sqrt {x}}{2 x^{2}+2}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {1+x +\sqrt {2}\, \sqrt {x}}{1+x -\sqrt {2}\, \sqrt {x}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{16}\)	\(69\)
default	\(\frac {\sqrt {x}}{2 x^{2}+2}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {1+x +\sqrt {2}\, \sqrt {x}}{1+x -\sqrt {2}\, \sqrt {x}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{16}\)	\(69\)
risch	\(\frac {\sqrt {x}}{2 x^{2}+2}+\frac {3 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right ) \sqrt {2}}{8}+\frac {3 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right ) \sqrt {2}}{8}+\frac {3 \sqrt {2}\, \ln \left (\frac {1+x +\sqrt {2}\, \sqrt {x}}{1+x -\sqrt {2}\, \sqrt {x}}\right )}{16}\)	\(74\)
meijerg	\(\frac {2 \sqrt {x}}{4 x^{2}+4}-\frac {3 \sqrt {x}\, \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{16 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {x}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{8 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {x}\, \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{16 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {x}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{8 \left (x^{2}\right )^{\frac {1}{4}}}\)	\(153\)
trager	\(\frac {\sqrt {x}}{2 x^{2}+2}+\frac {3 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{5} x -\RootOf \left (\textit {\_Z}^{4}+1\right )^{5}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-\RootOf \left (\textit {\_Z}^{4}+1\right ) x +4 \sqrt {x}-\RootOf \left (\textit {\_Z}^{4}+1\right )}{\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x -\RootOf \left (\textit {\_Z}^{4}+1\right )^{2}-x -1}\right )}{8}+\frac {3 \RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{5} x -\RootOf \left (\textit {\_Z}^{4}+1\right )^{5}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x +\RootOf \left (\textit {\_Z}^{4}+1\right ) x +4 \sqrt {x}+\RootOf \left (\textit {\_Z}^{4}+1\right )}{\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x -\RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+x +1}\right )}{8}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2+1)^2/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.55, size = 86, normalized size = 0.76 \begin {gather*} \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {3}{16} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {\sqrt {x}}{2 \, {\left (x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.05, size = 141, normalized size = 1.25 \begin {gather*} -\frac {12 \, \sqrt {2} {\left (x^{2} + 1\right )} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} \sqrt {x} + x + 1} - \sqrt {2} \sqrt {x} - 1\right ) + 12 \, \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 ) - 3 \, \sqrt {2} {\left (x^{2} + 1\right )} \log \left (4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4\right ) + 3 \, \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(12*sqrt(2)*(x^2 + 1)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) + 12*sqrt(2)*(
x^2 + 1)*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) - 3*sqrt(2)*(x^2 + 1)*lo
g(4*sqrt(2)*sqrt(x) + 4*x + 4) + 3*sqrt(2)*(x^2 + 1)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4) - 8*sqrt(x))/(x^2 + 1)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (102) = 204\).
time = 0.60, size = 264, normalized size = 2.34 \begin {gather*} \frac {8 \sqrt {x}}{16 x^{2} + 16} - \frac {3 \sqrt {2} x^{2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {3 \sqrt {2} x^{2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {6 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{16 x^{2} + 16} + \frac {6 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{16 x^{2} + 16} - \frac {3 \sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {3 \sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {6 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{16 x^{2} + 16} + \frac {6 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{16 x^{2} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*sqrt(x)/(16*x**2 + 16) - 3*sqrt(2)*x**2*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + 3*sqrt(2)*x**2*lo
g(4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + 6*sqrt(2)*x**2*atan(sqrt(2)*sqrt(x) - 1)/(16*x**2 + 16) + 6*sq
rt(2)*x**2*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16) - 3*sqrt(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 1
6) + 3*sqrt(2)*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + 6*sqrt(2)*atan(sqrt(2)*sqrt(x) - 1)/(16*x**2
+ 16) + 6*sqrt(2)*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16)

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Giac [A]
time = 3.20, size = 86, normalized size = 0.76 \begin {gather*} \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {3}{16} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {\sqrt {x}}{2 \, {\left (x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.66, size = 50, normalized size = 0.44 \begin {gather*} \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 {3}{8}+\frac {3}{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 {3}{8}-\frac {3}{8}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(x^2 + 1)^2),x)

[Out]

2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 - 1i/2))*(3/8 + 3i/8) + 2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 + 1i/2))*(3/8 - 3i
/8) + x^(1/2)/(2*(x^2 + 1))

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