3.317 \(\int \frac {1}{\sqrt {x} (1+x^2)} \, dx\)

Optimal. Leaf size=92 \[ -\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}+\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}} \]

[Out]

1/2*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)+1/2*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)-1/4*ln(1+x-2^(1/2)*x^(1/2))*2^(1/
2)+1/4*ln(1+x+2^(1/2)*x^(1/2))*2^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}+\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(1 + x^2)),x]

[Out]

-(ArcTan[1 - Sqrt[2]*Sqrt[x]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[x]]/Sqrt[2] - Log[1 - Sqrt[2]*Sqrt[x] + x]/(2
*Sqrt[2]) + Log[1 + Sqrt[2]*Sqrt[x] + x]/(2*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 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 {1}{\sqrt {x} \left (1+x^2\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {x}\right )\\ &=\operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {x}\right )+\operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {x}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {x}\right )+\frac {1}{2} \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 )}{2 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2}}\\ &=-\frac {\log \left (1-\sqrt {2} \sqrt {x}+x\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {x}+x\right )}{2 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\\ &=-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\log \left (1-\sqrt {2} \sqrt {x}+x\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {x}+x\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 76, normalized size = 0.83 \[ \frac {-\log \left (x-\sqrt {2} \sqrt {x}+1\right )+\log \left (x+\sqrt {2} \sqrt {x}+1\right )-2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )+2 \tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(1 + x^2)),x]

[Out]

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

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fricas [A]  time = 0.66, size = 107, normalized size = 1.16 \[ -\sqrt {2} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} \sqrt {x} + x + 1} - \sqrt {2} \sqrt {x} - 1\right ) - \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4} - \sqrt {2} \sqrt {x} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) - sqrt(2)*arctan(1/2*sqrt(2)*sqrt
(-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) + 1/4*sqrt(2)*log(4*sqrt(2)*sqrt(x) + 4*x + 4) - 1/4*sqr
t(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)

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giac [A]  time = 0.59, size = 74, normalized size = 0.80 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 62, normalized size = 0.67 \[ \frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, \sqrt {x}-1\right )}{2}+\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, \sqrt {x}+1\right )}{2}+\frac {\sqrt {2}\, \ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*2^(1/2)*ln((x+2^(1/2)*x^(1/2)+1)/(x-2^(1/2)*x^(1/2)+1))+1/2*2^(1/2)*arctan(2^(1/2)*x^(1/2)-1)+1/2*2^(1/2)*
arctan(2^(1/2)*x^(1/2)+1)

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maxima [A]  time = 2.89, size = 74, normalized size = 0.80 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 37, normalized size = 0.40 \[ \sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(x^2 + 1)),x)

[Out]

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

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sympy [A]  time = 0.73, size = 90, normalized size = 0.98 \[ - \frac {\sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{4} + \frac {\sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{4} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{2} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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