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

Optimal. Leaf size=65 \[ -\frac{1}{2} (1-i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{1-i} \sqrt{x}}{\sqrt{x+1}}\right )-\frac{1}{2} (1+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{1+i} \sqrt{x}}{\sqrt{x+1}}\right ) \]

[Out]

-((1 - I)^(3/2)*ArcTanh[(Sqrt[1 - I]*Sqrt[x])/Sqrt[1 + x]])/2 - ((1 + I)^(3/2)*ArcTanh[(Sqrt[1 + I]*Sqrt[x])/S
qrt[1 + x]])/2

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Rubi [A]  time = 0.048894, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {910, 93, 208} \[ -\frac{1}{2} (1-i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{1-i} \sqrt{x}}{\sqrt{x+1}}\right )-\frac{1}{2} (1+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{1+i} \sqrt{x}}{\sqrt{x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(Sqrt[1 + x]*(1 + x^2)),x]

[Out]

-((1 - I)^(3/2)*ArcTanh[(Sqrt[1 - I]*Sqrt[x])/Sqrt[1 + x]])/2 - ((1 + I)^(3/2)*ArcTanh[(Sqrt[1 + I]*Sqrt[x])/S
qrt[1 + x]])/2

Rule 910

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegr
and[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &
& NeQ[c*d^2 + a*e^2, 0] && IGtQ[m + 1/2, 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0614122, size = 63, normalized size = 0.97 \[ \frac{1}{2} \left (-(-1+i)^{3/2} \tan ^{-1}\left (\sqrt{-1+i} \sqrt{\frac{x}{x+1}}\right )-(1+i)^{3/2} \tanh ^{-1}\left (\sqrt{1+i} \sqrt{\frac{x}{x+1}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.231, size = 305, normalized size = 4.7 \begin{align*}{\frac{ \left ( \sqrt{2}-1+x \right ) \sqrt{2}}{ \left ( 12\,\sqrt{2}-16 \right ) \sqrt{1+\sqrt{2}}}\sqrt{{\frac{x \left ( 1+x \right ) }{ \left ( \sqrt{2}-1+x \right ) ^{2}}}} \left ( \sqrt{-2+2\,\sqrt{2}}\arctan \left ({\frac{\sqrt{-2+2\,\sqrt{2}} \left ( 3+2\,\sqrt{2} \right ) \left ( 3\,\sqrt{2}-4 \right ) \left ( \sqrt{2}+1-x \right ) \left ( \sqrt{2}-1+x \right ) }{ \left ( 4+4\,x \right ) x}\sqrt{{\frac{ \left ( 3\,\sqrt{2}-4 \right ) x \left ( 1+x \right ) \left ( 4+3\,\sqrt{2} \right ) }{ \left ( \sqrt{2}-1+x \right ) ^{2}}}}} \right ) \sqrt{1+\sqrt{2}}\sqrt{2}-2\,\sqrt{-2+2\,\sqrt{2}}\arctan \left ( 1/4\,{\frac{\sqrt{-2+2\,\sqrt{2}} \left ( 3+2\,\sqrt{2} \right ) \left ( 3\,\sqrt{2}-4 \right ) \left ( \sqrt{2}+1-x \right ) \left ( \sqrt{2}-1+x \right ) }{x \left ( 1+x \right ) }\sqrt{{\frac{ \left ( 3\,\sqrt{2}-4 \right ) x \left ( 1+x \right ) \left ( 4+3\,\sqrt{2} \right ) }{ \left ( \sqrt{2}-1+x \right ) ^{2}}}}} \right ) \sqrt{1+\sqrt{2}}+4\,{\it Artanh} \left ({\frac{\sqrt{2}}{\sqrt{1+\sqrt{2}}}\sqrt{{\frac{x \left ( 1+x \right ) }{ \left ( \sqrt{2}-1+x \right ) ^{2}}}}} \right ) \sqrt{2}-6\,{\it Artanh} \left ({\frac{\sqrt{2}}{\sqrt{1+\sqrt{2}}}\sqrt{{\frac{x \left ( 1+x \right ) }{ \left ( \sqrt{2}-1+x \right ) ^{2}}}}} \right ) \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{1+x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/x^(1/2)/(1+x)^(1/2)*((1+x)*x/(2^(1/2)-1+x)^2)^(1/2)*(2^(1/2)-1+x)*((-2+2*2^(1/2))^(1/2)*arctan(1/4*((3*2^(
1/2)-4)*x*(1+x)*(4+3*2^(1/2))/(2^(1/2)-1+x)^2)^(1/2)*(-2+2*2^(1/2))^(1/2)*(3+2*2^(1/2))*(3*2^(1/2)-4)*(2^(1/2)
+1-x)*(2^(1/2)-1+x)/(1+x)/x)*(1+2^(1/2))^(1/2)*2^(1/2)-2*(-2+2*2^(1/2))^(1/2)*arctan(1/4*((3*2^(1/2)-4)*x*(1+x
)*(4+3*2^(1/2))/(2^(1/2)-1+x)^2)^(1/2)*(-2+2*2^(1/2))^(1/2)*(3+2*2^(1/2))*(3*2^(1/2)-4)*(2^(1/2)+1-x)*(2^(1/2)
-1+x)/(1+x)/x)*(1+2^(1/2))^(1/2)+4*arctanh(2^(1/2)*((1+x)*x/(2^(1/2)-1+x)^2)^(1/2)/(1+2^(1/2))^(1/2))*2^(1/2)-
6*arctanh(2^(1/2)*((1+x)*x/(2^(1/2)-1+x)^2)^(1/2)/(1+2^(1/2))^(1/2)))*2^(1/2)/(3*2^(1/2)-4)/(1+2^(1/2))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{{\left (x^{2} + 1\right )} \sqrt{x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.76693, size = 2338, normalized size = 35.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*2^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 1)*log(-8*sqrt(x + 1)*x^(3/2) + 8*x^2 + 2*(2^(1/4)*sqrt(x + 1)*sqrt
(x)*(sqrt(2) - 2) - 2^(1/4)*(sqrt(2)*(x + 1) - 2*x))*sqrt(2*sqrt(2) + 4) + 4*x + 4*sqrt(2) + 4) - 1/8*2^(1/4)*
sqrt(2*sqrt(2) + 4)*(sqrt(2) - 1)*log(-8*sqrt(x + 1)*x^(3/2) + 8*x^2 - 2*(2^(1/4)*sqrt(x + 1)*sqrt(x)*(sqrt(2)
- 2) - 2^(1/4)*(sqrt(2)*(x + 1) - 2*x))*sqrt(2*sqrt(2) + 4) + 4*x + 4*sqrt(2) + 4) - 1/2*2^(1/4)*sqrt(2*sqrt(
2) + 4)*arctan(1/7*(sqrt(2)*(5*sqrt(2) + 6) + 8*sqrt(2) + 4)*sqrt(x + 1)*sqrt(x) - 1/7*sqrt(2)*(sqrt(2)*(5*x +
1) + 6*x + 4) - 1/28*sqrt(-8*sqrt(x + 1)*x^(3/2) + 8*x^2 - 2*(2^(1/4)*sqrt(x + 1)*sqrt(x)*(sqrt(2) - 2) - 2^(
1/4)*(sqrt(2)*(x + 1) - 2*x))*sqrt(2*sqrt(2) + 4) + 4*x + 4*sqrt(2) + 4)*(2*sqrt(2)*(5*sqrt(2) + 6) - (2^(3/4)
*(3*sqrt(2) + 5) + 2*2^(1/4)*(sqrt(2) + 4))*sqrt(2*sqrt(2) + 4) + 16*sqrt(2) + 8) - 1/7*sqrt(2)*(8*x + 3) - 1/
14*((2^(3/4)*(3*sqrt(2) + 5) + 2*2^(1/4)*(sqrt(2) + 4))*sqrt(x + 1)*sqrt(x) - 2^(3/4)*(sqrt(2)*(3*x + 2) + 5*x
+ 1) - 2*2^(1/4)*(sqrt(2)*(x + 3) + 4*x - 2))*sqrt(2*sqrt(2) + 4) - 4/7*x - 5/7) - 1/2*2^(1/4)*sqrt(2*sqrt(2)
+ 4)*arctan(-1/7*(sqrt(2)*(5*sqrt(2) + 6) + 8*sqrt(2) + 4)*sqrt(x + 1)*sqrt(x) + 1/7*sqrt(2)*(sqrt(2)*(5*x +
1) + 6*x + 4) + 1/28*sqrt(-8*sqrt(x + 1)*x^(3/2) + 8*x^2 + 2*(2^(1/4)*sqrt(x + 1)*sqrt(x)*(sqrt(2) - 2) - 2^(1
/4)*(sqrt(2)*(x + 1) - 2*x))*sqrt(2*sqrt(2) + 4) + 4*x + 4*sqrt(2) + 4)*(2*sqrt(2)*(5*sqrt(2) + 6) + (2^(3/4)*
(3*sqrt(2) + 5) + 2*2^(1/4)*(sqrt(2) + 4))*sqrt(2*sqrt(2) + 4) + 16*sqrt(2) + 8) + 1/7*sqrt(2)*(8*x + 3) - 1/1
4*((2^(3/4)*(3*sqrt(2) + 5) + 2*2^(1/4)*(sqrt(2) + 4))*sqrt(x + 1)*sqrt(x) - 2^(3/4)*(sqrt(2)*(3*x + 2) + 5*x
+ 1) - 2*2^(1/4)*(sqrt(2)*(x + 3) + 4*x - 2))*sqrt(2*sqrt(2) + 4) + 4/7*x + 5/7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{\sqrt{x + 1} \left (x^{2} + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.50335, size = 1, normalized size = 0.02 \begin{align*} 0 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0