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

Optimal. Leaf size=32 \[ \frac{2 \sqrt{\left (x^2+1\right ) \left (\sqrt{x^2+1}+1\right )}}{\sqrt{x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.100574, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {6715, 1588} \[ \frac{2 \sqrt{\left (x^2+1\right ) \left (\sqrt{x^2+1}+1\right )}}{\sqrt{x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[1 + x^2 + (1 + x^2)^(3/2)],x]

[Out]

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

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0484645, size = 37, normalized size = 1.16 \[ \frac{2 \left (x^2+\sqrt{x^2+1}+1\right )}{\sqrt{\left (x^2+1\right ) \left (\sqrt{x^2+1}+1\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[1 + x^2 + (1 + x^2)^(3/2)],x]

[Out]

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

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{x{\frac{1}{\sqrt{1+{x}^{2}+ \left ({x}^{2}+1 \right ) ^{{\frac{3}{2}}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{x^{2} +{\left (x^{2} + 1\right )}^{\frac{3}{2}} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(x^2 + (x^2 + 1)^(3/2) + 1), x)

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Fricas [A]  time = 0.555716, size = 66, normalized size = 2.06 \begin{align*} \frac{2 \, \sqrt{x^{2} +{\left (x^{2} + 1\right )}^{\frac{3}{2}} + 1}}{\sqrt{x^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x^2 + (x^2 + 1)^(3/2) + 1)/sqrt(x^2 + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11284, size = 20, normalized size = 0.62 \begin{align*} 2 \, \sqrt{\sqrt{x^{2} + 1} + 1} - 2 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(sqrt(x^2 + 1) + 1) - 2