3.11.7 \(\int \sqrt {\frac {x^2}{1+x^2}} \, dx\) [1007]

Optimal. Leaf size=20 \[ \frac {\sqrt {x^2} \sqrt {1+x^2}}{x} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1978, 15, 267} \begin {gather*} \frac {\sqrt {x^2} \sqrt {x^2+1}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x^2/(1 + x^2)],x]

[Out]

(Sqrt[x^2]*Sqrt[1 + x^2])/x

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1978

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*((a*e + b*e*
x^n)^p/(c + d*x^n)^p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - a*(d/b), 0]

Rubi steps

\begin {align*} \int \sqrt {\frac {x^2}{1+x^2}} \, dx &=\int \frac {\sqrt {x^2}}{\sqrt {1+x^2}} \, dx\\ &=\frac {\sqrt {x^2} \int \frac {x}{\sqrt {1+x^2}} \, dx}{x}\\ &=\frac {\sqrt {x^2} \sqrt {1+x^2}}{x}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 17, normalized size = 0.85 \begin {gather*} \frac {x}{\sqrt {\frac {x^2}{1+x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x^2/(1 + x^2)],x]

[Out]

x/Sqrt[x^2/(1 + x^2)]

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Maple [A]
time = 0.02, size = 23, normalized size = 1.15

method	result	size
gosper	\(\frac {\left (x^{2}+1\right ) \sqrt {\frac {x^{2}}{x^{2}+1}}}{x}\)	\(23\)
default	\(\frac {\left (x^{2}+1\right ) \sqrt {\frac {x^{2}}{x^{2}+1}}}{x}\)	\(23\)
trager	\(\frac {\left (x^{2}+1\right ) \sqrt {\frac {x^{2}}{x^{2}+1}}}{x}\)	\(23\)
risch	\(\frac {\left (x^{2}+1\right ) \sqrt {\frac {x^{2}}{x^{2}+1}}}{x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 7, normalized size = 0.35 \begin {gather*} \sqrt {x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 + 1)

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Fricas [A]
time = 0.38, size = 22, normalized size = 1.10 \begin {gather*} \frac {{\left (x^{2} + 1\right )} \sqrt {\frac {x^{2}}{x^{2} + 1}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2 + 1)*sqrt(x^2/(x^2 + 1))/x

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Sympy [A]
time = 0.16, size = 26, normalized size = 1.30 \begin {gather*} x \sqrt {\frac {x^{2}}{x^{2} + 1}} + \frac {\sqrt {\frac {x^{2}}{x^{2} + 1}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.92, size = 15, normalized size = 0.75 \begin {gather*} \sqrt {x^{2} + 1} \mathrm {sgn}\left (x\right ) - \mathrm {sgn}\left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 + 1)*sgn(x) - sgn(x)

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Mupad [B]
time = 3.41, size = 13, normalized size = 0.65 \begin {gather*} \frac {\sqrt {x^4+x^2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2/(x^2 + 1))^(1/2),x)

[Out]

(x^2 + x^4)^(1/2)/x

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