3.11.15 \(\int \frac {x}{\sqrt {a+b x^2+(2+2 c-2 (1+c)) x^4}} \, dx\) [1015]

Optimal. Leaf size=15 \[ \frac {\sqrt {a+b x^2}}{b} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {5, 267} \begin {gather*} \frac {\sqrt {a+b x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b*x^2 + (2 + 2*c - 2*(1 + c))*x^4],x]

[Out]

Sqrt[a + b*x^2]/b

Rule 5

Int[(u_.)*((a_.) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(a + b*x^n)^p, x] /; FreeQ[{
a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[c, 0]

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]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {a+b x^2+(2+2 c-2 (1+c)) x^4}} \, dx &=\int \frac {x}{\sqrt {a+b x^2}} \, dx\\ &=\frac {\sqrt {a+b x^2}}{b}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a+b x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a + b*x^2 + (2 + 2*c - 2*(1 + c))*x^4],x]

[Out]

Sqrt[a + b*x^2]/b

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Maple [A]
time = 0.13, size = 14, normalized size = 0.93

method	result	size
gosper	\(\frac {\sqrt {b \,x^{2}+a}}{b}\)	\(14\)
derivativedivides	\(\frac {\sqrt {b \,x^{2}+a}}{b}\)	\(14\)
default	\(\frac {\sqrt {b \,x^{2}+a}}{b}\)	\(14\)
trager	\(\frac {\sqrt {b \,x^{2}+a}}{b}\)	\(14\)
risch	\(\frac {\sqrt {b \,x^{2}+a}}{b}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 13, normalized size = 0.87 \begin {gather*} \frac {\sqrt {b x^{2} + a}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b*x^2 + a)/b

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Fricas [A]
time = 0.36, size = 13, normalized size = 0.87 \begin {gather*} \frac {\sqrt {b x^{2} + a}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b*x^2 + a)/b

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Sympy [A]
time = 0.17, size = 20, normalized size = 1.33 \begin {gather*} \begin {cases} \frac {\sqrt {a + b x^{2}}}{b} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 \sqrt {a}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sqrt(a + b*x**2)/b, Ne(b, 0)), (x**2/(2*sqrt(a)), True))

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Giac [A]
time = 5.70, size = 13, normalized size = 0.87 \begin {gather*} \frac {\sqrt {b x^{2} + a}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b*x^2 + a)/b

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Mupad [B]
time = 4.33, size = 13, normalized size = 0.87 \begin {gather*} \frac {\sqrt {b\,x^2+a}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x^2)^(1/2),x)

[Out]

(a + b*x^2)^(1/2)/b

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