3.1013 \(\int \frac{x^3}{\sqrt{a+b x^2+(2+2 c-2 (1+c)) x^4}} \, dx\)

Optimal. Leaf size=36 \[ \frac{\left (a+b x^2\right )^{3/2}}{3 b^2}-\frac{a \sqrt{a+b x^2}}{b^2} \]

[Out]

-((a*Sqrt[a + b*x^2])/b^2) + (a + b*x^2)^(3/2)/(3*b^2)

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Rubi [A]  time = 0.0225047, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {5, 266, 43} \[ \frac{\left (a+b x^2\right )^{3/2}}{3 b^2}-\frac{a \sqrt{a+b x^2}}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a + b*x^2 + (2 + 2*c - 2*(1 + c))*x^4],x]

[Out]

-((a*Sqrt[a + b*x^2])/b^2) + (a + b*x^2)^(3/2)/(3*b^2)

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 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^3}{\sqrt{a+b x^2+(2+2 c-2 (1+c)) x^4}} \, dx &=\int \frac{x^3}{\sqrt{a+b x^2}} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{\sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{a}{b \sqrt{a+b x}}+\frac{\sqrt{a+b x}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac{a \sqrt{a+b x^2}}{b^2}+\frac{\left (a+b x^2\right )^{3/2}}{3 b^2}\\ \end{align*}

Mathematica [A]  time = 0.013652, size = 27, normalized size = 0.75 \[ \frac{\left (b x^2-2 a\right ) \sqrt{a+b x^2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a + b*x^2 + (2 + 2*c - 2*(1 + c))*x^4],x]

[Out]

((-2*a + b*x^2)*Sqrt[a + b*x^2])/(3*b^2)

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Maple [A]  time = 0.043, size = 25, normalized size = 0.7 \begin{align*} -{\frac{-b{x}^{2}+2\,a}{3\,{b}^{2}}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)^(1/2),x)

[Out]

-1/3*(b*x^2+a)^(1/2)*(-b*x^2+2*a)/b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45639, size = 53, normalized size = 1.47 \begin{align*} \frac{\sqrt{b x^{2} + a}{\left (b x^{2} - 2 \, a\right )}}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(b*x^2 + a)*(b*x^2 - 2*a)/b^2

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Sympy [A]  time = 0.458261, size = 44, normalized size = 1.22 \begin{align*} \begin{cases} - \frac{2 a \sqrt{a + b x^{2}}}{3 b^{2}} + \frac{x^{2} \sqrt{a + b x^{2}}}{3 b} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 \sqrt{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*a*sqrt(a + b*x**2)/(3*b**2) + x**2*sqrt(a + b*x**2)/(3*b), Ne(b, 0)), (x**4/(4*sqrt(a)), True))

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Giac [A]  time = 1.11574, size = 36, normalized size = 1. \begin{align*} \frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}} - 3 \, \sqrt{b x^{2} + a} a}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((b*x^2 + a)^(3/2) - 3*sqrt(b*x^2 + a)*a)/b^2