3.252 \(\int x (c \sqrt{a+b x^2})^{3/2} \, dx\)

Optimal. Leaf size=36 \[ \frac{2 c \left (a+b x^2\right )^{3/2} \sqrt{c \sqrt{a+b x^2}}}{7 b} \]

[Out]

(2*c*Sqrt[c*Sqrt[a + b*x^2]]*(a + b*x^2)^(3/2))/(7*b)

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Rubi [A]  time = 0.0195566, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1591, 15, 30} \[ \frac{2 c \left (a+b x^2\right )^{3/2} \sqrt{c \sqrt{a+b x^2}}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[x*(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

(2*c*Sqrt[c*Sqrt[a + b*x^2]]*(a + b*x^2)^(3/2))/(7*b)

Rule 1591

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x \left (c \sqrt{a+b x^2}\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \left (c \sqrt{x}\right )^{3/2} \, dx,x,a+b x^2\right )}{2 b}\\ &=\frac{\left (c \sqrt{c \sqrt{a+b x^2}}\right ) \operatorname{Subst}\left (\int x^{3/4} \, dx,x,a+b x^2\right )}{2 b \sqrt [4]{a+b x^2}}\\ &=\frac{2 c \sqrt{c \sqrt{a+b x^2}} \left (a+b x^2\right )^{3/2}}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.0067952, size = 31, normalized size = 0.86 \[ \frac{2 \left (a+b x^2\right ) \left (c \sqrt{a+b x^2}\right )^{3/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2))/(7*b)

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Maple [A]  time = 0.003, size = 26, normalized size = 0.7 \begin{align*}{\frac{2\,b{x}^{2}+2\,a}{7\,b} \left ( c\sqrt{b{x}^{2}+a} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*x^2+a)*(c*(b*x^2+a)^(1/2))^(3/2)/b

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Maxima [A]  time = 0.993363, size = 34, normalized size = 0.94 \begin{align*} \frac{2 \,{\left (b x^{2} + a\right )} \left (\sqrt{b x^{2} + a} c\right )^{\frac{3}{2}}}{7 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*x^2 + a)*(sqrt(b*x^2 + a)*c)^(3/2)/b

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Fricas [A]  time = 1.6987, size = 85, normalized size = 2.36 \begin{align*} \frac{2 \,{\left (b c x^{2} + a c\right )} \sqrt{b x^{2} + a} \sqrt{\sqrt{b x^{2} + a} c}}{7 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*c*x^2 + a*c)*sqrt(b*x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)/b

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Sympy [A]  time = 20.1388, size = 58, normalized size = 1.61 \begin{align*} \begin{cases} \frac{2 a c^{\frac{3}{2}} \left (a + b x^{2}\right )^{\frac{3}{4}}}{7 b} + \frac{2 c^{\frac{3}{2}} x^{2} \left (a + b x^{2}\right )^{\frac{3}{4}}}{7} & \text{for}\: b \neq 0 \\\frac{x^{2} \left (\sqrt{a} c\right )^{\frac{3}{2}}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a*c**(3/2)*(a + b*x**2)**(3/4)/(7*b) + 2*c**(3/2)*x**2*(a + b*x**2)**(3/4)/7, Ne(b, 0)), (x**2*(s
qrt(a)*c)**(3/2)/2, True))

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Giac [A]  time = 1.13623, size = 23, normalized size = 0.64 \begin{align*} \frac{2 \,{\left (b x^{2} + a\right )}^{\frac{7}{4}} c^{\frac{3}{2}}}{7 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*x^2 + a)^(7/4)*c^(3/2)/b