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

Optimal. Leaf size=66 \[ \frac{2 \left (a+b x^2\right )^2 \left (c \sqrt{a+b x^2}\right )^{3/2}}{11 b^2}-\frac{2 a \left (a+b x^2\right ) \left (c \sqrt{a+b x^2}\right )^{3/2}}{7 b^2} \]

[Out]

(-2*a*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2))/(7*b^2) + (2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)^2)/(11*b^2)

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Rubi [A]  time = 0.136821, antiderivative size = 74, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {6720, 266, 43} \[ \frac{2 c \left (a+b x^2\right )^{5/2} \sqrt{c \sqrt{a+b x^2}}}{11 b^2}-\frac{2 a c \left (a+b x^2\right )^{3/2} \sqrt{c \sqrt{a+b x^2}}}{7 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

(-2*a*c*Sqrt[c*Sqrt[a + b*x^2]]*(a + b*x^2)^(3/2))/(7*b^2) + (2*c*Sqrt[c*Sqrt[a + b*x^2]]*(a + b*x^2)^(5/2))/(
11*b^2)

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

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 x^3 \left (c \sqrt{a+b x^2}\right )^{3/2} \, dx &=\frac{\left (c \sqrt{c \sqrt{a+b x^2}}\right ) \int x^3 \left (a+b x^2\right )^{3/4} \, dx}{\sqrt [4]{a+b x^2}}\\ &=\frac{\left (c \sqrt{c \sqrt{a+b x^2}}\right ) \operatorname{Subst}\left (\int x (a+b x)^{3/4} \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=\frac{\left (c \sqrt{c \sqrt{a+b x^2}}\right ) \operatorname{Subst}\left (\int \left (-\frac{a (a+b x)^{3/4}}{b}+\frac{(a+b x)^{7/4}}{b}\right ) \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=-\frac{2 a c \sqrt{c \sqrt{a+b x^2}} \left (a+b x^2\right )^{3/2}}{7 b^2}+\frac{2 c \sqrt{c \sqrt{a+b x^2}} \left (a+b x^2\right )^{5/2}}{11 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0202312, size = 41, normalized size = 0.62 \[ \frac{2 \left (a+b x^2\right ) \left (7 b x^2-4 a\right ) \left (c \sqrt{a+b x^2}\right )^{3/2}}{77 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)*(-4*a + 7*b*x^2))/(77*b^2)

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Maple [A]  time = 0.006, size = 36, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,b{x}^{2}+2\,a \right ) \left ( -7\,b{x}^{2}+4\,a \right ) }{77\,{b}^{2}} \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^3*(c*(b*x^2+a)^(1/2))^(3/2),x)

[Out]

-2/77*(b*x^2+a)*(-7*b*x^2+4*a)*(c*(b*x^2+a)^(1/2))^(3/2)/b^2

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Maxima [A]  time = 0.997918, size = 58, normalized size = 0.88 \begin{align*} -\frac{2 \,{\left (11 \, \left (\sqrt{b x^{2} + a} c\right )^{\frac{7}{2}} a c^{2} - 7 \, \left (\sqrt{b x^{2} + a} c\right )^{\frac{11}{2}}\right )}}{77 \, b^{2} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77*(11*(sqrt(b*x^2 + a)*c)^(7/2)*a*c^2 - 7*(sqrt(b*x^2 + a)*c)^(11/2))/(b^2*c^4)

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Fricas [A]  time = 1.68915, size = 119, normalized size = 1.8 \begin{align*} \frac{2 \,{\left (7 \, b^{2} c x^{4} + 3 \, a b c x^{2} - 4 \, a^{2} c\right )} \sqrt{b x^{2} + a} \sqrt{\sqrt{b x^{2} + a} c}}{77 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77*(7*b^2*c*x^4 + 3*a*b*c*x^2 - 4*a^2*c)*sqrt(b*x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)/b^2

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Sympy [A]  time = 52.6148, size = 87, normalized size = 1.32 \begin{align*} \begin{cases} - \frac{8 a^{2} c^{\frac{3}{2}} \left (a + b x^{2}\right )^{\frac{3}{4}}}{77 b^{2}} + \frac{6 a c^{\frac{3}{2}} x^{2} \left (a + b x^{2}\right )^{\frac{3}{4}}}{77 b} + \frac{2 c^{\frac{3}{2}} x^{4} \left (a + b x^{2}\right )^{\frac{3}{4}}}{11} & \text{for}\: b \neq 0 \\\frac{x^{4} \left (\sqrt{a} c\right )^{\frac{3}{2}}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-8*a**2*c**(3/2)*(a + b*x**2)**(3/4)/(77*b**2) + 6*a*c**(3/2)*x**2*(a + b*x**2)**(3/4)/(77*b) + 2*c
**(3/2)*x**4*(a + b*x**2)**(3/4)/11, Ne(b, 0)), (x**4*(sqrt(a)*c)**(3/2)/4, True))

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Giac [A]  time = 1.23085, size = 43, normalized size = 0.65 \begin{align*} \frac{2 \,{\left (7 \,{\left (b x^{2} + a\right )}^{\frac{11}{4}} - 11 \,{\left (b x^{2} + a\right )}^{\frac{7}{4}} a\right )} c^{\frac{3}{2}}}{77 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77*(7*(b*x^2 + a)^(11/4) - 11*(b*x^2 + a)^(7/4)*a)*c^(3/2)/b^2