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

Optimal. Leaf size=66 \[ \frac{c \left (a+b x^2\right )^4 \sqrt{c \left (a+b x^2\right )^2}}{10 b^2}-\frac{a c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^2}}{8 b^2} \]

[Out]

-(a*c*(a + b*x^2)^3*Sqrt[c*(a + b*x^2)^2])/(8*b^2) + (c*(a + b*x^2)^4*Sqrt[c*(a + b*x^2)^2])/(10*b^2)

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Rubi [A]  time = 0.113314, antiderivative size = 78, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1989, 1111, 640, 609} \[ \frac{\left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{5/2}}{10 b^2 c}-\frac{a \left (a+b x^2\right ) \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

-(a*(a + b*x^2)*(a^2*c + 2*a*b*c*x^2 + b^2*c*x^4)^(3/2))/(8*b^2) + (a^2*c + 2*a*b*c*x^2 + b^2*c*x^4)^(5/2)/(10
*b^2*c)

Rule 1989

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
TrinomialQ[u, x] &&  !TrinomialMatchQ[u, x]

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rubi steps

\begin{align*} \int x^3 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx &=\int x^3 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int x \left (a^2 c+2 a b c x+b^2 c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{\left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{5/2}}{10 b^2 c}-\frac{a \operatorname{Subst}\left (\int \left (a^2 c+2 a b c x+b^2 c x^2\right )^{3/2} \, dx,x,x^2\right )}{2 b}\\ &=-\frac{a \left (a+b x^2\right ) \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{8 b^2}+\frac{\left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{5/2}}{10 b^2 c}\\ \end{align*}

Mathematica [A]  time = 0.0213701, size = 63, normalized size = 0.95 \[ \frac{x^4 \left (20 a^2 b x^2+10 a^3+15 a b^2 x^4+4 b^3 x^6\right ) \left (c \left (a+b x^2\right )^2\right )^{3/2}}{40 \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(c*(a + b*x^2)^2)^(3/2)*(10*a^3 + 20*a^2*b*x^2 + 15*a*b^2*x^4 + 4*b^3*x^6))/(40*(a + b*x^2)^3)

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Maple [A]  time = 0.005, size = 60, normalized size = 0.9 \begin{align*}{\frac{{x}^{4} \left ( 4\,{b}^{3}{x}^{6}+15\,a{b}^{2}{x}^{4}+20\,{a}^{2}b{x}^{2}+10\,{a}^{3} \right ) }{40\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( c \left ( b{x}^{2}+a \right ) ^{2} \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)^2)^(3/2),x)

[Out]

1/40*x^4*(4*b^3*x^6+15*a*b^2*x^4+20*a^2*b*x^2+10*a^3)*(c*(b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.42047, size = 162, normalized size = 2.45 \begin{align*} \frac{{\left (4 \, b^{3} c x^{10} + 15 \, a b^{2} c x^{8} + 20 \, a^{2} b c x^{6} + 10 \, a^{3} c x^{4}\right )} \sqrt{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{40 \,{\left (b x^{2} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(4*b^3*c*x^10 + 15*a*b^2*c*x^8 + 20*a^2*b*c*x^6 + 10*a^3*c*x^4)*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)/(b*
x^2 + a)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.21654, size = 65, normalized size = 0.98 \begin{align*} \frac{1}{40} \,{\left (4 \, b^{3} x^{10} + 15 \, a b^{2} x^{8} + 20 \, a^{2} b x^{6} + 10 \, a^{3} x^{4}\right )} c^{\frac{3}{2}} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(4*b^3*x^10 + 15*a*b^2*x^8 + 20*a^2*b*x^6 + 10*a^3*x^4)*c^(3/2)*sgn(b*x^2 + a)