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]

-1/8*a*c*(b*x^2+a)^3*(c*(b*x^2+a)^2)^(1/2)/b^2+1/10*c*(b*x^2+a)^4*(c*(b*x^2+a)^2)^(1/2)/b^2

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Rubi [A]  time = 0.11, antiderivative size = 78, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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)]

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 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 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]

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*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.95 \[ \frac {x^4 \left (10 a^3+20 a^2 b x^2+15 a b^2 x^4+4 b^3 x^6\right ) \left (c \left (a+b x^2\right )^2\right )^{3/2}}{40 \left (a+b x^2\right )^3} \]

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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fricas [A]  time = 0.64, size = 74, normalized size = 1.12 \[ \frac {{\left (4 \, b^{3} c x^{10} + 15 \, a b^{2} c x^{8} + 20 \, a^{2} b c x^{6} + 10 \, a^{3} c x^{4}\right )} \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{40 \, {\left (b x^{2} + a\right )}} \]

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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giac [A]  time = 0.30, size = 48, normalized size = 0.73 \[ \frac {1}{40} \, {\left (4 \, b^{3} x^{10} + 15 \, a b^{2} x^{8} + 20 \, a^{2} b x^{6} + 10 \, a^{3} x^{4}\right )} c^{\frac {3}{2}} \mathrm {sgn}\left (b x^{2} + a\right ) \]

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)

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maple [A]  time = 0.01, size = 60, normalized size = 0.91 \[ \frac {\left (4 b^{3} x^{6}+15 a \,b^{2} x^{4}+20 a^{2} b \,x^{2}+10 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2} c \right )^{\frac {3}{2}} x^{4}}{40 \left (b \,x^{2}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 1.04, size = 98, normalized size = 1.48 \[ -\frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a x^{2}}{8 \, b} - \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a^{2}}{8 \, b^{2}} + \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {5}{2}}}{10 \, b^{2} c} \]

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]

-1/8*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(3/2)*a*x^2/b - 1/8*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(3/2)*a^2/b^2 + 1
/10*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(5/2)/(b^2*c)

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mupad [B]  time = 2.83, size = 50, normalized size = 0.76 \[ \frac {\left (-a^2+3\,a\,b\,x^2+4\,b^2\,x^4\right )\,{\left (c\,a^2+2\,c\,a\,b\,x^2+c\,b^2\,x^4\right )}^{3/2}}{40\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*(a + b*x^2)^2)^(3/2),x)

[Out]

((4*b^2*x^4 - a^2 + 3*a*b*x^2)*(a^2*c + b^2*c*x^4 + 2*a*b*c*x^2)^(3/2))/(40*b^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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