3.3.30 \(\int x^3 (c (a+b x^2)^2)^{3/2} \, dx\) [230]

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

[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 = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1973, 272, 45} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

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

Rule 45

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])

Rule 272

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 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

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} \text {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 \text {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 = 0.01, size = 63, normalized size = 0.95 \begin {gather*} \frac {x^4 \left (c \left (a+b x^2\right )^2\right )^{3/2} \left (10 a^3+20 a^2 b x^2+15 a b^2 x^4+4 b^3 x^6\right )}{40 \left (a+b x^2\right )^3} \end {gather*}

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 = 0.05, size = 60, normalized size = 0.91

method	result	size
gosper	\(\frac {x^{4} \left (4 b^{3} x^{6}+15 a \,b^{2} x^{4}+20 a^{2} b \,x^{2}+10 a^{3}\right ) \left (c \left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{40 \left (b \,x^{2}+a \right )^{3}}\)	\(60\)
default	\(\frac {x^{4} \left (4 b^{3} x^{6}+15 a \,b^{2} x^{4}+20 a^{2} b \,x^{2}+10 a^{3}\right ) \left (c \left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{40 \left (b \,x^{2}+a \right )^{3}}\)	\(60\)
trager	\(\frac {c \,x^{4} \left (4 b^{3} x^{6}+15 a \,b^{2} x^{4}+20 a^{2} b \,x^{2}+10 a^{3}\right ) \sqrt {b^{2} c \,x^{4}+2 a b c \,x^{2}+a^{2} c}}{40 b \,x^{2}+40 a}\)	\(72\)
risch	\(\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, b^{3} x^{10}}{10 b \,x^{2}+10 a}+\frac {3 c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, a \,b^{2} x^{8}}{8 \left (b \,x^{2}+a \right )}+\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, a^{2} b \,x^{6}}{2 b \,x^{2}+2 a}+\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, a^{3} x^{4}}{4 b \,x^{2}+4 a}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*(b*x^2+a)^2)^(3/2),x,method=_RETURNVERBOSE)

[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 [A]
time = 0.28, size = 98, normalized size = 1.48 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 0.32, size = 74, normalized size = 1.12 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (c \left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 5.80, size = 48, normalized size = 0.73 \begin {gather*} \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 {gather*}

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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Mupad [B]
time = 2.83, size = 50, normalized size = 0.76 \begin {gather*} \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} \end {gather*}

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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