3.3.32 \(\int x (c (a+b x^2)^2)^{3/2} \, dx\) [232]

Optimal. Leaf size=32 \[ \frac {c \left (a+b x^2\right )^3 \sqrt {c \left (a+b x^2\right )^2}}{8 b} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1605, 15, 30} \begin {gather*} \frac {c \left (a+b x^2\right )^3 \sqrt {c \left (a+b x^2\right )^2}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

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

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]

Rule 1605

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]

Rubi steps

\begin {align*} \int x \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx &=\frac {\text {Subst}\left (\int \left (c x^2\right )^{3/2} \, dx,x,a+b x^2\right )}{2 b}\\ &=\frac {\left (c \sqrt {c \left (a+b x^2\right )^2}\right ) \text {Subst}\left (\int x^3 \, dx,x,a+b x^2\right )}{2 b \left (a+b x^2\right )}\\ &=\frac {c \left (a+b x^2\right )^3 \sqrt {c \left (a+b x^2\right )^2}}{8 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 29, normalized size = 0.91 \begin {gather*} \frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^2\right )^{3/2}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)*(c*(a + b*x^2)^2)^(3/2))/(8*b)

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Maple [A]
time = 0.05, size = 26, normalized size = 0.81

method	result	size
default	\(\frac {\left (c \left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )}{8 b}\)	\(26\)
risch	\(\frac {c \left (b \,x^{2}+a \right )^{3} \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{8 b}\)	\(29\)
gosper	\(\frac {x^{2} \left (b^{3} x^{6}+4 a \,b^{2} x^{4}+6 a^{2} b \,x^{2}+4 a^{3}\right ) \left (c \left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{8 \left (b \,x^{2}+a \right )^{3}}\)	\(59\)
trager	\(\frac {c \,x^{2} \left (b^{3} x^{6}+4 a \,b^{2} x^{4}+6 a^{2} b \,x^{2}+4 a^{3}\right ) \sqrt {b^{2} c \,x^{4}+2 a b c \,x^{2}+a^{2} c}}{8 b \,x^{2}+8 a}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (28) = 56\).
time = 0.28, size = 60, normalized size = 1.88 \begin {gather*} \frac {1}{8} \, {\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} x^{2} + \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (28) = 56\).
time = 0.33, size = 73, normalized size = 2.28 \begin {gather*} \frac {{\left (b^{3} c x^{8} + 4 \, a b^{2} c x^{6} + 6 \, a^{2} b c x^{4} + 4 \, a^{3} c x^{2}\right )} \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{8 \, {\left (b x^{2} + a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^3*c*x^8 + 4*a*b^2*c*x^6 + 6*a^2*b*c*x^4 + 4*a^3*c*x^2)*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 \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*(c*(b*x**2+a)**2)**(3/2),x)

[Out]

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

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Giac [A]
time = 3.17, size = 25, normalized size = 0.78 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{4} c^{\frac {3}{2}} \mathrm {sgn}\left (b x^{2} + a\right )}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.84, size = 40, normalized size = 1.25 \begin {gather*} \frac {\left (b^2\,x^2+a\,b\right )\,{\left (c\,a^2+2\,c\,a\,b\,x^2+c\,b^2\,x^4\right )}^{3/2}}{8\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*(a + b*x^2)^2)^(3/2),x)

[Out]

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

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