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

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]

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

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Rubi [A]  time = 0.0223109, antiderivative size = 32, normalized size of antiderivative = 1., 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 = {1591, 15, 30} \[ \frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^2}}{8 b} \]

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 1591

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.0123346, size = 29, normalized size = 0.91 \[ \frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^2\right )^{3/2}}{8 b} \]

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 [B]  time = 0.003, size = 59, normalized size = 1.8 \begin{align*}{\frac{{x}^{2} \left ({b}^{3}{x}^{6}+4\,a{b}^{2}{x}^{4}+6\,{a}^{2}b{x}^{2}+4\,{a}^{3} \right ) }{8\, \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*(c*(b*x^2+a)^2)^(3/2),x)

[Out]

1/8*x^2*(b^3*x^6+4*a*b^2*x^4+6*a^2*b*x^2+4*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*(c*(b*x^2+a)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.39286, size = 153, normalized size = 4.78 \begin{align*} \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{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.23664, size = 34, normalized size = 1.06 \begin{align*} \frac{{\left (b x^{2} + a\right )}^{4} c^{\frac{3}{2}} \mathrm{sgn}\left (b x^{2} + a\right )}{8 \, b} \end{align*}

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