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

Optimal. Leaf size=32 \[ \frac{c \left (a+b x^2\right )^4 \sqrt{c \left (a+b x^2\right )^3}}{11 b} \]

[Out]

(c*(a + b*x^2)^4*Sqrt[c*(a + b*x^2)^3])/(11*b)

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Rubi [A]  time = 0.0265127, 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} \[ \frac{c \left (a+b x^2\right )^4 \sqrt{c \left (a+b x^2\right )^3}}{11 b} \]

Antiderivative was successfully verified.

[In]

Int[x*(c*(a + b*x^2)^3)^(3/2),x]

[Out]

(c*(a + b*x^2)^4*Sqrt[c*(a + b*x^2)^3])/(11*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 )^3\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \left (c x^3\right )^{3/2} \, dx,x,a+b x^2\right )}{2 b}\\ &=\frac{\left (c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int x^{9/2} \, dx,x,a+b x^2\right )}{2 b \left (a+b x^2\right )^{3/2}}\\ &=\frac{c \left (a+b x^2\right )^4 \sqrt{c \left (a+b x^2\right )^3}}{11 b}\\ \end{align*}

Mathematica [A]  time = 0.0176661, size = 29, normalized size = 0.91 \[ \frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^3\right )^{3/2}}{11 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)*(c*(a + b*x^2)^3)^(3/2))/(11*b)

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Maple [A]  time = 0.003, size = 26, normalized size = 0.8 \begin{align*}{\frac{b{x}^{2}+a}{11\,b} \left ( c \left ( b{x}^{2}+a \right ) ^{3} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*(b*x^2+a)/b*(c*(b*x^2+a)^3)^(3/2)

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Maxima [B]  time = 1.0748, size = 95, normalized size = 2.97 \begin{align*} \frac{{\left (b^{4} c^{\frac{3}{2}} x^{8} + 4 \, a b^{3} c^{\frac{3}{2}} x^{6} + 6 \, a^{2} b^{2} c^{\frac{3}{2}} x^{4} + 4 \, a^{3} b c^{\frac{3}{2}} x^{2} + a^{4} c^{\frac{3}{2}}\right )}{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{11 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*(b^4*c^(3/2)*x^8 + 4*a*b^3*c^(3/2)*x^6 + 6*a^2*b^2*c^(3/2)*x^4 + 4*a^3*b*c^(3/2)*x^2 + a^4*c^(3/2))*(b*x^
2 + a)^(3/2)/b

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Fricas [B]  time = 1.62584, size = 181, normalized size = 5.66 \begin{align*} \frac{{\left (b^{4} c x^{8} + 4 \, a b^{3} c x^{6} + 6 \, a^{2} b^{2} c x^{4} + 4 \, a^{3} b c x^{2} + a^{4} c\right )} \sqrt{b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{11 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*(b^4*c*x^8 + 4*a*b^3*c*x^6 + 6*a^2*b^2*c*x^4 + 4*a^3*b*c*x^2 + a^4*c)*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*
a^2*b*c*x^2 + a^3*c)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.19807, size = 450, normalized size = 14.06 \begin{align*} \frac{1155 \,{\left (b c x^{2} + a c\right )}^{\frac{3}{2}} a^{4} \mathrm{sgn}\left (b x^{2} + a\right ) - \frac{924 \,{\left (5 \,{\left (b c x^{2} + a c\right )}^{\frac{3}{2}} a c - 3 \,{\left (b c x^{2} + a c\right )}^{\frac{5}{2}}\right )} a^{3} \mathrm{sgn}\left (b x^{2} + a\right )}{c} + \frac{198 \,{\left (35 \,{\left (b c x^{2} + a c\right )}^{\frac{3}{2}} a^{2} c^{2} - 42 \,{\left (b c x^{2} + a c\right )}^{\frac{5}{2}} a c + 15 \,{\left (b c x^{2} + a c\right )}^{\frac{7}{2}}\right )} a^{2} \mathrm{sgn}\left (b x^{2} + a\right )}{c^{2}} - \frac{44 \,{\left (105 \,{\left (b c x^{2} + a c\right )}^{\frac{3}{2}} a^{3} c^{3} - 189 \,{\left (b c x^{2} + a c\right )}^{\frac{5}{2}} a^{2} c^{2} + 135 \,{\left (b c x^{2} + a c\right )}^{\frac{7}{2}} a c - 35 \,{\left (b c x^{2} + a c\right )}^{\frac{9}{2}}\right )} a \mathrm{sgn}\left (b x^{2} + a\right )}{c^{3}} + \frac{{\left (1155 \,{\left (b c x^{2} + a c\right )}^{\frac{3}{2}} a^{4} c^{4} - 2772 \,{\left (b c x^{2} + a c\right )}^{\frac{5}{2}} a^{3} c^{3} + 2970 \,{\left (b c x^{2} + a c\right )}^{\frac{7}{2}} a^{2} c^{2} - 1540 \,{\left (b c x^{2} + a c\right )}^{\frac{9}{2}} a c + 315 \,{\left (b c x^{2} + a c\right )}^{\frac{11}{2}}\right )} \mathrm{sgn}\left (b x^{2} + a\right )}{c^{4}}}{3465 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(1155*(b*c*x^2 + a*c)^(3/2)*a^4*sgn(b*x^2 + a) - 924*(5*(b*c*x^2 + a*c)^(3/2)*a*c - 3*(b*c*x^2 + a*c)^(
5/2))*a^3*sgn(b*x^2 + a)/c + 198*(35*(b*c*x^2 + a*c)^(3/2)*a^2*c^2 - 42*(b*c*x^2 + a*c)^(5/2)*a*c + 15*(b*c*x^
2 + a*c)^(7/2))*a^2*sgn(b*x^2 + a)/c^2 - 44*(105*(b*c*x^2 + a*c)^(3/2)*a^3*c^3 - 189*(b*c*x^2 + a*c)^(5/2)*a^2
*c^2 + 135*(b*c*x^2 + a*c)^(7/2)*a*c - 35*(b*c*x^2 + a*c)^(9/2))*a*sgn(b*x^2 + a)/c^3 + (1155*(b*c*x^2 + a*c)^
(3/2)*a^4*c^4 - 2772*(b*c*x^2 + a*c)^(5/2)*a^3*c^3 + 2970*(b*c*x^2 + a*c)^(7/2)*a^2*c^2 - 1540*(b*c*x^2 + a*c)
^(9/2)*a*c + 315*(b*c*x^2 + a*c)^(11/2))*sgn(b*x^2 + a)/c^4)/b