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]

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

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Rubi [A]  time = 0.03, 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 = {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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [B]  time = 0.48, size = 87, normalized size = 2.72 \[ \frac {{\left (b^{4} c x^{8} + 4 \, a b^{3} c x^{6} + 6 \, a^{2} b^{2} c x^{4} + 4 \, a^{3} b c x^{2} + a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{11 \, b} \]

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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giac [A]  time = 0.31, size = 28, normalized size = 0.88 \[ \frac {{\left (b c x^{2} + a c\right )}^{\frac {11}{2}} \mathrm {sgn}\left (b x^{2} + a\right )}{11 \, b c^{4}} \]

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]

1/11*(b*c*x^2 + a*c)^(11/2)*sgn(b*x^2 + a)/(b*c^4)

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maple [A]  time = 0.01, size = 26, normalized size = 0.81 \[ \frac {\left (b \,x^{2}+a \right ) \left (\left (b \,x^{2}+a \right )^{3} c \right )^{\frac {3}{2}}}{11 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.08, size = 70, normalized size = 2.19 \[ \frac {{\left (b^{4} c^{\frac {3}{2}} x^{8} + 4 \, a b^{3} c^{\frac {3}{2}} x^{6} + 6 \, a^{2} b^{2} c^{\frac {3}{2}} x^{4} + 4 \, a^{3} b c^{\frac {3}{2}} x^{2} + a^{4} c^{\frac {3}{2}}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}}}{11 \, b} \]

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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mupad [B]  time = 2.71, size = 62, normalized size = 1.94 \[ \sqrt {c\,{\left (b\,x^2+a\right )}^3}\,\left (\frac {a^4\,c}{11\,b}+\frac {4\,a^3\,c\,x^2}{11}+\frac {b^3\,c\,x^8}{11}+\frac {6\,a^2\,b\,c\,x^4}{11}+\frac {4\,a\,b^2\,c\,x^6}{11}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*(a + b*x^2)^3)^(3/2),x)

[Out]

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

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

[Out]

Timed out

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