3.24.75 \(\int x^5 \sqrt {a+b (c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=56 \[ \frac {2 \left (a+b \left (c x^2\right )^{3/2}\right )^{5/2}}{15 b^2 c^3}-\frac {2 a \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b^2 c^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {368, 266, 43} \begin {gather*} \frac {2 \left (a+b \left (c x^2\right )^{3/2}\right )^{5/2}}{15 b^2 c^3}-\frac {2 a \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b^2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

(-2*a*(a + b*(c*x^2)^(3/2))^(3/2))/(9*b^2*c^3) + (2*(a + b*(c*x^2)^(3/2))^(5/2))/(15*b^2*c^3)

Rule 43

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 266

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 368

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rubi steps

\begin {align*} \int x^5 \sqrt {a+b \left (c x^2\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int x^5 \sqrt {a+b x^3} \, dx,x,\sqrt {c x^2}\right )}{c^3}\\ &=\frac {\operatorname {Subst}\left (\int x \sqrt {a+b x} \, dx,x,\left (c x^2\right )^{3/2}\right )}{3 c^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a \sqrt {a+b x}}{b}+\frac {(a+b x)^{3/2}}{b}\right ) \, dx,x,\left (c x^2\right )^{3/2}\right )}{3 c^3}\\ &=-\frac {2 a \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b^2 c^3}+\frac {2 \left (a+b \left (c x^2\right )^{3/2}\right )^{5/2}}{15 b^2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.77 \begin {gather*} \frac {2 \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2} \left (3 b \left (c x^2\right )^{3/2}-2 a\right )}{45 b^2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

(2*(a + b*(c*x^2)^(3/2))^(3/2)*(-2*a + 3*b*(c*x^2)^(3/2)))/(45*b^2*c^3)

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IntegrateAlgebraic [A]  time = 2.72, size = 57, normalized size = 1.02 \begin {gather*} -\frac {2 \sqrt {a+b \left (c x^2\right )^{3/2}} \left (2 a^2-a b \left (c x^2\right )^{3/2}-3 b^2 c^3 x^6\right )}{45 b^2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5*Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

(-2*Sqrt[a + b*(c*x^2)^(3/2)]*(2*a^2 - 3*b^2*c^3*x^6 - a*b*(c*x^2)^(3/2)))/(45*b^2*c^3)

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fricas [A]  time = 0.82, size = 56, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} c^{3} x^{6} + \sqrt {c x^{2}} a b c x^{2} - 2 \, a^{2}\right )} \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{45 \, b^{2} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*(c*x^2)^(3/2))^(1/2),x, algorithm="fricas")

[Out]

2/45*(3*b^2*c^3*x^6 + sqrt(c*x^2)*a*b*c*x^2 - 2*a^2)*sqrt(sqrt(c*x^2)*b*c*x^2 + a)/(b^2*c^3)

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giac [A]  time = 0.18, size = 38, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (3 \, {\left (b c^{\frac {3}{2}} x^{3} + a\right )}^{\frac {5}{2}} - 5 \, {\left (b c^{\frac {3}{2}} x^{3} + a\right )}^{\frac {3}{2}} a\right )}}{45 \, b^{2} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*(c*x^2)^(3/2))^(1/2),x, algorithm="giac")

[Out]

2/45*(3*(b*c^(3/2)*x^3 + a)^(5/2) - 5*(b*c^(3/2)*x^3 + a)^(3/2)*a)/(b^2*c^3)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a +\left (c \,x^{2}\right )^{\frac {3}{2}} b}\, x^{5}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+(c*x^2)^(3/2)*b)^(1/2),x)

[Out]

int(x^5*(a+(c*x^2)^(3/2)*b)^(1/2),x)

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maxima [A]  time = 0.53, size = 43, normalized size = 0.77 \begin {gather*} \frac {2 \, {\left (\frac {3 \, {\left (\left (c x^{2}\right )^{\frac {3}{2}} b + a\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (\left (c x^{2}\right )^{\frac {3}{2}} b + a\right )}^{\frac {3}{2}} a}{b^{2}}\right )}}{45 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*(c*x^2)^(3/2))^(1/2),x, algorithm="maxima")

[Out]

2/45*(3*((c*x^2)^(3/2)*b + a)^(5/2)/b^2 - 5*((c*x^2)^(3/2)*b + a)^(3/2)*a/b^2)/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^5\,\sqrt {a+b\,{\left (c\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*(c*x^2)^(3/2))^(1/2),x)

[Out]

int(x^5*(a + b*(c*x^2)^(3/2))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \sqrt {a + b \left (c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*(c*x**2)**(3/2))**(1/2),x)

[Out]

Integral(x**5*sqrt(a + b*(c*x**2)**(3/2)), x)

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