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

Optimal. Leaf size=56 \[ \frac {4 \left (a+b \sqrt {c x^3}\right )^{5/2}}{15 b^2 c}-\frac {4 a \left (a+b \sqrt {c x^3}\right )^{3/2}}{9 b^2 c} \]

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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 = {369, 266, 43} \begin {gather*} \frac {4 \left (a+b \sqrt {c x^3}\right )^{5/2}}{15 b^2 c}-\frac {4 a \left (a+b \sqrt {c x^3}\right )^{3/2}}{9 b^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + b*Sqrt[c*x^3]],x]

[Out]

(-4*a*(a + b*Sqrt[c*x^3])^(3/2))/(9*b^2*c) + (4*(a + b*Sqrt[c*x^3])^(5/2))/(15*b^2*c)

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 369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b
, c, d, m, p, q}, x] && FractionQ[n]

Rubi steps

\begin {align*} \int x^2 \sqrt {a+b \sqrt {c x^3}} \, dx &=\operatorname {Subst}\left (\int x^2 \sqrt {a+b \sqrt {c} x^{3/2}} \, dx,\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=\operatorname {Subst}\left (\frac {2}{3} \operatorname {Subst}\left (\int x \sqrt {a+b \sqrt {c} x} \, dx,x,x^{3/2}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=\operatorname {Subst}\left (\frac {2}{3} \operatorname {Subst}\left (\int \left (-\frac {a \sqrt {a+b \sqrt {c} x}}{b \sqrt {c}}+\frac {\left (a+b \sqrt {c} x\right )^{3/2}}{b \sqrt {c}}\right ) \, dx,x,x^{3/2}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {4 a \left (a+b \sqrt {c x^3}\right )^{3/2}}{9 b^2 c}+\frac {4 \left (a+b \sqrt {c x^3}\right )^{5/2}}{15 b^2 c}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.77 \begin {gather*} \frac {4 \left (a+b \sqrt {c x^3}\right )^{3/2} \left (3 b \sqrt {c x^3}-2 a\right )}{45 b^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[a + b*Sqrt[c*x^3]],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.28, size = 50, normalized size = 0.89 \begin {gather*} -\frac {4 \left (5 a \left (a+b \sqrt {c x^3}\right )^{3/2}-3 \left (a+b \sqrt {c x^3}\right )^{5/2}\right )}{45 b^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*Sqrt[a + b*Sqrt[c*x^3]],x]

[Out]

(-4*(5*a*(a + b*Sqrt[c*x^3])^(3/2) - 3*(a + b*Sqrt[c*x^3])^(5/2)))/(45*b^2*c)

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fricas [A]  time = 0.91, size = 46, normalized size = 0.82 \begin {gather*} \frac {4 \, {\left (3 \, b^{2} c x^{3} + \sqrt {c x^{3}} a b - 2 \, a^{2}\right )} \sqrt {\sqrt {c x^{3}} b + a}}{45 \, b^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/45*(3*b^2*c*x^3 + sqrt(c*x^3)*a*b - 2*a^2)*sqrt(sqrt(c*x^3)*b + a)/(b^2*c)

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giac [A]  time = 0.18, size = 57, normalized size = 1.02 \begin {gather*} -\frac {4 \, {\left (5 \, {\left (\sqrt {c x} b c^{2} x + a c^{2}\right )}^{\frac {3}{2}} a c^{2} - 3 \, {\left (\sqrt {c x} b c^{2} x + a c^{2}\right )}^{\frac {5}{2}}\right )} {\left | c \right |}}{45 \, b^{2} c^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/45*(5*(sqrt(c*x)*b*c^2*x + a*c^2)^(3/2)*a*c^2 - 3*(sqrt(c*x)*b*c^2*x + a*c^2)^(5/2))*abs(c)/(b^2*c^7)

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maple [A]  time = 0.29, size = 65, normalized size = 1.16 \begin {gather*} \frac {4 \sqrt {a +\sqrt {c \,x^{3}}\, b}\, \left (a b c \,x^{3}+3 \sqrt {c \,x^{3}}\, b^{2} c \,x^{3}-2 \sqrt {c \,x^{3}}\, a^{2}\right )}{45 \sqrt {c \,x^{3}}\, b^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/45/c*(a+(c*x^3)^(1/2)*b)^(1/2)*(3*c*x^3*(c*x^3)^(1/2)*b^2+a*x^3*c*b-2*a^2*(c*x^3)^(1/2))/(c*x^3)^(1/2)/b^2

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maxima [A]  time = 0.45, size = 43, normalized size = 0.77 \begin {gather*} \frac {4 \, {\left (\frac {3 \, {\left (\sqrt {c x^{3}} b + a\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (\sqrt {c x^{3}} b + a\right )}^{\frac {3}{2}} a}{b^{2}}\right )}}{45 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/45*(3*(sqrt(c*x^3)*b + a)^(5/2)/b^2 - 5*(sqrt(c*x^3)*b + a)^(3/2)*a/b^2)/c

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mupad [B]  time = 1.38, size = 52, normalized size = 0.93 \begin {gather*} \frac {x^3\,\sqrt {a+b\,\sqrt {c\,x^3}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},2;\ 3;\ -\frac {b\,\sqrt {c\,x^3}}{a}\right )}{3\,\sqrt {\frac {b\,\sqrt {c\,x^3}}{a}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3*(a + b*(c*x^3)^(1/2))^(1/2)*hypergeom([-1/2, 2], 3, -(b*(c*x^3)^(1/2))/a))/(3*((b*(c*x^3)^(1/2))/a + 1)^(
1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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