3.2.77 \(\int x^2 \sqrt {a+b x^3} (A+B x^3) \, dx\)

Optimal. Leaf size=46 \[ \frac {2 \left (a+b x^3\right )^{3/2} (A b-a B)}{9 b^2}+\frac {2 B \left (a+b x^3\right )^{5/2}}{15 b^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {444, 43} \begin {gather*} \frac {2 \left (a+b x^3\right )^{3/2} (A b-a B)}{9 b^2}+\frac {2 B \left (a+b x^3\right )^{5/2}}{15 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + b*x^3]*(A + B*x^3),x]

[Out]

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

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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rubi steps

\begin {align*} \int x^2 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \sqrt {a+b x} (A+B x) \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {(A b-a B) \sqrt {a+b x}}{b}+\frac {B (a+b x)^{3/2}}{b}\right ) \, dx,x,x^3\right )\\ &=\frac {2 (A b-a B) \left (a+b x^3\right )^{3/2}}{9 b^2}+\frac {2 B \left (a+b x^3\right )^{5/2}}{15 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 34, normalized size = 0.74 \begin {gather*} \frac {2 \left (a+b x^3\right )^{3/2} \left (-2 a B+5 A b+3 b B x^3\right )}{45 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[a + b*x^3]*(A + B*x^3),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 34, normalized size = 0.74 \begin {gather*} \frac {2 \left (a+b x^3\right )^{3/2} \left (-2 a B+5 A b+3 b B x^3\right )}{45 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*Sqrt[a + b*x^3]*(A + B*x^3),x]

[Out]

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

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fricas [A]  time = 0.71, size = 50, normalized size = 1.09 \begin {gather*} \frac {2 \, {\left (3 \, B b^{2} x^{6} + {\left (B a b + 5 \, A b^{2}\right )} x^{3} - 2 \, B a^{2} + 5 \, A a b\right )} \sqrt {b x^{3} + a}}{45 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(3*B*b^2*x^6 + (B*a*b + 5*A*b^2)*x^3 - 2*B*a^2 + 5*A*a*b)*sqrt(b*x^3 + a)/b^2

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giac [A]  time = 0.15, size = 44, normalized size = 0.96 \begin {gather*} \frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B - 5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a + 5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A b\right )}}{45 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 31, normalized size = 0.67 \begin {gather*} \frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}} \left (3 B b \,x^{3}+5 A b -2 B a \right )}{45 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 49, normalized size = 1.07 \begin {gather*} \frac {2}{45} \, B {\left (\frac {3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a}{b^{2}}\right )} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A}{9 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*B*(3*(b*x^3 + a)^(5/2)/b^2 - 5*(b*x^3 + a)^(3/2)*a/b^2) + 2/9*(b*x^3 + a)^(3/2)*A/b

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mupad [B]  time = 2.60, size = 44, normalized size = 0.96 \begin {gather*} \frac {6\,B\,{\left (b\,x^3+a\right )}^{5/2}+10\,A\,b\,{\left (b\,x^3+a\right )}^{3/2}-10\,B\,a\,{\left (b\,x^3+a\right )}^{3/2}}{45\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*B*(a + b*x^3)^(5/2) + 10*A*b*(a + b*x^3)^(3/2) - 10*B*a*(a + b*x^3)^(3/2))/(45*b^2)

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sympy [A]  time = 0.74, size = 117, normalized size = 2.54 \begin {gather*} \begin {cases} \frac {2 A a \sqrt {a + b x^{3}}}{9 b} + \frac {2 A x^{3} \sqrt {a + b x^{3}}}{9} - \frac {4 B a^{2} \sqrt {a + b x^{3}}}{45 b^{2}} + \frac {2 B a x^{3} \sqrt {a + b x^{3}}}{45 b} + \frac {2 B x^{6} \sqrt {a + b x^{3}}}{15} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{3}}{3} + \frac {B x^{6}}{6}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*A*a*sqrt(a + b*x**3)/(9*b) + 2*A*x**3*sqrt(a + b*x**3)/9 - 4*B*a**2*sqrt(a + b*x**3)/(45*b**2) +
2*B*a*x**3*sqrt(a + b*x**3)/(45*b) + 2*B*x**6*sqrt(a + b*x**3)/15, Ne(b, 0)), (sqrt(a)*(A*x**3/3 + B*x**6/6),
True))

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