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

Optimal. Leaf size=46 \[ \frac {2 \left (a+b x^3\right )^{5/2} (A b-a B)}{15 b^2}+\frac {2 B \left (a+b x^3\right )^{7/2}}{21 b^2} \]

[Out]

2/15*(A*b-B*a)*(b*x^3+a)^(5/2)/b^2+2/21*B*(b*x^3+a)^(7/2)/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} \[ \frac {2 \left (a+b x^3\right )^{5/2} (A b-a B)}{15 b^2}+\frac {2 B \left (a+b x^3\right )^{7/2}}{21 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*(a + b*x^3)^(5/2))/(15*b^2) + (2*B*(a + b*x^3)^(7/2))/(21*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 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int (a+b x)^{3/2} (A+B x) \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {(A b-a B) (a+b x)^{3/2}}{b}+\frac {B (a+b x)^{5/2}}{b}\right ) \, dx,x,x^3\right )\\ &=\frac {2 (A b-a B) \left (a+b x^3\right )^{5/2}}{15 b^2}+\frac {2 B \left (a+b x^3\right )^{7/2}}{21 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 34, normalized size = 0.74 \[ \frac {2 \left (a+b x^3\right )^{5/2} \left (-2 a B+7 A b+5 b B x^3\right )}{105 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x^3)^(5/2)*(7*A*b - 2*a*B + 5*b*B*x^3))/(105*b^2)

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fricas [A]  time = 0.88, size = 73, normalized size = 1.59 \[ \frac {2 \, {\left (5 \, B b^{3} x^{9} + {\left (8 \, B a b^{2} + 7 \, A b^{3}\right )} x^{6} - 2 \, B a^{3} + 7 \, A a^{2} b + {\left (B a^{2} b + 14 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{105 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(5*B*b^3*x^9 + (8*B*a*b^2 + 7*A*b^3)*x^6 - 2*B*a^3 + 7*A*a^2*b + (B*a^2*b + 14*A*a*b^2)*x^3)*sqrt(b*x^3
+ a)/b^2

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giac [A]  time = 0.17, size = 44, normalized size = 0.96 \[ \frac {2 \, {\left (5 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} B - 7 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B a + 7 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A b\right )}}{105 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(5*(b*x^3 + a)^(7/2)*B - 7*(b*x^3 + a)^(5/2)*B*a + 7*(b*x^3 + a)^(5/2)*A*b)/b^2

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maple [A]  time = 0.05, size = 31, normalized size = 0.67 \[ \frac {2 \left (b \,x^{3}+a \right )^{\frac {5}{2}} \left (5 B b \,x^{3}+7 A b -2 B a \right )}{105 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(b*x^3+a)^(5/2)*(5*B*b*x^3+7*A*b-2*B*a)/b^2

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maxima [A]  time = 0.49, size = 49, normalized size = 1.07 \[ \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A}{15 \, b} + \frac {2}{105} \, {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}}}{b^{2}} - \frac {7 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a}{b^{2}}\right )} B \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(b*x^3 + a)^(5/2)*A/b + 2/105*(5*(b*x^3 + a)^(7/2)/b^2 - 7*(b*x^3 + a)^(5/2)*a/b^2)*B

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mupad [B]  time = 3.35, size = 150, normalized size = 3.26 \[ \frac {\left (2\,A\,a^2-\frac {2\,a\,\left (2\,B\,a^2+4\,A\,a\,b-\frac {4\,a\,\left (2\,A\,b^2+\frac {16\,B\,a\,b}{7}\right )}{5\,b}\right )}{3\,b}\right )\,\sqrt {b\,x^3+a}}{3\,b}+\frac {x^3\,\sqrt {b\,x^3+a}\,\left (2\,B\,a^2+4\,A\,a\,b-\frac {4\,a\,\left (2\,A\,b^2+\frac {16\,B\,a\,b}{7}\right )}{5\,b}\right )}{9\,b}+\frac {2\,B\,b\,x^9\,\sqrt {b\,x^3+a}}{21}+\frac {x^6\,\left (2\,A\,b^2+\frac {16\,B\,a\,b}{7}\right )\,\sqrt {b\,x^3+a}}{15\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*A*a^2 - (2*a*(2*B*a^2 + 4*A*a*b - (4*a*(2*A*b^2 + (16*B*a*b)/7))/(5*b)))/(3*b))*(a + b*x^3)^(1/2))/(3*b) +
(x^3*(a + b*x^3)^(1/2)*(2*B*a^2 + 4*A*a*b - (4*a*(2*A*b^2 + (16*B*a*b)/7))/(5*b)))/(9*b) + (2*B*b*x^9*(a + b*
x^3)^(1/2))/21 + (x^6*(2*A*b^2 + (16*B*a*b)/7)*(a + b*x^3)^(1/2))/(15*b)

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sympy [A]  time = 2.52, size = 165, normalized size = 3.59 \[ \begin {cases} \frac {2 A a^{2} \sqrt {a + b x^{3}}}{15 b} + \frac {4 A a x^{3} \sqrt {a + b x^{3}}}{15} + \frac {2 A b x^{6} \sqrt {a + b x^{3}}}{15} - \frac {4 B a^{3} \sqrt {a + b x^{3}}}{105 b^{2}} + \frac {2 B a^{2} x^{3} \sqrt {a + b x^{3}}}{105 b} + \frac {16 B a x^{6} \sqrt {a + b x^{3}}}{105} + \frac {2 B b x^{9} \sqrt {a + b x^{3}}}{21} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{3}}{3} + \frac {B x^{6}}{6}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*A*a**2*sqrt(a + b*x**3)/(15*b) + 4*A*a*x**3*sqrt(a + b*x**3)/15 + 2*A*b*x**6*sqrt(a + b*x**3)/15
- 4*B*a**3*sqrt(a + b*x**3)/(105*b**2) + 2*B*a**2*x**3*sqrt(a + b*x**3)/(105*b) + 16*B*a*x**6*sqrt(a + b*x**3)
/105 + 2*B*b*x**9*sqrt(a + b*x**3)/21, Ne(b, 0)), (a**(3/2)*(A*x**3/3 + B*x**6/6), True))

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