3.6.22 \(\int x^3 (a+b x^2)^{5/2} (A+B x^2) \, dx\)

Optimal. Leaf size=73 \[ \frac {\left (a+b x^2\right )^{9/2} (A b-2 a B)}{9 b^3}-\frac {a \left (a+b x^2\right )^{7/2} (A b-a B)}{7 b^3}+\frac {B \left (a+b x^2\right )^{11/2}}{11 b^3} \]

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Rubi [A]  time = 0.06, antiderivative size = 73, 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 = {446, 77} \begin {gather*} \frac {\left (a+b x^2\right )^{9/2} (A b-2 a B)}{9 b^3}-\frac {a \left (a+b x^2\right )^{7/2} (A b-a B)}{7 b^3}+\frac {B \left (a+b x^2\right )^{11/2}}{11 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^(5/2)*(A + B*x^2),x]

[Out]

-(a*(A*b - a*B)*(a + b*x^2)^(7/2))/(7*b^3) + ((A*b - 2*a*B)*(a + b*x^2)^(9/2))/(9*b^3) + (B*(a + b*x^2)^(11/2)
)/(11*b^3)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (a+b x)^{5/2} (A+B x) \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {a (-A b+a B) (a+b x)^{5/2}}{b^2}+\frac {(A b-2 a B) (a+b x)^{7/2}}{b^2}+\frac {B (a+b x)^{9/2}}{b^2}\right ) \, dx,x,x^2\right )\\ &=-\frac {a (A b-a B) \left (a+b x^2\right )^{7/2}}{7 b^3}+\frac {(A b-2 a B) \left (a+b x^2\right )^{9/2}}{9 b^3}+\frac {B \left (a+b x^2\right )^{11/2}}{11 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.78 \begin {gather*} \frac {\left (a+b x^2\right )^{7/2} \left (8 a^2 B-2 a b \left (11 A+14 B x^2\right )+7 b^2 x^2 \left (11 A+9 B x^2\right )\right )}{693 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(7/2)*(8*a^2*B + 7*b^2*x^2*(11*A + 9*B*x^2) - 2*a*b*(11*A + 14*B*x^2)))/(693*b^3)

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IntegrateAlgebraic [A]  time = 0.04, size = 56, normalized size = 0.77 \begin {gather*} \frac {\left (a+b x^2\right )^{7/2} \left (8 a^2 B-22 a A b-28 a b B x^2+77 A b^2 x^2+63 b^2 B x^4\right )}{693 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(7/2)*(-22*a*A*b + 8*a^2*B + 77*A*b^2*x^2 - 28*a*b*B*x^2 + 63*b^2*B*x^4))/(693*b^3)

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fricas [A]  time = 0.72, size = 122, normalized size = 1.67 \begin {gather*} \frac {{\left (63 \, B b^{5} x^{10} + 7 \, {\left (23 \, B a b^{4} + 11 \, A b^{5}\right )} x^{8} + {\left (113 \, B a^{2} b^{3} + 209 \, A a b^{4}\right )} x^{6} + 8 \, B a^{5} - 22 \, A a^{4} b + 3 \, {\left (B a^{3} b^{2} + 55 \, A a^{2} b^{3}\right )} x^{4} - {\left (4 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{693 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/693*(63*B*b^5*x^10 + 7*(23*B*a*b^4 + 11*A*b^5)*x^8 + (113*B*a^2*b^3 + 209*A*a*b^4)*x^6 + 8*B*a^5 - 22*A*a^4*
b + 3*(B*a^3*b^2 + 55*A*a^2*b^3)*x^4 - (4*B*a^4*b - 11*A*a^3*b^2)*x^2)*sqrt(b*x^2 + a)/b^3

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giac [A]  time = 0.41, size = 73, normalized size = 1.00 \begin {gather*} \frac {63 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} B - 154 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B a + 99 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2} + 77 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} A b - 99 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a b}{693 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/693*(63*(b*x^2 + a)^(11/2)*B - 154*(b*x^2 + a)^(9/2)*B*a + 99*(b*x^2 + a)^(7/2)*B*a^2 + 77*(b*x^2 + a)^(9/2)
*A*b - 99*(b*x^2 + a)^(7/2)*A*a*b)/b^3

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maple [A]  time = 0.01, size = 53, normalized size = 0.73 \begin {gather*} -\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \left (-63 B \,b^{2} x^{4}-77 A \,b^{2} x^{2}+28 B a b \,x^{2}+22 a b A -8 a^{2} B \right )}{693 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^(5/2)*(B*x^2+A),x)

[Out]

-1/693*(b*x^2+a)^(7/2)*(-63*B*b^2*x^4-77*A*b^2*x^2+28*B*a*b*x^2+22*A*a*b-8*B*a^2)/b^3

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maxima [A]  time = 1.06, size = 90, normalized size = 1.23 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x^{4}}{11 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a x^{2}}{99 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A x^{2}}{9 \, b} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2}}{693 \, b^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a}{63 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*(b*x^2 + a)^(7/2)*B*x^4/b - 4/99*(b*x^2 + a)^(7/2)*B*a*x^2/b^2 + 1/9*(b*x^2 + a)^(7/2)*A*x^2/b + 8/693*(b
*x^2 + a)^(7/2)*B*a^2/b^3 - 2/63*(b*x^2 + a)^(7/2)*A*a/b^2

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mupad [B]  time = 0.69, size = 115, normalized size = 1.58 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {8\,B\,a^5-22\,A\,a^4\,b}{693\,b^3}+\frac {B\,b^2\,x^{10}}{11}+\frac {x^8\,\left (77\,A\,b^5+161\,B\,a\,b^4\right )}{693\,b^3}+\frac {a\,x^6\,\left (209\,A\,b+113\,B\,a\right )}{693}+\frac {a^3\,x^2\,\left (11\,A\,b-4\,B\,a\right )}{693\,b^2}+\frac {a^2\,x^4\,\left (55\,A\,b+B\,a\right )}{231\,b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(A + B*x^2)*(a + b*x^2)^(5/2),x)

[Out]

(a + b*x^2)^(1/2)*((8*B*a^5 - 22*A*a^4*b)/(693*b^3) + (B*b^2*x^10)/11 + (x^8*(77*A*b^5 + 161*B*a*b^4))/(693*b^
3) + (a*x^6*(209*A*b + 113*B*a))/693 + (a^3*x^2*(11*A*b - 4*B*a))/(693*b^2) + (a^2*x^4*(55*A*b + B*a))/(231*b)
)

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sympy [A]  time = 8.85, size = 260, normalized size = 3.56 \begin {gather*} \begin {cases} - \frac {2 A a^{4} \sqrt {a + b x^{2}}}{63 b^{2}} + \frac {A a^{3} x^{2} \sqrt {a + b x^{2}}}{63 b} + \frac {5 A a^{2} x^{4} \sqrt {a + b x^{2}}}{21} + \frac {19 A a b x^{6} \sqrt {a + b x^{2}}}{63} + \frac {A b^{2} x^{8} \sqrt {a + b x^{2}}}{9} + \frac {8 B a^{5} \sqrt {a + b x^{2}}}{693 b^{3}} - \frac {4 B a^{4} x^{2} \sqrt {a + b x^{2}}}{693 b^{2}} + \frac {B a^{3} x^{4} \sqrt {a + b x^{2}}}{231 b} + \frac {113 B a^{2} x^{6} \sqrt {a + b x^{2}}}{693} + \frac {23 B a b x^{8} \sqrt {a + b x^{2}}}{99} + \frac {B b^{2} x^{10} \sqrt {a + b x^{2}}}{11} & \text {for}\: b \neq 0 \\a^{\frac {5}{2}} \left (\frac {A x^{4}}{4} + \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**3*(b*x**2+a)**(5/2)*(B*x**2+A),x)

[Out]

Piecewise((-2*A*a**4*sqrt(a + b*x**2)/(63*b**2) + A*a**3*x**2*sqrt(a + b*x**2)/(63*b) + 5*A*a**2*x**4*sqrt(a +
b*x**2)/21 + 19*A*a*b*x**6*sqrt(a + b*x**2)/63 + A*b**2*x**8*sqrt(a + b*x**2)/9 + 8*B*a**5*sqrt(a + b*x**2)/(
693*b**3) - 4*B*a**4*x**2*sqrt(a + b*x**2)/(693*b**2) + B*a**3*x**4*sqrt(a + b*x**2)/(231*b) + 113*B*a**2*x**6
*sqrt(a + b*x**2)/693 + 23*B*a*b*x**8*sqrt(a + b*x**2)/99 + B*b**2*x**10*sqrt(a + b*x**2)/11, Ne(b, 0)), (a**(
5/2)*(A*x**4/4 + B*x**6/6), True))

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