3.584 \(\int \frac{x^7 (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=128 \[ \frac{a^3 (A b-a B)}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac{a^2 (3 A b-4 a B)}{b^5 \sqrt{a+b x^2}}-\frac{3 a \sqrt{a+b x^2} (A b-2 a B)}{b^5}+\frac{\left (a+b x^2\right )^{3/2} (A b-4 a B)}{3 b^5}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^5} \]

[Out]

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

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Rubi [A]  time = 0.101529, antiderivative size = 128, normalized size of antiderivative = 1., 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} \[ \frac{a^3 (A b-a B)}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac{a^2 (3 A b-4 a B)}{b^5 \sqrt{a+b x^2}}-\frac{3 a \sqrt{a+b x^2} (A b-2 a B)}{b^5}+\frac{\left (a+b x^2\right )^{3/2} (A b-4 a B)}{3 b^5}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

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

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]]

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]))))

Rubi steps

\begin{align*} \int \frac{x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3 (A+B x)}{(a+b x)^{5/2}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{a^3 (-A b+a B)}{b^4 (a+b x)^{5/2}}-\frac{a^2 (-3 A b+4 a B)}{b^4 (a+b x)^{3/2}}+\frac{3 a (-A b+2 a B)}{b^4 \sqrt{a+b x}}+\frac{(A b-4 a B) \sqrt{a+b x}}{b^4}+\frac{B (a+b x)^{3/2}}{b^4}\right ) \, dx,x,x^2\right )\\ &=\frac{a^3 (A b-a B)}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac{a^2 (3 A b-4 a B)}{b^5 \sqrt{a+b x^2}}-\frac{3 a (A b-2 a B) \sqrt{a+b x^2}}{b^5}+\frac{(A b-4 a B) \left (a+b x^2\right )^{3/2}}{3 b^5}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^5}\\ \end{align*}

Mathematica [A]  time = 0.0659652, size = 98, normalized size = 0.77 \[ \frac{24 a^2 b^2 x^2 \left (2 B x^2-5 A\right )+a^3 \left (192 b B x^2-80 A b\right )+128 a^4 B-2 a b^3 x^4 \left (15 A+4 B x^2\right )+b^4 x^6 \left (5 A+3 B x^2\right )}{15 b^5 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

(128*a^4*B + 24*a^2*b^2*x^2*(-5*A + 2*B*x^2) + b^4*x^6*(5*A + 3*B*x^2) - 2*a*b^3*x^4*(15*A + 4*B*x^2) + a^3*(-
80*A*b + 192*b*B*x^2))/(15*b^5*(a + b*x^2)^(3/2))

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Maple [A]  time = 0.006, size = 101, normalized size = 0.8 \begin{align*} -{\frac{-3\,{x}^{8}B{b}^{4}-5\,A{b}^{4}{x}^{6}+8\,Ba{b}^{3}{x}^{6}+30\,Aa{b}^{3}{x}^{4}-48\,B{a}^{2}{b}^{2}{x}^{4}+120\,A{a}^{2}{b}^{2}{x}^{2}-192\,B{a}^{3}b{x}^{2}+80\,A{a}^{3}b-128\,B{a}^{4}}{15\,{b}^{5}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(B*x^2+A)/(b*x^2+a)^(5/2),x)

[Out]

-1/15*(-3*B*b^4*x^8-5*A*b^4*x^6+8*B*a*b^3*x^6+30*A*a*b^3*x^4-48*B*a^2*b^2*x^4+120*A*a^2*b^2*x^2-192*B*a^3*b*x^
2+80*A*a^3*b-128*B*a^4)/(b*x^2+a)^(3/2)/b^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62329, size = 259, normalized size = 2.02 \begin{align*} \frac{{\left (3 \, B b^{4} x^{8} -{\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} x^{6} + 128 \, B a^{4} - 80 \, A a^{3} b + 6 \,{\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 24 \,{\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{15 \,{\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

1/15*(3*B*b^4*x^8 - (8*B*a*b^3 - 5*A*b^4)*x^6 + 128*B*a^4 - 80*A*a^3*b + 6*(8*B*a^2*b^2 - 5*A*a*b^3)*x^4 + 24*
(8*B*a^3*b - 5*A*a^2*b^2)*x^2)*sqrt(b*x^2 + a)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5)

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Sympy [A]  time = 2.93182, size = 437, normalized size = 3.41 \begin{align*} \begin{cases} - \frac{80 A a^{3} b}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} - \frac{120 A a^{2} b^{2} x^{2}}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} - \frac{30 A a b^{3} x^{4}}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} + \frac{5 A b^{4} x^{6}}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} + \frac{128 B a^{4}}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} + \frac{192 B a^{3} b x^{2}}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} + \frac{48 B a^{2} b^{2} x^{4}}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} - \frac{8 B a b^{3} x^{6}}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} + \frac{3 B b^{4} x^{8}}{15 a b^{5} \sqrt{a + b x^{2}} + 15 b^{6} x^{2} \sqrt{a + b x^{2}}} & \text{for}\: b \neq 0 \\\frac{\frac{A x^{8}}{8} + \frac{B x^{10}}{10}}{a^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(B*x**2+A)/(b*x**2+a)**(5/2),x)

[Out]

Piecewise((-80*A*a**3*b/(15*a*b**5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) - 120*A*a**2*b**2*x**2/(1
5*a*b**5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) - 30*A*a*b**3*x**4/(15*a*b**5*sqrt(a + b*x**2) + 15
*b**6*x**2*sqrt(a + b*x**2)) + 5*A*b**4*x**6/(15*a*b**5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) + 12
8*B*a**4/(15*a*b**5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) + 192*B*a**3*b*x**2/(15*a*b**5*sqrt(a +
b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) + 48*B*a**2*b**2*x**4/(15*a*b**5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt
(a + b*x**2)) - 8*B*a*b**3*x**6/(15*a*b**5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) + 3*B*b**4*x**8/(
15*a*b**5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)), Ne(b, 0)), ((A*x**8/8 + B*x**10/10)/a**(5/2), Tru
e))

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Giac [A]  time = 1.13117, size = 167, normalized size = 1.3 \begin{align*} \frac{3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B - 20 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a + 90 \, \sqrt{b x^{2} + a} B a^{2} + 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A b - 45 \, \sqrt{b x^{2} + a} A a b + \frac{5 \,{\left (12 \,{\left (b x^{2} + a\right )} B a^{3} - B a^{4} - 9 \,{\left (b x^{2} + a\right )} A a^{2} b + A a^{3} b\right )}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/15*(3*(b*x^2 + a)^(5/2)*B - 20*(b*x^2 + a)^(3/2)*B*a + 90*sqrt(b*x^2 + a)*B*a^2 + 5*(b*x^2 + a)^(3/2)*A*b -
45*sqrt(b*x^2 + a)*A*a*b + 5*(12*(b*x^2 + a)*B*a^3 - B*a^4 - 9*(b*x^2 + a)*A*a^2*b + A*a^3*b)/(b*x^2 + a)^(3/2
))/b^5