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

Optimal. Leaf size=77 \[ -\frac{2 x (4 A b-a B)}{3 a^3 \sqrt{a+b x^2}}-\frac{x (4 A b-a B)}{3 a^2 \left (a+b x^2\right )^{3/2}}-\frac{A}{a x \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0265301, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {453, 192, 191} \[ -\frac{2 x (4 A b-a B)}{3 a^3 \sqrt{a+b x^2}}-\frac{x (4 A b-a B)}{3 a^2 \left (a+b x^2\right )^{3/2}}-\frac{A}{a x \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0203443, size = 60, normalized size = 0.78 \[ \frac{-3 a^2 \left (A-B x^2\right )+2 a b x^2 \left (B x^2-6 A\right )-8 A b^2 x^4}{3 a^3 x \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*A*b^2*x^4 - 3*a^2*(A - B*x^2) + 2*a*b*x^2*(-6*A + B*x^2))/(3*a^3*x*(a + b*x^2)^(3/2))

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Maple [A]  time = 0.004, size = 59, normalized size = 0.8 \begin{align*} -{\frac{8\,A{b}^{2}{x}^{4}-2\,B{x}^{4}ab+12\,aAb{x}^{2}-3\,B{x}^{2}{a}^{2}+3\,A{a}^{2}}{3\,x{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(8*A*b^2*x^4-2*B*a*b*x^4+12*A*a*b*x^2-3*B*a^2*x^2+3*A*a^2)/(b*x^2+a)^(3/2)/x/a^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83957, size = 161, normalized size = 2.09 \begin{align*} \frac{{\left (2 \,{\left (B a b - 4 \, A b^{2}\right )} x^{4} - 3 \, A a^{2} + 3 \,{\left (B a^{2} - 4 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{3 \,{\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*(B*a*b - 4*A*b^2)*x^4 - 3*A*a^2 + 3*(B*a^2 - 4*A*a*b)*x^2)*sqrt(b*x^2 + a)/(a^3*b^2*x^5 + 2*a^4*b*x^3 +
a^5*x)

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Sympy [B]  time = 23.1413, size = 265, normalized size = 3.44 \begin{align*} A \left (- \frac{3 a^{2} b^{\frac{9}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac{12 a b^{\frac{11}{2}} x^{2} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac{8 b^{\frac{13}{2}} x^{4} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}}\right ) + B \left (\frac{3 a x}{3 a^{\frac{7}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{5}{2}} b x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{2 b x^{3}}{3 a^{\frac{7}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{5}{2}} b x^{2} \sqrt{1 + \frac{b x^{2}}{a}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-3*a**2*b**(9/2)*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4) - 12*a*b**(11/2)*
x**2*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4) - 8*b**(13/2)*x**4*sqrt(a/(b*x**
2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4)) + B*(3*a*x/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**
(5/2)*b*x**2*sqrt(1 + b*x**2/a)) + 2*b*x**3/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2
/a)))

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Giac [A]  time = 1.14633, size = 136, normalized size = 1.77 \begin{align*} \frac{x{\left (\frac{{\left (2 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{2}}{a^{5} b} + \frac{3 \,{\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )}}{a^{5} b}\right )}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} + \frac{2 \, A \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x*((2*B*a^3*b^2 - 5*A*a^2*b^3)*x^2/(a^5*b) + 3*(B*a^4*b - 2*A*a^3*b^2)/(a^5*b))/(b*x^2 + a)^(3/2) + 2*A*sq
rt(b)/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)*a^2)