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

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

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Rubi [A]  time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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]

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 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]

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*}

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Mathematica [A]  time = 0.04, size = 60, normalized size = 0.78 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.13, size = 62, normalized size = 0.81 \begin {gather*} \frac {-3 a^2 A+3 a^2 B x^2-12 a A b x^2+2 a b B x^4-8 A b^2 x^4}{3 a^3 x \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.95, size = 77, normalized size = 1.00 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.42, size = 101, normalized size = 1.31 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.00, size = 59, normalized size = 0.77 \begin {gather*} -\frac {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^{2} A}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3} x} \end {gather*}

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 [A]  time = 1.13, size = 85, normalized size = 1.10 \begin {gather*} \frac {2 \, B x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {B x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {8 \, A b x}{3 \, \sqrt {b x^{2} + a} a^{3}} - \frac {4 \, A b x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {A}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x} \end {gather*}

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]

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

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mupad [B]  time = 0.62, size = 68, normalized size = 0.88 \begin {gather*} \frac {A\,a^2-8\,A\,{\left (b\,x^2+a\right )}^2+B\,a^2\,x^2+4\,A\,a\,\left (b\,x^2+a\right )+2\,B\,a\,x^2\,\left (b\,x^2+a\right )}{3\,a^3\,x\,{\left (b\,x^2+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 21.74, size = 265, normalized size = 3.44 \begin {gather*} 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 {gather*}

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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