3.595 \(\int \frac{A+B x^2}{x^4 (a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=108 \[ \frac{8 b x (2 A b-a B)}{3 a^4 \sqrt{a+b x^2}}+\frac{4 b x (2 A b-a B)}{3 a^3 \left (a+b x^2\right )^{3/2}}+\frac{2 A b-a B}{a^2 x \left (a+b x^2\right )^{3/2}}-\frac{A}{3 a x^3 \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0487297, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {453, 271, 192, 191} \[ \frac{8 b x (2 A b-a B)}{3 a^4 \sqrt{a+b x^2}}+\frac{4 b x (2 A b-a B)}{3 a^3 \left (a+b x^2\right )^{3/2}}+\frac{2 A b-a B}{a^2 x \left (a+b x^2\right )^{3/2}}-\frac{A}{3 a x^3 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^4*(a + b*x^2)^(5/2)),x]

[Out]

-A/(3*a*x^3*(a + b*x^2)^(3/2)) + (2*A*b - a*B)/(a^2*x*(a + b*x^2)^(3/2)) + (4*b*(2*A*b - a*B)*x)/(3*a^3*(a + b
*x^2)^(3/2)) + (8*b*(2*A*b - a*B)*x)/(3*a^4*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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -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^4 \left (a+b x^2\right )^{5/2}} \, dx &=-\frac{A}{3 a x^3 \left (a+b x^2\right )^{3/2}}-\frac{(6 A b-3 a B) \int \frac{1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx}{3 a}\\ &=-\frac{A}{3 a x^3 \left (a+b x^2\right )^{3/2}}+\frac{2 A b-a B}{a^2 x \left (a+b x^2\right )^{3/2}}+\frac{(4 b (2 A b-a B)) \int \frac{1}{\left (a+b x^2\right )^{5/2}} \, dx}{a^2}\\ &=-\frac{A}{3 a x^3 \left (a+b x^2\right )^{3/2}}+\frac{2 A b-a B}{a^2 x \left (a+b x^2\right )^{3/2}}+\frac{4 b (2 A b-a B) x}{3 a^3 \left (a+b x^2\right )^{3/2}}+\frac{(8 b (2 A b-a B)) \int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx}{3 a^3}\\ &=-\frac{A}{3 a x^3 \left (a+b x^2\right )^{3/2}}+\frac{2 A b-a B}{a^2 x \left (a+b x^2\right )^{3/2}}+\frac{4 b (2 A b-a B) x}{3 a^3 \left (a+b x^2\right )^{3/2}}+\frac{8 b (2 A b-a B) x}{3 a^4 \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0245709, size = 79, normalized size = 0.73 \[ \frac{6 a^2 b x^2 \left (A-2 B x^2\right )-a^3 \left (A+3 B x^2\right )-8 a b^2 x^4 \left (B x^2-3 A\right )+16 A b^3 x^6}{3 a^4 x^3 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^4*(a + b*x^2)^(5/2)),x]

[Out]

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

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Maple [A]  time = 0.006, size = 82, normalized size = 0.8 \begin{align*} -{\frac{-16\,A{b}^{3}{x}^{6}+8\,Ba{b}^{2}{x}^{6}-24\,Aa{b}^{2}{x}^{4}+12\,B{a}^{2}b{x}^{4}-6\,A{a}^{2}b{x}^{2}+3\,B{a}^{3}{x}^{2}+A{a}^{3}}{3\,{x}^{3}{a}^{4}} \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^4/(b*x^2+a)^(5/2),x)

[Out]

-1/3*(-16*A*b^3*x^6+8*B*a*b^2*x^6-24*A*a*b^2*x^4+12*B*a^2*b*x^4-6*A*a^2*b*x^2+3*B*a^3*x^2+A*a^3)/(b*x^2+a)^(3/
2)/x^3/a^4

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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^4/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63431, size = 209, normalized size = 1.94 \begin{align*} -\frac{{\left (8 \,{\left (B a b^{2} - 2 \, A b^{3}\right )} x^{6} + 12 \,{\left (B a^{2} b - 2 \, A a b^{2}\right )} x^{4} + A a^{3} + 3 \,{\left (B a^{3} - 2 \, A a^{2} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{3 \,{\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 43.6028, size = 524, normalized size = 4.85 \begin{align*} A \left (- \frac{a^{4} b^{\frac{19}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}} + \frac{5 a^{3} b^{\frac{21}{2}} x^{2} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}} + \frac{30 a^{2} b^{\frac{23}{2}} x^{4} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}} + \frac{40 a b^{\frac{25}{2}} x^{6} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}} + \frac{16 b^{\frac{27}{2}} x^{8} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}}\right ) + B \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**4/(b*x**2+a)**(5/2),x)

[Out]

A*(-a**4*b**(19/2)*sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b**10*x**4 + 9*a**5*b**11*x**6 + 3*a**4*b**
12*x**8) + 5*a**3*b**(21/2)*x**2*sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b**10*x**4 + 9*a**5*b**11*x**
6 + 3*a**4*b**12*x**8) + 30*a**2*b**(23/2)*x**4*sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b**10*x**4 + 9
*a**5*b**11*x**6 + 3*a**4*b**12*x**8) + 40*a*b**(25/2)*x**6*sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b*
*10*x**4 + 9*a**5*b**11*x**6 + 3*a**4*b**12*x**8) + 16*b**(27/2)*x**8*sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 +
9*a**6*b**10*x**4 + 9*a**5*b**11*x**6 + 3*a**4*b**12*x**8)) + B*(-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))

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Giac [B]  time = 1.15275, size = 302, normalized size = 2.8 \begin{align*} -\frac{x{\left (\frac{{\left (5 \, B a^{4} b^{3} - 8 \, A a^{3} b^{4}\right )} x^{2}}{a^{7} b} + \frac{3 \,{\left (2 \, B a^{5} b^{2} - 3 \, A a^{4} b^{3}\right )}}{a^{7} b}\right )}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} + \frac{2 \,{\left (3 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} B a \sqrt{b} - 6 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} A b^{\frac{3}{2}} - 6 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} B a^{2} \sqrt{b} + 18 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} A a b^{\frac{3}{2}} + 3 \, B a^{3} \sqrt{b} - 8 \, A a^{2} b^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{3} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

-1/3*x*((5*B*a^4*b^3 - 8*A*a^3*b^4)*x^2/(a^7*b) + 3*(2*B*a^5*b^2 - 3*A*a^4*b^3)/(a^7*b))/(b*x^2 + a)^(3/2) + 2
/3*(3*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a*sqrt(b) - 6*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*b^(3/2) - 6*(sqrt(b)*x
- sqrt(b*x^2 + a))^2*B*a^2*sqrt(b) + 18*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a*b^(3/2) + 3*B*a^3*sqrt(b) - 8*A*a
^2*b^(3/2))/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^3*a^3)