3.6.48 \(\int \frac {A+B x^2}{x^6 \sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac {2 b \sqrt {a+b x^2} (4 A b-5 a B)}{15 a^3 x}+\frac {\sqrt {a+b x^2} (4 A b-5 a B)}{15 a^2 x^3}-\frac {A \sqrt {a+b x^2}}{5 a x^5} \]

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Rubi [A]  time = 0.03, antiderivative size = 84, 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, 271, 264} \begin {gather*} -\frac {2 b \sqrt {a+b x^2} (4 A b-5 a B)}{15 a^3 x}+\frac {\sqrt {a+b x^2} (4 A b-5 a B)}{15 a^2 x^3}-\frac {A \sqrt {a+b x^2}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^6*Sqrt[a + b*x^2]),x]

[Out]

-(A*Sqrt[a + b*x^2])/(5*a*x^5) + ((4*A*b - 5*a*B)*Sqrt[a + b*x^2])/(15*a^2*x^3) - (2*b*(4*A*b - 5*a*B)*Sqrt[a
+ b*x^2])/(15*a^3*x)

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -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 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^6 \sqrt {a+b x^2}} \, dx &=-\frac {A \sqrt {a+b x^2}}{5 a x^5}-\frac {(4 A b-5 a B) \int \frac {1}{x^4 \sqrt {a+b x^2}} \, dx}{5 a}\\ &=-\frac {A \sqrt {a+b x^2}}{5 a x^5}+\frac {(4 A b-5 a B) \sqrt {a+b x^2}}{15 a^2 x^3}+\frac {(2 b (4 A b-5 a B)) \int \frac {1}{x^2 \sqrt {a+b x^2}} \, dx}{15 a^2}\\ &=-\frac {A \sqrt {a+b x^2}}{5 a x^5}+\frac {(4 A b-5 a B) \sqrt {a+b x^2}}{15 a^2 x^3}-\frac {2 b (4 A b-5 a B) \sqrt {a+b x^2}}{15 a^3 x}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 62, normalized size = 0.74 \begin {gather*} -\frac {\sqrt {a+b x^2} \left (a^2 \left (3 A+5 B x^2\right )-2 a b x^2 \left (2 A+5 B x^2\right )+8 A b^2 x^4\right )}{15 a^3 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^6*Sqrt[a + b*x^2]),x]

[Out]

-1/15*(Sqrt[a + b*x^2]*(8*A*b^2*x^4 - 2*a*b*x^2*(2*A + 5*B*x^2) + a^2*(3*A + 5*B*x^2)))/(a^3*x^5)

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IntegrateAlgebraic [A]  time = 0.12, size = 62, normalized size = 0.74 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-3 a^2 A-5 a^2 B x^2+4 a A b x^2+10 a b B x^4-8 A b^2 x^4\right )}{15 a^3 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x^2)/(x^6*Sqrt[a + b*x^2]),x]

[Out]

(Sqrt[a + b*x^2]*(-3*a^2*A + 4*a*A*b*x^2 - 5*a^2*B*x^2 - 8*A*b^2*x^4 + 10*a*b*B*x^4))/(15*a^3*x^5)

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fricas [A]  time = 1.16, size = 58, normalized size = 0.69 \begin {gather*} \frac {{\left (2 \, {\left (5 \, B a b - 4 \, A b^{2}\right )} x^{4} - 3 \, A a^{2} - {\left (5 \, B a^{2} - 4 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, a^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/15*(2*(5*B*a*b - 4*A*b^2)*x^4 - 3*A*a^2 - (5*B*a^2 - 4*A*a*b)*x^2)*sqrt(b*x^2 + a)/(a^3*x^5)

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giac [B]  time = 0.42, size = 176, normalized size = 2.10 \begin {gather*} \frac {4 \, {\left (15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B b^{\frac {3}{2}} - 35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a b^{\frac {3}{2}} + 40 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A b^{\frac {5}{2}} + 25 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} b^{\frac {3}{2}} - 20 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a b^{\frac {5}{2}} - 5 \, B a^{3} b^{\frac {3}{2}} + 4 \, A a^{2} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

4/15*(15*(sqrt(b)*x - sqrt(b*x^2 + a))^6*B*b^(3/2) - 35*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a*b^(3/2) + 40*(sqrt
(b)*x - sqrt(b*x^2 + a))^4*A*b^(5/2) + 25*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^2*b^(3/2) - 20*(sqrt(b)*x - sqrt
(b*x^2 + a))^2*A*a*b^(5/2) - 5*B*a^3*b^(3/2) + 4*A*a^2*b^(5/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5

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maple [A]  time = 0.01, size = 59, normalized size = 0.70 \begin {gather*} -\frac {\sqrt {b \,x^{2}+a}\, \left (8 A \,b^{2} x^{4}-10 B a b \,x^{4}-4 A a b \,x^{2}+5 B \,a^{2} x^{2}+3 a^{2} A \right )}{15 a^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^6/(b*x^2+a)^(1/2),x)

[Out]

-1/15*(b*x^2+a)^(1/2)*(8*A*b^2*x^4-10*B*a*b*x^4-4*A*a*b*x^2+5*B*a^2*x^2+3*A*a^2)/x^5/a^3

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maxima [A]  time = 1.02, size = 96, normalized size = 1.14 \begin {gather*} \frac {2 \, \sqrt {b x^{2} + a} B b}{3 \, a^{2} x} - \frac {8 \, \sqrt {b x^{2} + a} A b^{2}}{15 \, a^{3} x} - \frac {\sqrt {b x^{2} + a} B}{3 \, a x^{3}} + \frac {4 \, \sqrt {b x^{2} + a} A b}{15 \, a^{2} x^{3}} - \frac {\sqrt {b x^{2} + a} A}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

2/3*sqrt(b*x^2 + a)*B*b/(a^2*x) - 8/15*sqrt(b*x^2 + a)*A*b^2/(a^3*x) - 1/3*sqrt(b*x^2 + a)*B/(a*x^3) + 4/15*sq
rt(b*x^2 + a)*A*b/(a^2*x^3) - 1/5*sqrt(b*x^2 + a)*A/(a*x^5)

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mupad [B]  time = 0.68, size = 58, normalized size = 0.69 \begin {gather*} -\frac {\sqrt {b\,x^2+a}\,\left (5\,B\,a^2\,x^2+3\,A\,a^2-10\,B\,a\,b\,x^4-4\,A\,a\,b\,x^2+8\,A\,b^2\,x^4\right )}{15\,a^3\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^6*(a + b*x^2)^(1/2)),x)

[Out]

-((a + b*x^2)^(1/2)*(3*A*a^2 + 5*B*a^2*x^2 + 8*A*b^2*x^4 - 4*A*a*b*x^2 - 10*B*a*b*x^4))/(15*a^3*x^5)

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sympy [B]  time = 3.17, size = 355, normalized size = 4.23 \begin {gather*} - \frac {3 A a^{4} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {2 A a^{3} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {3 A a^{2} b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {12 A a b^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {8 A b^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a x^{2}} + \frac {2 B b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**6/(b*x**2+a)**(1/2),x)

[Out]

-3*A*a**4*b**(9/2)*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 2*A*a**3
*b**(11/2)*x**2*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 3*A*a**2*b*
*(13/2)*x**4*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 12*A*a*b**(15/
2)*x**6*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 8*A*b**(17/2)*x**8*
sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - B*sqrt(b)*sqrt(a/(b*x**2) +
1)/(3*a*x**2) + 2*B*b**(3/2)*sqrt(a/(b*x**2) + 1)/(3*a**2)

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