\(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{x^6} \, dx\) [515]

Optimal result

Integrand size = 22, antiderivative size = 53 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^6} \, dx=-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^5}+\frac {(2 A b-5 a B) \left (a+b x^2\right )^{3/2}}{15 a^2 x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {464, 270} \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^6} \, dx=\frac {\left (a+b x^2\right )^{3/2} (2 A b-5 a B)}{15 a^2 x^3}-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^5} \]

[In]

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

[Out]

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

Rule 270

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 464

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*} \text {integral}& = -\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^5}-\frac {(2 A b-5 a B) \int \frac {\sqrt {a+b x^2}}{x^4} \, dx}{5 a} \\ & = -\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^5}+\frac {(2 A b-5 a B) \left (a+b x^2\right )^{3/2}}{15 a^2 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^6} \, dx=\frac {\left (a+b x^2\right )^{3/2} \left (-3 a A+2 A b x^2-5 a B x^2\right )}{15 a^2 x^5} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.74 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(-\frac {\left (\left (\frac {5 x^{2} B}{3}+A \right ) a -\frac {2 A b \,x^{2}}{3}\right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 x^{5} a^{2}}\)	\(36\)
gosper	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (-2 A b \,x^{2}+5 B a \,x^{2}+3 A a \right )}{15 x^{5} a^{2}}\)	\(37\)
default	\(-\frac {B \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 a \,x^{3}}+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 a \,x^{5}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 a^{2} x^{3}}\right )\)	\(58\)
trager	\(-\frac {\left (-2 A \,b^{2} x^{4}+5 B a b \,x^{4}+a A b \,x^{2}+5 a^{2} B \,x^{2}+3 a^{2} A \right ) \sqrt {b \,x^{2}+a}}{15 x^{5} a^{2}}\)	\(58\)
risch	\(-\frac {\left (-2 A \,b^{2} x^{4}+5 B a b \,x^{4}+a A b \,x^{2}+5 a^{2} B \,x^{2}+3 a^{2} A \right ) \sqrt {b \,x^{2}+a}}{15 x^{5} a^{2}}\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^6} \, dx=-\frac {{\left ({\left (5 \, B a b - 2 \, A b^{2}\right )} x^{4} + 3 \, A a^{2} + {\left (5 \, B a^{2} + A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, a^{2} x^{5}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (46) = 92\).

Time = 1.14 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.25 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^6} \, dx=- \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a x^{2}} + \frac {2 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{2}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {B b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^6} \, dx=-\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{3 \, a x^{3}} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{5 \, a x^{5}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (45) = 90\).

Time = 0.30 (sec) , antiderivative size = 232, normalized size of antiderivative = 4.38 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^6} \, dx=\frac {2 \, {\left (15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B b^{\frac {3}{2}} - 30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a b^{\frac {3}{2}} + 30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} A b^{\frac {5}{2}} + 20 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{2} b^{\frac {3}{2}} + 10 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a b^{\frac {5}{2}} - 10 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{3} b^{\frac {3}{2}} + 10 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{2} b^{\frac {5}{2}} + 5 \, B a^{4} b^{\frac {3}{2}} - 2 \, A a^{3} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \]

[In]

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

[Out]

2/15*(15*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*b^(3/2) - 30*(sqrt(b)*x - sqrt(b*x^2 + a))^6*B*a*b^(3/2) + 30*(sqrt
(b)*x - sqrt(b*x^2 + a))^6*A*b^(5/2) + 20*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^2*b^(3/2) + 10*(sqrt(b)*x - sqrt
(b*x^2 + a))^4*A*a*b^(5/2) - 10*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^3*b^(3/2) + 10*(sqrt(b)*x - sqrt(b*x^2 + a
))^2*A*a^2*b^(5/2) + 5*B*a^4*b^(3/2) - 2*A*a^3*b^(5/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5

Mupad [B] (verification not implemented)

Time = 5.41 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.83 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^6} \, dx=\frac {\left (A\,b^2+B\,a\,b\right )\,\sqrt {b\,x^2+a}}{5\,a^2\,x}-\frac {\left (5\,B\,a^2+A\,b\,a\right )\,\sqrt {b\,x^2+a}}{15\,a^2\,x^3}-\frac {A\,\sqrt {b\,x^2+a}}{5\,x^5}-\frac {b\,\sqrt {b\,x^2+a}\,\left (A\,b+8\,B\,a\right )}{15\,a^2\,x} \]

[In]

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

[Out]

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