3.330 \(\int \frac {\sqrt {a+\frac {b}{c+d x^2}}}{x^6} \, dx\)

Optimal. Leaf size=466 \[ -\frac {d^{5/2} \left (3 a^2 c^2+13 a b c+8 b^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 c^{5/2} (a c+b)^2 \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {d^3 x \left (3 a^2 c^2+13 a b c+8 b^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{15 c^3 (a c+b)^2}-\frac {d^2 \left (3 a^2 c^2+13 a b c+8 b^2\right ) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{15 c^3 x (a c+b)^2}+\frac {a d^{5/2} (3 a c+4 b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 c^{3/2} (a c+b)^2 \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {d (3 a c+4 b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{15 c^2 x^3 (a c+b)}-\frac {\left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{5 c x^5} \]

[Out]

1/15*(3*a^2*c^2+13*a*b*c+8*b^2)*d^3*x*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^3/(a*c+b)^2-1/5*(d*x^2+c)*((a*d*x^2+
a*c+b)/(d*x^2+c))^(1/2)/c/x^5+1/15*(3*a*c+4*b)*d*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^2/(a*c+b)/x^3-1
/15*(3*a^2*c^2+13*a*b*c+8*b^2)*d^2*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^3/(a*c+b)^2/x-1/15*(3*a^2*c^2
+13*a*b*c+8*b^2)*d^(5/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2)
,(b/(a*c+b))^(1/2))*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^(5/2)/(a*c+b)^2/(c*(a*d*x^2+a*c+b)/(a*c+b)/(d*x^2+c))^
(1/2)+1/15*a*(3*a*c+4*b)*d^(5/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/
c)^(1/2),(b/(a*c+b))^(1/2))*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^(3/2)/(a*c+b)^2/(c*(a*d*x^2+a*c+b)/(a*c+b)/(d*
x^2+c))^(1/2)

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Rubi [A]  time = 0.81, antiderivative size = 598, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6722, 1975, 475, 583, 531, 418, 492, 411} \[ \frac {d^3 x \left (3 a^2 c^2+13 a b c+8 b^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^3 (a c+b)^2 \sqrt {a \left (c+d x^2\right )+b}}-\frac {d^2 \left (3 a^2 c^2+13 a b c+8 b^2\right ) \left (c+d x^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^3 x (a c+b)^2 \sqrt {a \left (c+d x^2\right )+b}}-\frac {d^{5/2} \left (3 a^2 c^2+13 a b c+8 b^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 c^{5/2} (a c+b)^2 \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a \left (c+d x^2\right )+b}}+\frac {a d^{5/2} (3 a c+4 b) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 c^{3/2} (a c+b)^2 \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a \left (c+d x^2\right )+b}}+\frac {d (3 a c+4 b) \left (c+d x^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^2 x^3 (a c+b) \sqrt {a \left (c+d x^2\right )+b}}-\frac {\left (c+d x^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{5 c x^5 \sqrt {a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/(c + d*x^2)]/x^6,x]

[Out]

((8*b^2 + 13*a*b*c + 3*a^2*c^2)*d^3*x*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(15*c^3*(b + a*c)^2*Sqr
t[b + a*(c + d*x^2)]) - ((c + d*x^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(5*c*x^5*Sqrt[b + a*(c +
d*x^2)]) + ((4*b + 3*a*c)*d*(c + d*x^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(15*c^2*(b + a*c)*x^
3*Sqrt[b + a*(c + d*x^2)]) - ((8*b^2 + 13*a*b*c + 3*a^2*c^2)*d^2*(c + d*x^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a +
b/(c + d*x^2)])/(15*c^3*(b + a*c)^2*x*Sqrt[b + a*(c + d*x^2)]) - ((8*b^2 + 13*a*b*c + 3*a^2*c^2)*d^(5/2)*Sqrt[
b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(15*c^(5/2)*(b
+ a*c)^2*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[b + a*(c + d*x^2)]) + (a*(4*b + 3*a*c)*d^
(5/2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(15
*c^(3/2)*(b + a*c)^2*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[b + a*(c + d*x^2)])

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 475

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+\frac {b}{c+d x^2}}}{x^6} \, dx &=\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {\sqrt {b+a \left (c+d x^2\right )}}{x^6 \sqrt {c+d x^2}} \, dx}{\sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {\sqrt {b+a c+a d x^2}}{x^6 \sqrt {c+d x^2}} \, dx}{\sqrt {b+a \left (c+d x^2\right )}}\\ &=-\frac {\left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{5 c x^5 \sqrt {b+a \left (c+d x^2\right )}}+\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {-(4 b+3 a c) d-3 a d^2 x^2}{x^4 \sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{5 c \sqrt {b+a \left (c+d x^2\right )}}\\ &=-\frac {\left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{5 c x^5 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(4 b+3 a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^2 (b+a c) x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {-\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^2-a (4 b+3 a c) d^3 x^2}{x^2 \sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{15 c^2 (b+a c) \sqrt {b+a \left (c+d x^2\right )}}\\ &=-\frac {\left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{5 c x^5 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(4 b+3 a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^2 (b+a c) x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^2 \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^3 (b+a c)^2 x \sqrt {b+a \left (c+d x^2\right )}}+\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {a c (b+a c) (4 b+3 a c) d^3+a \left (8 b^2+13 a b c+3 a^2 c^2\right ) d^4 x^2}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{15 c^3 (b+a c)^2 \sqrt {b+a \left (c+d x^2\right )}}\\ &=-\frac {\left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{5 c x^5 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(4 b+3 a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^2 (b+a c) x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^2 \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^3 (b+a c)^2 x \sqrt {b+a \left (c+d x^2\right )}}+\frac {\left (a (4 b+3 a c) d^3 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {1}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{15 c^2 (b+a c) \sqrt {b+a \left (c+d x^2\right )}}+\frac {\left (a \left (8 b^2+13 a b c+3 a^2 c^2\right ) d^4 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{15 c^3 (b+a c)^2 \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^3 x \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^3 (b+a c)^2 \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{5 c x^5 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(4 b+3 a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^2 (b+a c) x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^2 \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^3 (b+a c)^2 x \sqrt {b+a \left (c+d x^2\right )}}+\frac {a (4 b+3 a c) d^{5/2} \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 c^{3/2} (b+a c)^2 \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^3 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {\sqrt {b+a c+a d x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{15 c^2 (b+a c)^2 \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^3 x \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^3 (b+a c)^2 \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{5 c x^5 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(4 b+3 a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^2 (b+a c) x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^2 \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{15 c^3 (b+a c)^2 x \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^{5/2} \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 c^{5/2} (b+a c)^2 \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {b+a \left (c+d x^2\right )}}+\frac {a (4 b+3 a c) d^{5/2} \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 c^{3/2} (b+a c)^2 \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {b+a \left (c+d x^2\right )}}\\ \end {align*}

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Mathematica [C]  time = 1.09, size = 402, normalized size = 0.86 \[ -\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (i a c d^3 x^5 \left (3 a^2 c^2+13 a b c+8 b^2\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )+\left (c+d x^2\right ) \sqrt {\frac {a d}{a c+b}} \left (3 a^3 c^2 \left (c^3+d^3 x^6\right )+a^2 b c \left (9 c^3-4 c^2 d x^2+9 c d^2 x^4+13 d^3 x^6\right )+a b^2 \left (9 c^3-8 c^2 d x^2+17 c d^2 x^4+8 d^3 x^6\right )+b^3 \left (3 c^2-4 c d x^2+8 d^2 x^4\right )\right )-2 i a b c d^3 x^5 (3 a c+2 b) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )\right )}{15 c^3 x^5 (a c+b)^2 \sqrt {\frac {a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(Sqrt[(a*d)/(b + a*c)]*(c + d*x^2)*(b^3*(3*c^2 - 4*c*d*x^2 + 8*d^
2*x^4) + 3*a^3*c^2*(c^3 + d^3*x^6) + a*b^2*(9*c^3 - 8*c^2*d*x^2 + 17*c*d^2*x^4 + 8*d^3*x^6) + a^2*b*c*(9*c^3 -
4*c^2*d*x^2 + 9*c*d^2*x^4 + 13*d^3*x^6)) + I*a*c*(8*b^2 + 13*a*b*c + 3*a^2*c^2)*d^3*x^5*Sqrt[(b + a*c + a*d*x
^2)/(b + a*c)]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[(a*d)/(b + a*c)]*x], 1 + b/(a*c)] - (2*I)*a*b*c*(2
*b + 3*a*c)*d^3*x^5*Sqrt[(b + a*c + a*d*x^2)/(b + a*c)]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[(a*d)/(b
+ a*c)]*x], 1 + b/(a*c)]))/(c^3*(b + a*c)^2*Sqrt[(a*d)/(b + a*c)]*x^5*(b + a*(c + d*x^2)))

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/x^6, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{d x^{2} + c}}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/(d*x^2 + c))/x^6, x)

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maple [A]  time = 0.05, size = 955, normalized size = 2.05 \[ -\frac {\left (3 \sqrt {-\frac {a d}{a c +b}}\, a^{3} c^{2} d^{4} x^{8}+13 \sqrt {-\frac {a d}{a c +b}}\, a^{2} b c \,d^{4} x^{8}+3 \sqrt {-\frac {a d}{a c +b}}\, a^{3} c^{3} d^{3} x^{6}-3 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{3} c^{3} d^{3} x^{5} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+8 \sqrt {-\frac {a d}{a c +b}}\, a \,b^{2} d^{4} x^{8}+22 \sqrt {-\frac {a d}{a c +b}}\, a^{2} b \,c^{2} d^{3} x^{6}-13 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} b \,c^{2} d^{3} x^{5} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+6 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} b \,c^{2} d^{3} x^{5} \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+25 \sqrt {-\frac {a d}{a c +b}}\, a \,b^{2} c \,d^{3} x^{6}-8 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a \,b^{2} c \,d^{3} x^{5} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+4 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a \,b^{2} c \,d^{3} x^{5} \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+5 \sqrt {-\frac {a d}{a c +b}}\, a^{2} b \,c^{3} d^{2} x^{4}+8 \sqrt {-\frac {a d}{a c +b}}\, b^{3} d^{3} x^{6}+3 \sqrt {-\frac {a d}{a c +b}}\, a^{3} c^{5} d \,x^{2}+9 \sqrt {-\frac {a d}{a c +b}}\, a \,b^{2} c^{2} d^{2} x^{4}+5 \sqrt {-\frac {a d}{a c +b}}\, a^{2} b \,c^{4} d \,x^{2}+4 \sqrt {-\frac {a d}{a c +b}}\, b^{3} c \,d^{2} x^{4}+3 \sqrt {-\frac {a d}{a c +b}}\, a^{3} c^{6}+\sqrt {-\frac {a d}{a c +b}}\, a \,b^{2} c^{3} d \,x^{2}+9 \sqrt {-\frac {a d}{a c +b}}\, a^{2} b \,c^{5}-\sqrt {-\frac {a d}{a c +b}}\, b^{3} c^{2} d \,x^{2}+9 \sqrt {-\frac {a d}{a c +b}}\, a \,b^{2} c^{4}+3 \sqrt {-\frac {a d}{a c +b}}\, b^{3} c^{3}\right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{15 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \left (a c +b \right )^{2} \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, c^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/(d*x^2+c))^(1/2)/x^6,x)

[Out]

-1/15*(3*(-1/(a*c+b)*a*d)^(1/2)*x^8*a^3*c^2*d^4+13*(-1/(a*c+b)*a*d)^(1/2)*x^8*a^2*b*c*d^4-3*((a*d*x^2+a*c+b)/(
a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*x^5*a^3*c^3*d^3+8*(-
1/(a*c+b)*a*d)^(1/2)*x^8*a*b^2*d^4+3*(-1/(a*c+b)*a*d)^(1/2)*x^6*a^3*c^3*d^3+6*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*
((d*x^2+c)/c)^(1/2)*EllipticF((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*x^5*a^2*b*c^2*d^3-13*((a*d*x^2+a*c
+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*x^5*a^2*b*c^2*d
^3+22*(-1/(a*c+b)*a*d)^(1/2)*x^6*a^2*b*c^2*d^3+4*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF
((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*x^5*a*b^2*c*d^3-8*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)
^(1/2)*EllipticE((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*x^5*a*b^2*c*d^3+25*(-1/(a*c+b)*a*d)^(1/2)*x^6*a
*b^2*c*d^3+8*(-1/(a*c+b)*a*d)^(1/2)*x^6*b^3*d^3+5*(-1/(a*c+b)*a*d)^(1/2)*x^4*a^2*b*c^3*d^2+9*(-1/(a*c+b)*a*d)^
(1/2)*x^4*a*b^2*c^2*d^2+3*(-1/(a*c+b)*a*d)^(1/2)*x^2*a^3*c^5*d+4*(-1/(a*c+b)*a*d)^(1/2)*x^4*b^3*c*d^2+5*(-1/(a
*c+b)*a*d)^(1/2)*x^2*a^2*b*c^4*d+(-1/(a*c+b)*a*d)^(1/2)*x^2*a*b^2*c^3*d+3*(-1/(a*c+b)*a*d)^(1/2)*a^3*c^6-(-1/(
a*c+b)*a*d)^(1/2)*x^2*b^3*c^2*d+9*(-1/(a*c+b)*a*d)^(1/2)*a^2*b*c^5+9*(-1/(a*c+b)*a*d)^(1/2)*a*b^2*c^4+3*(-1/(a
*c+b)*a*d)^(1/2)*b^3*c^3)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c
)^(1/2)/(-1/(a*c+b)*a*d)^(1/2)/(a*c+b)^2/x^5/c^3/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{d x^{2} + c}}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/(d*x^2 + c))/x^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a+\frac {b}{d\,x^2+c}}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/(c + d*x^2))^(1/2)/x^6,x)

[Out]

int((a + b/(c + d*x^2))^(1/2)/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/(d*x**2+c))**(1/2)/x**6,x)

[Out]

Integral(sqrt((a*c + a*d*x**2 + b)/(c + d*x**2))/x**6, x)

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