3.353 \(\int \frac {1}{x^2 \sqrt {a+\frac {b}{c+d x^2}}} \, dx\)

Optimal. Leaf size=343 \[ -\frac {a c+a d x^2+b}{x (a c+b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}+\frac {d x \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}+\frac {\sqrt {c} \sqrt {d} \left (a c+a d x^2+b\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{(a c+b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {\sqrt {c} \sqrt {d} \left (a c+a d x^2+b\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{(a c+b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}} \]

[Out]

(-a*d*x^2-a*c-b)/(a*c+b)/x/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)+d*x*(a*d*x^2+a*c+b)/(a*c+b)/(d*x^2+c)/((a*d*x^2+a
*c+b)/(d*x^2+c))^(1/2)-(a*d*x^2+a*c+b)*(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))*c^(1/2)*d^(1/2)/(a*c+b)/(d*x^2+c)/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(c*(a*d*
x^2+a*c+b)/(a*c+b)/(d*x^2+c))^(1/2)+(a*d*x^2+a*c+b)*(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))*c^(1/2)*d^(1/2)/(a*c+b)/(d*x^2+c)/((a*d*x^2+a*c+b)/(d*x^2+c))^(
1/2)/(c*(a*d*x^2+a*c+b)/(a*c+b)/(d*x^2+c))^(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 387, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6722, 1975, 475, 21, 422, 418, 492, 411} \[ \frac {d x \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{(a c+b) \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{x (a c+b) \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\sqrt {c} \sqrt {d} \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{(a c+b) \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\sqrt {c} \sqrt {d} \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{(a c+b) \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[a + b/(c + d*x^2)]),x]

[Out]

-((Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/((b + a*c)*x*Sqrt[a + b/(c + d*x^2)])) + (d*x*Sqrt[b + a*c
+ a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/((b + a*c)*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)]) - (Sqrt[c]*Sqrt[d]*Sqrt[b
+ a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/((b + a*c)*(c +
d*x^2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[a + b/(c + d*x^2)]) + (Sqrt[c]*Sqrt[d]*Sqrt
[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/((b + a*c)*(c
+ d*x^2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[a + b/(c + d*x^2)])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 422

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

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

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Mathematica [A]  time = 0.18, size = 151, normalized size = 0.44 \[ \frac {d \sqrt {\frac {c+d x^2}{c}} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} E\left (\sin ^{-1}\left (\sqrt {-\frac {d}{c}} x\right )|\frac {a c}{b+a c}\right )}{\sqrt {-\frac {d}{c}} (a c+b) \sqrt {\frac {a c+a d x^2+b}{a c+b}}}-\frac {\left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{x (a c+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[a + b/(c + d*x^2)]),x]

[Out]

-(((c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/((b + a*c)*x)) + (d*Sqrt[(c + d*x^2)/c]*Sqrt[(b + a*c +
a*d*x^2)/(c + d*x^2)]*EllipticE[ArcSin[Sqrt[-(d/c)]*x], (a*c)/(b + a*c)])/((b + a*c)*Sqrt[-(d/c)]*Sqrt[(b + a*
c + a*d*x^2)/(b + a*c)])

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d x^{2} + c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{a d x^{4} + {\left (a c + b\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x^2 + c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a*d*x^4 + (a*c + b)*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + \frac {b}{d x^{2} + c}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 345, normalized size = 1.01 \[ -\frac {\left (\sqrt {-\frac {a d}{a c +b}}\, a \,d^{2} x^{4}+2 \sqrt {-\frac {a d}{a c +b}}\, a c d \,x^{2}-\sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a c d x \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+\sqrt {-\frac {a d}{a c +b}}\, b d \,x^{2}-\sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b d x \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+\sqrt {-\frac {a d}{a c +b}}\, a \,c^{2}+\sqrt {-\frac {a d}{a c +b}}\, b c \right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \left (a c +b \right ) \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((-1/(a*c+b)*a*d)^(1/2)*a*d^2*x^4-a*d*c*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*x*EllipticE((-1/(
a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))+2*(-1/(a*c+b)*a*d)^(1/2)*a*c*d*x^2-((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*b*d+(-1/(a*c+b)*a*d)^(1/2)*b*d*x^2
+(-1/(a*c+b)*a*d)^(1/2)*a*c^2+(-1/(a*c+b)*a*d)^(1/2)*b*c)*(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)/x/(a*c+b)/((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 \[ \int \frac {1}{\sqrt {a + \frac {b}{d x^{2} + c}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*sqrt((a*c + a*d*x**2 + b)/(c + d*x**2))), x)

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