3.348 \(\int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {\left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 x^2 (a c+b)}-\frac {b d \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a c+b}}\right )}{2 \sqrt {c} (a c+b)^{3/2}} \]

[Out]

-1/2*b*d*arctanh(c^(1/2)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(a*c+b)^(1/2))/(a*c+b)^(3/2)/c^(1/2)-1/2*(d*x^2+c)*
((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(a*c+b)/x^2

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Rubi [A]  time = 0.39, antiderivative size = 148, normalized size of antiderivative = 1.37, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6722, 1975, 446, 94, 93, 208} \[ -\frac {a \left (c+d x^2\right )+b}{2 x^2 (a c+b) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {b d \sqrt {a \left (c+d x^2\right )+b} \tanh ^{-1}\left (\frac {\sqrt {a c+b} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a \left (c+d x^2\right )+b}}\right )}{2 \sqrt {c} (a c+b)^{3/2} \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a + b/(c + d*x^2)]),x]

[Out]

-(b + a*(c + d*x^2))/(2*(b + a*c)*x^2*Sqrt[a + b/(c + d*x^2)]) - (b*d*Sqrt[b + a*(c + d*x^2)]*ArcTanh[(Sqrt[b
+ a*c]*Sqrt[c + d*x^2])/(Sqrt[c]*Sqrt[b + a*(c + d*x^2)])])/(2*Sqrt[c]*(b + a*c)^(3/2)*Sqrt[c + d*x^2]*Sqrt[a
+ b/(c + d*x^2)])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 446

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

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^3 \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^3 \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^3 \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 \left (c+d x^2\right )} \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x^2 \sqrt {b+a c+a d x}} \, dx,x,x^2\right )}{2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {b+a \left (c+d x^2\right )}{2 (b+a c) x^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (b d \sqrt {b+a \left (c+d x^2\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x} \sqrt {b+a c+a d x}} \, dx,x,x^2\right )}{4 (b+a c) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {b+a \left (c+d x^2\right )}{2 (b+a c) x^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (b d \sqrt {b+a \left (c+d x^2\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{-c-(-b-a c) x^2} \, dx,x,\frac {\sqrt {c+d x^2}}{\sqrt {b+a \left (c+d x^2\right )}}\right )}{2 (b+a c) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {b+a \left (c+d x^2\right )}{2 (b+a c) x^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {b d \sqrt {b+a \left (c+d x^2\right )} \tanh ^{-1}\left (\frac {\sqrt {b+a c} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {b+a \left (c+d x^2\right )}}\right )}{2 \sqrt {c} (b+a c)^{3/2} \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 210, normalized size = 1.94 \[ -\frac {c \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (2 \sqrt {c (a c+b)} \left (c+d x^2\right ) \left (a c+a d x^2+b\right )-2 b d x^2 \log (x) \sqrt {\left (c+d x^2\right ) \left (a \left (c+d x^2\right )+b\right )}+b d x^2 \sqrt {\left (c+d x^2\right ) \left (a c+a d x^2+b\right )} \log \left (2 \sqrt {c (a c+b)} \sqrt {\left (c+d x^2\right ) \left (a c+a d x^2+b\right )}+2 a c \left (c+d x^2\right )+b \left (2 c+d x^2\right )\right )\right )}{4 x^2 (c (a c+b))^{3/2} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a + b/(c + d*x^2)]),x]

[Out]

-1/4*(c*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(2*Sqrt[c*(b + a*c)]*(c + d*x^2)*(b + a*c + a*d*x^2) - 2*b*d*x^2
*Sqrt[(c + d*x^2)*(b + a*(c + d*x^2))]*Log[x] + b*d*x^2*Sqrt[(c + d*x^2)*(b + a*c + a*d*x^2)]*Log[2*a*c*(c + d
*x^2) + b*(2*c + d*x^2) + 2*Sqrt[c*(b + a*c)]*Sqrt[(c + d*x^2)*(b + a*c + a*d*x^2)]]))/((c*(b + a*c))^(3/2)*x^
2*(b + a*(c + d*x^2)))

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fricas [B]  time = 0.62, size = 451, normalized size = 4.18 \[ \left [\frac {\sqrt {a c^{2} + b c} b d x^{2} \log \left (\frac {{\left (8 \, a^{2} c^{2} + 8 \, a b c + b^{2}\right )} d^{2} x^{4} + 8 \, a^{2} c^{4} + 16 \, a b c^{3} + 8 \, b^{2} c^{2} + 8 \, {\left (2 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c\right )} d x^{2} - 4 \, {\left ({\left (2 \, a c + b\right )} d^{2} x^{4} + 2 \, a c^{3} + {\left (4 \, a c^{2} + 3 \, b c\right )} d x^{2} + 2 \, b c^{2}\right )} \sqrt {a c^{2} + b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{x^{4}}\right ) - 4 \, {\left (a c^{3} + {\left (a c^{2} + b c\right )} d x^{2} + b c^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + b^{2} c\right )} x^{2}}, \frac {\sqrt {-a c^{2} - b c} b d x^{2} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {-a c^{2} - b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + {\left (a^{2} c^{2} + a b c\right )} d x^{2} + b^{2} c\right )}}\right ) - 2 \, {\left (a c^{3} + {\left (a c^{2} + b c\right )} d x^{2} + b c^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + b^{2} c\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b/(d*x^2+c))^(1/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(a*c^2 + b*c)*b*d*x^2*log(((8*a^2*c^2 + 8*a*b*c + b^2)*d^2*x^4 + 8*a^2*c^4 + 16*a*b*c^3 + 8*b^2*c^2
+ 8*(2*a^2*c^3 + 3*a*b*c^2 + b^2*c)*d*x^2 - 4*((2*a*c + b)*d^2*x^4 + 2*a*c^3 + (4*a*c^2 + 3*b*c)*d*x^2 + 2*b*c
^2)*sqrt(a*c^2 + b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/x^4) - 4*(a*c^3 + (a*c^2 + b*c)*d*x^2 + b*c^2)*sq
rt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a^2*c^3 + 2*a*b*c^2 + b^2*c)*x^2), 1/4*(sqrt(-a*c^2 - b*c)*b*d*x^2*arct
an(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt(-a*c^2 - b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*c^3
+ 2*a*b*c^2 + (a^2*c^2 + a*b*c)*d*x^2 + b^2*c)) - 2*(a*c^3 + (a*c^2 + b*c)*d*x^2 + b*c^2)*sqrt((a*d*x^2 + a*c
+ b)/(d*x^2 + c)))/((a^2*c^3 + 2*a*b*c^2 + b^2*c)*x^2)]

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giac [B]  time = 0.70, size = 315, normalized size = 2.92 \[ \frac {b d \arctan \left (-\frac {\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}}{\sqrt {-a c^{2} - b c}}\right )}{2 \, \sqrt {-a c^{2} - b c} {\left (a c \mathrm {sgn}\left (d x^{2} + c\right ) + b \mathrm {sgn}\left (d x^{2} + c\right )\right )}} - \frac {2 \, a^{\frac {3}{2}} c^{2} {\left | d \right |} + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a c d + 2 \, \sqrt {a} b c {\left | d \right |} + {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} b d}{2 \, {\left (a c^{2} - {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} + b c\right )} {\left (a c \mathrm {sgn}\left (d x^{2} + c\right ) + b \mathrm {sgn}\left (d x^{2} + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b/(d*x^2+c))^(1/2),x, algorithm="giac")

[Out]

1/2*b*d*arctan(-(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))/sqrt(-a*c^2 - b*c))/
(sqrt(-a*c^2 - b*c)*(a*c*sgn(d*x^2 + c) + b*sgn(d*x^2 + c))) - 1/2*(2*a^(3/2)*c^2*abs(d) + 2*(sqrt(a*d^2)*x^2
- sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a*c*d + 2*sqrt(a)*b*c*abs(d) + (sqrt(a*d^2)*x^2 - sqr
t(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*b*d)/((a*c^2 - (sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d
*x^2 + b*d*x^2 + a*c^2 + b*c))^2 + b*c)*(a*c*sgn(d*x^2 + c) + b*sgn(d*x^2 + c)))

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maple [B]  time = 0.04, size = 452, normalized size = 4.19 \[ -\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (a b \,c^{2} d \,x^{2} \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 a \,c^{2}+2 b c +2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}{x^{2}}\right )-2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c}\, a \,d^{2} x^{4}+b^{2} c d \,x^{2} \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 a \,c^{2}+2 b c +2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}{x^{2}}\right )-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c}\, a c d \,x^{2}-2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c}\, b d \,x^{2}+2 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c \right )^{\frac {3}{2}} \sqrt {a \,c^{2}+b c}\right )}{4 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \left (a c +b \right )^{2} \sqrt {a \,c^{2}+b c}\, c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b/(d*x^2+c))^(1/2),x)

[Out]

-1/4*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-2*a*d^2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*x^4
*(a*c^2+b*c)^(1/2)+ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^2*a*b*c^2*d-4*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*a*c*d*x^2*(a*c^2+b*c)^(1/
2)+ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/
x^2)*x^2*b^2*c*d-2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*b*d*x^2*(a*c^2+b*c)^(1/2)+2*(a*d^2*x^4+2*a*
c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(1/2))/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)/(a*c+b)^2/c/x^2/(a*c^2+b
*c)^(1/2)

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maxima [A]  time = 2.78, size = 173, normalized size = 1.60 \[ -\frac {b d \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} c^{2} + 2 \, a b c + b^{2} - \frac {{\left (a d x^{2} + a c + b\right )} {\left (a c^{2} + b c\right )}}{d x^{2} + c}\right )}} + \frac {b d \log \left (\frac {c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - \sqrt {{\left (a c + b\right )} c}}{c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} + \sqrt {{\left (a c + b\right )} c}}\right )}{4 \, \sqrt {{\left (a c + b\right )} c} {\left (a c + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b/(d*x^2+c))^(1/2),x, algorithm="maxima")

[Out]

-1/2*b*d*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*c^2 + 2*a*b*c + b^2 - (a*d*x^2 + a*c + b)*(a*c^2 + b*c)/(d
*x^2 + c)) + 1/4*b*d*log((c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)) - sqrt((a*c + b)*c))/(c*sqrt((a*d*x^2 + a*c
+ b)/(d*x^2 + c)) + sqrt((a*c + b)*c)))/(sqrt((a*c + b)*c)*(a*c + b))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^3\,\sqrt {a+\frac {b}{d\,x^2+c}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b/(c + d*x^2))^(1/2)),x)

[Out]

int(1/(x^3*(a + b/(c + d*x^2))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(a+b/(d*x**2+c))**(1/2),x)

[Out]

Integral(1/(x**3*sqrt((a*c + a*d*x**2 + b)/(c + d*x**2))), x)

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