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]

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

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Rubi [A]  time = 0.387278, 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 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]

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 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 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 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 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]

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*}

Mathematica [A]  time = 0.213945, size = 158, normalized size = 1.46 \[ -\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (\sqrt{c} \sqrt{a c+b} \left (c+d x^2\right ) \sqrt{a \left (c+d x^2\right )+b}+b d x^2 \sqrt{c+d x^2} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a c+a d x^2+b}}\right )\right )}{2 \sqrt{c} x^2 (a c+b)^{3/2} \sqrt{a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.013, size = 452, normalized size = 4.2 \begin{align*} -{\frac{d{x}^{2}+c}{4\, \left ( ac+b \right ) ^{2}c{x}^{2}}\sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}} \left ( -2\,a{d}^{2}\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}{x}^{4}\sqrt{{c}^{2}a+bc}+\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,acd{x}^{2}+bd{x}^{2}+2\,{c}^{2}a+2\,\sqrt{{c}^{2}a+bc}\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}+2\,bc \right ) } \right ){x}^{2}ab{c}^{2}d-4\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}acd{x}^{2}\sqrt{{c}^{2}a+bc}+\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,acd{x}^{2}+bd{x}^{2}+2\,{c}^{2}a+2\,\sqrt{{c}^{2}a+bc}\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}+2\,bc \right ) } \right ){x}^{2}{b}^{2}cd-2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}bd{x}^{2}\sqrt{{c}^{2}a+bc}+2\, \left ( a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc \right ) ^{3/2}\sqrt{{c}^{2}a+bc} \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }}}{\frac{1}{\sqrt{{c}^{2}a+bc}}}} \end{align*}

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*c^2*a+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)+2*b*c)/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*c^2*a+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)+2*b*c)/
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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.76838, size = 972, normalized size = 9. \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{\frac{a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + \frac{b}{d x^{2} + c}} x^{3}}\,{d x} \end{align*}

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]

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