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

Optimal. Leaf size=134 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a c+b}}\right )}{(a c+b)^{3/2}}-\frac{b}{a (a c+b) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.51397, antiderivative size = 214, normalized size of antiderivative = 1.6, number of steps used = 10, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {6722, 1975, 446, 98, 157, 63, 217, 206, 93, 208} \[ \frac{\sqrt{a \left (c+d x^2\right )+b} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c+d x^2}}{\sqrt{a \left (c+d x^2\right )+b}}\right )}{a^{3/2} \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}-\frac{c^{3/2} \sqrt{a \left (c+d x^2\right )+b} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a \left (c+d x^2\right )+b}}\right )}{(a c+b)^{3/2} \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}-\frac{b}{a (a c+b) \sqrt{a+\frac{b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b/(a*(b + a*c)*Sqrt[a + b/(c + d*x^2)])) + (Sqrt[b + a*(c + d*x^2)]*ArcTanh[(Sqrt[a]*Sqrt[c + d*x^2])/Sqrt[b
+ a*(c + d*x^2)]])/(a^(3/2)*Sqrt[c + d*x^2]*Sqrt[a + b/(c + d*x^2)]) - (c^(3/2)*Sqrt[b + a*(c + d*x^2)]*ArcTa
nh[(Sqrt[b + a*c]*Sqrt[c + d*x^2])/(Sqrt[c]*Sqrt[b + a*(c + d*x^2)])])/((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 98

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

Rule 157

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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

Mathematica [A]  time = 0.472127, size = 110, normalized size = 0.82 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a c+b}}\right )}{(a c+b)^{3/2}}-\frac{b}{a (a c+b) \sqrt{a+\frac{b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.017, size = 1014, normalized size = 7.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.9179, size = 3229, normalized size = 24.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)*sqrt(a)*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2*c + a*b
)*d*x^2 + 8*a*b*c + b^2 + 4*(2*a*d^2*x^4 + (4*a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)
/(d*x^2 + c))) + (a^3*c*d*x^2 + a^3*c^2 + a^2*b*c)*sqrt(c/(a*c + b))*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^2*c^2 + 3*a*b*c + b^2
)*d^2*x^4 + 2*a^2*c^4 + 4*a*b*c^3 + 2*b^2*c^2 + (4*a^2*c^3 + 7*a*b*c^2 + 3*b^2*c)*d*x^2)*sqrt((a*d*x^2 + a*c +
b)/(d*x^2 + c))*sqrt(c/(a*c + b)))/x^4) - 4*(a*b*d*x^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*c
^2 + 2*a^3*b*c + a^2*b^2 + (a^4*c + a^3*b)*d*x^2), -1/4*(2*(a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)*sqr
t(-a)*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c + a
*b)) - (a^3*c*d*x^2 + a^3*c^2 + a^2*b*c)*sqrt(c/(a*c + b))*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^2*c^2 + 3*a*b*c + b^2)*d^2*x^4
+ 2*a^2*c^4 + 4*a*b*c^3 + 2*b^2*c^2 + (4*a^2*c^3 + 7*a*b*c^2 + 3*b^2*c)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2
+ c))*sqrt(c/(a*c + b)))/x^4) + 4*(a*b*d*x^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*c^2 + 2*a^3
*b*c + a^2*b^2 + (a^4*c + a^3*b)*d*x^2), 1/4*(2*(a^3*c*d*x^2 + a^3*c^2 + a^2*b*c)*sqrt(-c/(a*c + b))*arctan(1/
2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))*sqrt(-c/(a*c + b))/(a*c*d*x^2 +
a*c^2 + b*c)) + (a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)*sqrt(a)*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a
^2*c + a*b)*d*x^2 + 8*a*b*c + b^2 + 4*(2*a*d^2*x^4 + (4*a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2
+ a*c + b)/(d*x^2 + c))) - 4*(a*b*d*x^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*c^2 + 2*a^3*b*c +
a^2*b^2 + (a^4*c + a^3*b)*d*x^2), -1/2*((a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)*sqrt(-a)*arctan(1/2*(
2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c + a*b)) - (a^3*c*d*x^
2 + a^3*c^2 + a^2*b*c)*sqrt(-c/(a*c + b))*arctan(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt((a*d*x^2 + a*c
+ b)/(d*x^2 + c))*sqrt(-c/(a*c + b))/(a*c*d*x^2 + a*c^2 + b*c)) + 2*(a*b*d*x^2 + a*b*c)*sqrt((a*d*x^2 + a*c +
b)/(d*x^2 + c)))/(a^4*c^2 + 2*a^3*b*c + a^2*b^2 + (a^4*c + a^3*b)*d*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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