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

Optimal. Leaf size=126 \[ a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a}}\right )-\frac{(a c+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a c+b}}\right )}{c^{3/2}}+\frac{b \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{c} \]

[Out]

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

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Rubi [A]  time = 0.490115, antiderivative size = 206, normalized size of antiderivative = 1.63, 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} \[ \frac{a^{3/2} \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c+d x^2}}{\sqrt{a \left (c+d x^2\right )+b}}\right )}{\sqrt{a \left (c+d x^2\right )+b}}-\frac{(a c+b)^{3/2} \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a \left (c+d x^2\right )+b}}\right )}{c^{3/2} \sqrt{a \left (c+d x^2\right )+b}}+\frac{b \sqrt{a+\frac{b}{c+d x^2}}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2)/x,x]

[Out]

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

Mathematica [A]  time = 0.233806, size = 118, normalized size = 0.94 \[ \frac{a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )+b \sqrt{c} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}-(a c+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a c+b}}\right )}{c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.016, size = 652, normalized size = 5.2 \begin{align*}{\frac{1}{2\,{c}^{2}} \left ( \ln \left ({\frac{1}{2} \left ( 2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd \right ){\frac{1}{\sqrt{a{d}^{2}}}}} \right ){x}^{2}{a}^{2}{c}^{2}{d}^{2}-\sqrt{a{d}^{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}acd+\ln \left ({\frac{1}{2} \left ( 2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd \right ){\frac{1}{\sqrt{a{d}^{2}}}}} \right ){a}^{2}{c}^{3}d-\sqrt{a{d}^{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}bd-\sqrt{a{d}^{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 ) a{c}^{2}-\sqrt{a{d}^{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 ) bc+2\,\sqrt{a{d}^{2}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }bc \right ) \sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}}{\frac{1}{\sqrt{a{d}^{2}}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.82807, size = 2414, normalized size = 19.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(a^(3/2)*c*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*c + b)*sqrt((a*c + b)/c)*l
og(((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^2 + b*c)*d^2*x^4 + 2*a*c^4 + 2*b*c^3 + (4*a*c^3 + 3*b*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b
)/(d*x^2 + c))*sqrt((a*c + b)/c))/x^4) + 4*b*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/c, -1/4*(2*sqrt(-a)*a*c*ar
ctan(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*c + b)*sqrt((a*c + b)/c)*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^2 + b*c)*d^2*x^4 + 2*a*c^4 + 2*b*c^3 + (4*a*c^3 + 3*b*c^2)*d*
x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))*sqrt((a*c + b)/c))/x^4) - 4*b*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))
)/c, 1/4*(a^(3/2)*c*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))) + 2*(a*c + b)*sqrt(-(a*c +
b)/c)*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(-(a*c + b)/c
)/(a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)) + 4*b*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/c, -1/2*(sqrt(
-a)*a*c*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*c + b)*sqrt(-(a*c + b)/c)*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(-(a*c + b)/c)/(a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)) - 2*b*sqrt((a*d*x^2 + a*c +
b)/(d*x^2 + c)))/c]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((a+b/(d*x^2+c))^(3/2)/x,x, algorithm="giac")

[Out]

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