3.360 \(\int \frac{1}{x^5 (a+\frac{b}{c+d x^2})^{3/2}} \, dx\)

Optimal. Leaf size=212 \[ -\frac{a b d^2}{(a c+b)^3 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}-\frac{3 b d^2 (b-4 a c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a c+b}}\right )}{8 \sqrt{c} (a c+b)^{7/2}}-\frac{d (3 b-4 a c) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{8 x^2 (a c+b)^3}-\frac{\left (c+d x^2\right )^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{4 x^4 (a c+b)^2} \]

[Out]

-((a*b*d^2)/((b + a*c)^3*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])) - ((3*b - 4*a*c)*d*(c + d*x^2)*Sqrt[(b + a*c
+ a*d*x^2)/(c + d*x^2)])/(8*(b + a*c)^3*x^2) - ((c + d*x^2)^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*(b + a
*c)^2*x^4) - (3*b*(b - 4*a*c)*d^2*ArcTanh[(Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[b + a*c]])/(8*S
qrt[c]*(b + a*c)^(7/2))

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Rubi [A]  time = 0.584397, antiderivative size = 246, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6722, 1975, 446, 96, 94, 93, 208} \[ \frac{3 b d^2 (b-4 a c)}{8 c (a c+b)^3 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{3 b d^2 (b-4 a c) \sqrt{a \left (c+d x^2\right )+b} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a \left (c+d x^2\right )+b}}\right )}{8 \sqrt{c} (a c+b)^{7/2} \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}-\frac{d (b-4 a c) \left (c+d x^2\right )}{8 c x^2 (a c+b)^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{\left (c+d x^2\right )^2}{4 c x^4 (a c+b) \sqrt{a+\frac{b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b/(c + d*x^2))^(3/2)),x]

[Out]

(3*b*(b - 4*a*c)*d^2)/(8*c*(b + a*c)^3*Sqrt[a + b/(c + d*x^2)]) - ((b - 4*a*c)*d*(c + d*x^2))/(8*c*(b + a*c)^2
*x^2*Sqrt[a + b/(c + d*x^2)]) - (c + d*x^2)^2/(4*c*(b + a*c)*x^4*Sqrt[a + b/(c + d*x^2)]) - (3*b*(b - 4*a*c)*d
^2*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)])])/(8*Sqrt
[c]*(b + a*c)^(7/2)*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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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^5 \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^5 \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^5 \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^3 (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{\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{\left ((b-4 a c) d \sqrt{b+a \left (c+d x^2\right )}\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x^2 (b+a c+a d x)^{3/2}} \, dx,x,x^2\right )}{8 c (b+a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=-\frac{(b-4 a c) d \left (c+d x^2\right )}{8 c (b+a c)^2 x^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{\left (3 b (b-4 a c) d^2 \sqrt{b+a \left (c+d x^2\right )}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x (b+a c+a d x)^{3/2}} \, dx,x,x^2\right )}{16 c (b+a c)^2 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=\frac{3 b (b-4 a c) d^2}{8 c (b+a c)^3 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{(b-4 a c) d \left (c+d x^2\right )}{8 c (b+a c)^2 x^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{\left (3 b (b-4 a c) d^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 )}{16 (b+a c)^3 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=\frac{3 b (b-4 a c) d^2}{8 c (b+a c)^3 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{(b-4 a c) d \left (c+d x^2\right )}{8 c (b+a c)^2 x^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{\left (3 b (b-4 a c) d^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 )}{8 (b+a c)^3 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=\frac{3 b (b-4 a c) d^2}{8 c (b+a c)^3 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{(b-4 a c) d \left (c+d x^2\right )}{8 c (b+a c)^2 x^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{3 b (b-4 a c) d^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 )}{8 \sqrt{c} (b+a c)^{7/2} \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ \end{align*}

Mathematica [A]  time = 0.560356, size = 202, normalized size = 0.95 \[ \frac{1}{8} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (\frac{8 b^2 d^2}{(a c+b)^3 \left (a \left (c+d x^2\right )+b\right )}-\frac{2 c^2}{x^4 (a c+b)^2}-\frac{3 b d^2 (b-4 a c) \sqrt{c+d x^2} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a c+a d x^2+b}}\right )}{\sqrt{c} (a c+b)^{7/2} \sqrt{a \left (c+d x^2\right )+b}}-\frac{d^2 (13 b-2 a c)}{(a c+b)^3}-\frac{7 b c d}{x^2 (a c+b)^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b/(c + d*x^2))^(3/2)),x]

[Out]

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(-(((13*b - 2*a*c)*d^2)/(b + a*c)^3) - (2*c^2)/((b + a*c)^2*x^4) - (7*b
*c*d)/((b + a*c)^3*x^2) + (8*b^2*d^2)/((b + a*c)^3*(b + a*(c + d*x^2))) - (3*b*(b - 4*a*c)*d^2*Sqrt[c + d*x^2]
*ArcTanh[(Sqrt[b + a*c]*Sqrt[c + d*x^2])/(Sqrt[c]*Sqrt[b + a*c + a*d*x^2])])/(Sqrt[c]*(b + a*c)^(7/2)*Sqrt[b +
a*(c + d*x^2)])))/8

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Maple [B]  time = 0.018, size = 1947, normalized size = 9.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(a+b/(d*x^2+c))^(3/2),x)

[Out]

-1/16*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-12*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^6*a^4*b*c^5*d^3-6*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*
c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^8*a^2*b*d^4-21*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^6*a^3*b^2*c^4*d^3+32*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^
2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^6*a^3*c^2*d^3-6*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^6*a^2*b^3*c^3*d^3-12*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^4*a^4*b*c^6*d^2+3*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^6*a*b^4*c^
2*d^3-33*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^4*a^3*b^2*c^5*d^2-12*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^6*a*b^2*
d^3+20*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^4*a^3*c^3*d^2-27*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^4*a^2*b^3*c^4*
d^2-12*(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)^(3/2)*x^4*a^2*c*d^2-3*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^4*a*b^4*c^3*d^2+6
*(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)^(3/2)*x^4*a*b*d^2-8*(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)^(3/2)*x^2*a^2*c^2*d+4*(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)^(3/2)*a^2*c^3+4*(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)^(3/2)*b^2*c+12*(a*d^2*x^4+2*a*c
*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^8*a^3*c*d^4+16*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(
1/2)*(a*c^2+b*c)^(3/2)*x^4*a^2*b*c^2*d^2+16*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(a*c^2+b*c)^(3/2)*x^4*a*b^2*c*d^
2-10*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^4*a*b^2*c*d^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)^(3/2)*x^2*a*b*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*
c^2+b*c)^(3/2)*x^6*a^2*b*c*d^3+16*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(a*c^2+b*c)^(3/2)*x^4*a^2*b*c^2*d^2+3*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^
4*b^5*c^2*d^2-6*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^4*b^3*d^2+6*(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)^(3/2)*x^2*b^2*d+8*(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)^(3/2)*a*b*c^2)/c/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)/(a*c+b)^4/x^4/(a*c^2+b*c)^(3/2)/(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^5/(a+b/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.74492, size = 1997, normalized size = 9.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(a+b/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

[1/32*(3*((4*a^2*b*c - a*b^2)*d^3*x^6 + (4*a^2*b*c^2 + 3*a*b^2*c - b^3)*d^2*x^4)*sqrt(a*c^2 + 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 + 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*((2*a^3*c^3 - 11*a^2*b*c^2 - 13*a*b^2*c)*d^3*x^6 - 2*a^3*c^6 - 6*a^2*b*c^5 - 6*a*b^
2*c^4 + (2*a^3*c^4 - 16*a^2*b*c^3 - 23*a*b^2*c^2 - 5*b^3*c)*d^2*x^4 - 2*b^3*c^3 - (2*a^3*c^5 + 11*a^2*b*c^4 +
16*a*b^2*c^3 + 7*b^3*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a^5*c^5 + 4*a^4*b*c^4 + 6*a^3*b^2*c^
3 + 4*a^2*b^3*c^2 + a*b^4*c)*d*x^6 + (a^5*c^6 + 5*a^4*b*c^5 + 10*a^3*b^2*c^4 + 10*a^2*b^3*c^3 + 5*a*b^4*c^2 +
b^5*c)*x^4), -1/16*(3*((4*a^2*b*c - a*b^2)*d^3*x^6 + (4*a^2*b*c^2 + 3*a*b^2*c - b^3)*d^2*x^4)*sqrt(-a*c^2 - b*
c)*arctan(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*((2*a^3*c^3 - 11*a^2*b*c^2 - 13*a*b^2*c)*d^3*x^6 -
2*a^3*c^6 - 6*a^2*b*c^5 - 6*a*b^2*c^4 + (2*a^3*c^4 - 16*a^2*b*c^3 - 23*a*b^2*c^2 - 5*b^3*c)*d^2*x^4 - 2*b^3*c
^3 - (2*a^3*c^5 + 11*a^2*b*c^4 + 16*a*b^2*c^3 + 7*b^3*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a^5
*c^5 + 4*a^4*b*c^4 + 6*a^3*b^2*c^3 + 4*a^2*b^3*c^2 + a*b^4*c)*d*x^6 + (a^5*c^6 + 5*a^4*b*c^5 + 10*a^3*b^2*c^4
+ 10*a^2*b^3*c^3 + 5*a*b^4*c^2 + b^5*c)*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{5} \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**5/(a+b/(d*x**2+c))**(3/2),x)

[Out]

Integral(1/(x**5*((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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(a+b/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a + b/(d*x^2 + c))^(3/2)*x^5), x)