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

Optimal. Leaf size=205 \[ \frac{b d^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{c^3}-\frac{3 b d^2 (4 a c+5 b) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a c+b}}\right )}{8 c^{7/2} \sqrt{a c+b}}+\frac{d (4 a c+9 b) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{8 c^3 x^2}-\frac{(a c+b) \left (c+d x^2\right )^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{4 c^3 x^4} \]

[Out]

(b*d^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/c^3 + ((9*b + 4*a*c)*d*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c +
d*x^2)])/(8*c^3*x^2) - ((b + a*c)*(c + d*x^2)^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*c^3*x^4) - (3*b*(5*
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*c^(7/2)*Sqrt[b + a*c
])

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Rubi [A]  time = 0.591398, antiderivative size = 260, normalized size of antiderivative = 1.27, 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} \[ \frac{3 b d^2 (4 a c+5 b) \sqrt{a+\frac{b}{c+d x^2}}}{8 c^3 (a c+b)}-\frac{3 b d^2 (4 a c+5 b) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a \left (c+d x^2\right )+b}}\right )}{8 c^{7/2} \sqrt{a c+b} \sqrt{a \left (c+d x^2\right )+b}}+\frac{d (4 a c+5 b) \sqrt{a+\frac{b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )}{8 c^2 x^2 (a c+b)}-\frac{\sqrt{a+\frac{b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )^2}{4 c x^4 (a c+b)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2)/x^5,x]

[Out]

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

Mathematica [A]  time = 0.28848, size = 202, normalized size = 0.99 \[ -\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (\sqrt{c} \sqrt{a c+b} \sqrt{a \left (c+d x^2\right )+b} \left (2 a c \left (c^2-d^2 x^4\right )+b \left (2 c^2-5 c d x^2-15 d^2 x^4\right )\right )+3 b d^2 x^4 (4 a c+5 b) \sqrt{c+d x^2} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a c+a d x^2+b}}\right )\right )}{8 c^{7/2} x^4 \sqrt{a c+b} \sqrt{a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(Sqrt[c]*Sqrt[b + a*c]*Sqrt[b + a*(c + d*x^2)]*(b*(2*c^2 - 5*c*d*x^2 -
15*d^2*x^4) + 2*a*c*(c^2 - d^2*x^4)) + 3*b*(5*b + 4*a*c)*d^2*x^4*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])]))/(8*c^(7/2)*Sqrt[b + a*c]*x^4*Sqrt[b + a*(c + d*x^2)])

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Maple [B]  time = 0.018, size = 1653, normalized size = 8.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(-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^8*a^2*c*d^4-12*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^3*b*c^5*d^3-18*(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*b*d^4-39*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^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^2*
c^2*d^3-42*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^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^3*b*c^6*d^2-62*(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*c*d^3-15*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*b^4*c^2*d^3-39*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^2*c^5*d^2-18*(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*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^2*c^3*d^2-42*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^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*c*d^2-44*(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*c^2*d^2+16*(a*c^2+b*
c)^(3/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^4*a*b*c^2*d^2-15*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^4*c^3*d^2+18*(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*b*d^2-18*(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^2*c*d^2+16*(a*c^2+b*c)^(3/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^4*b^2*c*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*c^2*d+14*(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*c*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*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*c^2)/(a*c^2+b*c)^(3/2)/x^4/(a*c+b)/c^4/((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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.98366, size = 1191, normalized size = 5.81 \begin{align*} \left [\frac{3 \,{\left (4 \, a b c + 5 \, b^{2}\right )} \sqrt{a c^{2} + b c} d^{2} x^{4} \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 (2 \, a^{2} c^{5} -{\left (2 \, a^{2} c^{3} + 17 \, a b c^{2} + 15 \, b^{2} c\right )} d^{2} x^{4} + 4 \, a b c^{4} + 2 \, b^{2} c^{3} - 5 \,{\left (a b c^{3} + b^{2} c^{2}\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{32 \,{\left (a c^{5} + b c^{4}\right )} x^{4}}, \frac{3 \,{\left (4 \, a b c + 5 \, b^{2}\right )} \sqrt{-a c^{2} - b c} d^{2} x^{4} \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 (2 \, a^{2} c^{5} -{\left (2 \, a^{2} c^{3} + 17 \, a b c^{2} + 15 \, b^{2} c\right )} d^{2} x^{4} + 4 \, a b c^{4} + 2 \, b^{2} c^{3} - 5 \,{\left (a b c^{3} + b^{2} c^{2}\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{16 \,{\left (a c^{5} + b c^{4}\right )} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(4*a*b*c + 5*b^2)*sqrt(a*c^2 + b*c)*d^2*x^4*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^2*c^5 - (2*a^2
*c^3 + 17*a*b*c^2 + 15*b^2*c)*d^2*x^4 + 4*a*b*c^4 + 2*b^2*c^3 - 5*(a*b*c^3 + b^2*c^2)*d*x^2)*sqrt((a*d*x^2 + a
*c + b)/(d*x^2 + c)))/((a*c^5 + b*c^4)*x^4), 1/16*(3*(4*a*b*c + 5*b^2)*sqrt(-a*c^2 - b*c)*d^2*x^4*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^2*c^5 - (2*a^2*c^3 + 17*a*b*c^2 + 15*b^2*c)*d^2*x^4 + 4*a*b*
c^4 + 2*b^2*c^3 - 5*(a*b*c^3 + b^2*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a*c^5 + b*c^4)*x^4)]

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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^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(((a*c + a*d*x**2 + b)/(c + d*x**2))**(3/2)/x**5, 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^{5}}\,{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^5,x, algorithm="giac")

[Out]

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