3.342 \(\int \frac {(a+\frac {b}{c+d x^2})^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=388 \[ -\frac {a d^{3/2} (a c+4 b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 c^{3/2} (a c+b) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {d^{3/2} (a c+8 b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 c^{5/2} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {d^2 x (a c+8 b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{3 c^3}+\frac {d (a c+8 b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{3 c^3 x}-\frac {(a c+4 b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{3 c^2 x^3}+\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{c x^3} \]

[Out]

b*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c/x^3-1/3*(a*c+8*b)*d^2*x*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^3-1/3*(a*c+4
*b)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^2/x^3+1/3*(a*c+8*b)*d*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^
(1/2)/c^3/x+1/3*(a*c+8*b)*d^(3/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2
/c)^(1/2),(b/(a*c+b))^(1/2))*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^(5/2)/(c*(a*d*x^2+a*c+b)/(a*c+b)/(d*x^2+c))^(
1/2)-1/3*a*(a*c+4*b)*d^(3/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(
1/2),(b/(a*c+b))^(1/2))*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^(3/2)/(a*c+b)/(c*(a*d*x^2+a*c+b)/(a*c+b)/(d*x^2+c)
)^(1/2)

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Rubi [A]  time = 0.87, antiderivative size = 520, normalized size of antiderivative = 1.34, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6722, 1975, 468, 583, 531, 418, 492, 411} \[ -\frac {d^2 x (a c+8 b) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^3 \sqrt {a \left (c+d x^2\right )+b}}-\frac {a d^{3/2} (a c+4 b) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 c^{3/2} (a c+b) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a \left (c+d x^2\right )+b}}+\frac {d^{3/2} (a c+8 b) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 c^{5/2} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a \left (c+d x^2\right )+b}}+\frac {d (a c+8 b) \left (c+d x^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^3 x \sqrt {a \left (c+d x^2\right )+b}}-\frac {(a c+4 b) \left (c+d x^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^2 x^3 \sqrt {a \left (c+d x^2\right )+b}}+\frac {b \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{c x^3 \sqrt {a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2)/x^4,x]

[Out]

(b*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(c*x^3*Sqrt[b + a*(c + d*x^2)]) - ((8*b + a*c)*d^2*x*Sqrt[
b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(3*c^3*Sqrt[b + a*(c + d*x^2)]) - ((4*b + a*c)*(c + d*x^2)*Sqrt[b
+ a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(3*c^2*x^3*Sqrt[b + a*(c + d*x^2)]) + ((8*b + a*c)*d*(c + d*x^2)*Sqr
t[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(3*c^3*x*Sqrt[b + a*(c + d*x^2)]) + ((8*b + a*c)*d^(3/2)*Sqrt[b
+ a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(3*c^(5/2)*Sqrt[
(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[b + a*(c + d*x^2)]) - (a*(4*b + a*c)*d^(3/2)*Sqrt[b + a*
c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(3*c^(3/2)*(b + a*c)
*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[b + a*(c + d*x^2)])

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((c*b -
a*d)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*b*e*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), I
nt[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(p
+ 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 531

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

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{c+d x^2}\right )^{3/2}}{x^4} \, 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^4 \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^4 \left (c+d x^2\right )^{3/2}} \, dx}{\sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {b \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{c x^3 \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 {-(b+a c) (4 b+a c) d-a (3 b+a c) d^2 x^2}{x^4 \sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{c d \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {b \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{c x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {(4 b+a c) \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^2 x^3 \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 {-(b+a c)^2 (8 b+a c) d^2-a (b+a c) (4 b+a c) d^3 x^2}{x^2 \sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{3 c^2 (b+a c) d \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {b \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{c x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {(4 b+a c) \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^2 x^3 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(8 b+a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^3 x \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 {a c (b+a c)^2 (4 b+a c) d^3+a (b+a c)^2 (8 b+a c) d^4 x^2}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{3 c^3 (b+a c)^2 d \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {b \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{c x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {(4 b+a c) \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^2 x^3 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(8 b+a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^3 x \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (a (4 b+a c) d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {1}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{3 c^2 \sqrt {b+a \left (c+d x^2\right )}}-\frac {\left (a (8 b+a c) d^3 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{3 c^3 \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {b \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{c x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {(8 b+a c) d^2 x \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {(4 b+a c) \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^2 x^3 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(8 b+a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^3 x \sqrt {b+a \left (c+d x^2\right )}}-\frac {a (4 b+a c) d^{3/2} \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 c^{3/2} (b+a c) \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {b+a \left (c+d x^2\right )}}+\frac {\left ((8 b+a c) d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {\sqrt {b+a c+a d x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c^2 \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {b \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{c x^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {(8 b+a c) d^2 x \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^3 \sqrt {b+a \left (c+d x^2\right )}}-\frac {(4 b+a c) \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^2 x^3 \sqrt {b+a \left (c+d x^2\right )}}+\frac {(8 b+a c) d \left (c+d x^2\right ) \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{3 c^3 x \sqrt {b+a \left (c+d x^2\right )}}+\frac {(8 b+a c) d^{3/2} \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 c^{5/2} \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {b+a \left (c+d x^2\right )}}-\frac {a (4 b+a c) d^{3/2} \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 c^{3/2} (b+a c) \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {b+a \left (c+d x^2\right )}}\\ \end {align*}

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Mathematica [C]  time = 0.83, size = 329, normalized size = 0.85 \[ -\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (\sqrt {\frac {a d}{a c+b}} \left (a^2 c \left (c-d x^2\right ) \left (c+d x^2\right )^2+a b \left (2 c^3-3 c^2 d x^2-13 c d^2 x^4-8 d^3 x^6\right )+b^2 \left (c^2-4 c d x^2-8 d^2 x^4\right )\right )+4 i a b c d^2 x^3 \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )-i a c d^2 x^3 (a c+8 b) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )\right )}{3 c^3 x^3 \sqrt {\frac {a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/(c + d*x^2))^(3/2)/x^4,x]

[Out]

-1/3*(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(Sqrt[(a*d)/(b + a*c)]*(a^2*c*(c - d*x^2)*(c + d*x^2)^2 + b^2*(c^2
- 4*c*d*x^2 - 8*d^2*x^4) + a*b*(2*c^3 - 3*c^2*d*x^2 - 13*c*d^2*x^4 - 8*d^3*x^6)) - I*a*c*(8*b + a*c)*d^2*x^3*
Sqrt[(b + a*c + a*d*x^2)/(b + a*c)]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[(a*d)/(b + a*c)]*x], 1 + b/(a
*c)] + (4*I)*a*b*c*d^2*x^3*Sqrt[(b + a*c + a*d*x^2)/(b + a*c)]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[(a
*d)/(b + a*c)]*x], 1 + b/(a*c)]))/(c^3*Sqrt[(a*d)/(b + a*c)]*x^3*(b + a*(c + d*x^2)))

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a d x^{2} + a c + b\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{d x^{6} + c x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/(d*x^2+c))^(3/2)/x^4,x, algorithm="fricas")

[Out]

integral((a*d*x^2 + a*c + b)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(d*x^6 + c*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{d x^{2} + c}\right )}^{\frac {3}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/(d*x^2+c))^(3/2)/x^4,x, algorithm="giac")

[Out]

integrate((a + b/(d*x^2 + c))^(3/2)/x^4, x)

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maple [B]  time = 0.04, size = 1039, normalized size = 2.68 \[ \frac {\left (\sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a^{2} c \,d^{3} x^{6}+3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, a b \,d^{3} x^{6}+5 \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a b \,d^{3} x^{6}+\sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a^{2} c^{2} d^{2} x^{4}-\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c^{2} d^{2} x^{3} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, a b c \,d^{2} x^{4}+10 \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a b c \,d^{2} x^{4}-8 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c \,d^{2} x^{3} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+4 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c \,d^{2} x^{3} \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-\sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a^{2} c^{3} d \,x^{2}+3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, b^{2} d^{2} x^{4}+5 \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, b^{2} d^{2} x^{4}+3 \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a b \,c^{2} d \,x^{2}-\sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a^{2} c^{4}+4 \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, b^{2} c d \,x^{2}-2 \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a b \,c^{3}-\sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, b^{2} c^{2}\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \left (a d \,x^{2}+a c +b \right ) c^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/(d*x^2+c))^(3/2)/x^4,x)

[Out]

1/3*((-1/(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^6*a^2*c*d^3+3*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a
*c^2+b*c)^(1/2)*(-1/(a*c+b)*a*d)^(1/2)*x^6*a*b*d^3+5*(-1/(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*
x^6*a*b*d^3-((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-
1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*x^3*a^2*c^2*d^2+(-1/(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))
^(1/2)*x^4*a^2*c^2*d^2+4*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)
*EllipticF((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*x^3*a*b*c*d^2-8*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*((a
*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*x^3*a
*b*c*d^2+3*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(-1/(a*c+b)*a*d)^(1/2)*x^4*a*b*c*d^2+10*(-1/(a*c+b)
*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^4*a*b*c*d^2+3*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*
(-1/(a*c+b)*a*d)^(1/2)*x^4*b^2*d^2+5*(-1/(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^4*b^2*d^2-(-1/
(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^2*a^2*c^3*d+3*(-1/(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^
2+a*c+b))^(1/2)*x^2*a*b*c^2*d+4*(-1/(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^2*b^2*c*d-(-1/(a*c+
b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*a^2*c^4-2*(-1/(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(
1/2)*a*b*c^3-(-1/(a*c+b)*a*d)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*b^2*c^2)*((a*d*x^2+a*c+b)/(d*x^2+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)/(-1/(a*c+b)*a*d)^(1/2)/x^3/c^3/(a*d*x^2+a*c+b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{d x^{2} + c}\right )}^{\frac {3}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/(d*x^2+c))^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate((a + b/(d*x^2 + c))^(3/2)/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/(c + d*x^2))^(3/2)/x^4,x)

[Out]

int((a + b/(c + d*x^2))^(3/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/(d*x**2+c))**(3/2)/x**4,x)

[Out]

Integral(((a*c + a*d*x**2 + b)/(c + d*x**2))**(3/2)/x**4, x)

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