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

Optimal. Leaf size=142 \[ \frac {b (b c-4 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{5/2}}+\frac {d x (2 a d+b c)}{2 a c \sqrt {c+d x^2} (b c-a d)^2}+\frac {b x}{2 a \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)} \]

[Out]

1/2*b*(-4*a*d+b*c)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))/a^(3/2)/(-a*d+b*c)^(5/2)+1/2*d*(2*a*d+b*
c)*x/a/c/(-a*d+b*c)^2/(d*x^2+c)^(1/2)+1/2*b*x/a/(-a*d+b*c)/(b*x^2+a)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {414, 527, 12, 377, 205} \[ \frac {b (b c-4 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{5/2}}+\frac {d x (2 a d+b c)}{2 a c \sqrt {c+d x^2} (b c-a d)^2}+\frac {b x}{2 a \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

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

Rubi steps

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

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Mathematica [A]  time = 5.25, size = 120, normalized size = 0.85 \[ \frac {1}{2} \left (\frac {b (b c-4 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} (b c-a d)^{5/2}}+\frac {x \sqrt {c+d x^2} \left (\frac {b^2}{a^2+a b x^2}+\frac {2 d^2}{c^2+c d x^2}\right )}{(b c-a d)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.66, size = 854, normalized size = 6.01 \[ \left [\frac {{\left (a b^{2} c^{3} - 4 \, a^{2} b c^{2} d + {\left (b^{3} c^{2} d - 4 \, a b^{2} c d^{2}\right )} x^{4} + {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 4 \, a^{2} b c d^{2}\right )} x^{2}\right )} \sqrt {-a b c + a^{2} d} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left ({\left (a b^{3} c^{2} d + a^{2} b^{2} c d^{2} - 2 \, a^{3} b d^{3}\right )} x^{3} + {\left (a b^{3} c^{3} - a^{2} b^{2} c^{2} d + 2 \, a^{3} b c d^{2} - 2 \, a^{4} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{3} b^{3} c^{5} - 3 \, a^{4} b^{2} c^{4} d + 3 \, a^{5} b c^{3} d^{2} - a^{6} c^{2} d^{3} + {\left (a^{2} b^{4} c^{4} d - 3 \, a^{3} b^{3} c^{3} d^{2} + 3 \, a^{4} b^{2} c^{2} d^{3} - a^{5} b c d^{4}\right )} x^{4} + {\left (a^{2} b^{4} c^{5} - 2 \, a^{3} b^{3} c^{4} d + 2 \, a^{5} b c^{2} d^{3} - a^{6} c d^{4}\right )} x^{2}\right )}}, \frac {{\left (a b^{2} c^{3} - 4 \, a^{2} b c^{2} d + {\left (b^{3} c^{2} d - 4 \, a b^{2} c d^{2}\right )} x^{4} + {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 4 \, a^{2} b c d^{2}\right )} x^{2}\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left ({\left (a b^{3} c^{2} d + a^{2} b^{2} c d^{2} - 2 \, a^{3} b d^{3}\right )} x^{3} + {\left (a b^{3} c^{3} - a^{2} b^{2} c^{2} d + 2 \, a^{3} b c d^{2} - 2 \, a^{4} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{3} b^{3} c^{5} - 3 \, a^{4} b^{2} c^{4} d + 3 \, a^{5} b c^{3} d^{2} - a^{6} c^{2} d^{3} + {\left (a^{2} b^{4} c^{4} d - 3 \, a^{3} b^{3} c^{3} d^{2} + 3 \, a^{4} b^{2} c^{2} d^{3} - a^{5} b c d^{4}\right )} x^{4} + {\left (a^{2} b^{4} c^{5} - 2 \, a^{3} b^{3} c^{4} d + 2 \, a^{5} b c^{2} d^{3} - a^{6} c d^{4}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((a*b^2*c^3 - 4*a^2*b*c^2*d + (b^3*c^2*d - 4*a*b^2*c*d^2)*x^4 + (b^3*c^3 - 3*a*b^2*c^2*d - 4*a^2*b*c*d^2)
*x^2)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^
2 + 4*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*((a*b
^3*c^2*d + a^2*b^2*c*d^2 - 2*a^3*b*d^3)*x^3 + (a*b^3*c^3 - a^2*b^2*c^2*d + 2*a^3*b*c*d^2 - 2*a^4*d^3)*x)*sqrt(
d*x^2 + c))/(a^3*b^3*c^5 - 3*a^4*b^2*c^4*d + 3*a^5*b*c^3*d^2 - a^6*c^2*d^3 + (a^2*b^4*c^4*d - 3*a^3*b^3*c^3*d^
2 + 3*a^4*b^2*c^2*d^3 - a^5*b*c*d^4)*x^4 + (a^2*b^4*c^5 - 2*a^3*b^3*c^4*d + 2*a^5*b*c^2*d^3 - a^6*c*d^4)*x^2),
1/4*((a*b^2*c^3 - 4*a^2*b*c^2*d + (b^3*c^2*d - 4*a*b^2*c*d^2)*x^4 + (b^3*c^3 - 3*a*b^2*c^2*d - 4*a^2*b*c*d^2)
*x^2)*sqrt(a*b*c - a^2*d)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d -
a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) + 2*((a*b^3*c^2*d + a^2*b^2*c*d^2 - 2*a^3*b*d^3)*x^3 + (a*b^3*c^3 - a^
2*b^2*c^2*d + 2*a^3*b*c*d^2 - 2*a^4*d^3)*x)*sqrt(d*x^2 + c))/(a^3*b^3*c^5 - 3*a^4*b^2*c^4*d + 3*a^5*b*c^3*d^2
- a^6*c^2*d^3 + (a^2*b^4*c^4*d - 3*a^3*b^3*c^3*d^2 + 3*a^4*b^2*c^2*d^3 - a^5*b*c*d^4)*x^4 + (a^2*b^4*c^5 - 2*a
^3*b^3*c^4*d + 2*a^5*b*c^2*d^3 - a^6*c*d^4)*x^2)]

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giac [B]  time = 4.53, size = 318, normalized size = 2.24 \[ \frac {d^{2} x}{{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {d x^{2} + c}} - \frac {{\left (b^{2} c \sqrt {d} - 4 \, a b d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b d^{\frac {3}{2}} - b^{2} c^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*x/((b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(d*x^2 + c)) - 1/2*(b^2*c*sqrt(d) - 4*a*b*d^(3/2))*arctan(1/2*(
(sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/((a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)
*sqrt(a*b*c*d - a^2*d^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2
*a*b*d^(3/2) - b^2*c^2*sqrt(d))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c +
4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2))

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maple [B]  time = 0.01, size = 1461, normalized size = 10.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a/(a*d-b*c)/(x-(-a*b)^(1/2)/b)/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)
^(1/2)+3/4/a*(-a*b)^(1/2)*d/(a*d-b*c)^2/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c
)/b)^(1/2)+3/4*d^2/(a*d-b*c)^2/c/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1
/2)*x-3/4/a*(-a*b)^(1/2)*d/(a*d-b*c)^2/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b
*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))
/(x-(-a*b)^(1/2)/b))-1/4/a/(a*d-b*c)/c/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)
/b)^(1/2)*d*x+1/4/a/(-a*b)^(1/2)/(a*d-b*c)*b/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*
d-b*c)/b)^(1/2)-1/4/a/(a*d-b*c)/c/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(
1/2)*d*x-1/4/a/(-a*b)^(1/2)/(a*d-b*c)*b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d
-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2
))/(x+(-a*b)^(1/2)/b))-1/4/a/(a*d-b*c)/(x+(-a*b)^(1/2)/b)/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/
2)/b)/b*d-(a*d-b*c)/b)^(1/2)-3/4/a*(-a*b)^(1/2)*d/(a*d-b*c)^2/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)
^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)+3/4*d^2/(a*d-b*c)^2/c/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/
b)/b*d-(a*d-b*c)/b)^(1/2)*x+3/4/a*(-a*b)^(1/2)*d/(a*d-b*c)^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b
)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/
b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))-1/4/a/(-a*b)^(1/2)/(a*d-b*c)*b/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1
/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)+1/4/a/(-a*b)^(1/2)/(a*d-b*c)*b/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)
^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(
-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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