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

Optimal. Leaf size=123 \[ -\frac {x}{2 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}-\frac {3 d x}{2 \sqrt {c+d x^2} (b c-a d)^2}+\frac {(2 a d+b c) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} (b c-a d)^{5/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {471, 527, 12, 377, 205} \begin {gather*} -\frac {x}{2 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}-\frac {3 d x}{2 \sqrt {c+d x^2} (b c-a d)^2}+\frac {(2 a d+b c) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} (b c-a d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d*x)/(2*(b*c - a*d)^2*Sqrt[c + d*x^2]) - x/(2*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) + ((b*c + 2*a*d)*Ar
cTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*Sqrt[a]*(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 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, 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 {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx &=-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {\int \frac {c-2 d x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{2 (b c-a d)}\\ &=-\frac {3 d x}{2 (b c-a d)^2 \sqrt {c+d x^2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {\int \frac {c (b c+2 a d)}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 c (b c-a d)^2}\\ &=-\frac {3 d x}{2 (b c-a d)^2 \sqrt {c+d x^2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {(b c+2 a d) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 (b c-a d)^2}\\ &=-\frac {3 d x}{2 (b c-a d)^2 \sqrt {c+d x^2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {(b c+2 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 (b c-a d)^2}\\ &=-\frac {3 d x}{2 (b c-a d)^2 \sqrt {c+d x^2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {(b c+2 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 1.18, size = 133, normalized size = 1.08 \begin {gather*} \frac {x^3 \left (\frac {8 x^2 \left (c+d x^2\right ) (b c-a d) \, _2F_1\left (2,3;\frac {9}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )}{a+b x^2}+7 c \left (5 c+2 d x^2\right ) \, _2F_1\left (1,2;\frac {7}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{105 c^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^3*(7*c*(5*c + 2*d*x^2)*Hypergeometric2F1[1, 2, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + (8*(b*c - a*d)*x^2
*(c + d*x^2)*Hypergeometric2F1[2, 3, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(a + b*x^2)))/(105*c^3*(a + b*x^
2)^2*Sqrt[c + d*x^2])

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IntegrateAlgebraic [A]  time = 0.73, size = 135, normalized size = 1.10 \begin {gather*} \frac {(-2 a d-b c) \tan ^{-1}\left (\frac {a \sqrt {d}-b x \sqrt {c+d x^2}+b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 \sqrt {a} (b c-a d)^{5/2}}+\frac {-2 a d x-b c x-3 b d x^3}{2 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.73, size = 744, normalized size = 6.05 \begin {gather*} \left [-\frac {{\left ({\left (b^{2} c d + 2 \, a b d^{2}\right )} x^{4} + a b c^{2} + 2 \, a^{2} c d + {\left (b^{2} c^{2} + 3 \, a b c d + 2 \, a^{2} 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 (3 \, {\left (a b^{2} c d - a^{2} b d^{2}\right )} x^{3} + {\left (a b^{2} c^{2} + a^{2} b c d - 2 \, a^{3} d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 3 \, a^{3} b^{2} c d^{3} - a^{4} b d^{4}\right )} x^{4} + {\left (a b^{4} c^{4} - 2 \, a^{2} b^{3} c^{3} d + 2 \, a^{4} b c d^{3} - a^{5} d^{4}\right )} x^{2}\right )}}, \frac {{\left ({\left (b^{2} c d + 2 \, a b d^{2}\right )} x^{4} + a b c^{2} + 2 \, a^{2} c d + {\left (b^{2} c^{2} + 3 \, a b c d + 2 \, a^{2} 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 (3 \, {\left (a b^{2} c d - a^{2} b d^{2}\right )} x^{3} + {\left (a b^{2} c^{2} + a^{2} b c d - 2 \, a^{3} d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 3 \, a^{3} b^{2} c d^{3} - a^{4} b d^{4}\right )} x^{4} + {\left (a b^{4} c^{4} - 2 \, a^{2} b^{3} c^{3} d + 2 \, a^{4} b c d^{3} - a^{5} d^{4}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 4.45, size = 299, normalized size = 2.43 \begin {gather*} -\frac {d x}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {d x^{2} + c}} - \frac {{\left (b c \sqrt {d} + 2 \, a 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 (b^{2} c^{2} - 2 \, a b c d + a^{2} 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 c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d^{\frac {3}{2}} - b 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 (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 1453, normalized size = 11.81

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/((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 \begin {gather*} \int \frac {x^2}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2/((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 \begin {gather*} \int \frac {x^{2}}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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