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

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

[Out]

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

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Rubi [A]  time = 0.268332, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {472, 579, 583, 12, 377, 205} \[ -\frac{\sqrt{c+d x^2} \left (4 a^2 d^2-4 a b c d+3 b^2 c^2\right )}{2 a^2 c^2 x (b c-a d)^2}-\frac{3 b^2 (b c-2 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{5/2} (b c-a d)^{5/2}}+\frac{b}{2 a x \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)}+\frac{d (2 a d+b c)}{2 a c x \sqrt{c+d x^2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 472

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

Rule 579

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

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 12

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

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 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]

Rubi steps

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

Mathematica [A]  time = 5.35102, size = 145, normalized size = 0.71 \[ \sqrt{c+d x^2} \left (-\frac{\frac{b^3 x}{2 \left (a+b x^2\right ) (b c-a d)^2}+\frac{1}{c^2 x}}{a^2}-\frac{d^3 x}{c^2 \left (c+d x^2\right ) (b c-a d)^2}\right )-\frac{3 b^2 (b c-2 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{5/2} (b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^2]*(-((d^3*x)/(c^2*(b*c - a*d)^2*(c + d*x^2))) - (1/(c^2*x) + (b^3*x)/(2*(b*c - a*d)^2*(a + b*x^2
)))/a^2) - (3*b^2*(b*c - 2*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(5/2)*(b*c - a*d)^
(5/2))

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Maple [B]  time = 0.015, size = 1524, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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^2), x)

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Fricas [B]  time = 6.08287, size = 2020, normalized size = 9.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*((b^4*c^3*d - 2*a*b^3*c^2*d^2)*x^5 + (b^4*c^4 - a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2)*x^3 + (a*b^3*c^4 - 2*
a^2*b^2*c^3*d)*x)*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*(2*a^2*b^3*c^4 - 6*a^3*b^2*c^3*d + 6*a^4*b*c^2*d^2 - 2*a^5*c*d^3 + (3*a*b^4*c^3*d - 7*a^2*b^3*c^2*d^2 +
8*a^3*b^2*c*d^3 - 4*a^4*b*d^4)*x^4 + (3*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 6*a^4*b*c*d^3 - 4*a^5*d^4)*x^2)*sqrt(d*
x^2 + c))/((a^3*b^4*c^5*d - 3*a^4*b^3*c^4*d^2 + 3*a^5*b^2*c^3*d^3 - a^6*b*c^2*d^4)*x^5 + (a^3*b^4*c^6 - 2*a^4*
b^3*c^5*d + 2*a^6*b*c^3*d^3 - a^7*c^2*d^4)*x^3 + (a^4*b^3*c^6 - 3*a^5*b^2*c^5*d + 3*a^6*b*c^4*d^2 - a^7*c^3*d^
3)*x), -1/4*(3*((b^4*c^3*d - 2*a*b^3*c^2*d^2)*x^5 + (b^4*c^4 - a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2)*x^3 + (a*b^3*c
^4 - 2*a^2*b^2*c^3*d)*x)*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*(2*a^2*b^3*c^4 - 6*a^3*b^2*c^3*d + 6*a^4*b*c^2*
d^2 - 2*a^5*c*d^3 + (3*a*b^4*c^3*d - 7*a^2*b^3*c^2*d^2 + 8*a^3*b^2*c*d^3 - 4*a^4*b*d^4)*x^4 + (3*a*b^4*c^4 - 5
*a^2*b^3*c^3*d + 6*a^4*b*c*d^3 - 4*a^5*d^4)*x^2)*sqrt(d*x^2 + c))/((a^3*b^4*c^5*d - 3*a^4*b^3*c^4*d^2 + 3*a^5*
b^2*c^3*d^3 - a^6*b*c^2*d^4)*x^5 + (a^3*b^4*c^6 - 2*a^4*b^3*c^5*d + 2*a^6*b*c^3*d^3 - a^7*c^2*d^4)*x^3 + (a^4*
b^3*c^6 - 3*a^5*b^2*c^5*d + 3*a^6*b*c^4*d^2 - a^7*c^3*d^3)*x)]

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

[Out]

Exception raised: ValueError

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Giac [B]  time = 6.96892, size = 748, normalized size = 3.65 \begin{align*} -\frac{d^{3} x}{{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \sqrt{d x^{2} + c}} + \frac{3 \,{\left (b^{3} c \sqrt{d} - 2 \, a b^{2} 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^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt{a b c d - a^{2} d^{2}}} + \frac{3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b^{3} c^{2} \sqrt{d} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b^{2} c d^{\frac{3}{2}} + 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} b d^{\frac{5}{2}} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{3} c^{3} \sqrt{d} + 18 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b^{2} c^{2} d^{\frac{3}{2}} - 20 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} b c d^{\frac{5}{2}} + 8 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{3} d^{\frac{7}{2}} + 3 \, b^{3} c^{4} \sqrt{d} - 4 \, a b^{2} c^{3} d^{\frac{3}{2}} + 2 \, a^{2} b c^{2} d^{\frac{5}{2}}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} b - 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a d + 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c^{2} - 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a c d - b c^{3}\right )}{\left (a^{2} b^{2} c^{3} - 2 \, a^{3} b c^{2} d + a^{4} c d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d^3*x/((b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*sqrt(d*x^2 + c)) + 3/2*(b^3*c*sqrt(d) - 2*a*b^2*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^2*b^2*c^2 - 2*a^3*b*c*d + a
^4*d^2)*sqrt(a*b*c*d - a^2*d^2)) + (3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^3*c^2*sqrt(d) - 6*(sqrt(d)*x - sqrt(d*
x^2 + c))^4*a*b^2*c*d^(3/2) + 2*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*b*d^(5/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c)
)^2*b^3*c^3*sqrt(d) + 18*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b^2*c^2*d^(3/2) - 20*(sqrt(d)*x - sqrt(d*x^2 + c))^
2*a^2*b*c*d^(5/2) + 8*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^3*d^(7/2) + 3*b^3*c^4*sqrt(d) - 4*a*b^2*c^3*d^(3/2) +
2*a^2*b*c^2*d^(5/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^6*b - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c + 4*(sqrt(d)*
x - sqrt(d*x^2 + c))^4*a*d + 3*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^2 - 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c*d
- b*c^3)*(a^2*b^2*c^3 - 2*a^3*b*c^2*d + a^4*c*d^2))