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

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

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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]

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*}

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Mathematica [A]  time = 5.37, 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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fricas [B]  time = 2.24, size = 1018, normalized size = 4.97 \[ \left [\frac {3 \, {\left ({\left (b^{4} c^{3} d - 2 \, a b^{3} c^{2} d^{2}\right )} x^{5} + {\left (b^{4} c^{4} - a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2}\right )} x^{3} + {\left (a b^{3} c^{4} - 2 \, a^{2} b^{2} c^{3} d\right )} x\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 (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} + {\left (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}\right )} x^{4} + {\left (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}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left ({\left (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}\right )} x^{5} + {\left (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}\right )} x^{3} + {\left (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}\right )} x\right )}}, -\frac {3 \, {\left ({\left (b^{4} c^{3} d - 2 \, a b^{3} c^{2} d^{2}\right )} x^{5} + {\left (b^{4} c^{4} - a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2}\right )} x^{3} + {\left (a b^{3} c^{4} - 2 \, a^{2} b^{2} c^{3} d\right )} x\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 (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} + {\left (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}\right )} x^{4} + {\left (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}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left ({\left (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}\right )} x^{5} + {\left (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}\right )} x^{3} + {\left (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}\right )} x\right )}}\right ] \]

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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giac [B]  time = 4.77, size = 554, normalized size = 2.70 \[ -\frac {d^{3} x}{{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \sqrt {d x^{2} + c}} + \frac {3 \, {\left (b^{3} c \sqrt {d} - 2 \, a b^{2} d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{3} c^{2} \sqrt {d} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b^{2} c d^{\frac {3}{2}} + 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} b d^{\frac {5}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{3} c^{3} \sqrt {d} + 18 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{2} c^{2} d^{\frac {3}{2}} - 20 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} b c d^{\frac {5}{2}} + 8 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{3} d^{\frac {7}{2}} + 3 \, b^{3} c^{4} \sqrt {d} - 4 \, a b^{2} c^{3} d^{\frac {3}{2}} + 2 \, a^{2} b c^{2} d^{\frac {5}{2}}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a d + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{2} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c d - b c^{3}\right )} {\left (a^{2} b^{2} c^{3} - 2 \, a^{3} b c^{2} d + a^{4} c d^{2}\right )}} \]

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))

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maple [B]  time = 0.02, size = 1524, normalized size = 7.43 \[ \text {result too large to display} \]

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-(-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^2*(-a*b)^(1/2)*d/(a*d-b*c)^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*b*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^2*(-a*b)^(1/2)*d/(a*d-b*c)^2*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^2/(a*d-b*c)*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)*d*x-3/4*b^2/a^2/(-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)-1/4*b/a^2/(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+3/4*b^2/a^2/(-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/a^2/(a*d-b*c)*b/(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^2*(-a*b)^(1/2)*d/(a*d-b*c)^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*b*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^2*(-a*b)^(1/2)*d/(a*d-b*c)^2*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))+3/4*b^2/a^2/(-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^2/a^2/(
-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/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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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