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

Optimal. Leaf size=211 \[ -\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (b c-a d)^3}+\frac {d^{3/2} \left (15 a^2 d^2-42 a b c d+35 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} (b c-a d)^3}-\frac {15 a^2 d^2-27 a b c d+8 b^2 c^2}{8 a c^3 x (b c-a d)^2}-\frac {d (9 b c-5 a d)}{8 c^2 x \left (c+d x^2\right ) (b c-a d)^2}-\frac {d}{4 c x \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

1/8*(-15*a^2*d^2+27*a*b*c*d-8*b^2*c^2)/a/c^3/(-a*d+b*c)^2/x-1/4*d/c/(-a*d+b*c)/x/(d*x^2+c)^2-1/8*d*(-5*a*d+9*b
*c)/c^2/(-a*d+b*c)^2/x/(d*x^2+c)-b^(7/2)*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/(-a*d+b*c)^3+1/8*d^(3/2)*(15*a^2*d^
2-42*a*b*c*d+35*b^2*c^2)*arctan(x*d^(1/2)/c^(1/2))/c^(7/2)/(-a*d+b*c)^3

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Rubi [A]  time = 0.31, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {472, 579, 583, 522, 205} \[ -\frac {15 a^2 d^2-27 a b c d+8 b^2 c^2}{8 a c^3 x (b c-a d)^2}+\frac {d^{3/2} \left (15 a^2 d^2-42 a b c d+35 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} (b c-a d)^3}-\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (b c-a d)^3}-\frac {d (9 b c-5 a d)}{8 c^2 x \left (c+d x^2\right ) (b c-a d)^2}-\frac {d}{4 c x \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(8*b^2*c^2 - 27*a*b*c*d + 15*a^2*d^2)/(8*a*c^3*(b*c - a*d)^2*x) - d/(4*c*(b*c - a*d)*x*(c + d*x^2)^2) - (d*(9
*b*c - 5*a*d))/(8*c^2*(b*c - a*d)^2*x*(c + d*x^2)) - (b^(7/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3/2)*(b*c - a*d
)^3) + (d^(3/2)*(35*b^2*c^2 - 42*a*b*c*d + 15*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(7/2)*(b*c - a*d)^3)

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 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 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, 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 ) \left (c+d x^2\right )^3} \, dx &=-\frac {d}{4 c (b c-a d) x \left (c+d x^2\right )^2}+\frac {\int \frac {4 b c-5 a d-5 b d x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{4 c (b c-a d)}\\ &=-\frac {d}{4 c (b c-a d) x \left (c+d x^2\right )^2}-\frac {d (9 b c-5 a d)}{8 c^2 (b c-a d)^2 x \left (c+d x^2\right )}+\frac {\int \frac {8 b^2 c^2-27 a b c d+15 a^2 d^2-3 b d (9 b c-5 a d) x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{8 c^2 (b c-a d)^2}\\ &=-\frac {\frac {8 b^2 c}{a}-27 b d+\frac {15 a d^2}{c}}{8 c^2 (b c-a d)^2 x}-\frac {d}{4 c (b c-a d) x \left (c+d x^2\right )^2}-\frac {d (9 b c-5 a d)}{8 c^2 (b c-a d)^2 x \left (c+d x^2\right )}-\frac {\int \frac {8 b^3 c^3+8 a b^2 c^2 d-27 a^2 b c d^2+15 a^3 d^3+b d \left (8 b^2 c^2-27 a b c d+15 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{8 a c^3 (b c-a d)^2}\\ &=-\frac {\frac {8 b^2 c}{a}-27 b d+\frac {15 a d^2}{c}}{8 c^2 (b c-a d)^2 x}-\frac {d}{4 c (b c-a d) x \left (c+d x^2\right )^2}-\frac {d (9 b c-5 a d)}{8 c^2 (b c-a d)^2 x \left (c+d x^2\right )}-\frac {b^4 \int \frac {1}{a+b x^2} \, dx}{a (b c-a d)^3}+\frac {\left (d^2 \left (35 b^2 c^2-42 a b c d+15 a^2 d^2\right )\right ) \int \frac {1}{c+d x^2} \, dx}{8 c^3 (b c-a d)^3}\\ &=-\frac {\frac {8 b^2 c}{a}-27 b d+\frac {15 a d^2}{c}}{8 c^2 (b c-a d)^2 x}-\frac {d}{4 c (b c-a d) x \left (c+d x^2\right )^2}-\frac {d (9 b c-5 a d)}{8 c^2 (b c-a d)^2 x \left (c+d x^2\right )}-\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (b c-a d)^3}+\frac {d^{3/2} \left (35 b^2 c^2-42 a b c d+15 a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} (b c-a d)^3}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 172, normalized size = 0.82 \[ \frac {1}{8} \left (\frac {8 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (a d-b c)^3}+\frac {d^{3/2} \left (15 a^2 d^2-42 a b c d+35 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2} (b c-a d)^3}+\frac {d^2 x (11 b c-7 a d)}{c^3 \left (c+d x^2\right ) (b c-a d)^2}+\frac {2 d^2 x}{c^2 \left (c+d x^2\right )^2 (b c-a d)}-\frac {8}{a c^3 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8/(a*c^3*x) + (2*d^2*x)/(c^2*(b*c - a*d)*(c + d*x^2)^2) + (d^2*(11*b*c - 7*a*d)*x)/(c^3*(b*c - a*d)^2*(c + d
*x^2)) + (8*b^(7/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3/2)*(-(b*c) + a*d)^3) + (d^(3/2)*(35*b^2*c^2 - 42*a*b*c*
d + 15*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(7/2)*(b*c - a*d)^3))/8

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fricas [B]  time = 2.40, size = 1991, normalized size = 9.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(16*b^3*c^5 - 48*a*b^2*c^4*d + 48*a^2*b*c^3*d^2 - 16*a^3*c^2*d^3 + 2*(8*b^3*c^3*d^2 - 35*a*b^2*c^2*d^3
+ 42*a^2*b*c*d^4 - 15*a^3*d^5)*x^4 + 2*(16*b^3*c^4*d - 61*a*b^2*c^3*d^2 + 70*a^2*b*c^2*d^3 - 25*a^3*c*d^4)*x^2
+ 8*(b^3*c^3*d^2*x^5 + 2*b^3*c^4*d*x^3 + b^3*c^5*x)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)
) + ((35*a*b^2*c^2*d^3 - 42*a^2*b*c*d^4 + 15*a^3*d^5)*x^5 + 2*(35*a*b^2*c^3*d^2 - 42*a^2*b*c^2*d^3 + 15*a^3*c*
d^4)*x^3 + (35*a*b^2*c^4*d - 42*a^2*b*c^3*d^2 + 15*a^3*c^2*d^3)*x)*sqrt(-d/c)*log((d*x^2 - 2*c*x*sqrt(-d/c) -
c)/(d*x^2 + c)))/((a*b^3*c^6*d^2 - 3*a^2*b^2*c^5*d^3 + 3*a^3*b*c^4*d^4 - a^4*c^3*d^5)*x^5 + 2*(a*b^3*c^7*d - 3
*a^2*b^2*c^6*d^2 + 3*a^3*b*c^5*d^3 - a^4*c^4*d^4)*x^3 + (a*b^3*c^8 - 3*a^2*b^2*c^7*d + 3*a^3*b*c^6*d^2 - a^4*c
^5*d^3)*x), -1/8*(8*b^3*c^5 - 24*a*b^2*c^4*d + 24*a^2*b*c^3*d^2 - 8*a^3*c^2*d^3 + (8*b^3*c^3*d^2 - 35*a*b^2*c^
2*d^3 + 42*a^2*b*c*d^4 - 15*a^3*d^5)*x^4 + (16*b^3*c^4*d - 61*a*b^2*c^3*d^2 + 70*a^2*b*c^2*d^3 - 25*a^3*c*d^4)
*x^2 - ((35*a*b^2*c^2*d^3 - 42*a^2*b*c*d^4 + 15*a^3*d^5)*x^5 + 2*(35*a*b^2*c^3*d^2 - 42*a^2*b*c^2*d^3 + 15*a^3
*c*d^4)*x^3 + (35*a*b^2*c^4*d - 42*a^2*b*c^3*d^2 + 15*a^3*c^2*d^3)*x)*sqrt(d/c)*arctan(x*sqrt(d/c)) + 4*(b^3*c
^3*d^2*x^5 + 2*b^3*c^4*d*x^3 + b^3*c^5*x)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/((a*b^3*
c^6*d^2 - 3*a^2*b^2*c^5*d^3 + 3*a^3*b*c^4*d^4 - a^4*c^3*d^5)*x^5 + 2*(a*b^3*c^7*d - 3*a^2*b^2*c^6*d^2 + 3*a^3*
b*c^5*d^3 - a^4*c^4*d^4)*x^3 + (a*b^3*c^8 - 3*a^2*b^2*c^7*d + 3*a^3*b*c^6*d^2 - a^4*c^5*d^3)*x), -1/16*(16*b^3
*c^5 - 48*a*b^2*c^4*d + 48*a^2*b*c^3*d^2 - 16*a^3*c^2*d^3 + 2*(8*b^3*c^3*d^2 - 35*a*b^2*c^2*d^3 + 42*a^2*b*c*d
^4 - 15*a^3*d^5)*x^4 + 2*(16*b^3*c^4*d - 61*a*b^2*c^3*d^2 + 70*a^2*b*c^2*d^3 - 25*a^3*c*d^4)*x^2 + 16*(b^3*c^3
*d^2*x^5 + 2*b^3*c^4*d*x^3 + b^3*c^5*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) + ((35*a*b^2*c^2*d^3 - 42*a^2*b*c*d^4 +
15*a^3*d^5)*x^5 + 2*(35*a*b^2*c^3*d^2 - 42*a^2*b*c^2*d^3 + 15*a^3*c*d^4)*x^3 + (35*a*b^2*c^4*d - 42*a^2*b*c^3*
d^2 + 15*a^3*c^2*d^3)*x)*sqrt(-d/c)*log((d*x^2 - 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a*b^3*c^6*d^2 - 3*a^2*b
^2*c^5*d^3 + 3*a^3*b*c^4*d^4 - a^4*c^3*d^5)*x^5 + 2*(a*b^3*c^7*d - 3*a^2*b^2*c^6*d^2 + 3*a^3*b*c^5*d^3 - a^4*c
^4*d^4)*x^3 + (a*b^3*c^8 - 3*a^2*b^2*c^7*d + 3*a^3*b*c^6*d^2 - a^4*c^5*d^3)*x), -1/8*(8*b^3*c^5 - 24*a*b^2*c^4
*d + 24*a^2*b*c^3*d^2 - 8*a^3*c^2*d^3 + (8*b^3*c^3*d^2 - 35*a*b^2*c^2*d^3 + 42*a^2*b*c*d^4 - 15*a^3*d^5)*x^4 +
(16*b^3*c^4*d - 61*a*b^2*c^3*d^2 + 70*a^2*b*c^2*d^3 - 25*a^3*c*d^4)*x^2 + 8*(b^3*c^3*d^2*x^5 + 2*b^3*c^4*d*x^
3 + b^3*c^5*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) - ((35*a*b^2*c^2*d^3 - 42*a^2*b*c*d^4 + 15*a^3*d^5)*x^5 + 2*(35*a
*b^2*c^3*d^2 - 42*a^2*b*c^2*d^3 + 15*a^3*c*d^4)*x^3 + (35*a*b^2*c^4*d - 42*a^2*b*c^3*d^2 + 15*a^3*c^2*d^3)*x)*
sqrt(d/c)*arctan(x*sqrt(d/c)))/((a*b^3*c^6*d^2 - 3*a^2*b^2*c^5*d^3 + 3*a^3*b*c^4*d^4 - a^4*c^3*d^5)*x^5 + 2*(a
*b^3*c^7*d - 3*a^2*b^2*c^6*d^2 + 3*a^3*b*c^5*d^3 - a^4*c^4*d^4)*x^3 + (a*b^3*c^8 - 3*a^2*b^2*c^7*d + 3*a^3*b*c
^6*d^2 - a^4*c^5*d^3)*x)]

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giac [A]  time = 0.33, size = 236, normalized size = 1.12 \[ -\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {a b}} + \frac {{\left (35 \, b^{2} c^{2} d^{2} - 42 \, a b c d^{3} + 15 \, a^{2} d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{6} - 3 \, a b^{2} c^{5} d + 3 \, a^{2} b c^{4} d^{2} - a^{3} c^{3} d^{3}\right )} \sqrt {c d}} + \frac {11 \, b c d^{3} x^{3} - 7 \, a d^{4} x^{3} + 13 \, b c^{2} d^{2} x - 9 \, a c d^{3} x}{8 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} {\left (d x^{2} + c\right )}^{2}} - \frac {1}{a c^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^4*arctan(b*x/sqrt(a*b))/((a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*sqrt(a*b)) + 1/8*(35*b^2*c
^2*d^2 - 42*a*b*c*d^3 + 15*a^2*d^4)*arctan(d*x/sqrt(c*d))/((b^3*c^6 - 3*a*b^2*c^5*d + 3*a^2*b*c^4*d^2 - a^3*c^
3*d^3)*sqrt(c*d)) + 1/8*(11*b*c*d^3*x^3 - 7*a*d^4*x^3 + 13*b*c^2*d^2*x - 9*a*c*d^3*x)/((b^2*c^5 - 2*a*b*c^4*d
+ a^2*c^3*d^2)*(d*x^2 + c)^2) - 1/(a*c^3*x)

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maple [A]  time = 0.02, size = 335, normalized size = 1.59 \[ -\frac {7 a^{2} d^{5} x^{3}}{8 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{3}}+\frac {9 a b \,d^{4} x^{3}}{4 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{2}}-\frac {11 b^{2} d^{3} x^{3}}{8 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c}-\frac {9 a^{2} d^{4} x}{8 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{2}}+\frac {11 a b \,d^{3} x}{4 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c}-\frac {13 b^{2} d^{2} x}{8 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2}}-\frac {15 a^{2} d^{4} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \left (a d -b c \right )^{3} \sqrt {c d}\, c^{3}}+\frac {21 a b \,d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{4 \left (a d -b c \right )^{3} \sqrt {c d}\, c^{2}}+\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\left (a d -b c \right )^{3} \sqrt {a b}\, a}-\frac {35 b^{2} d^{2} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \left (a d -b c \right )^{3} \sqrt {c d}\, c}-\frac {1}{a \,c^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*b^4/(a*d-b*c)^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)-7/8*d^5/c^3/(a*d-b*c)^3/(d*x^2+c)^2*x^3*a^2+9/4*d^4/
c^2/(a*d-b*c)^3/(d*x^2+c)^2*x^3*a*b-11/8*d^3/c/(a*d-b*c)^3/(d*x^2+c)^2*x^3*b^2-9/8*d^4/c^2/(a*d-b*c)^3/(d*x^2+
c)^2*a^2*x+11/4*d^3/c/(a*d-b*c)^3/(d*x^2+c)^2*a*b*x-13/8*d^2/(a*d-b*c)^3/(d*x^2+c)^2*b^2*x-15/8*d^4/c^3/(a*d-b
*c)^3/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*a^2+21/4*d^3/c^2/(a*d-b*c)^3/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)
*a*b-35/8*d^2/c/(a*d-b*c)^3/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*b^2-1/a/c^3/x

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maxima [A]  time = 2.54, size = 352, normalized size = 1.67 \[ -\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {a b}} + \frac {{\left (35 \, b^{2} c^{2} d^{2} - 42 \, a b c d^{3} + 15 \, a^{2} d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{6} - 3 \, a b^{2} c^{5} d + 3 \, a^{2} b c^{4} d^{2} - a^{3} c^{3} d^{3}\right )} \sqrt {c d}} - \frac {8 \, b^{2} c^{4} - 16 \, a b c^{3} d + 8 \, a^{2} c^{2} d^{2} + {\left (8 \, b^{2} c^{2} d^{2} - 27 \, a b c d^{3} + 15 \, a^{2} d^{4}\right )} x^{4} + {\left (16 \, b^{2} c^{3} d - 45 \, a b c^{2} d^{2} + 25 \, a^{2} c d^{3}\right )} x^{2}}{8 \, {\left ({\left (a b^{2} c^{5} d^{2} - 2 \, a^{2} b c^{4} d^{3} + a^{3} c^{3} d^{4}\right )} x^{5} + 2 \, {\left (a b^{2} c^{6} d - 2 \, a^{2} b c^{5} d^{2} + a^{3} c^{4} d^{3}\right )} x^{3} + {\left (a b^{2} c^{7} - 2 \, a^{2} b c^{6} d + a^{3} c^{5} d^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^4*arctan(b*x/sqrt(a*b))/((a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*sqrt(a*b)) + 1/8*(35*b^2*c
^2*d^2 - 42*a*b*c*d^3 + 15*a^2*d^4)*arctan(d*x/sqrt(c*d))/((b^3*c^6 - 3*a*b^2*c^5*d + 3*a^2*b*c^4*d^2 - a^3*c^
3*d^3)*sqrt(c*d)) - 1/8*(8*b^2*c^4 - 16*a*b*c^3*d + 8*a^2*c^2*d^2 + (8*b^2*c^2*d^2 - 27*a*b*c*d^3 + 15*a^2*d^4
)*x^4 + (16*b^2*c^3*d - 45*a*b*c^2*d^2 + 25*a^2*c*d^3)*x^2)/((a*b^2*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3*c^3*d^4)*x
^5 + 2*(a*b^2*c^6*d - 2*a^2*b*c^5*d^2 + a^3*c^4*d^3)*x^3 + (a*b^2*c^7 - 2*a^2*b*c^6*d + a^3*c^5*d^2)*x)

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mupad [B]  time = 1.30, size = 738, normalized size = 3.50 \[ -\frac {\frac {1}{a\,c}+\frac {x^4\,\left (15\,a^2\,d^4-27\,a\,b\,c\,d^3+8\,b^2\,c^2\,d^2\right )}{8\,a\,c^3\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {x^2\,\left (25\,a^2\,d^3-45\,a\,b\,c\,d^2+16\,b^2\,c^2\,d\right )}{8\,a\,c^2\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}}{c^2\,x+2\,c\,d\,x^3+d^2\,x^5}-\frac {\mathrm {atan}\left (\frac {b\,c^7\,x\,{\left (-a^3\,b^7\right )}^{3/2}\,64{}\mathrm {i}+a^{10}\,b\,d^7\,x\,\sqrt {-a^3\,b^7}\,225{}\mathrm {i}+a^6\,b^5\,c^4\,d^3\,x\,\sqrt {-a^3\,b^7}\,1225{}\mathrm {i}-a^7\,b^4\,c^3\,d^4\,x\,\sqrt {-a^3\,b^7}\,2940{}\mathrm {i}+a^8\,b^3\,c^2\,d^5\,x\,\sqrt {-a^3\,b^7}\,2814{}\mathrm {i}-a^9\,b^2\,c\,d^6\,x\,\sqrt {-a^3\,b^7}\,1260{}\mathrm {i}}{a^3\,b^7\,\left (2940\,a^6\,c^3\,d^4-1225\,a^5\,b\,c^4\,d^3+64\,a^2\,b^4\,c^7\right )-225\,a^{12}\,b^4\,d^7+1260\,a^{11}\,b^5\,c\,d^6-2814\,a^{10}\,b^6\,c^2\,d^5}\right )\,\sqrt {-a^3\,b^7}\,1{}\mathrm {i}}{a^6\,d^3-3\,a^5\,b\,c\,d^2+3\,a^4\,b^2\,c^2\,d-a^3\,b^3\,c^3}-\frac {\mathrm {atan}\left (\frac {a^7\,d^5\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,225{}\mathrm {i}+b^7\,c^{14}\,d\,x\,\sqrt {-c^7\,d^3}\,64{}\mathrm {i}-a^4\,b^3\,c^3\,d^2\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,2940{}\mathrm {i}+a^5\,b^2\,c^2\,d^3\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,2814{}\mathrm {i}-a^6\,b\,c\,d^4\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,1260{}\mathrm {i}+a^3\,b^4\,c^4\,d\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,1225{}\mathrm {i}}{225\,a^7\,c^{11}\,d^9-1260\,a^6\,b\,c^{12}\,d^8+2814\,a^5\,b^2\,c^{13}\,d^7-2940\,a^4\,b^3\,c^{14}\,d^6+1225\,a^3\,b^4\,c^{15}\,d^5-64\,b^7\,c^{18}\,d^2}\right )\,\sqrt {-c^7\,d^3}\,\left (15\,a^2\,d^2-42\,a\,b\,c\,d+35\,b^2\,c^2\right )\,1{}\mathrm {i}}{8\,\left (-a^3\,c^7\,d^3+3\,a^2\,b\,c^8\,d^2-3\,a\,b^2\,c^9\,d+b^3\,c^{10}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (1/(a*c) + (x^4*(15*a^2*d^4 + 8*b^2*c^2*d^2 - 27*a*b*c*d^3))/(8*a*c^3*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (x^
2*(25*a^2*d^3 + 16*b^2*c^2*d - 45*a*b*c*d^2))/(8*a*c^2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(c^2*x + d^2*x^5 + 2*
c*d*x^3) - (atan((b*c^7*x*(-a^3*b^7)^(3/2)*64i + a^10*b*d^7*x*(-a^3*b^7)^(1/2)*225i + a^6*b^5*c^4*d^3*x*(-a^3*
b^7)^(1/2)*1225i - a^7*b^4*c^3*d^4*x*(-a^3*b^7)^(1/2)*2940i + a^8*b^3*c^2*d^5*x*(-a^3*b^7)^(1/2)*2814i - a^9*b
^2*c*d^6*x*(-a^3*b^7)^(1/2)*1260i)/(a^3*b^7*(64*a^2*b^4*c^7 + 2940*a^6*c^3*d^4 - 1225*a^5*b*c^4*d^3) - 225*a^1
2*b^4*d^7 + 1260*a^11*b^5*c*d^6 - 2814*a^10*b^6*c^2*d^5))*(-a^3*b^7)^(1/2)*1i)/(a^6*d^3 - a^3*b^3*c^3 + 3*a^4*
b^2*c^2*d - 3*a^5*b*c*d^2) - (atan((a^7*d^5*x*(-c^7*d^3)^(3/2)*225i + b^7*c^14*d*x*(-c^7*d^3)^(1/2)*64i - a^4*
b^3*c^3*d^2*x*(-c^7*d^3)^(3/2)*2940i + a^5*b^2*c^2*d^3*x*(-c^7*d^3)^(3/2)*2814i - a^6*b*c*d^4*x*(-c^7*d^3)^(3/
2)*1260i + a^3*b^4*c^4*d*x*(-c^7*d^3)^(3/2)*1225i)/(225*a^7*c^11*d^9 - 64*b^7*c^18*d^2 - 1260*a^6*b*c^12*d^8 +
1225*a^3*b^4*c^15*d^5 - 2940*a^4*b^3*c^14*d^6 + 2814*a^5*b^2*c^13*d^7))*(-c^7*d^3)^(1/2)*(15*a^2*d^2 + 35*b^2
*c^2 - 42*a*b*c*d)*1i)/(8*(b^3*c^10 - a^3*c^7*d^3 + 3*a^2*b*c^8*d^2 - 3*a*b^2*c^9*d))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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