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

Optimal. Leaf size=297 \[ \frac{d \left (-5 a^2 d^2+13 a b c d+4 b^2 c^2\right )}{8 a c^2 x \left (c+d x^2\right ) (b c-a d)^3}-\frac{3 (2 b c-a d) \left (5 a^2 d^2-3 a b c d+2 b^2 c^2\right )}{8 a^2 c^3 x (b c-a d)^3}-\frac{3 d^{5/2} \left (5 a^2 d^2-18 a b c d+21 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{7/2} (b c-a d)^4}-\frac{3 b^{7/2} (b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} (b c-a d)^4}+\frac{b}{2 a x \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}+\frac{d (a d+2 b c)}{4 a c x \left (c+d x^2\right )^2 (b c-a d)^2} \]

[Out]

(-3*(2*b*c - a*d)*(2*b^2*c^2 - 3*a*b*c*d + 5*a^2*d^2))/(8*a^2*c^3*(b*c - a*d)^3*x) + (d*(2*b*c + a*d))/(4*a*c*
(b*c - a*d)^2*x*(c + d*x^2)^2) + b/(2*a*(b*c - a*d)*x*(a + b*x^2)*(c + d*x^2)^2) + (d*(4*b^2*c^2 + 13*a*b*c*d
- 5*a^2*d^2))/(8*a*c^2*(b*c - a*d)^3*x*(c + d*x^2)) - (3*b^(7/2)*(b*c - 3*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2
*a^(5/2)*(b*c - a*d)^4) - (3*d^(5/2)*(21*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(
7/2)*(b*c - a*d)^4)

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Rubi [A]  time = 0.496568, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{d \left (-5 a^2 d^2+13 a b c d+4 b^2 c^2\right )}{8 a c^2 x \left (c+d x^2\right ) (b c-a d)^3}-\frac{3 (2 b c-a d) \left (5 a^2 d^2-3 a b c d+2 b^2 c^2\right )}{8 a^2 c^3 x (b c-a d)^3}-\frac{3 d^{5/2} \left (5 a^2 d^2-18 a b c d+21 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{7/2} (b c-a d)^4}-\frac{3 b^{7/2} (b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} (b c-a d)^4}+\frac{b}{2 a x \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}+\frac{d (a d+2 b c)}{4 a c x \left (c+d x^2\right )^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(2*b*c - a*d)*(2*b^2*c^2 - 3*a*b*c*d + 5*a^2*d^2))/(8*a^2*c^3*(b*c - a*d)^3*x) + (d*(2*b*c + a*d))/(4*a*c*
(b*c - a*d)^2*x*(c + d*x^2)^2) + b/(2*a*(b*c - a*d)*x*(a + b*x^2)*(c + d*x^2)^2) + (d*(4*b^2*c^2 + 13*a*b*c*d
- 5*a^2*d^2))/(8*a*c^2*(b*c - a*d)^3*x*(c + d*x^2)) - (3*b^(7/2)*(b*c - 3*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2
*a^(5/2)*(b*c - a*d)^4) - (3*d^(5/2)*(21*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(
7/2)*(b*c - a*d)^4)

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 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 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} \, dx &=\frac{b}{2 a (b c-a d) x \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac{\int \frac{-3 b c+2 a d-7 b d x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx}{2 a (b c-a d)}\\ &=\frac{d (2 b c+a d)}{4 a c (b c-a d)^2 x \left (c+d x^2\right )^2}+\frac{b}{2 a (b c-a d) x \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac{\int \frac{-2 \left (6 b^2 c^2-8 a b c d+5 a^2 d^2\right )-10 b d (2 b c+a d) x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{8 a c (b c-a d)^2}\\ &=\frac{d (2 b c+a d)}{4 a c (b c-a d)^2 x \left (c+d x^2\right )^2}+\frac{b}{2 a (b c-a d) x \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d \left (4 b^2 c^2+13 a b c d-5 a^2 d^2\right )}{8 a c^2 (b c-a d)^3 x \left (c+d x^2\right )}-\frac{\int \frac{-6 (2 b c-a d) \left (2 b^2 c^2-3 a b c d+5 a^2 d^2\right )-6 b d \left (4 b^2 c^2+13 a b c d-5 a^2 d^2\right ) x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a c^2 (b c-a d)^3}\\ &=-\frac{3 (2 b c-a d) \left (2 b^2 c^2-3 a b c d+5 a^2 d^2\right )}{8 a^2 c^3 (b c-a d)^3 x}+\frac{d (2 b c+a d)}{4 a c (b c-a d)^2 x \left (c+d x^2\right )^2}+\frac{b}{2 a (b c-a d) x \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d \left (4 b^2 c^2+13 a b c d-5 a^2 d^2\right )}{8 a c^2 (b c-a d)^3 x \left (c+d x^2\right )}+\frac{\int \frac{-6 \left (4 b^4 c^4-8 a b^3 c^3 d-8 a^2 b^2 c^2 d^2+13 a^3 b c d^3-5 a^4 d^4\right )-6 b d (2 b c-a d) \left (2 b^2 c^2-3 a b c d+5 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a^2 c^3 (b c-a d)^3}\\ &=-\frac{3 (2 b c-a d) \left (2 b^2 c^2-3 a b c d+5 a^2 d^2\right )}{8 a^2 c^3 (b c-a d)^3 x}+\frac{d (2 b c+a d)}{4 a c (b c-a d)^2 x \left (c+d x^2\right )^2}+\frac{b}{2 a (b c-a d) x \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d \left (4 b^2 c^2+13 a b c d-5 a^2 d^2\right )}{8 a c^2 (b c-a d)^3 x \left (c+d x^2\right )}-\frac{\left (3 b^4 (b c-3 a d)\right ) \int \frac{1}{a+b x^2} \, dx}{2 a^2 (b c-a d)^4}-\frac{\left (3 d^3 \left (21 b^2 c^2-18 a b c d+5 a^2 d^2\right )\right ) \int \frac{1}{c+d x^2} \, dx}{8 c^3 (b c-a d)^4}\\ &=-\frac{3 (2 b c-a d) \left (2 b^2 c^2-3 a b c d+5 a^2 d^2\right )}{8 a^2 c^3 (b c-a d)^3 x}+\frac{d (2 b c+a d)}{4 a c (b c-a d)^2 x \left (c+d x^2\right )^2}+\frac{b}{2 a (b c-a d) x \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d \left (4 b^2 c^2+13 a b c d-5 a^2 d^2\right )}{8 a c^2 (b c-a d)^3 x \left (c+d x^2\right )}-\frac{3 b^{7/2} (b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} (b c-a d)^4}-\frac{3 d^{5/2} \left (21 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{7/2} (b c-a d)^4}\\ \end{align*}

Mathematica [A]  time = 0.439648, size = 210, normalized size = 0.71 \[ \frac{1}{8} \left (-\frac{3 d^{5/2} \left (5 a^2 d^2-18 a b c d+21 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{7/2} (b c-a d)^4}+\frac{4 b^4 x}{a^2 \left (a+b x^2\right ) (a d-b c)^3}+\frac{12 b^{7/2} (3 a d-b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (b c-a d)^4}-\frac{8}{a^2 c^3 x}+\frac{d^3 x (7 a d-15 b c)}{c^3 \left (c+d x^2\right ) (b c-a d)^3}-\frac{2 d^3 x}{c^2 \left (c+d x^2\right )^2 (b c-a d)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8/(a^2*c^3*x) + (4*b^4*x)/(a^2*(-(b*c) + a*d)^3*(a + b*x^2)) - (2*d^3*x)/(c^2*(b*c - a*d)^2*(c + d*x^2)^2) +
(d^3*(-15*b*c + 7*a*d)*x)/(c^3*(b*c - a*d)^3*(c + d*x^2)) + (12*b^(7/2)*(-(b*c) + 3*a*d)*ArcTan[(Sqrt[b]*x)/S
qrt[a]])/(a^(5/2)*(b*c - a*d)^4) - (3*d^(5/2)*(21*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]
])/(c^(7/2)*(b*c - a*d)^4))/8

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Maple [A]  time = 0.02, size = 428, normalized size = 1.4 \begin{align*} -{\frac{7\,{d}^{6}{x}^{3}{a}^{2}}{8\,{c}^{3} \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{11\,{d}^{5}{x}^{3}ab}{4\,{c}^{2} \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{15\,{d}^{4}{x}^{3}{b}^{2}}{8\,c \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{9\,{a}^{2}{d}^{5}x}{8\,{c}^{2} \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{13\,{d}^{4}abx}{4\,c \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{17\,{b}^{2}{d}^{3}x}{8\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{15\,{a}^{2}{d}^{5}}{8\,{c}^{3} \left ( ad-bc \right ) ^{4}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{27\,{d}^{4}ab}{4\,{c}^{2} \left ( ad-bc \right ) ^{4}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{63\,{b}^{2}{d}^{3}}{8\,c \left ( ad-bc \right ) ^{4}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{1}{{a}^{2}{c}^{3}x}}+{\frac{{b}^{4}xd}{2\,a \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) }}-{\frac{{b}^{5}xc}{2\,{a}^{2} \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) }}+{\frac{9\,{b}^{4}d}{2\,a \left ( ad-bc \right ) ^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,{b}^{5}c}{2\,{a}^{2} \left ( ad-bc \right ) ^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \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,x)

[Out]

-7/8*d^6/c^3/(a*d-b*c)^4/(d*x^2+c)^2*x^3*a^2+11/4*d^5/c^2/(a*d-b*c)^4/(d*x^2+c)^2*x^3*a*b-15/8*d^4/c/(a*d-b*c)
^4/(d*x^2+c)^2*x^3*b^2-9/8*d^5/c^2/(a*d-b*c)^4/(d*x^2+c)^2*a^2*x+13/4*d^4/c/(a*d-b*c)^4/(d*x^2+c)^2*a*b*x-17/8
*d^3/(a*d-b*c)^4/(d*x^2+c)^2*b^2*x-15/8*d^5/c^3/(a*d-b*c)^4/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a^2+27/4*d^4/c
^2/(a*d-b*c)^4/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a*b-63/8*d^3/c/(a*d-b*c)^4/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/
2))*b^2-1/a^2/c^3/x+1/2*b^4/a/(a*d-b*c)^4*x/(b*x^2+a)*d-1/2*b^5/a^2/(a*d-b*c)^4*x/(b*x^2+a)*c+9/2*b^4/a/(a*d-b
*c)^4/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*d-3/2*b^5/a^2/(a*d-b*c)^4/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c

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Maxima [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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 23.7778, size = 7524, normalized size = 25.33 \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,x, algorithm="fricas")

[Out]

[-1/16*(16*a*b^4*c^6 - 64*a^2*b^3*c^5*d + 96*a^3*b^2*c^4*d^2 - 64*a^4*b*c^3*d^3 + 16*a^5*c^2*d^4 + 6*(4*b^5*c^
4*d^2 - 12*a*b^4*c^3*d^3 + 21*a^2*b^3*c^2*d^4 - 18*a^3*b^2*c*d^5 + 5*a^4*b*d^6)*x^6 + 2*(24*b^5*c^5*d - 64*a*b
^4*c^4*d^2 + 81*a^2*b^3*c^3*d^3 - 27*a^3*b^2*c^2*d^4 - 29*a^4*b*c*d^5 + 15*a^5*d^6)*x^4 + 2*(12*b^5*c^6 - 20*a
*b^4*c^5*d - 16*a^2*b^3*c^4*d^2 + 81*a^3*b^2*c^3*d^3 - 82*a^4*b*c^2*d^4 + 25*a^5*c*d^5)*x^2 + 12*((b^5*c^4*d^2
- 3*a*b^4*c^3*d^3)*x^7 + (2*b^5*c^5*d - 5*a*b^4*c^4*d^2 - 3*a^2*b^3*c^3*d^3)*x^5 + (b^5*c^6 - a*b^4*c^5*d - 6
*a^2*b^3*c^4*d^2)*x^3 + (a*b^4*c^6 - 3*a^2*b^3*c^5*d)*x)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2
+ a)) - 3*((21*a^2*b^3*c^2*d^4 - 18*a^3*b^2*c*d^5 + 5*a^4*b*d^6)*x^7 + (42*a^2*b^3*c^3*d^3 - 15*a^3*b^2*c^2*d^
4 - 8*a^4*b*c*d^5 + 5*a^5*d^6)*x^5 + (21*a^2*b^3*c^4*d^2 + 24*a^3*b^2*c^3*d^3 - 31*a^4*b*c^2*d^4 + 10*a^5*c*d^
5)*x^3 + (21*a^3*b^2*c^4*d^2 - 18*a^4*b*c^3*d^3 + 5*a^5*c^2*d^4)*x)*sqrt(-d/c)*log((d*x^2 - 2*c*x*sqrt(-d/c) -
c)/(d*x^2 + c)))/((a^2*b^5*c^7*d^2 - 4*a^3*b^4*c^6*d^3 + 6*a^4*b^3*c^5*d^4 - 4*a^5*b^2*c^4*d^5 + a^6*b*c^3*d^
6)*x^7 + (2*a^2*b^5*c^8*d - 7*a^3*b^4*c^7*d^2 + 8*a^4*b^3*c^6*d^3 - 2*a^5*b^2*c^5*d^4 - 2*a^6*b*c^4*d^5 + a^7*
c^3*d^6)*x^5 + (a^2*b^5*c^9 - 2*a^3*b^4*c^8*d - 2*a^4*b^3*c^7*d^2 + 8*a^5*b^2*c^6*d^3 - 7*a^6*b*c^5*d^4 + 2*a^
7*c^4*d^5)*x^3 + (a^3*b^4*c^9 - 4*a^4*b^3*c^8*d + 6*a^5*b^2*c^7*d^2 - 4*a^6*b*c^6*d^3 + a^7*c^5*d^4)*x), -1/8*
(8*a*b^4*c^6 - 32*a^2*b^3*c^5*d + 48*a^3*b^2*c^4*d^2 - 32*a^4*b*c^3*d^3 + 8*a^5*c^2*d^4 + 3*(4*b^5*c^4*d^2 - 1
2*a*b^4*c^3*d^3 + 21*a^2*b^3*c^2*d^4 - 18*a^3*b^2*c*d^5 + 5*a^4*b*d^6)*x^6 + (24*b^5*c^5*d - 64*a*b^4*c^4*d^2
+ 81*a^2*b^3*c^3*d^3 - 27*a^3*b^2*c^2*d^4 - 29*a^4*b*c*d^5 + 15*a^5*d^6)*x^4 + (12*b^5*c^6 - 20*a*b^4*c^5*d -
16*a^2*b^3*c^4*d^2 + 81*a^3*b^2*c^3*d^3 - 82*a^4*b*c^2*d^4 + 25*a^5*c*d^5)*x^2 + 3*((21*a^2*b^3*c^2*d^4 - 18*a
^3*b^2*c*d^5 + 5*a^4*b*d^6)*x^7 + (42*a^2*b^3*c^3*d^3 - 15*a^3*b^2*c^2*d^4 - 8*a^4*b*c*d^5 + 5*a^5*d^6)*x^5 +
(21*a^2*b^3*c^4*d^2 + 24*a^3*b^2*c^3*d^3 - 31*a^4*b*c^2*d^4 + 10*a^5*c*d^5)*x^3 + (21*a^3*b^2*c^4*d^2 - 18*a^4
*b*c^3*d^3 + 5*a^5*c^2*d^4)*x)*sqrt(d/c)*arctan(x*sqrt(d/c)) + 6*((b^5*c^4*d^2 - 3*a*b^4*c^3*d^3)*x^7 + (2*b^5
*c^5*d - 5*a*b^4*c^4*d^2 - 3*a^2*b^3*c^3*d^3)*x^5 + (b^5*c^6 - a*b^4*c^5*d - 6*a^2*b^3*c^4*d^2)*x^3 + (a*b^4*c
^6 - 3*a^2*b^3*c^5*d)*x)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/((a^2*b^5*c^7*d^2 - 4*a^3
*b^4*c^6*d^3 + 6*a^4*b^3*c^5*d^4 - 4*a^5*b^2*c^4*d^5 + a^6*b*c^3*d^6)*x^7 + (2*a^2*b^5*c^8*d - 7*a^3*b^4*c^7*d
^2 + 8*a^4*b^3*c^6*d^3 - 2*a^5*b^2*c^5*d^4 - 2*a^6*b*c^4*d^5 + a^7*c^3*d^6)*x^5 + (a^2*b^5*c^9 - 2*a^3*b^4*c^8
*d - 2*a^4*b^3*c^7*d^2 + 8*a^5*b^2*c^6*d^3 - 7*a^6*b*c^5*d^4 + 2*a^7*c^4*d^5)*x^3 + (a^3*b^4*c^9 - 4*a^4*b^3*c
^8*d + 6*a^5*b^2*c^7*d^2 - 4*a^6*b*c^6*d^3 + a^7*c^5*d^4)*x), -1/16*(16*a*b^4*c^6 - 64*a^2*b^3*c^5*d + 96*a^3*
b^2*c^4*d^2 - 64*a^4*b*c^3*d^3 + 16*a^5*c^2*d^4 + 6*(4*b^5*c^4*d^2 - 12*a*b^4*c^3*d^3 + 21*a^2*b^3*c^2*d^4 - 1
8*a^3*b^2*c*d^5 + 5*a^4*b*d^6)*x^6 + 2*(24*b^5*c^5*d - 64*a*b^4*c^4*d^2 + 81*a^2*b^3*c^3*d^3 - 27*a^3*b^2*c^2*
d^4 - 29*a^4*b*c*d^5 + 15*a^5*d^6)*x^4 + 2*(12*b^5*c^6 - 20*a*b^4*c^5*d - 16*a^2*b^3*c^4*d^2 + 81*a^3*b^2*c^3*
d^3 - 82*a^4*b*c^2*d^4 + 25*a^5*c*d^5)*x^2 + 24*((b^5*c^4*d^2 - 3*a*b^4*c^3*d^3)*x^7 + (2*b^5*c^5*d - 5*a*b^4*
c^4*d^2 - 3*a^2*b^3*c^3*d^3)*x^5 + (b^5*c^6 - a*b^4*c^5*d - 6*a^2*b^3*c^4*d^2)*x^3 + (a*b^4*c^6 - 3*a^2*b^3*c^
5*d)*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) - 3*((21*a^2*b^3*c^2*d^4 - 18*a^3*b^2*c*d^5 + 5*a^4*b*d^6)*x^7 + (42*a^2
*b^3*c^3*d^3 - 15*a^3*b^2*c^2*d^4 - 8*a^4*b*c*d^5 + 5*a^5*d^6)*x^5 + (21*a^2*b^3*c^4*d^2 + 24*a^3*b^2*c^3*d^3
- 31*a^4*b*c^2*d^4 + 10*a^5*c*d^5)*x^3 + (21*a^3*b^2*c^4*d^2 - 18*a^4*b*c^3*d^3 + 5*a^5*c^2*d^4)*x)*sqrt(-d/c)
*log((d*x^2 - 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^5*c^7*d^2 - 4*a^3*b^4*c^6*d^3 + 6*a^4*b^3*c^5*d^4 -
4*a^5*b^2*c^4*d^5 + a^6*b*c^3*d^6)*x^7 + (2*a^2*b^5*c^8*d - 7*a^3*b^4*c^7*d^2 + 8*a^4*b^3*c^6*d^3 - 2*a^5*b^2*
c^5*d^4 - 2*a^6*b*c^4*d^5 + a^7*c^3*d^6)*x^5 + (a^2*b^5*c^9 - 2*a^3*b^4*c^8*d - 2*a^4*b^3*c^7*d^2 + 8*a^5*b^2*
c^6*d^3 - 7*a^6*b*c^5*d^4 + 2*a^7*c^4*d^5)*x^3 + (a^3*b^4*c^9 - 4*a^4*b^3*c^8*d + 6*a^5*b^2*c^7*d^2 - 4*a^6*b*
c^6*d^3 + a^7*c^5*d^4)*x), -1/8*(8*a*b^4*c^6 - 32*a^2*b^3*c^5*d + 48*a^3*b^2*c^4*d^2 - 32*a^4*b*c^3*d^3 + 8*a^
5*c^2*d^4 + 3*(4*b^5*c^4*d^2 - 12*a*b^4*c^3*d^3 + 21*a^2*b^3*c^2*d^4 - 18*a^3*b^2*c*d^5 + 5*a^4*b*d^6)*x^6 + (
24*b^5*c^5*d - 64*a*b^4*c^4*d^2 + 81*a^2*b^3*c^3*d^3 - 27*a^3*b^2*c^2*d^4 - 29*a^4*b*c*d^5 + 15*a^5*d^6)*x^4 +
(12*b^5*c^6 - 20*a*b^4*c^5*d - 16*a^2*b^3*c^4*d^2 + 81*a^3*b^2*c^3*d^3 - 82*a^4*b*c^2*d^4 + 25*a^5*c*d^5)*x^2
+ 12*((b^5*c^4*d^2 - 3*a*b^4*c^3*d^3)*x^7 + (2*b^5*c^5*d - 5*a*b^4*c^4*d^2 - 3*a^2*b^3*c^3*d^3)*x^5 + (b^5*c^
6 - a*b^4*c^5*d - 6*a^2*b^3*c^4*d^2)*x^3 + (a*b^4*c^6 - 3*a^2*b^3*c^5*d)*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 3*
((21*a^2*b^3*c^2*d^4 - 18*a^3*b^2*c*d^5 + 5*a^4*b*d^6)*x^7 + (42*a^2*b^3*c^3*d^3 - 15*a^3*b^2*c^2*d^4 - 8*a^4*
b*c*d^5 + 5*a^5*d^6)*x^5 + (21*a^2*b^3*c^4*d^2 + 24*a^3*b^2*c^3*d^3 - 31*a^4*b*c^2*d^4 + 10*a^5*c*d^5)*x^3 + (
21*a^3*b^2*c^4*d^2 - 18*a^4*b*c^3*d^3 + 5*a^5*c^2*d^4)*x)*sqrt(d/c)*arctan(x*sqrt(d/c)))/((a^2*b^5*c^7*d^2 - 4
*a^3*b^4*c^6*d^3 + 6*a^4*b^3*c^5*d^4 - 4*a^5*b^2*c^4*d^5 + a^6*b*c^3*d^6)*x^7 + (2*a^2*b^5*c^8*d - 7*a^3*b^4*c
^7*d^2 + 8*a^4*b^3*c^6*d^3 - 2*a^5*b^2*c^5*d^4 - 2*a^6*b*c^4*d^5 + a^7*c^3*d^6)*x^5 + (a^2*b^5*c^9 - 2*a^3*b^4
*c^8*d - 2*a^4*b^3*c^7*d^2 + 8*a^5*b^2*c^6*d^3 - 7*a^6*b*c^5*d^4 + 2*a^7*c^4*d^5)*x^3 + (a^3*b^4*c^9 - 4*a^4*b
^3*c^8*d + 6*a^5*b^2*c^7*d^2 - 4*a^6*b*c^6*d^3 + a^7*c^5*d^4)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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,x)

[Out]

Timed out

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Giac [A]  time = 1.20822, size = 581, normalized size = 1.96 \begin{align*} -\frac{3 \,{\left (b^{5} c - 3 \, a b^{4} d\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \,{\left (a^{2} b^{4} c^{4} - 4 \, a^{3} b^{3} c^{3} d + 6 \, a^{4} b^{2} c^{2} d^{2} - 4 \, a^{5} b c d^{3} + a^{6} d^{4}\right )} \sqrt{a b}} - \frac{3 \,{\left (21 \, b^{2} c^{2} d^{3} - 18 \, a b c d^{4} + 5 \, a^{2} d^{5}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{8 \,{\left (b^{4} c^{7} - 4 \, a b^{3} c^{6} d + 6 \, a^{2} b^{2} c^{5} d^{2} - 4 \, a^{3} b c^{4} d^{3} + a^{4} c^{3} d^{4}\right )} \sqrt{c d}} - \frac{3 \, b^{4} c^{3} x^{2} - 6 \, a b^{3} c^{2} d x^{2} + 6 \, a^{2} b^{2} c d^{2} x^{2} - 2 \, a^{3} b d^{3} x^{2} + 2 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 6 \, a^{3} b c d^{2} - 2 \, a^{4} d^{3}}{2 \,{\left (a^{2} b^{3} c^{6} - 3 \, a^{3} b^{2} c^{5} d + 3 \, a^{4} b c^{4} d^{2} - a^{5} c^{3} d^{3}\right )}{\left (b x^{3} + a x\right )}} - \frac{15 \, b c d^{4} x^{3} - 7 \, a d^{5} x^{3} + 17 \, b c^{2} d^{3} x - 9 \, a c d^{4} x}{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 )}{\left (d x^{2} + c\right )}^{2}} \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,x, algorithm="giac")

[Out]

-3/2*(b^5*c - 3*a*b^4*d)*arctan(b*x/sqrt(a*b))/((a^2*b^4*c^4 - 4*a^3*b^3*c^3*d + 6*a^4*b^2*c^2*d^2 - 4*a^5*b*c
*d^3 + a^6*d^4)*sqrt(a*b)) - 3/8*(21*b^2*c^2*d^3 - 18*a*b*c*d^4 + 5*a^2*d^5)*arctan(d*x/sqrt(c*d))/((b^4*c^7 -
4*a*b^3*c^6*d + 6*a^2*b^2*c^5*d^2 - 4*a^3*b*c^4*d^3 + a^4*c^3*d^4)*sqrt(c*d)) - 1/2*(3*b^4*c^3*x^2 - 6*a*b^3*
c^2*d*x^2 + 6*a^2*b^2*c*d^2*x^2 - 2*a^3*b*d^3*x^2 + 2*a*b^3*c^3 - 6*a^2*b^2*c^2*d + 6*a^3*b*c*d^2 - 2*a^4*d^3)
/((a^2*b^3*c^6 - 3*a^3*b^2*c^5*d + 3*a^4*b*c^4*d^2 - a^5*c^3*d^3)*(b*x^3 + a*x)) - 1/8*(15*b*c*d^4*x^3 - 7*a*d
^5*x^3 + 17*b*c^2*d^3*x - 9*a*c*d^4*x)/((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)*(d*x^2 + c)^
2)