3.250 \(\int \frac{1}{x^4 (a+b x^2) (c+d x^2)^2} \, dx\)

Optimal. Leaf size=189 \[ \frac{-5 a^2 d^2+2 a b c d+2 b^2 c^2}{2 a^2 c^3 x (b c-a d)}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (b c-a d)^2}-\frac{d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{7/2} (b c-a d)^2}-\frac{2 b c-5 a d}{6 a c^2 x^3 (b c-a d)}-\frac{d}{2 c x^3 \left (c+d x^2\right ) (b c-a d)} \]

[Out]

-(2*b*c - 5*a*d)/(6*a*c^2*(b*c - a*d)*x^3) + (2*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)/(2*a^2*c^3*(b*c - a*d)*x) - d
/(2*c*(b*c - a*d)*x^3*(c + d*x^2)) + (b^(7/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(5/2)*(b*c - a*d)^2) - (d^(5/2)*
(7*b*c - 5*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(7/2)*(b*c - a*d)^2)

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Rubi [A]  time = 0.272061, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {472, 583, 522, 205} \[ \frac{-5 a^2 d^2+2 a b c d+2 b^2 c^2}{2 a^2 c^3 x (b c-a d)}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (b c-a d)^2}-\frac{d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{7/2} (b c-a d)^2}-\frac{2 b c-5 a d}{6 a c^2 x^3 (b c-a d)}-\frac{d}{2 c x^3 \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)*(c + d*x^2)^2),x]

[Out]

-(2*b*c - 5*a*d)/(6*a*c^2*(b*c - a*d)*x^3) + (2*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)/(2*a^2*c^3*(b*c - a*d)*x) - d
/(2*c*(b*c - a*d)*x^3*(c + d*x^2)) + (b^(7/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(5/2)*(b*c - a*d)^2) - (d^(5/2)*
(7*b*c - 5*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(7/2)*(b*c - a*d)^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 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^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx &=-\frac{d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac{\int \frac{2 b c-5 a d-5 b d x^2}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{2 c (b c-a d)}\\ &=-\frac{2 b c-5 a d}{6 a c^2 (b c-a d) x^3}-\frac{d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}-\frac{\int \frac{3 \left (2 b^2 c^2+2 a b c d-5 a^2 d^2\right )+3 b d (2 b c-5 a d) x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{6 a c^2 (b c-a d)}\\ &=-\frac{2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac{2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac{d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac{\int \frac{3 \left (2 b^3 c^3+2 a b^2 c^2 d+2 a^2 b c d^2-5 a^3 d^3\right )+3 b d \left (2 b^2 c^2+2 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}{6 a^2 c^3 (b c-a d)}\\ &=-\frac{2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac{2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac{d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac{b^4 \int \frac{1}{a+b x^2} \, dx}{a^2 (b c-a d)^2}-\frac{\left (d^3 (7 b c-5 a d)\right ) \int \frac{1}{c+d x^2} \, dx}{2 c^3 (b c-a d)^2}\\ &=-\frac{2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac{2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac{d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (b c-a d)^2}-\frac{d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{7/2} (b c-a d)^2}\\ \end{align*}

Mathematica [A]  time = 0.372498, size = 142, normalized size = 0.75 \[ \frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (a d-b c)^2}+\frac{2 a d+b c}{a^2 c^3 x}-\frac{d^3 x}{2 c^3 \left (c+d x^2\right ) (b c-a d)}-\frac{d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{7/2} (b c-a d)^2}-\frac{1}{3 a c^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)*(c + d*x^2)^2),x]

[Out]

-1/(3*a*c^2*x^3) + (b*c + 2*a*d)/(a^2*c^3*x) - (d^3*x)/(2*c^3*(b*c - a*d)*(c + d*x^2)) + (b^(7/2)*ArcTan[(Sqrt
[b]*x)/Sqrt[a]])/(a^(5/2)*(-(b*c) + a*d)^2) - (d^(5/2)*(7*b*c - 5*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(7/2)
*(b*c - a*d)^2)

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Maple [A]  time = 0.016, size = 191, normalized size = 1. \begin{align*}{\frac{{d}^{4}xa}{2\,{c}^{3} \left ( ad-bc \right ) ^{2} \left ( d{x}^{2}+c \right ) }}-{\frac{{d}^{3}xb}{2\,{c}^{2} \left ( ad-bc \right ) ^{2} \left ( d{x}^{2}+c \right ) }}+{\frac{5\,{d}^{4}a}{2\,{c}^{3} \left ( ad-bc \right ) ^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{7\,{d}^{3}b}{2\,{c}^{2} \left ( ad-bc \right ) ^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{1}{3\,a{c}^{2}{x}^{3}}}+2\,{\frac{d}{a{c}^{3}x}}+{\frac{b}{{a}^{2}{c}^{2}x}}+{\frac{{b}^{4}}{{a}^{2} \left ( ad-bc \right ) ^{2}}\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^4/(b*x^2+a)/(d*x^2+c)^2,x)

[Out]

1/2*d^4/c^3/(a*d-b*c)^2*x/(d*x^2+c)*a-1/2*d^3/c^2/(a*d-b*c)^2*x/(d*x^2+c)*b+5/2*d^4/c^3/(a*d-b*c)^2/(c*d)^(1/2
)*arctan(x*d/(c*d)^(1/2))*a-7/2*d^3/c^2/(a*d-b*c)^2/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*b-1/3/a/c^2/x^3+2/a/c^
3/x*d+1/a^2/c^2/x*b+1/a^2*b^4/(a*d-b*c)^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))

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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^4/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.90387, size = 2558, normalized size = 13.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

[-1/12*(4*a*b^2*c^4 - 8*a^2*b*c^3*d + 4*a^3*c^2*d^2 - 6*(2*b^3*c^3*d - 7*a^2*b*c*d^3 + 5*a^3*d^4)*x^4 - 4*(3*b
^3*c^4 - a*b^2*c^3*d - 7*a^2*b*c^2*d^2 + 5*a^3*c*d^3)*x^2 - 6*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt(-b/a)*log((b*
x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 3*((7*a^2*b*c*d^3 - 5*a^3*d^4)*x^5 + (7*a^2*b*c^2*d^2 - 5*a^3*c*d^3
)*x^3)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^2*c^5*d - 2*a^3*b*c^4*d^2 + a^4*c^3
*d^3)*x^5 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a^4*c^4*d^2)*x^3), -1/6*(2*a*b^2*c^4 - 4*a^2*b*c^3*d + 2*a^3*c^2*d^
2 - 3*(2*b^3*c^3*d - 7*a^2*b*c*d^3 + 5*a^3*d^4)*x^4 - 2*(3*b^3*c^4 - a*b^2*c^3*d - 7*a^2*b*c^2*d^2 + 5*a^3*c*d
^3)*x^2 + 3*((7*a^2*b*c*d^3 - 5*a^3*d^4)*x^5 + (7*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*x^3)*sqrt(d/c)*arctan(x*sqrt(d/
c)) - 3*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/((a^2*b^2*c^
5*d - 2*a^3*b*c^4*d^2 + a^4*c^3*d^3)*x^5 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a^4*c^4*d^2)*x^3), -1/12*(4*a*b^2*c^
4 - 8*a^2*b*c^3*d + 4*a^3*c^2*d^2 - 6*(2*b^3*c^3*d - 7*a^2*b*c*d^3 + 5*a^3*d^4)*x^4 - 4*(3*b^3*c^4 - a*b^2*c^3
*d - 7*a^2*b*c^2*d^2 + 5*a^3*c*d^3)*x^2 - 12*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 3*(
(7*a^2*b*c*d^3 - 5*a^3*d^4)*x^5 + (7*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*x^3)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c
) - c)/(d*x^2 + c)))/((a^2*b^2*c^5*d - 2*a^3*b*c^4*d^2 + a^4*c^3*d^3)*x^5 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a^4
*c^4*d^2)*x^3), -1/6*(2*a*b^2*c^4 - 4*a^2*b*c^3*d + 2*a^3*c^2*d^2 - 3*(2*b^3*c^3*d - 7*a^2*b*c*d^3 + 5*a^3*d^4
)*x^4 - 2*(3*b^3*c^4 - a*b^2*c^3*d - 7*a^2*b*c^2*d^2 + 5*a^3*c*d^3)*x^2 - 6*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt
(b/a)*arctan(x*sqrt(b/a)) + 3*((7*a^2*b*c*d^3 - 5*a^3*d^4)*x^5 + (7*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*x^3)*sqrt(d/c
)*arctan(x*sqrt(d/c)))/((a^2*b^2*c^5*d - 2*a^3*b*c^4*d^2 + a^4*c^3*d^3)*x^5 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a
^4*c^4*d^2)*x^3)]

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

[Out]

Timed out

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Giac [A]  time = 1.17051, size = 223, normalized size = 1.18 \begin{align*} \frac{b^{4} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt{a b}} - \frac{d^{3} x}{2 \,{\left (b c^{4} - a c^{3} d\right )}{\left (d x^{2} + c\right )}} - \frac{{\left (7 \, b c d^{3} - 5 \, a d^{4}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{2 \,{\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} \sqrt{c d}} + \frac{3 \, b c x^{2} + 6 \, a d x^{2} - a c}{3 \, a^{2} c^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="giac")

[Out]

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