3.260 \(\int \frac {1}{x^4 (a+b x^2) (c+d x^2)^3} \, dx\)

Optimal. Leaf size=270 \[ \frac {b^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^3}-\frac {d^{5/2} \left (35 a^2 d^2-90 a b c d+63 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{9/2} (b c-a d)^3}-\frac {35 a^2 d^2-55 a b c d+8 b^2 c^2}{24 a c^3 x^3 (b c-a d)^2}+\frac {35 a^3 d^3-55 a^2 b c d^2+8 a b^2 c^2 d+8 b^3 c^3}{8 a^2 c^4 x (b c-a d)^2}-\frac {d (11 b c-7 a d)}{8 c^2 x^3 \left (c+d x^2\right ) (b c-a d)^2}-\frac {d}{4 c x^3 \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

1/24*(-35*a^2*d^2+55*a*b*c*d-8*b^2*c^2)/a/c^3/(-a*d+b*c)^2/x^3+1/8*(35*a^3*d^3-55*a^2*b*c*d^2+8*a*b^2*c^2*d+8*
b^3*c^3)/a^2/c^4/(-a*d+b*c)^2/x-1/4*d/c/(-a*d+b*c)/x^3/(d*x^2+c)^2-1/8*d*(-7*a*d+11*b*c)/c^2/(-a*d+b*c)^2/x^3/
(d*x^2+c)+b^(9/2)*arctan(x*b^(1/2)/a^(1/2))/a^(5/2)/(-a*d+b*c)^3-1/8*d^(5/2)*(35*a^2*d^2-90*a*b*c*d+63*b^2*c^2
)*arctan(x*d^(1/2)/c^(1/2))/c^(9/2)/(-a*d+b*c)^3

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Rubi [A]  time = 0.43, antiderivative size = 270, normalized size of antiderivative = 1.00, 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} \[ -\frac {35 a^2 d^2-55 a b c d+8 b^2 c^2}{24 a c^3 x^3 (b c-a d)^2}+\frac {-55 a^2 b c d^2+35 a^3 d^3+8 a b^2 c^2 d+8 b^3 c^3}{8 a^2 c^4 x (b c-a d)^2}-\frac {d^{5/2} \left (35 a^2 d^2-90 a b c d+63 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{9/2} (b c-a d)^3}+\frac {b^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^3}-\frac {d (11 b c-7 a d)}{8 c^2 x^3 \left (c+d x^2\right ) (b c-a d)^2}-\frac {d}{4 c x^3 \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)*(c + d*x^2)^3),x]

[Out]

-(8*b^2*c^2 - 55*a*b*c*d + 35*a^2*d^2)/(24*a*c^3*(b*c - a*d)^2*x^3) + (8*b^3*c^3 + 8*a*b^2*c^2*d - 55*a^2*b*c*
d^2 + 35*a^3*d^3)/(8*a^2*c^4*(b*c - a*d)^2*x) - d/(4*c*(b*c - a*d)*x^3*(c + d*x^2)^2) - (d*(11*b*c - 7*a*d))/(
8*c^2*(b*c - a*d)^2*x^3*(c + d*x^2)) + (b^(9/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(5/2)*(b*c - a*d)^3) - (d^(5/2
)*(63*b^2*c^2 - 90*a*b*c*d + 35*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(9/2)*(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^4 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx &=-\frac {d}{4 c (b c-a d) x^3 \left (c+d x^2\right )^2}+\frac {\int \frac {4 b c-7 a d-7 b d x^2}{x^4 \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^3 \left (c+d x^2\right )^2}-\frac {d (11 b c-7 a d)}{8 c^2 (b c-a d)^2 x^3 \left (c+d x^2\right )}+\frac {\int \frac {8 b^2 c^2-55 a b c d+35 a^2 d^2-5 b d (11 b c-7 a d) x^2}{x^4 \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}-55 b d+\frac {35 a d^2}{c}}{24 c^2 (b c-a d)^2 x^3}-\frac {d}{4 c (b c-a d) x^3 \left (c+d x^2\right )^2}-\frac {d (11 b c-7 a d)}{8 c^2 (b c-a d)^2 x^3 \left (c+d x^2\right )}-\frac {\int \frac {3 \left (8 b^3 c^3+8 a b^2 c^2 d-55 a^2 b c d^2+35 a^3 d^3\right )+3 b d \left (8 b^2 c^2-55 a b c d+35 a^2 d^2\right ) x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{24 a c^3 (b c-a d)^2}\\ &=-\frac {\frac {8 b^2 c}{a}-55 b d+\frac {35 a d^2}{c}}{24 c^2 (b c-a d)^2 x^3}+\frac {8 b^3 c^3+8 a b^2 c^2 d-55 a^2 b c d^2+35 a^3 d^3}{8 a^2 c^4 (b c-a d)^2 x}-\frac {d}{4 c (b c-a d) x^3 \left (c+d x^2\right )^2}-\frac {d (11 b c-7 a d)}{8 c^2 (b c-a d)^2 x^3 \left (c+d x^2\right )}+\frac {\int \frac {3 \left (8 b^4 c^4+8 a b^3 c^3 d+8 a^2 b^2 c^2 d^2-55 a^3 b c d^3+35 a^4 d^4\right )+3 b d \left (8 b^3 c^3+8 a b^2 c^2 d-55 a^2 b c d^2+35 a^3 d^3\right ) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{24 a^2 c^4 (b c-a d)^2}\\ &=-\frac {\frac {8 b^2 c}{a}-55 b d+\frac {35 a d^2}{c}}{24 c^2 (b c-a d)^2 x^3}+\frac {8 b^3 c^3+8 a b^2 c^2 d-55 a^2 b c d^2+35 a^3 d^3}{8 a^2 c^4 (b c-a d)^2 x}-\frac {d}{4 c (b c-a d) x^3 \left (c+d x^2\right )^2}-\frac {d (11 b c-7 a d)}{8 c^2 (b c-a d)^2 x^3 \left (c+d x^2\right )}+\frac {b^5 \int \frac {1}{a+b x^2} \, dx}{a^2 (b c-a d)^3}-\frac {\left (d^3 \left (63 b^2 c^2-90 a b c d+35 a^2 d^2\right )\right ) \int \frac {1}{c+d x^2} \, dx}{8 c^4 (b c-a d)^3}\\ &=-\frac {\frac {8 b^2 c}{a}-55 b d+\frac {35 a d^2}{c}}{24 c^2 (b c-a d)^2 x^3}+\frac {8 b^3 c^3+8 a b^2 c^2 d-55 a^2 b c d^2+35 a^3 d^3}{8 a^2 c^4 (b c-a d)^2 x}-\frac {d}{4 c (b c-a d) x^3 \left (c+d x^2\right )^2}-\frac {d (11 b c-7 a d)}{8 c^2 (b c-a d)^2 x^3 \left (c+d x^2\right )}+\frac {b^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^3}-\frac {d^{5/2} \left (63 b^2 c^2-90 a b c d+35 a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{9/2} (b c-a d)^3}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 196, normalized size = 0.73 \[ -\frac {b^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (a d-b c)^3}-\frac {d^{5/2} \left (35 a^2 d^2-90 a b c d+63 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{9/2} (b c-a d)^3}+\frac {3 a d+b c}{a^2 c^4 x}-\frac {d^3 x (15 b c-11 a d)}{8 c^4 \left (c+d x^2\right ) (b c-a d)^2}-\frac {d^3 x}{4 c^3 \left (c+d x^2\right )^2 (b c-a d)}-\frac {1}{3 a c^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)*(c + d*x^2)^3),x]

[Out]

-1/3*1/(a*c^3*x^3) + (b*c + 3*a*d)/(a^2*c^4*x) - (d^3*x)/(4*c^3*(b*c - a*d)*(c + d*x^2)^2) - (d^3*(15*b*c - 11
*a*d)*x)/(8*c^4*(b*c - a*d)^2*(c + d*x^2)) - (b^(9/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(5/2)*(-(b*c) + a*d)^3)
- (d^(5/2)*(63*b^2*c^2 - 90*a*b*c*d + 35*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(9/2)*(b*c - a*d)^3)

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fricas [B]  time = 6.54, size = 2397, normalized size = 8.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

[-1/48*(16*a*b^3*c^6 - 48*a^2*b^2*c^5*d + 48*a^3*b*c^4*d^2 - 16*a^4*c^3*d^3 - 6*(8*b^4*c^4*d^2 - 63*a^2*b^2*c^
2*d^4 + 90*a^3*b*c*d^5 - 35*a^4*d^6)*x^6 - 2*(48*b^4*c^5*d - 8*a*b^3*c^4*d^2 - 315*a^2*b^2*c^3*d^3 + 450*a^3*b
*c^2*d^4 - 175*a^4*c*d^5)*x^4 - 16*(3*b^4*c^6 - 2*a*b^3*c^5*d - 12*a^2*b^2*c^4*d^2 + 18*a^3*b*c^3*d^3 - 7*a^4*
c^2*d^4)*x^2 + 24*(b^4*c^4*d^2*x^7 + 2*b^4*c^5*d*x^5 + b^4*c^6*x^3)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) -
a)/(b*x^2 + a)) + 3*((63*a^2*b^2*c^2*d^4 - 90*a^3*b*c*d^5 + 35*a^4*d^6)*x^7 + 2*(63*a^2*b^2*c^3*d^3 - 90*a^3*
b*c^2*d^4 + 35*a^4*c*d^5)*x^5 + (63*a^2*b^2*c^4*d^2 - 90*a^3*b*c^3*d^3 + 35*a^4*c^2*d^4)*x^3)*sqrt(-d/c)*log((
d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^3*c^7*d^2 - 3*a^3*b^2*c^6*d^3 + 3*a^4*b*c^5*d^4 - a^5*c^4*
d^5)*x^7 + 2*(a^2*b^3*c^8*d - 3*a^3*b^2*c^7*d^2 + 3*a^4*b*c^6*d^3 - a^5*c^5*d^4)*x^5 + (a^2*b^3*c^9 - 3*a^3*b^
2*c^8*d + 3*a^4*b*c^7*d^2 - a^5*c^6*d^3)*x^3), -1/24*(8*a*b^3*c^6 - 24*a^2*b^2*c^5*d + 24*a^3*b*c^4*d^2 - 8*a^
4*c^3*d^3 - 3*(8*b^4*c^4*d^2 - 63*a^2*b^2*c^2*d^4 + 90*a^3*b*c*d^5 - 35*a^4*d^6)*x^6 - (48*b^4*c^5*d - 8*a*b^3
*c^4*d^2 - 315*a^2*b^2*c^3*d^3 + 450*a^3*b*c^2*d^4 - 175*a^4*c*d^5)*x^4 - 8*(3*b^4*c^6 - 2*a*b^3*c^5*d - 12*a^
2*b^2*c^4*d^2 + 18*a^3*b*c^3*d^3 - 7*a^4*c^2*d^4)*x^2 + 3*((63*a^2*b^2*c^2*d^4 - 90*a^3*b*c*d^5 + 35*a^4*d^6)*
x^7 + 2*(63*a^2*b^2*c^3*d^3 - 90*a^3*b*c^2*d^4 + 35*a^4*c*d^5)*x^5 + (63*a^2*b^2*c^4*d^2 - 90*a^3*b*c^3*d^3 +
35*a^4*c^2*d^4)*x^3)*sqrt(d/c)*arctan(x*sqrt(d/c)) + 12*(b^4*c^4*d^2*x^7 + 2*b^4*c^5*d*x^5 + b^4*c^6*x^3)*sqrt
(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/((a^2*b^3*c^7*d^2 - 3*a^3*b^2*c^6*d^3 + 3*a^4*b*c^5*d^
4 - a^5*c^4*d^5)*x^7 + 2*(a^2*b^3*c^8*d - 3*a^3*b^2*c^7*d^2 + 3*a^4*b*c^6*d^3 - a^5*c^5*d^4)*x^5 + (a^2*b^3*c^
9 - 3*a^3*b^2*c^8*d + 3*a^4*b*c^7*d^2 - a^5*c^6*d^3)*x^3), -1/48*(16*a*b^3*c^6 - 48*a^2*b^2*c^5*d + 48*a^3*b*c
^4*d^2 - 16*a^4*c^3*d^3 - 6*(8*b^4*c^4*d^2 - 63*a^2*b^2*c^2*d^4 + 90*a^3*b*c*d^5 - 35*a^4*d^6)*x^6 - 2*(48*b^4
*c^5*d - 8*a*b^3*c^4*d^2 - 315*a^2*b^2*c^3*d^3 + 450*a^3*b*c^2*d^4 - 175*a^4*c*d^5)*x^4 - 16*(3*b^4*c^6 - 2*a*
b^3*c^5*d - 12*a^2*b^2*c^4*d^2 + 18*a^3*b*c^3*d^3 - 7*a^4*c^2*d^4)*x^2 - 48*(b^4*c^4*d^2*x^7 + 2*b^4*c^5*d*x^5
+ b^4*c^6*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 3*((63*a^2*b^2*c^2*d^4 - 90*a^3*b*c*d^5 + 35*a^4*d^6)*x^7 + 2*
(63*a^2*b^2*c^3*d^3 - 90*a^3*b*c^2*d^4 + 35*a^4*c*d^5)*x^5 + (63*a^2*b^2*c^4*d^2 - 90*a^3*b*c^3*d^3 + 35*a^4*c
^2*d^4)*x^3)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^3*c^7*d^2 - 3*a^3*b^2*c^6*d^3
+ 3*a^4*b*c^5*d^4 - a^5*c^4*d^5)*x^7 + 2*(a^2*b^3*c^8*d - 3*a^3*b^2*c^7*d^2 + 3*a^4*b*c^6*d^3 - a^5*c^5*d^4)*
x^5 + (a^2*b^3*c^9 - 3*a^3*b^2*c^8*d + 3*a^4*b*c^7*d^2 - a^5*c^6*d^3)*x^3), -1/24*(8*a*b^3*c^6 - 24*a^2*b^2*c^
5*d + 24*a^3*b*c^4*d^2 - 8*a^4*c^3*d^3 - 3*(8*b^4*c^4*d^2 - 63*a^2*b^2*c^2*d^4 + 90*a^3*b*c*d^5 - 35*a^4*d^6)*
x^6 - (48*b^4*c^5*d - 8*a*b^3*c^4*d^2 - 315*a^2*b^2*c^3*d^3 + 450*a^3*b*c^2*d^4 - 175*a^4*c*d^5)*x^4 - 8*(3*b^
4*c^6 - 2*a*b^3*c^5*d - 12*a^2*b^2*c^4*d^2 + 18*a^3*b*c^3*d^3 - 7*a^4*c^2*d^4)*x^2 - 24*(b^4*c^4*d^2*x^7 + 2*b
^4*c^5*d*x^5 + b^4*c^6*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 3*((63*a^2*b^2*c^2*d^4 - 90*a^3*b*c*d^5 + 35*a^4*d
^6)*x^7 + 2*(63*a^2*b^2*c^3*d^3 - 90*a^3*b*c^2*d^4 + 35*a^4*c*d^5)*x^5 + (63*a^2*b^2*c^4*d^2 - 90*a^3*b*c^3*d^
3 + 35*a^4*c^2*d^4)*x^3)*sqrt(d/c)*arctan(x*sqrt(d/c)))/((a^2*b^3*c^7*d^2 - 3*a^3*b^2*c^6*d^3 + 3*a^4*b*c^5*d^
4 - a^5*c^4*d^5)*x^7 + 2*(a^2*b^3*c^8*d - 3*a^3*b^2*c^7*d^2 + 3*a^4*b*c^6*d^3 - a^5*c^5*d^4)*x^5 + (a^2*b^3*c^
9 - 3*a^3*b^2*c^8*d + 3*a^4*b*c^7*d^2 - a^5*c^6*d^3)*x^3)]

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giac [A]  time = 0.31, size = 256, normalized size = 0.95 \[ \frac {b^{5} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt {a b}} - \frac {{\left (63 \, b^{2} c^{2} d^{3} - 90 \, a b c d^{4} + 35 \, a^{2} d^{5}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{7} - 3 \, a b^{2} c^{6} d + 3 \, a^{2} b c^{5} d^{2} - a^{3} c^{4} d^{3}\right )} \sqrt {c d}} - \frac {15 \, b c d^{4} x^{3} - 11 \, a d^{5} x^{3} + 17 \, b c^{2} d^{3} x - 13 \, a c d^{4} x}{8 \, {\left (b^{2} c^{6} - 2 \, a b c^{5} d + a^{2} c^{4} d^{2}\right )} {\left (d x^{2} + c\right )}^{2}} + \frac {3 \, b c x^{2} + 9 \, a d x^{2} - a c}{3 \, a^{2} c^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="giac")

[Out]

b^5*arctan(b*x/sqrt(a*b))/((a^2*b^3*c^3 - 3*a^3*b^2*c^2*d + 3*a^4*b*c*d^2 - a^5*d^3)*sqrt(a*b)) - 1/8*(63*b^2*
c^2*d^3 - 90*a*b*c*d^4 + 35*a^2*d^5)*arctan(d*x/sqrt(c*d))/((b^3*c^7 - 3*a*b^2*c^6*d + 3*a^2*b*c^5*d^2 - a^3*c
^4*d^3)*sqrt(c*d)) - 1/8*(15*b*c*d^4*x^3 - 11*a*d^5*x^3 + 17*b*c^2*d^3*x - 13*a*c*d^4*x)/((b^2*c^6 - 2*a*b*c^5
*d + a^2*c^4*d^2)*(d*x^2 + c)^2) + 1/3*(3*b*c*x^2 + 9*a*d*x^2 - a*c)/(a^2*c^4*x^3)

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maple [A]  time = 0.02, size = 362, normalized size = 1.34 \[ \frac {11 a^{2} d^{6} x^{3}}{8 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{4}}-\frac {13 a b \,d^{5} x^{3}}{4 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{3}}+\frac {15 b^{2} d^{4} x^{3}}{8 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{2}}+\frac {13 a^{2} d^{5} x}{8 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{3}}-\frac {15 a b \,d^{4} x}{4 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{2}}+\frac {17 b^{2} d^{3} x}{8 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c}+\frac {35 a^{2} d^{5} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \left (a d -b c \right )^{3} \sqrt {c d}\, c^{4}}-\frac {45 a b \,d^{4} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{4 \left (a d -b c \right )^{3} \sqrt {c d}\, c^{3}}-\frac {b^{5} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\left (a d -b c \right )^{3} \sqrt {a b}\, a^{2}}+\frac {63 b^{2} d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \left (a d -b c \right )^{3} \sqrt {c d}\, c^{2}}+\frac {3 d}{a \,c^{4} x}+\frac {b}{a^{2} c^{3} x}-\frac {1}{3 a \,c^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^2+a)/(d*x^2+c)^3,x)

[Out]

-1/a^2*b^5/(a*d-b*c)^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)+11/8*d^6/c^4/(a*d-b*c)^3/(d*x^2+c)^2*x^3*a^2-13/4
*d^5/c^3/(a*d-b*c)^3/(d*x^2+c)^2*x^3*a*b+15/8*d^4/c^2/(a*d-b*c)^3/(d*x^2+c)^2*x^3*b^2+13/8*d^5/c^3/(a*d-b*c)^3
/(d*x^2+c)^2*a^2*x-15/4*d^4/c^2/(a*d-b*c)^3/(d*x^2+c)^2*a*b*x+17/8*d^3/c/(a*d-b*c)^3/(d*x^2+c)^2*b^2*x+35/8*d^
5/c^4/(a*d-b*c)^3/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*a^2-45/4*d^4/c^3/(a*d-b*c)^3/(c*d)^(1/2)*arctan(1/(c*d
)^(1/2)*d*x)*a*b+63/8*d^3/c^2/(a*d-b*c)^3/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*b^2-1/3/a/c^3/x^3+3/a/c^4/x*d+
1/a^2/c^3/x*b

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maxima [A]  time = 2.58, size = 440, normalized size = 1.63 \[ \frac {b^{5} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt {a b}} - \frac {{\left (63 \, b^{2} c^{2} d^{3} - 90 \, a b c d^{4} + 35 \, a^{2} d^{5}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{7} - 3 \, a b^{2} c^{6} d + 3 \, a^{2} b c^{5} d^{2} - a^{3} c^{4} d^{3}\right )} \sqrt {c d}} - \frac {8 \, a b^{2} c^{5} - 16 \, a^{2} b c^{4} d + 8 \, a^{3} c^{3} d^{2} - 3 \, {\left (8 \, b^{3} c^{3} d^{2} + 8 \, a b^{2} c^{2} d^{3} - 55 \, a^{2} b c d^{4} + 35 \, a^{3} d^{5}\right )} x^{6} - {\left (48 \, b^{3} c^{4} d + 40 \, a b^{2} c^{3} d^{2} - 275 \, a^{2} b c^{2} d^{3} + 175 \, a^{3} c d^{4}\right )} x^{4} - 8 \, {\left (3 \, b^{3} c^{5} + a b^{2} c^{4} d - 11 \, a^{2} b c^{3} d^{2} + 7 \, a^{3} c^{2} d^{3}\right )} x^{2}}{24 \, {\left ({\left (a^{2} b^{2} c^{6} d^{2} - 2 \, a^{3} b c^{5} d^{3} + a^{4} c^{4} d^{4}\right )} x^{7} + 2 \, {\left (a^{2} b^{2} c^{7} d - 2 \, a^{3} b c^{6} d^{2} + a^{4} c^{5} d^{3}\right )} x^{5} + {\left (a^{2} b^{2} c^{8} - 2 \, a^{3} b c^{7} d + a^{4} c^{6} d^{2}\right )} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

b^5*arctan(b*x/sqrt(a*b))/((a^2*b^3*c^3 - 3*a^3*b^2*c^2*d + 3*a^4*b*c*d^2 - a^5*d^3)*sqrt(a*b)) - 1/8*(63*b^2*
c^2*d^3 - 90*a*b*c*d^4 + 35*a^2*d^5)*arctan(d*x/sqrt(c*d))/((b^3*c^7 - 3*a*b^2*c^6*d + 3*a^2*b*c^5*d^2 - a^3*c
^4*d^3)*sqrt(c*d)) - 1/24*(8*a*b^2*c^5 - 16*a^2*b*c^4*d + 8*a^3*c^3*d^2 - 3*(8*b^3*c^3*d^2 + 8*a*b^2*c^2*d^3 -
55*a^2*b*c*d^4 + 35*a^3*d^5)*x^6 - (48*b^3*c^4*d + 40*a*b^2*c^3*d^2 - 275*a^2*b*c^2*d^3 + 175*a^3*c*d^4)*x^4
- 8*(3*b^3*c^5 + a*b^2*c^4*d - 11*a^2*b*c^3*d^2 + 7*a^3*c^2*d^3)*x^2)/((a^2*b^2*c^6*d^2 - 2*a^3*b*c^5*d^3 + a^
4*c^4*d^4)*x^7 + 2*(a^2*b^2*c^7*d - 2*a^3*b*c^6*d^2 + a^4*c^5*d^3)*x^5 + (a^2*b^2*c^8 - 2*a^3*b*c^7*d + a^4*c^
6*d^2)*x^3)

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mupad [B]  time = 1.42, size = 785, normalized size = 2.91 \[ \frac {\frac {x^2\,\left (7\,a\,d+3\,b\,c\right )}{3\,a^2\,c^2}-\frac {1}{3\,a\,c}+\frac {x^4\,\left (175\,a^3\,d^4-275\,a^2\,b\,c\,d^3+40\,a\,b^2\,c^2\,d^2+48\,b^3\,c^3\,d\right )}{24\,a^2\,c^3\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {x^6\,\left (35\,a^3\,d^5-55\,a^2\,b\,c\,d^4+8\,a\,b^2\,c^2\,d^3+8\,b^3\,c^3\,d^2\right )}{8\,a^2\,c^4\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}}{c^2\,x^3+2\,c\,d\,x^5+d^2\,x^7}+\frac {\mathrm {atan}\left (\frac {b\,c^9\,x\,{\left (-a^5\,b^9\right )}^{3/2}\,64{}\mathrm {i}+a^{14}\,b\,d^9\,x\,\sqrt {-a^5\,b^9}\,1225{}\mathrm {i}+a^{10}\,b^5\,c^4\,d^5\,x\,\sqrt {-a^5\,b^9}\,3969{}\mathrm {i}-a^{11}\,b^4\,c^3\,d^6\,x\,\sqrt {-a^5\,b^9}\,11340{}\mathrm {i}+a^{12}\,b^3\,c^2\,d^7\,x\,\sqrt {-a^5\,b^9}\,12510{}\mathrm {i}-a^{13}\,b^2\,c\,d^8\,x\,\sqrt {-a^5\,b^9}\,6300{}\mathrm {i}}{-1225\,a^{17}\,b^5\,d^9+6300\,a^{16}\,b^6\,c\,d^8-12510\,a^{15}\,b^7\,c^2\,d^7+11340\,a^{14}\,b^8\,c^3\,d^6-3969\,a^{13}\,b^9\,c^4\,d^5+64\,a^8\,b^{14}\,c^9}\right )\,\sqrt {-a^5\,b^9}\,1{}\mathrm {i}}{a^8\,d^3-3\,a^7\,b\,c\,d^2+3\,a^6\,b^2\,c^2\,d-a^5\,b^3\,c^3}+\frac {\mathrm {atan}\left (\frac {a^9\,d^5\,x\,{\left (-c^9\,d^5\right )}^{3/2}\,1225{}\mathrm {i}+b^9\,c^{18}\,d\,x\,\sqrt {-c^9\,d^5}\,64{}\mathrm {i}-a^6\,b^3\,c^3\,d^2\,x\,{\left (-c^9\,d^5\right )}^{3/2}\,11340{}\mathrm {i}+a^7\,b^2\,c^2\,d^3\,x\,{\left (-c^9\,d^5\right )}^{3/2}\,12510{}\mathrm {i}-a^8\,b\,c\,d^4\,x\,{\left (-c^9\,d^5\right )}^{3/2}\,6300{}\mathrm {i}+a^5\,b^4\,c^4\,d\,x\,{\left (-c^9\,d^5\right )}^{3/2}\,3969{}\mathrm {i}}{1225\,a^9\,c^{14}\,d^{12}-6300\,a^8\,b\,c^{15}\,d^{11}+12510\,a^7\,b^2\,c^{16}\,d^{10}-11340\,a^6\,b^3\,c^{17}\,d^9+3969\,a^5\,b^4\,c^{18}\,d^8-64\,b^9\,c^{23}\,d^3}\right )\,\sqrt {-c^9\,d^5}\,\left (35\,a^2\,d^2-90\,a\,b\,c\,d+63\,b^2\,c^2\right )\,1{}\mathrm {i}}{8\,\left (-a^3\,c^9\,d^3+3\,a^2\,b\,c^{10}\,d^2-3\,a\,b^2\,c^{11}\,d+b^3\,c^{12}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^2)*(c + d*x^2)^3),x)

[Out]

((x^2*(7*a*d + 3*b*c))/(3*a^2*c^2) - 1/(3*a*c) + (x^4*(175*a^3*d^4 + 48*b^3*c^3*d + 40*a*b^2*c^2*d^2 - 275*a^2
*b*c*d^3))/(24*a^2*c^3*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (x^6*(35*a^3*d^5 + 8*b^3*c^3*d^2 + 8*a*b^2*c^2*d^3 -
55*a^2*b*c*d^4))/(8*a^2*c^4*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(c^2*x^3 + d^2*x^7 + 2*c*d*x^5) + (atan((b*c^9*
x*(-a^5*b^9)^(3/2)*64i + a^14*b*d^9*x*(-a^5*b^9)^(1/2)*1225i + a^10*b^5*c^4*d^5*x*(-a^5*b^9)^(1/2)*3969i - a^1
1*b^4*c^3*d^6*x*(-a^5*b^9)^(1/2)*11340i + a^12*b^3*c^2*d^7*x*(-a^5*b^9)^(1/2)*12510i - a^13*b^2*c*d^8*x*(-a^5*
b^9)^(1/2)*6300i)/(64*a^8*b^14*c^9 - 1225*a^17*b^5*d^9 + 6300*a^16*b^6*c*d^8 - 3969*a^13*b^9*c^4*d^5 + 11340*a
^14*b^8*c^3*d^6 - 12510*a^15*b^7*c^2*d^7))*(-a^5*b^9)^(1/2)*1i)/(a^8*d^3 - a^5*b^3*c^3 + 3*a^6*b^2*c^2*d - 3*a
^7*b*c*d^2) + (atan((a^9*d^5*x*(-c^9*d^5)^(3/2)*1225i + b^9*c^18*d*x*(-c^9*d^5)^(1/2)*64i - a^6*b^3*c^3*d^2*x*
(-c^9*d^5)^(3/2)*11340i + a^7*b^2*c^2*d^3*x*(-c^9*d^5)^(3/2)*12510i - a^8*b*c*d^4*x*(-c^9*d^5)^(3/2)*6300i + a
^5*b^4*c^4*d*x*(-c^9*d^5)^(3/2)*3969i)/(1225*a^9*c^14*d^12 - 64*b^9*c^23*d^3 - 6300*a^8*b*c^15*d^11 + 3969*a^5
*b^4*c^18*d^8 - 11340*a^6*b^3*c^17*d^9 + 12510*a^7*b^2*c^16*d^10))*(-c^9*d^5)^(1/2)*(35*a^2*d^2 + 63*b^2*c^2 -
90*a*b*c*d)*1i)/(8*(b^3*c^12 - a^3*c^9*d^3 + 3*a^2*b*c^10*d^2 - 3*a*b^2*c^11*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**4/(b*x**2+a)/(d*x**2+c)**3,x)

[Out]

Timed out

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