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

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

[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)

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Rubi [A]  time = 0.433726, antiderivative size = 270, 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} \[ -\frac{35 a^2 d^2-55 a b c d+8 b^2 c^2}{24 a c^3 x^3 (b c-a d)^2}+\frac{-55 a^2 b c d^2+35 a^3 d^3+8 a b^2 c^2 d+8 b^3 c^3}{8 a^2 c^4 x (b c-a d)^2}-\frac{d^{5/2} \left (35 a^2 d^2-90 a b c d+63 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{9/2} (b c-a d)^3}+\frac{b^{9/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (b c-a d)^3}-\frac{d (11 b c-7 a d)}{8 c^2 x^3 \left (c+d x^2\right ) (b c-a d)^2}-\frac{d}{4 c x^3 \left (c+d x^2\right )^2 (b c-a d)} \]

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 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^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*}

Mathematica [A]  time = 0.442556, size = 196, normalized size = 0.73 \[ -\frac{d^{5/2} \left (35 a^2 d^2-90 a b c d+63 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{9/2} (b c-a d)^3}-\frac{b^{9/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (a d-b c)^3}+\frac{3 a d+b c}{a^2 c^4 x}-\frac{d^3 x (15 b c-11 a d)}{8 c^4 \left (c+d x^2\right ) (b c-a d)^2}-\frac{d^3 x}{4 c^3 \left (c+d x^2\right )^2 (b c-a d)}-\frac{1}{3 a c^3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*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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Maple [A]  time = 0.017, size = 362, normalized size = 1.3 \begin{align*}{\frac{11\,{d}^{6}{x}^{3}{a}^{2}}{8\,{c}^{4} \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{13\,{d}^{5}{x}^{3}ab}{4\,{c}^{3} \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{15\,{d}^{4}{x}^{3}{b}^{2}}{8\,{c}^{2} \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{13\,{a}^{2}{d}^{5}x}{8\,{c}^{3} \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{15\,ab{d}^{4}x}{4\,{c}^{2} \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{17\,{b}^{2}{d}^{3}x}{8\,c \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{35\,{a}^{2}{d}^{5}}{8\,{c}^{4} \left ( ad-bc \right ) ^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{45\,ab{d}^{4}}{4\,{c}^{3} \left ( ad-bc \right ) ^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{63\,{b}^{2}{d}^{3}}{8\,{c}^{2} \left ( ad-bc \right ) ^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{1}{3\,a{c}^{3}{x}^{3}}}+3\,{\frac{d}{a{c}^{4}x}}+{\frac{b}{{a}^{2}{c}^{3}x}}-{\frac{{b}^{5}}{{a}^{2} \left ( ad-bc \right ) ^{3}}\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)^3,x)

[Out]

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(x*d/(c*d)^(1/2))*a^2-45/
4*d^4/c^3/(a*d-b*c)^3/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a*b+63/8*d^3/c^2/(a*d-b*c)^3/(c*d)^(1/2)*arctan(x*d/
(c*d)^(1/2))*b^2-1/3/a/c^3/x^3+3/a/c^4/x*d+1/a^2/c^3/x*b-1/a^2*b^5/(a*d-b*c)^3/(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 32.6831, size = 4841, normalized size = 17.93 \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)^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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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15645, size = 346, normalized size = 1.28 \begin{align*} \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}} \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)^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)