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

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

[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)

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Rubi [A]  time = 0.309268, antiderivative size = 211, normalized size of antiderivative = 1., 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 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 ) \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*}

Mathematica [A]  time = 0.389394, size = 172, normalized size = 0.82 \[ \frac{1}{8} \left (\frac{d^{3/2} \left (15 a^2 d^2-42 a b c d+35 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{7/2} (b c-a d)^3}+\frac{8 b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2} (a d-b c)^3}+\frac{d^2 x (11 b c-7 a d)}{c^3 \left (c+d x^2\right ) (b c-a d)^2}+\frac{2 d^2 x}{c^2 \left (c+d x^2\right )^2 (b c-a d)}-\frac{8}{a c^3 x}\right ) \]

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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Maple [A]  time = 0.017, size = 335, normalized size = 1.6 \begin{align*} -{\frac{7\,{d}^{5}{x}^{3}{a}^{2}}{8\,{c}^{3} \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{9\,{d}^{4}{x}^{3}ab}{4\,{c}^{2} \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{11\,{d}^{3}{x}^{3}{b}^{2}}{8\,c \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{9\,{a}^{2}{d}^{4}x}{8\,{c}^{2} \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{11\,a{d}^{3}bx}{4\,c \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{13\,{d}^{2}{b}^{2}x}{8\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{15\,{a}^{2}{d}^{4}}{8\,{c}^{3} \left ( ad-bc \right ) ^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{21\,a{d}^{3}b}{4\,{c}^{2} \left ( ad-bc \right ) ^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{35\,{d}^{2}{b}^{2}}{8\,c \left ( ad-bc \right ) ^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{1}{a{c}^{3}x}}+{\frac{{b}^{4}}{a \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^2/(b*x^2+a)/(d*x^2+c)^3,x)

[Out]

-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(x*d/(c*d)^(1/2))*a^2+21/4*d^3/c^
2/(a*d-b*c)^3/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a*b-35/8*d^2/c/(a*d-b*c)^3/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2
))*b^2-1/a/c^3/x+1/a*b^4/(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^2/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 12.5784, size = 4054, normalized size = 19.21 \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)/(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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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)/(d*x**2+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.17243, size = 319, normalized size = 1.51 \begin{align*} -\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} \end{align*}

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)