Integrand size = 22, antiderivative size = 211 \[ \int \frac {1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=-\frac {8 b^2 c^2-27 a b c d+15 a^2 d^2}{8 a c^3 (b c-a d)^2 x}-\frac {d}{4 c (b c-a d) x \left (c+d x^2\right )^2}-\frac {d (9 b c-5 a d)}{8 c^2 (b c-a d)^2 x \left (c+d x^2\right )}-\frac {b^{7/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (b c-a d)^3}+\frac {d^{3/2} \left (35 b^2 c^2-42 a b c d+15 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} (b c-a d)^3} \]
1/8*(-15*a^2*d^2+27*a*b*c*d-8*b^2*c^2)/a/c^3/(-a*d+b*c)^2/x-1/4*d/c/(-a*d+ b*c)/x/(d*x^2+c)^2-1/8*d*(-5*a*d+9*b*c)/c^2/(-a*d+b*c)^2/x/(d*x^2+c)-b^(7/ 2)*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/(-a*d+b*c)^3+1/8*d^(3/2)*(15*a^2*d^2- 42*a*b*c*d+35*b^2*c^2)*arctan(x*d^(1/2)/c^(1/2))/c^(7/2)/(-a*d+b*c)^3
Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {1}{8} \left (-\frac {8}{a c^3 x}+\frac {2 d^2 x}{c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac {d^2 (11 b c-7 a d) x}{c^3 (b c-a d)^2 \left (c+d x^2\right )}+\frac {8 b^{7/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (-b c+a d)^3}+\frac {d^{3/2} \left (35 b^2 c^2-42 a b c d+15 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2} (b c-a d)^3}\right ) \]
(-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
Time = 0.43 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {374, 441, 445, 397, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 374 |
| \(\displaystyle \frac {\int \frac {-5 b d x^2+4 b c-5 a d}{x^2 \left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}-\frac {d}{4 c x \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 441 |
| \(\displaystyle \frac {\frac {\int \frac {8 b^2 c^2-27 a b d c+15 a^2 d^2-3 b d (9 b c-5 a d) x^2}{x^2 \left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {d (9 b c-5 a d)}{2 c x \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d}{4 c x \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {8 b^3 c^3+8 a b^2 d c^2-27 a^2 b d^2 c+15 a^3 d^3+b d \left (8 b^2 c^2-27 a b d c+15 a^2 d^2\right ) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{a c}-\frac {\frac {8 b^2 c}{a}+\frac {15 a d^2}{c}-27 b d}{x}}{2 c (b c-a d)}-\frac {d (9 b c-5 a d)}{2 c x \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d}{4 c x \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {-\frac {\frac {8 b^4 c^3 \int \frac {1}{b x^2+a}dx}{b c-a d}-\frac {a d^2 \left (15 a^2 d^2-42 a b c d+35 b^2 c^2\right ) \int \frac {1}{d x^2+c}dx}{b c-a d}}{a c}-\frac {\frac {8 b^2 c}{a}+\frac {15 a d^2}{c}-27 b d}{x}}{2 c (b c-a d)}-\frac {d (9 b c-5 a d)}{2 c x \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d}{4 c x \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {-\frac {\frac {8 b^{7/2} c^3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} (b c-a d)}-\frac {a d^{3/2} \left (15 a^2 d^2-42 a b c d+35 b^2 c^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)}}{a c}-\frac {\frac {8 b^2 c}{a}+\frac {15 a d^2}{c}-27 b d}{x}}{2 c (b c-a d)}-\frac {d (9 b c-5 a d)}{2 c x \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d}{4 c x \left (c+d x^2\right )^2 (b c-a d)}\) |
-1/4*d/(c*(b*c - a*d)*x*(c + d*x^2)^2) + (-1/2*(d*(9*b*c - 5*a*d))/(c*(b*c - a*d)*x*(c + d*x^2)) + (-(((8*b^2*c)/a - 27*b*d + (15*a*d^2)/c)/x) - ((8 *b^(7/2)*c^3*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*(b*c - a*d)) - (a*d^(3/ 2)*(35*b^2*c^2 - 42*a*b*c*d + 15*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sq rt[c]*(b*c - a*d)))/(a*c))/(2*c*(b*c - a*d)))/(4*c*(b*c - a*d))
3.3.58.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*g*2*(b*c - a*d)*(p + 1))), x] + Si mp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2 )^q*Simp[c*(b*e - a*f)*(m + 1) + e*2*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, x] && LtQ[p, -1]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Time = 2.89 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.81
| method | result | size |
| default | \(-\frac {1}{a \,c^{3} x}+\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a \left (a d -b c \right )^{3} \sqrt {a b}}-\frac {d^{2} \left (\frac {\left (\frac {7}{8} a^{2} d^{3}-\frac {9}{4} a b c \,d^{2}+\frac {11}{8} b^{2} c^{2} d \right ) x^{3}+\frac {c \left (9 a^{2} d^{2}-22 a b c d +13 b^{2} c^{2}\right ) x}{8}}{\left (d \,x^{2}+c \right )^{2}}+\frac {\left (15 a^{2} d^{2}-42 a b c d +35 b^{2} c^{2}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \sqrt {c d}}\right )}{\left (a d -b c \right )^{3} c^{3}}\) | \(170\) |
| risch | \(\text {Expression too large to display}\) | \(1521\) |
-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))-d^2/(a* d-b*c)^3/c^3*(((7/8*a^2*d^3-9/4*a*b*c*d^2+11/8*b^2*c^2*d)*x^3+1/8*c*(9*a^2 *d^2-22*a*b*c*d+13*b^2*c^2)*x)/(d*x^2+c)^2+1/8*(15*a^2*d^2-42*a*b*c*d+35*b ^2*c^2)/(c*d)^(1/2)*arctan(d*x/(c*d)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (187) = 374\).
Time = 1.22 (sec) , antiderivative size = 1991, normalized size of antiderivative = 9.44 \[ \int \frac {1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \]
[-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 + 1 5*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)*arct an(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...
Timed out. \[ \int \frac {1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.67 \[ \int \frac {1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=-\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {a b}} + \frac {{\left (35 \, b^{2} c^{2} d^{2} - 42 \, a b c d^{3} + 15 \, a^{2} d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{6} - 3 \, a b^{2} c^{5} d + 3 \, a^{2} b c^{4} d^{2} - a^{3} c^{3} d^{3}\right )} \sqrt {c d}} - \frac {8 \, b^{2} c^{4} - 16 \, a b c^{3} d + 8 \, a^{2} c^{2} d^{2} + {\left (8 \, b^{2} c^{2} d^{2} - 27 \, a b c d^{3} + 15 \, a^{2} d^{4}\right )} x^{4} + {\left (16 \, b^{2} c^{3} d - 45 \, a b c^{2} d^{2} + 25 \, a^{2} c d^{3}\right )} x^{2}}{8 \, {\left ({\left (a b^{2} c^{5} d^{2} - 2 \, a^{2} b c^{4} d^{3} + a^{3} c^{3} d^{4}\right )} x^{5} + 2 \, {\left (a b^{2} c^{6} d - 2 \, a^{2} b c^{5} d^{2} + a^{3} c^{4} d^{3}\right )} x^{3} + {\left (a b^{2} c^{7} - 2 \, a^{2} b c^{6} d + a^{3} c^{5} d^{2}\right )} x\right )}} \]
-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)*ar ctan(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*(8*b^2*c^4 - 16*a*b*c^3*d + 8*a^2*c^2*d^2 + (8*b^2*c ^2*d^2 - 27*a*b*c*d^3 + 15*a^2*d^4)*x^4 + (16*b^2*c^3*d - 45*a*b*c^2*d^2 + 25*a^2*c*d^3)*x^2)/((a*b^2*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3*c^3*d^4)*x^5 + 2*(a*b^2*c^6*d - 2*a^2*b*c^5*d^2 + a^3*c^4*d^3)*x^3 + (a*b^2*c^7 - 2*a^2* b*c^6*d + a^3*c^5*d^2)*x)
Time = 0.29 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=-\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {a b}} + \frac {{\left (35 \, b^{2} c^{2} d^{2} - 42 \, a b c d^{3} + 15 \, a^{2} d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{6} - 3 \, a b^{2} c^{5} d + 3 \, a^{2} b c^{4} d^{2} - a^{3} c^{3} d^{3}\right )} \sqrt {c d}} + \frac {11 \, b c d^{3} x^{3} - 7 \, a d^{4} x^{3} + 13 \, b c^{2} d^{2} x - 9 \, a c d^{3} x}{8 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} {\left (d x^{2} + c\right )}^{2}} - \frac {1}{a c^{3} x} \]
-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)*ar ctan(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)
Time = 6.44 (sec) , antiderivative size = 738, normalized size of antiderivative = 3.50 \[ \int \frac {1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=-\frac {\frac {1}{a\,c}+\frac {x^4\,\left (15\,a^2\,d^4-27\,a\,b\,c\,d^3+8\,b^2\,c^2\,d^2\right )}{8\,a\,c^3\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {x^2\,\left (25\,a^2\,d^3-45\,a\,b\,c\,d^2+16\,b^2\,c^2\,d\right )}{8\,a\,c^2\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}}{c^2\,x+2\,c\,d\,x^3+d^2\,x^5}-\frac {\mathrm {atan}\left (\frac {b\,c^7\,x\,{\left (-a^3\,b^7\right )}^{3/2}\,64{}\mathrm {i}+a^{10}\,b\,d^7\,x\,\sqrt {-a^3\,b^7}\,225{}\mathrm {i}+a^6\,b^5\,c^4\,d^3\,x\,\sqrt {-a^3\,b^7}\,1225{}\mathrm {i}-a^7\,b^4\,c^3\,d^4\,x\,\sqrt {-a^3\,b^7}\,2940{}\mathrm {i}+a^8\,b^3\,c^2\,d^5\,x\,\sqrt {-a^3\,b^7}\,2814{}\mathrm {i}-a^9\,b^2\,c\,d^6\,x\,\sqrt {-a^3\,b^7}\,1260{}\mathrm {i}}{a^3\,b^7\,\left (2940\,a^6\,c^3\,d^4-1225\,a^5\,b\,c^4\,d^3+64\,a^2\,b^4\,c^7\right )-225\,a^{12}\,b^4\,d^7+1260\,a^{11}\,b^5\,c\,d^6-2814\,a^{10}\,b^6\,c^2\,d^5}\right )\,\sqrt {-a^3\,b^7}\,1{}\mathrm {i}}{a^6\,d^3-3\,a^5\,b\,c\,d^2+3\,a^4\,b^2\,c^2\,d-a^3\,b^3\,c^3}-\frac {\mathrm {atan}\left (\frac {a^7\,d^5\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,225{}\mathrm {i}+b^7\,c^{14}\,d\,x\,\sqrt {-c^7\,d^3}\,64{}\mathrm {i}-a^4\,b^3\,c^3\,d^2\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,2940{}\mathrm {i}+a^5\,b^2\,c^2\,d^3\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,2814{}\mathrm {i}-a^6\,b\,c\,d^4\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,1260{}\mathrm {i}+a^3\,b^4\,c^4\,d\,x\,{\left (-c^7\,d^3\right )}^{3/2}\,1225{}\mathrm {i}}{225\,a^7\,c^{11}\,d^9-1260\,a^6\,b\,c^{12}\,d^8+2814\,a^5\,b^2\,c^{13}\,d^7-2940\,a^4\,b^3\,c^{14}\,d^6+1225\,a^3\,b^4\,c^{15}\,d^5-64\,b^7\,c^{18}\,d^2}\right )\,\sqrt {-c^7\,d^3}\,\left (15\,a^2\,d^2-42\,a\,b\,c\,d+35\,b^2\,c^2\right )\,1{}\mathrm {i}}{8\,\left (-a^3\,c^7\,d^3+3\,a^2\,b\,c^8\,d^2-3\,a\,b^2\,c^9\,d+b^3\,c^{10}\right )} \]
- (1/(a*c) + (x^4*(15*a^2*d^4 + 8*b^2*c^2*d^2 - 27*a*b*c*d^3))/(8*a*c^3*(a ^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (x^2*(25*a^2*d^3 + 16*b^2*c^2*d - 45*a*b* c*d^2))/(8*a*c^2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(c^2*x + d^2*x^5 + 2*c* d*x^3) - (atan((b*c^7*x*(-a^3*b^7)^(3/2)*64i + a^10*b*d^7*x*(-a^3*b^7)^(1/ 2)*225i + a^6*b^5*c^4*d^3*x*(-a^3*b^7)^(1/2)*1225i - a^7*b^4*c^3*d^4*x*(-a ^3*b^7)^(1/2)*2940i + a^8*b^3*c^2*d^5*x*(-a^3*b^7)^(1/2)*2814i - a^9*b^2*c *d^6*x*(-a^3*b^7)^(1/2)*1260i)/(a^3*b^7*(64*a^2*b^4*c^7 + 2940*a^6*c^3*d^4 - 1225*a^5*b*c^4*d^3) - 225*a^12*b^4*d^7 + 1260*a^11*b^5*c*d^6 - 2814*a^1 0*b^6*c^2*d^5))*(-a^3*b^7)^(1/2)*1i)/(a^6*d^3 - a^3*b^3*c^3 + 3*a^4*b^2*c^ 2*d - 3*a^5*b*c*d^2) - (atan((a^7*d^5*x*(-c^7*d^3)^(3/2)*225i + b^7*c^14*d *x*(-c^7*d^3)^(1/2)*64i - a^4*b^3*c^3*d^2*x*(-c^7*d^3)^(3/2)*2940i + a^5*b ^2*c^2*d^3*x*(-c^7*d^3)^(3/2)*2814i - a^6*b*c*d^4*x*(-c^7*d^3)^(3/2)*1260i + a^3*b^4*c^4*d*x*(-c^7*d^3)^(3/2)*1225i)/(225*a^7*c^11*d^9 - 64*b^7*c^18 *d^2 - 1260*a^6*b*c^12*d^8 + 1225*a^3*b^4*c^15*d^5 - 2940*a^4*b^3*c^14*d^6 + 2814*a^5*b^2*c^13*d^7))*(-c^7*d^3)^(1/2)*(15*a^2*d^2 + 35*b^2*c^2 - 42* a*b*c*d)*1i)/(8*(b^3*c^10 - a^3*c^7*d^3 + 3*a^2*b*c^8*d^2 - 3*a*b^2*c^9*d) )