Integrand size = 19, antiderivative size = 291 \[ \int \frac {\left (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx=\frac {b^2 x}{d^2}+\frac {(b c-a d)^2 x}{4 c d^2 \left (c+d x^4\right )}+\frac {(b c-a d) (5 b c+3 a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt {2} c^{7/4} d^{9/4}}-\frac {(b c-a d) (5 b c+3 a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt {2} c^{7/4} d^{9/4}}+\frac {(b c-a d) (5 b c+3 a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{16 \sqrt {2} c^{7/4} d^{9/4}}-\frac {(b c-a d) (5 b c+3 a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{16 \sqrt {2} c^{7/4} d^{9/4}} \]
b^2*x/d^2+1/4*(-a*d+b*c)^2*x/c/d^2/(d*x^4+c)-1/16*(-a*d+b*c)*(3*a*d+5*b*c) *arctan(-1+d^(1/4)*x*2^(1/2)/c^(1/4))/c^(7/4)/d^(9/4)*2^(1/2)-1/16*(-a*d+b *c)*(3*a*d+5*b*c)*arctan(1+d^(1/4)*x*2^(1/2)/c^(1/4))/c^(7/4)/d^(9/4)*2^(1 /2)+1/32*(-a*d+b*c)*(3*a*d+5*b*c)*ln(-c^(1/4)*d^(1/4)*x*2^(1/2)+c^(1/2)+x^ 2*d^(1/2))/c^(7/4)/d^(9/4)*2^(1/2)-1/32*(-a*d+b*c)*(3*a*d+5*b*c)*ln(c^(1/4 )*d^(1/4)*x*2^(1/2)+c^(1/2)+x^2*d^(1/2))/c^(7/4)/d^(9/4)*2^(1/2)
Time = 0.17 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx=\frac {32 b^2 \sqrt [4]{d} x+\frac {8 \sqrt [4]{d} (b c-a d)^2 x}{c \left (c+d x^4\right )}+\frac {2 \sqrt {2} \left (5 b^2 c^2-2 a b c d-3 a^2 d^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{7/4}}-\frac {2 \sqrt {2} \left (5 b^2 c^2-2 a b c d-3 a^2 d^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{7/4}}+\frac {\sqrt {2} \left (5 b^2 c^2-2 a b c d-3 a^2 d^2\right ) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{c^{7/4}}-\frac {\sqrt {2} \left (5 b^2 c^2-2 a b c d-3 a^2 d^2\right ) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{c^{7/4}}}{32 d^{9/4}} \]
(32*b^2*d^(1/4)*x + (8*d^(1/4)*(b*c - a*d)^2*x)/(c*(c + d*x^4)) + (2*Sqrt[ 2]*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1 /4)])/c^(7/4) - (2*Sqrt[2]*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(7/4) + (Sqrt[2]*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/c^(7/4) - (Sqrt[2]*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1 /4)*d^(1/4)*x + Sqrt[d]*x^2])/c^(7/4))/(32*d^(9/4))
Time = 0.52 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {915, 2009}
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.
| \(\displaystyle \int \frac {\left (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 915 |
| \(\displaystyle \int \left (\frac {b^2}{d^2}-\frac {-a^2 d^2+2 b d x^4 (b c-a d)+b^2 c^2}{d^2 \left (c+d x^4\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(b c-a d) (3 a d+5 b c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt {2} c^{7/4} d^{9/4}}-\frac {(b c-a d) (3 a d+5 b c) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{8 \sqrt {2} c^{7/4} d^{9/4}}+\frac {(b c-a d) (3 a d+5 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{16 \sqrt {2} c^{7/4} d^{9/4}}-\frac {(b c-a d) (3 a d+5 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{16 \sqrt {2} c^{7/4} d^{9/4}}+\frac {x (b c-a d)^2}{4 c d^2 \left (c+d x^4\right )}+\frac {b^2 x}{d^2}\) |
(b^2*x)/d^2 + ((b*c - a*d)^2*x)/(4*c*d^2*(c + d*x^4)) + ((b*c - a*d)*(5*b* c + 3*a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(8*Sqrt[2]*c^(7/4)*d^( 9/4)) - ((b*c - a*d)*(5*b*c + 3*a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4 )])/(8*Sqrt[2]*c^(7/4)*d^(9/4)) + ((b*c - a*d)*(5*b*c + 3*a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/(16*Sqrt[2]*c^(7/4)*d^(9/4)) - ((b*c - a*d)*(5*b*c + 3*a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x + S qrt[d]*x^2])/(16*Sqrt[2]*c^(7/4)*d^(9/4))
3.2.58.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a , b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.96 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.35
| method | result | size |
| risch | \(\frac {b^{2} x}{d^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x}{4 c \,d^{2} \left (d \,x^{4}+c \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{4}+c \right )}{\sum }\frac {\left (3 a^{2} d^{2}+2 a b c d -5 b^{2} c^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{16 d^{3} c}\) | \(101\) |
| default | \(\frac {b^{2} x}{d^{2}}+\frac {\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x}{4 c \left (d \,x^{4}+c \right )}+\frac {\left (3 a^{2} d^{2}+2 a b c d -5 b^{2} c^{2}\right ) \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}{x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c^{2}}}{d^{2}}\) | \(175\) |
b^2*x/d^2+1/4*(a^2*d^2-2*a*b*c*d+b^2*c^2)/c*x/d^2/(d*x^4+c)+1/16/d^3/c*sum ((3*a^2*d^2+2*a*b*c*d-5*b^2*c^2)/_R^3*ln(x-_R),_R=RootOf(_Z^4*d+c))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 1210, normalized size of antiderivative = 4.16 \[ \int \frac {\left (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx=\text {Too large to display} \]
1/16*(16*b^2*c*d*x^5 + (c*d^3*x^4 + c^2*d^2)*(-(625*b^8*c^8 - 1000*a*b^7*c ^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/ (c^7*d^9))^(1/4)*log(c^2*d^2*(-(625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b ^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3* d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4) - (5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*x) - (-I*c*d^3*x^4 - I*c^2*d^2)*(-( 625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^ 3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216* a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4)*log(I*c^2*d^2*(-(625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^ 4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4) - (5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*x) - (I *c*d^3*x^4 + I*c^2*d^2)*(-(625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^ 6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4)*log( -I*c^2*d^2*(-(625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640* a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2* c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4) - (5*b^2*c^2 - 2* a*b*c*d - 3*a^2*d^2)*x) - (c*d^3*x^4 + c^2*d^2)*(-(625*b^8*c^8 - 1000*a...
Time = 2.02 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx=\frac {b^{2} x}{d^{2}} + \frac {x \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{4 c^{2} d^{2} + 4 c d^{3} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} c^{7} d^{9} + 81 a^{8} d^{8} + 216 a^{7} b c d^{7} - 324 a^{6} b^{2} c^{2} d^{6} - 984 a^{5} b^{3} c^{3} d^{5} + 646 a^{4} b^{4} c^{4} d^{4} + 1640 a^{3} b^{5} c^{5} d^{3} - 900 a^{2} b^{6} c^{6} d^{2} - 1000 a b^{7} c^{7} d + 625 b^{8} c^{8}, \left ( t \mapsto t \log {\left (\frac {16 t c^{2} d^{2}}{3 a^{2} d^{2} + 2 a b c d - 5 b^{2} c^{2}} + x \right )} \right )\right )} \]
b**2*x/d**2 + x*(a**2*d**2 - 2*a*b*c*d + b**2*c**2)/(4*c**2*d**2 + 4*c*d** 3*x**4) + RootSum(65536*_t**4*c**7*d**9 + 81*a**8*d**8 + 216*a**7*b*c*d**7 - 324*a**6*b**2*c**2*d**6 - 984*a**5*b**3*c**3*d**5 + 646*a**4*b**4*c**4* d**4 + 1640*a**3*b**5*c**5*d**3 - 900*a**2*b**6*c**6*d**2 - 1000*a*b**7*c* *7*d + 625*b**8*c**8, Lambda(_t, _t*log(16*_t*c**2*d**2/(3*a**2*d**2 + 2*a *b*c*d - 5*b**2*c**2) + x)))
Time = 0.32 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx=\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{4 \, {\left (c d^{3} x^{4} + c^{2} d^{2}\right )}} + \frac {b^{2} x}{d^{2}} - \frac {\frac {2 \, \sqrt {2} {\left (5 \, b^{2} c^{2} - 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (5 \, b^{2} c^{2} - 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (5 \, b^{2} c^{2} - 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (5 \, b^{2} c^{2} - 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}}{32 \, c d^{2}} \]
1/4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x/(c*d^3*x^4 + c^2*d^2) + b^2*x/d^2 - 1/32*(2*sqrt(2)*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*arctan(1/2*sqrt(2)*(2* sqrt(d)*x + sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt( sqrt(c)*sqrt(d))) + 2*sqrt(2)*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*arctan(1 /2*sqrt(2)*(2*sqrt(d)*x - sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqrt(d)))/ (sqrt(c)*sqrt(sqrt(c)*sqrt(d))) + sqrt(2)*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d ^2)*log(sqrt(d)*x^2 + sqrt(2)*c^(1/4)*d^(1/4)*x + sqrt(c))/(c^(3/4)*d^(1/4 )) - sqrt(2)*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*log(sqrt(d)*x^2 - sqrt(2) *c^(1/4)*d^(1/4)*x + sqrt(c))/(c^(3/4)*d^(1/4)))/(c*d^2)
Time = 0.30 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx=\frac {b^{2} x}{d^{2}} - \frac {\sqrt {2} {\left (5 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{16 \, c^{2} d^{3}} - \frac {\sqrt {2} {\left (5 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{16 \, c^{2} d^{3}} - \frac {\sqrt {2} {\left (5 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{32 \, c^{2} d^{3}} + \frac {\sqrt {2} {\left (5 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{32 \, c^{2} d^{3}} + \frac {b^{2} c^{2} x - 2 \, a b c d x + a^{2} d^{2} x}{4 \, {\left (d x^{4} + c\right )} c d^{2}} \]
b^2*x/d^2 - 1/16*sqrt(2)*(5*(c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c* d - 3*(c*d^3)^(1/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(c/d)^(1/4) )/(c/d)^(1/4))/(c^2*d^3) - 1/16*sqrt(2)*(5*(c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^ 3)^(1/4)*a*b*c*d - 3*(c*d^3)^(1/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(2*x - sqrt (2)*(c/d)^(1/4))/(c/d)^(1/4))/(c^2*d^3) - 1/32*sqrt(2)*(5*(c*d^3)^(1/4)*b^ 2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d - 3*(c*d^3)^(1/4)*a^2*d^2)*log(x^2 + sqrt( 2)*x*(c/d)^(1/4) + sqrt(c/d))/(c^2*d^3) + 1/32*sqrt(2)*(5*(c*d^3)^(1/4)*b^ 2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d - 3*(c*d^3)^(1/4)*a^2*d^2)*log(x^2 - sqrt( 2)*x*(c/d)^(1/4) + sqrt(c/d))/(c^2*d^3) + 1/4*(b^2*c^2*x - 2*a*b*c*d*x + a ^2*d^2*x)/((d*x^4 + c)*c*d^2)
Time = 5.95 (sec) , antiderivative size = 1254, normalized size of antiderivative = 4.31 \[ \int \frac {\left (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx=\text {Too large to display} \]
(b^2*x)/d^2 + (x*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(4*c*(c*d^2 + d^3*x^4)) + (atan(((((x*(9*a^4*d^4 + 25*b^4*c^4 - 26*a^2*b^2*c^2*d^2 - 20*a*b^3*c^3* d + 12*a^3*b*c*d^3))/(4*c^2*d) - ((a*d - b*c)*(3*a*d + 5*b*c)*(12*a^2*d^3 - 20*b^2*c^2*d + 8*a*b*c*d^2))/(16*(-c)^(7/4)*d^(9/4)))*(a*d - b*c)*(3*a*d + 5*b*c)*1i)/(16*(-c)^(7/4)*d^(9/4)) + (((x*(9*a^4*d^4 + 25*b^4*c^4 - 26* a^2*b^2*c^2*d^2 - 20*a*b^3*c^3*d + 12*a^3*b*c*d^3))/(4*c^2*d) + ((a*d - b* c)*(3*a*d + 5*b*c)*(12*a^2*d^3 - 20*b^2*c^2*d + 8*a*b*c*d^2))/(16*(-c)^(7/ 4)*d^(9/4)))*(a*d - b*c)*(3*a*d + 5*b*c)*1i)/(16*(-c)^(7/4)*d^(9/4)))/(((( x*(9*a^4*d^4 + 25*b^4*c^4 - 26*a^2*b^2*c^2*d^2 - 20*a*b^3*c^3*d + 12*a^3*b *c*d^3))/(4*c^2*d) - ((a*d - b*c)*(3*a*d + 5*b*c)*(12*a^2*d^3 - 20*b^2*c^2 *d + 8*a*b*c*d^2))/(16*(-c)^(7/4)*d^(9/4)))*(a*d - b*c)*(3*a*d + 5*b*c))/( 16*(-c)^(7/4)*d^(9/4)) - (((x*(9*a^4*d^4 + 25*b^4*c^4 - 26*a^2*b^2*c^2*d^2 - 20*a*b^3*c^3*d + 12*a^3*b*c*d^3))/(4*c^2*d) + ((a*d - b*c)*(3*a*d + 5*b *c)*(12*a^2*d^3 - 20*b^2*c^2*d + 8*a*b*c*d^2))/(16*(-c)^(7/4)*d^(9/4)))*(a *d - b*c)*(3*a*d + 5*b*c))/(16*(-c)^(7/4)*d^(9/4))))*(a*d - b*c)*(3*a*d + 5*b*c)*1i)/(8*(-c)^(7/4)*d^(9/4)) + (atan(((((x*(9*a^4*d^4 + 25*b^4*c^4 - 26*a^2*b^2*c^2*d^2 - 20*a*b^3*c^3*d + 12*a^3*b*c*d^3))/(4*c^2*d) - ((a*d - b*c)*(3*a*d + 5*b*c)*(12*a^2*d^3 - 20*b^2*c^2*d + 8*a*b*c*d^2)*1i)/(16*(- c)^(7/4)*d^(9/4)))*(a*d - b*c)*(3*a*d + 5*b*c))/(16*(-c)^(7/4)*d^(9/4)) + (((x*(9*a^4*d^4 + 25*b^4*c^4 - 26*a^2*b^2*c^2*d^2 - 20*a*b^3*c^3*d + 12...