Integrand size = 23, antiderivative size = 129 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\left (3 a^2-10 a b+15 b^2\right ) x}{8 (a-b)^3}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^3 d}+\frac {(3 a-7 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d} \]
1/8*(3*a^2-10*a*b+15*b^2)*x/(a-b)^3+1/8*(3*a-7*b)*cos(d*x+c)*sin(d*x+c)/(a -b)^2/d+1/4*cos(d*x+c)^3*sin(d*x+c)/(a-b)/d-b^(5/2)*arctan(b^(1/2)*tan(d*x +c)/a^(1/2))/(a-b)^3/d/a^(1/2)
Time = 0.95 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {-32 b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a} \left (4 \left (3 a^2-10 a b+15 b^2\right ) (c+d x)+8 \left (a^2-3 a b+2 b^2\right ) \sin (2 (c+d x))+(a-b)^2 \sin (4 (c+d x))\right )}{32 \sqrt {a} (a-b)^3 d} \]
(-32*b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a]*(4*(3*a^2 - 10*a*b + 15*b^2)*(c + d*x) + 8*(a^2 - 3*a*b + 2*b^2)*Sin[2*(c + d*x)] + (a - b)^2*Sin[4*(c + d*x)]))/(32*Sqrt[a]*(a - b)^3*d)
Time = 0.35 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4158, 316, 25, 402, 25, 397, 216, 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.
| \(\displaystyle \int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^4 \left (a+b \tan (c+d x)^2\right )}dx\) |
| \(\Big \downarrow \) 4158 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right )^3 \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2}-\frac {\int -\frac {3 b \tan ^2(c+d x)+3 a-4 b}{\left (\tan ^2(c+d x)+1\right )^2 \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{4 (a-b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {3 b \tan ^2(c+d x)+3 a-4 b}{\left (\tan ^2(c+d x)+1\right )^2 \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {(3 a-7 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}-\frac {\int -\frac {3 a^2-7 b a+8 b^2+(3 a-7 b) b \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 (a-b)}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 a^2-7 b a+8 b^2+(3 a-7 b) b \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 (a-b)}+\frac {(3 a-7 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-10 a b+15 b^2\right ) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{a-b}-\frac {8 b^3 \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{2 (a-b)}+\frac {(3 a-7 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-10 a b+15 b^2\right ) \arctan (\tan (c+d x))}{a-b}-\frac {8 b^3 \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{2 (a-b)}+\frac {(3 a-7 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-10 a b+15 b^2\right ) \arctan (\tan (c+d x))}{a-b}-\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{2 (a-b)}+\frac {(3 a-7 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2}}{d}\) |
(Tan[c + d*x]/(4*(a - b)*(1 + Tan[c + d*x]^2)^2) + ((((3*a^2 - 10*a*b + 15 *b^2)*ArcTan[Tan[c + d*x]])/(a - b) - (8*b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d *x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*(a - b)) + ((3*a - 7*b)*Tan[c + d*x]) /(2*(a - b)*(1 + Tan[c + d*x]^2)))/(4*(a - b)))/d
3.5.61.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 5.00 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(\frac {-\frac {b^{3} \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{3} \sqrt {a b}}+\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {5}{4} a b +\frac {7}{8} b^{2}\right ) \tan \left (d x +c \right )^{3}+\left (-\frac {7}{4} a b +\frac {9}{8} b^{2}+\frac {5}{8} a^{2}\right ) \tan \left (d x +c \right )}{\left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-10 a b +15 b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a -b \right )^{3}}}{d}\) | \(130\) |
| default | \(\frac {-\frac {b^{3} \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{3} \sqrt {a b}}+\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {5}{4} a b +\frac {7}{8} b^{2}\right ) \tan \left (d x +c \right )^{3}+\left (-\frac {7}{4} a b +\frac {9}{8} b^{2}+\frac {5}{8} a^{2}\right ) \tan \left (d x +c \right )}{\left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-10 a b +15 b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a -b \right )^{3}}}{d}\) | \(130\) |
| risch | \(\frac {3 x \,a^{2}}{8 \left (a -b \right )^{3}}-\frac {5 x a b}{4 \left (a -b \right )^{3}}+\frac {15 x \,b^{2}}{8 \left (a -b \right )^{3}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 \left (a -b \right )^{2} d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b}{4 \left (a -b \right )^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b}{4 \left (a^{2}-2 a b +b^{2}\right ) d}+\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 a \left (a -b \right )^{3} d}-\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 a \left (a -b \right )^{3} d}+\frac {\sin \left (4 d x +4 c \right )}{32 \left (a -b \right ) d}\) | \(280\) |
1/d*(-b^3/(a-b)^3/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))+1/(a-b)^3*( ((3/8*a^2-5/4*a*b+7/8*b^2)*tan(d*x+c)^3+(-7/4*a*b+9/8*b^2+5/8*a^2)*tan(d*x +c))/(1+tan(d*x+c)^2)^2+1/8*(3*a^2-10*a*b+15*b^2)*arctan(tan(d*x+c))))
Time = 0.32 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.11 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [-\frac {2 \, b^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - {\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} d x - {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d}, \frac {4 \, b^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + {\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} d x + {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d}\right ] \]
[-1/8*(2*b^2*sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b^2)*cos(d*x + c)^2 - 4*((a^2 + a*b)*cos(d*x + c)^3 - a*b*cos(d*x + c)) *sqrt(-b/a)*sin(d*x + c) + b^2)/((a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(a *b - b^2)*cos(d*x + c)^2 + b^2)) - (3*a^2 - 10*a*b + 15*b^2)*d*x - (2*(a^2 - 2*a*b + b^2)*cos(d*x + c)^3 + (3*a^2 - 10*a*b + 7*b^2)*cos(d*x + c))*si n(d*x + c))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d), 1/8*(4*b^2*sqrt(b/a)*arct an(1/2*((a + b)*cos(d*x + c)^2 - b)*sqrt(b/a)/(b*cos(d*x + c)*sin(d*x + c) )) + (3*a^2 - 10*a*b + 15*b^2)*d*x + (2*(a^2 - 2*a*b + b^2)*cos(d*x + c)^3 + (3*a^2 - 10*a*b + 7*b^2)*cos(d*x + c))*sin(d*x + c))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d)]
Timed out. \[ \int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {\frac {8 \, b^{3} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (3 \, a - 7 \, b\right )} \tan \left (d x + c\right )^{3} + {\left (5 \, a - 9 \, b\right )} \tan \left (d x + c\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2} - 2 \, a b + b^{2}}}{8 \, d} \]
-1/8*(8*b^3*arctan(b*tan(d*x + c)/sqrt(a*b))/((a^3 - 3*a^2*b + 3*a*b^2 - b ^3)*sqrt(a*b)) - (3*a^2 - 10*a*b + 15*b^2)*(d*x + c)/(a^3 - 3*a^2*b + 3*a* b^2 - b^3) - ((3*a - 7*b)*tan(d*x + c)^3 + (5*a - 9*b)*tan(d*x + c))/((a^2 - 2*a*b + b^2)*tan(d*x + c)^4 + 2*(a^2 - 2*a*b + b^2)*tan(d*x + c)^2 + a^ 2 - 2*a*b + b^2))/d
Time = 0.52 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} b^{3}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {3 \, a \tan \left (d x + c\right )^{3} - 7 \, b \tan \left (d x + c\right )^{3} + 5 \, a \tan \left (d x + c\right ) - 9 \, b \tan \left (d x + c\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \]
-1/8*(8*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt( a*b)))*b^3/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)) - (3*a^2 - 10*a*b + 15*b^2)*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*a*tan(d*x + c)^3 - 7*b*tan(d*x + c)^3 + 5*a*tan(d*x + c) - 9*b*tan(d*x + c))/((a^2 - 2*a*b + b^2)*(tan(d*x + c)^2 + 1)^2))/d
Time = 15.98 (sec) , antiderivative size = 3681, normalized size of antiderivative = 28.53 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Too large to display} \]
((tan(c + d*x)*(5*a - 9*b))/(8*(a^2 - 2*a*b + b^2)) + (tan(c + d*x)^3*(3*a - 7*b))/(8*(a^2 - 2*a*b + b^2)))/(d*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1)) - (atan(((((tan(c + d*x)*(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a^3*b ^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (((256 *b^10 - 1760*a*b^9 + 5280*a^2*b^8 - 9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5 *b^5 + 3040*a^6*b^4 - 800*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b ^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - (tan(c + d*x)*(3*a^2 - 10*a*b + 15*b^2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 12 80*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))* (3*a^2 - 10*a*b + 15*b^2))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)))*( 3*a^2 - 10*a*b + 15*b^2)*1i)/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) + (((tan(c + d*x)*(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a^3*b^4 + 9*a^4* b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - (((256*b^10 - 176 0*a*b^9 + 5280*a^2*b^8 - 9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5*b^5 + 3040 *a^6*b^4 - 800*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(c + d*x)*(3*a^2 - 10*a*b + 1 5*b^2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(3*a^2 -...