Integrand size = 24, antiderivative size = 197 \[ \int \frac {\csc ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {b^2 \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{11/4} \sqrt {\sqrt {a}-\sqrt {b}} d}+\frac {b^2 \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{11/4} \sqrt {\sqrt {a}+\sqrt {b}} d}-\frac {(a+b) \cot (c+d x)}{a^2 d}-\frac {(3 a+b) \cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot ^5(c+d x)}{5 a d}-\frac {\cot ^7(c+d x)}{7 a d} \]
-(a+b)*cot(d*x+c)/a^2/d-1/3*(3*a+b)*cot(d*x+c)^3/a^2/d-3/5*cot(d*x+c)^5/a/ d-1/7*cot(d*x+c)^7/a/d+1/2*b^2*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a ^(1/4))/a^(11/4)/d/(a^(1/2)-b^(1/2))^(1/2)+1/2*b^2*arctan((a^(1/2)+b^(1/2) )^(1/2)*tan(d*x+c)/a^(1/4))/a^(11/4)/d/(a^(1/2)+b^(1/2))^(1/2)
Time = 6.33 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.98 \[ \int \frac {\csc ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {105 b^2 \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {105 b^2 \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}-2 \sqrt {a} \cot (c+d x) \left (48 a+70 b+(24 a+35 b) \csc ^2(c+d x)+18 a \csc ^4(c+d x)+15 a \csc ^6(c+d x)\right )}{210 a^{5/2} d} \]
((105*b^2*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[ b]]])/Sqrt[a + Sqrt[a]*Sqrt[b]] - (105*b^2*ArcTanh[((Sqrt[a] - Sqrt[b])*Ta n[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/Sqrt[-a + Sqrt[a]*Sqrt[b]] - 2*Sq rt[a]*Cot[c + d*x]*(48*a + 70*b + (24*a + 35*b)*Csc[c + d*x]^2 + 18*a*Csc[ c + d*x]^4 + 15*a*Csc[c + d*x]^6))/(210*a^(5/2)*d)
Time = 0.41 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3696, 1610, 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.
\(\displaystyle \int \frac {\csc ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^8 \left (a-b \sin (c+d x)^4\right )}dx\) |
\(\Big \downarrow \) 3696 |
\(\displaystyle \frac {\int \frac {\cot ^8(c+d x) \left (\tan ^2(c+d x)+1\right )^5}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 1610 |
\(\displaystyle \frac {\int \left (\frac {\cot ^8(c+d x)}{a}+\frac {3 \cot ^6(c+d x)}{a}+\frac {(3 a+b) \cot ^4(c+d x)}{a^2}+\frac {(a+b) \cot ^2(c+d x)}{a^2}+\frac {b^2 \left (\tan ^2(c+d x)+1\right )}{a^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^2 \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{11/4} \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {b^2 \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{11/4} \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {(3 a+b) \cot ^3(c+d x)}{3 a^2}-\frac {(a+b) \cot (c+d x)}{a^2}-\frac {\cot ^7(c+d x)}{7 a}-\frac {3 \cot ^5(c+d x)}{5 a}}{d}\) |
((b^2*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(11/4)* Sqrt[Sqrt[a] - Sqrt[b]]) + (b^2*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d* x])/a^(1/4)])/(2*a^(11/4)*Sqrt[Sqrt[a] + Sqrt[b]]) - ((a + b)*Cot[c + d*x] )/a^2 - ((3*a + b)*Cot[c + d*x]^3)/(3*a^2) - (3*Cot[c + d*x]^5)/(5*a) - Co t[c + d*x]^7/(7*a))/d
3.3.11.3.1 Defintions of rubi rules used
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e*x^2)^q/(a + b*x^2 + c*x^4)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4 *a*c, 0] && IntegerQ[q] && IntegerQ[m]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2) ^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] & & IntegerQ[m/2] && IntegerQ[p]
Time = 2.95 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {-\frac {1}{7 a \tan \left (d x +c \right )^{7}}-\frac {a +b}{a^{2} \tan \left (d x +c \right )}-\frac {3 a +b}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {3}{5 a \tan \left (d x +c \right )^{5}}+\frac {b^{2} \left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{a^{2}}}{d}\) | \(213\) |
default | \(\frac {-\frac {1}{7 a \tan \left (d x +c \right )^{7}}-\frac {a +b}{a^{2} \tan \left (d x +c \right )}-\frac {3 a +b}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {3}{5 a \tan \left (d x +c \right )^{5}}+\frac {b^{2} \left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{a^{2}}}{d}\) | \(213\) |
risch | \(\frac {4 i \left (105 b \,{\mathrm e}^{10 i \left (d x +c \right )}-455 \,{\mathrm e}^{8 i \left (d x +c \right )} b +840 a \,{\mathrm e}^{6 i \left (d x +c \right )}+770 b \,{\mathrm e}^{6 i \left (d x +c \right )}-504 a \,{\mathrm e}^{4 i \left (d x +c \right )}-630 b \,{\mathrm e}^{4 i \left (d x +c \right )}+168 a \,{\mathrm e}^{2 i \left (d x +c \right )}+245 b \,{\mathrm e}^{2 i \left (d x +c \right )}-24 a -35 b \right )}{105 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{7}}+256 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (1099511627776 a^{12} d^{4}-1099511627776 a^{11} b \,d^{4}\right ) \textit {\_Z}^{4}+2097152 a^{6} b^{4} d^{2} \textit {\_Z}^{2}+b^{8}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {2147483648 i d^{3} a^{10}}{b^{7}}-\frac {2147483648 i d^{3} a^{9}}{b^{6}}\right ) \textit {\_R}^{3}+\left (-\frac {2097152 d^{2} a^{7}}{b^{5}}+\frac {2097152 d^{2} a^{6}}{b^{4}}\right ) \textit {\_R}^{2}+\left (\frac {2048 i d \,a^{4}}{b^{3}}+\frac {2048 i d \,a^{3}}{b^{2}}\right ) \textit {\_R} -\frac {2 a}{b}-1\right )\right )\) | \(272\) |
1/d*(-1/7/a/tan(d*x+c)^7-(a+b)/a^2/tan(d*x+c)-1/3*(3*a+b)/a^2/tan(d*x+c)^3 -3/5/a/tan(d*x+c)^5+b^2/a^2*(a-b)*(1/2*((a*b)^(1/2)+b)/(a*b)^(1/2)/(a-b)/( ((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b ))^(1/2))+1/2*((a*b)^(1/2)-b)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1 /2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 1585 vs. \(2 (155) = 310\).
Time = 0.44 (sec) , antiderivative size = 1585, normalized size of antiderivative = 8.05 \[ \int \frac {\csc ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
-1/840*(16*(24*a + 35*b)*cos(d*x + c)^7 - 56*(24*a + 35*b)*cos(d*x + c)^5 + 560*(3*a + 4*b)*cos(d*x + c)^3 + 105*(a^2*d*cos(d*x + c)^6 - 3*a^2*d*cos (d*x + c)^4 + 3*a^2*d*cos(d*x + c)^2 - a^2*d)*sqrt(-(b^4 + (a^6 - a^5*b)*s qrt(b^9/((a^13 - 2*a^12*b + a^11*b^2)*d^4))*d^2)/((a^6 - a^5*b)*d^2))*log( 1/4*b^7*cos(d*x + c)^2 - 1/4*b^7 - 1/4*(2*(a^7*b^2 - a^6*b^3)*d^2*cos(d*x + c)^2 - (a^7*b^2 - a^6*b^3)*d^2)*sqrt(b^9/((a^13 - 2*a^12*b + a^11*b^2)*d ^4)) + 1/2*(a^3*b^5*d*cos(d*x + c)*sin(d*x + c) - (a^10 - a^9*b)*sqrt(b^9/ ((a^13 - 2*a^12*b + a^11*b^2)*d^4))*d^3*cos(d*x + c)*sin(d*x + c))*sqrt(-( b^4 + (a^6 - a^5*b)*sqrt(b^9/((a^13 - 2*a^12*b + a^11*b^2)*d^4))*d^2)/((a^ 6 - a^5*b)*d^2)))*sin(d*x + c) - 105*(a^2*d*cos(d*x + c)^6 - 3*a^2*d*cos(d *x + c)^4 + 3*a^2*d*cos(d*x + c)^2 - a^2*d)*sqrt(-(b^4 + (a^6 - a^5*b)*sqr t(b^9/((a^13 - 2*a^12*b + a^11*b^2)*d^4))*d^2)/((a^6 - a^5*b)*d^2))*log(1/ 4*b^7*cos(d*x + c)^2 - 1/4*b^7 - 1/4*(2*(a^7*b^2 - a^6*b^3)*d^2*cos(d*x + c)^2 - (a^7*b^2 - a^6*b^3)*d^2)*sqrt(b^9/((a^13 - 2*a^12*b + a^11*b^2)*d^4 )) - 1/2*(a^3*b^5*d*cos(d*x + c)*sin(d*x + c) - (a^10 - a^9*b)*sqrt(b^9/(( a^13 - 2*a^12*b + a^11*b^2)*d^4))*d^3*cos(d*x + c)*sin(d*x + c))*sqrt(-(b^ 4 + (a^6 - a^5*b)*sqrt(b^9/((a^13 - 2*a^12*b + a^11*b^2)*d^4))*d^2)/((a^6 - a^5*b)*d^2)))*sin(d*x + c) - 105*(a^2*d*cos(d*x + c)^6 - 3*a^2*d*cos(d*x + c)^4 + 3*a^2*d*cos(d*x + c)^2 - a^2*d)*sqrt(-(b^4 - (a^6 - a^5*b)*sqrt( b^9/((a^13 - 2*a^12*b + a^11*b^2)*d^4))*d^2)/((a^6 - a^5*b)*d^2))*log(-...
Timed out. \[ \int \frac {\csc ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\csc \left (d x + c\right )^{8}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \]
4/105*(735*b*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) - 7*(15*b*sin(10*d*x + 10*c ) - 65*b*sin(8*d*x + 8*c) + 10*(12*a + 11*b)*sin(6*d*x + 6*c) - 18*(4*a + 5*b)*sin(4*d*x + 4*c) + (24*a + 35*b)*sin(2*d*x + 2*c))*cos(14*d*x + 14*c) + 49*(15*b*sin(10*d*x + 10*c) - 65*b*sin(8*d*x + 8*c) + 10*(12*a + 11*b)* sin(6*d*x + 6*c) - 18*(4*a + 5*b)*sin(4*d*x + 4*c) + (24*a + 35*b)*sin(2*d *x + 2*c))*cos(12*d*x + 12*c) + 147*(40*b*sin(8*d*x + 8*c) - 5*(24*a + 17* b)*sin(6*d*x + 6*c) + 3*(24*a + 25*b)*sin(4*d*x + 4*c) - 6*(4*a + 5*b)*sin (2*d*x + 2*c))*cos(10*d*x + 10*c) + 245*(15*(8*a + 3*b)*sin(6*d*x + 6*c) - 3*(24*a + 17*b)*sin(4*d*x + 4*c) + 2*(12*a + 11*b)*sin(2*d*x + 2*c))*cos( 8*d*x + 8*c) + 245*(24*b*sin(4*d*x + 4*c) - 13*b*sin(2*d*x + 2*c))*cos(6*d *x + 6*c) - 420*(a^2*b^2*d*cos(14*d*x + 14*c)^2 + 49*a^2*b^2*d*cos(12*d*x + 12*c)^2 + 441*a^2*b^2*d*cos(10*d*x + 10*c)^2 + 1225*a^2*b^2*d*cos(8*d*x + 8*c)^2 + 1225*a^2*b^2*d*cos(6*d*x + 6*c)^2 + 441*a^2*b^2*d*cos(4*d*x + 4 *c)^2 + 49*a^2*b^2*d*cos(2*d*x + 2*c)^2 + a^2*b^2*d*sin(14*d*x + 14*c)^2 + 49*a^2*b^2*d*sin(12*d*x + 12*c)^2 + 441*a^2*b^2*d*sin(10*d*x + 10*c)^2 + 1225*a^2*b^2*d*sin(8*d*x + 8*c)^2 + 1225*a^2*b^2*d*sin(6*d*x + 6*c)^2 + 44 1*a^2*b^2*d*sin(4*d*x + 4*c)^2 - 294*a^2*b^2*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 49*a^2*b^2*d*sin(2*d*x + 2*c)^2 - 14*a^2*b^2*d*cos(2*d*x + 2*c) + a^2*b^2*d - 2*(7*a^2*b^2*d*cos(12*d*x + 12*c) - 21*a^2*b^2*d*cos(10*d*x + 10*c) + 35*a^2*b^2*d*cos(8*d*x + 8*c) - 35*a^2*b^2*d*cos(6*d*x + 6*c) ...
Leaf count of result is larger than twice the leaf count of optimal. 467 vs. \(2 (155) = 310\).
Time = 0.81 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.37 \[ \int \frac {\csc ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {105 \, {\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{2} b^{2} - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a b^{3} - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (d x + c\right )}{\sqrt {\frac {a^{3} + \sqrt {a^{6} - {\left (a^{3} - a^{2} b\right )} a^{3}}}{a^{3} - a^{2} b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{7} - 12 \, a^{6} b + 14 \, a^{5} b^{2} - 4 \, a^{4} b^{3} - a^{3} b^{4}} + \frac {105 \, {\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{2} b^{2} - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a b^{3} - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (d x + c\right )}{\sqrt {\frac {a^{3} - \sqrt {a^{6} - {\left (a^{3} - a^{2} b\right )} a^{3}}}{a^{3} - a^{2} b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{7} - 12 \, a^{6} b + 14 \, a^{5} b^{2} - 4 \, a^{4} b^{3} - a^{3} b^{4}} - \frac {2 \, {\left (105 \, a \tan \left (d x + c\right )^{6} + 105 \, b \tan \left (d x + c\right )^{6} + 105 \, a \tan \left (d x + c\right )^{4} + 35 \, b \tan \left (d x + c\right )^{4} + 63 \, a \tan \left (d x + c\right )^{2} + 15 \, a\right )}}{a^{2} \tan \left (d x + c\right )^{7}}}{210 \, d} \]
1/210*(105*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^2 - 6*sqrt(a^2 - a *b + sqrt(a*b)*(a - b))*a*b^3 - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*b^4)*( pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/sqrt((a^3 + sqrt(a^6 - (a^3 - a^2*b)*a^3))/(a^3 - a^2*b))))*abs(a - b)/(3*a^7 - 12*a^6*b + 14*a^5 *b^2 - 4*a^4*b^3 - a^3*b^4) + 105*(3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a ^2*b^2 - 6*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a*b^3 - sqrt(a^2 - a*b - sq rt(a*b)*(a - b))*b^4)*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/ sqrt((a^3 - sqrt(a^6 - (a^3 - a^2*b)*a^3))/(a^3 - a^2*b))))*abs(a - b)/(3* a^7 - 12*a^6*b + 14*a^5*b^2 - 4*a^4*b^3 - a^3*b^4) - 2*(105*a*tan(d*x + c) ^6 + 105*b*tan(d*x + c)^6 + 105*a*tan(d*x + c)^4 + 35*b*tan(d*x + c)^4 + 6 3*a*tan(d*x + c)^2 + 15*a)/(a^2*tan(d*x + c)^7))/d
Time = 16.82 (sec) , antiderivative size = 1704, normalized size of antiderivative = 8.65 \[ \int \frac {\csc ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
(atan((((((a^11*b^9)^(1/2) + a^6*b^4)/(16*(a^11*b - a^12)))^(1/2)*(16*a^9* b^5 - 32*a^10*b^4 + 16*a^11*b^3 + tan(c + d*x)*(((a^11*b^9)^(1/2) + a^6*b^ 4)/(16*(a^11*b - a^12)))^(1/2)*(64*a^14*b + 64*a^12*b^3 - 128*a^13*b^2)) - tan(c + d*x)*(4*a^6*b^7 - 4*a^8*b^5))*(((a^11*b^9)^(1/2) + a^6*b^4)/(16*( a^11*b - a^12)))^(1/2)*1i - ((((a^11*b^9)^(1/2) + a^6*b^4)/(16*(a^11*b - a ^12)))^(1/2)*(16*a^9*b^5 - 32*a^10*b^4 + 16*a^11*b^3 - tan(c + d*x)*(((a^1 1*b^9)^(1/2) + a^6*b^4)/(16*(a^11*b - a^12)))^(1/2)*(64*a^14*b + 64*a^12*b ^3 - 128*a^13*b^2)) + tan(c + d*x)*(4*a^6*b^7 - 4*a^8*b^5))*(((a^11*b^9)^( 1/2) + a^6*b^4)/(16*(a^11*b - a^12)))^(1/2)*1i)/(((((a^11*b^9)^(1/2) + a^6 *b^4)/(16*(a^11*b - a^12)))^(1/2)*(16*a^9*b^5 - 32*a^10*b^4 + 16*a^11*b^3 + tan(c + d*x)*(((a^11*b^9)^(1/2) + a^6*b^4)/(16*(a^11*b - a^12)))^(1/2)*( 64*a^14*b + 64*a^12*b^3 - 128*a^13*b^2)) - tan(c + d*x)*(4*a^6*b^7 - 4*a^8 *b^5))*(((a^11*b^9)^(1/2) + a^6*b^4)/(16*(a^11*b - a^12)))^(1/2) + ((((a^1 1*b^9)^(1/2) + a^6*b^4)/(16*(a^11*b - a^12)))^(1/2)*(16*a^9*b^5 - 32*a^10* b^4 + 16*a^11*b^3 - tan(c + d*x)*(((a^11*b^9)^(1/2) + a^6*b^4)/(16*(a^11*b - a^12)))^(1/2)*(64*a^14*b + 64*a^12*b^3 - 128*a^13*b^2)) + tan(c + d*x)* (4*a^6*b^7 - 4*a^8*b^5))*(((a^11*b^9)^(1/2) + a^6*b^4)/(16*(a^11*b - a^12) ))^(1/2) - 2*a^4*b^8 + 2*a^5*b^7))*(((a^11*b^9)^(1/2) + a^6*b^4)/(16*(a^11 *b - a^12)))^(1/2)*2i)/d + (atan((((-((a^11*b^9)^(1/2) - a^6*b^4)/(16*(a^1 1*b - a^12)))^(1/2)*(16*a^9*b^5 - 32*a^10*b^4 + 16*a^11*b^3 + tan(c + d...