Integrand size = 24, antiderivative size = 177 \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{9/4} d}-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{9/4} d}+\frac {(a+b) \cos (c+d x)}{b^2 d}-\frac {2 \cos ^3(c+d x)}{3 b d}+\frac {\cos ^5(c+d x)}{5 b d} \]
(a+b)*cos(d*x+c)/b^2/d-2/3*cos(d*x+c)^3/b/d+1/5*cos(d*x+c)^5/b/d-1/2*a^(3/ 2)*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))/b^(9/4)/d/(a^(1/2)-b ^(1/2))^(1/2)-1/2*a^(3/2)*arctanh(b^(1/4)*cos(d*x+c)/(a^(1/2)+b^(1/2))^(1/ 2))/b^(9/4)/d/(a^(1/2)+b^(1/2))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 2.67 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.29 \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\cos (c+d x) (120 a+89 b-28 b \cos (2 (c+d x))+3 b \cos (4 (c+d x)))+60 i a^2 \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^3-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3}{-b-8 a \text {$\#$1}^2+3 b \text {$\#$1}^2-3 b \text {$\#$1}^4+b \text {$\#$1}^6}\&\right ]}{120 b^2 d} \]
(Cos[c + d*x]*(120*a + 89*b - 28*b*Cos[2*(c + d*x)] + 3*b*Cos[4*(c + d*x)] ) + (60*I)*a^2*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b* #1^8 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 + I*Log[1 - 2*Cos [c + d*x]*#1 + #1^2]*#1 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^3)/(-b - 8*a*#1^2 + 3*b*#1^2 - 3* b*#1^4 + b*#1^6) & ])/(120*b^2*d)
Time = 0.41 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3694, 1484, 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.
| \(\displaystyle \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^9}{a-b \sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3694 |
| \(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(c+d x)\right )^4}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 1484 |
| \(\displaystyle -\frac {\int \left (-\frac {\cos ^4(c+d x)}{b}+\frac {2 \cos ^2(c+d x)}{b}-\frac {a+b}{b^2}+\frac {a^2}{b^2 \left (-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b\right )}\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {a^{3/2} \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{9/4} \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{9/4} \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {(a+b) \cos (c+d x)}{b^2}-\frac {\cos ^5(c+d x)}{5 b}+\frac {2 \cos ^3(c+d x)}{3 b}}{d}\) |
-(((a^(3/2)*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqr t[Sqrt[a] - Sqrt[b]]*b^(9/4)) + (a^(3/2)*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sq rt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(9/4)) - ((a + b)*Cos [c + d*x])/b^2 + (2*Cos[c + d*x]^3)/(3*b) - Cos[c + d*x]^5/(5*b))/d)
3.2.95.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 2.97 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.77
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{5}-\frac {2 b \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right ) a +\cos \left (d x +c \right ) b}{b^{2}}+\frac {a^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {\arctan \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{b}}{d}\) | \(137\) |
| default | \(\frac {\frac {\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{5}-\frac {2 b \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right ) a +\cos \left (d x +c \right ) b}{b^{2}}+\frac {a^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {\arctan \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{b}}{d}\) | \(137\) |
| risch | \(\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d \,b^{2}}+\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}}{16 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} a}{2 b^{2} d}+\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 b d}-\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a \,b^{9} d^{4}-b^{10} d^{4}\right ) \textit {\_Z}^{4}-32768 a^{3} b^{5} d^{2} \textit {\_Z}^{2}-268435456 a^{6}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {i b^{7} d^{3}}{1048576 a^{4}}+\frac {i b^{8} d^{3}}{1048576 a^{5}}\right ) \textit {\_R}^{3}+\left (\frac {i b^{2} d}{64 a}+\frac {i b^{3} d}{64 a^{2}}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{512}+\frac {\cos \left (5 d x +5 c \right )}{80 b d}-\frac {5 \cos \left (3 d x +3 c \right )}{48 b d}\) | \(231\) |
1/d*(1/b^2*(1/5*b*cos(d*x+c)^5-2/3*b*cos(d*x+c)^3+cos(d*x+c)*a+cos(d*x+c)* b)+a^2/b*(-1/2/(a*b)^(1/2)/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/ (((a*b)^(1/2)+b)*b)^(1/2))-1/2/(a*b)^(1/2)/(((a*b)^(1/2)-b)*b)^(1/2)*arcta n(cos(d*x+c)*b/(((a*b)^(1/2)-b)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 872 vs. \(2 (133) = 266\).
Time = 0.34 (sec) , antiderivative size = 872, normalized size of antiderivative = 4.93 \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {12 \, b \cos \left (d x + c\right )^{5} - 15 \, b^{2} d \sqrt {-\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} + a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}} \log \left (a^{5} \cos \left (d x + c\right ) + {\left (a^{4} b^{2} d - {\left (a b^{7} - b^{8}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{3}\right )} \sqrt {-\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} + a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}}\right ) + 15 \, b^{2} d \sqrt {\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} - a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}} \log \left (a^{5} \cos \left (d x + c\right ) - {\left (a^{4} b^{2} d + {\left (a b^{7} - b^{8}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{3}\right )} \sqrt {\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} - a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}}\right ) + 15 \, b^{2} d \sqrt {-\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} + a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}} \log \left (-a^{5} \cos \left (d x + c\right ) + {\left (a^{4} b^{2} d - {\left (a b^{7} - b^{8}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{3}\right )} \sqrt {-\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} + a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}}\right ) - 15 \, b^{2} d \sqrt {\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} - a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}} \log \left (-a^{5} \cos \left (d x + c\right ) - {\left (a^{4} b^{2} d + {\left (a b^{7} - b^{8}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{3}\right )} \sqrt {\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} - a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}}\right ) - 40 \, b \cos \left (d x + c\right )^{3} + 60 \, {\left (a + b\right )} \cos \left (d x + c\right )}{60 \, b^{2} d} \]
1/60*(12*b*cos(d*x + c)^5 - 15*b^2*d*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b ^9 - 2*a*b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2))*log(a^5*cos(d* x + c) + (a^4*b^2*d - (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)* d^4))*d^3)*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4) )*d^2 + a^3)/((a*b^4 - b^5)*d^2))) + 15*b^2*d*sqrt(((a*b^4 - b^5)*sqrt(a^7 /((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))*log(a^ 5*cos(d*x + c) - (a^4*b^2*d + (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^1 1)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))) + 15*b^2*d*sqrt(-((a*b^4 - b^5)* sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2) )*log(-a^5*cos(d*x + c) + (a^4*b^2*d - (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a* b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2))) - 15*b^2*d*sqrt(((a*b^ 4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))*log(-a^5*cos(d*x + c) - (a^4*b^2*d + (a*b^7 - b^8)*sqrt(a^7/((a ^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(((a*b^4 - b^5)*sqrt(a^7/((a^2*b^ 9 - 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))) - 40*b*cos(d*x + c)^3 + 60*(a + b)*cos(d*x + c))/(b^2*d)
Timed out. \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sin \left (d x + c\right )^{9}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \]
1/240*(240*b^2*d*integrate(8*(4*a^2*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 2*(8*a^3 - 3*a^2*b)*cos(3*d*x + 3*c)*sin(4*d*x + 4*c) - 2*(8*a^3 - 3*a^2*b )*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) - (a^2*b*sin(5*d*x + 5*c) - a^2*b*sin( 3*d*x + 3*c))*cos(8*d*x + 8*c) + 4*(a^2*b*sin(5*d*x + 5*c) - a^2*b*sin(3*d *x + 3*c))*cos(6*d*x + 6*c) - 2*(2*a^2*b*sin(2*d*x + 2*c) + (8*a^3 - 3*a^2 *b)*sin(4*d*x + 4*c))*cos(5*d*x + 5*c) + (a^2*b*cos(5*d*x + 5*c) - a^2*b*c os(3*d*x + 3*c))*sin(8*d*x + 8*c) - 4*(a^2*b*cos(5*d*x + 5*c) - a^2*b*cos( 3*d*x + 3*c))*sin(6*d*x + 6*c) + (4*a^2*b*cos(2*d*x + 2*c) - a^2*b + 2*(8* a^3 - 3*a^2*b)*cos(4*d*x + 4*c))*sin(5*d*x + 5*c) - (4*a^2*b*cos(2*d*x + 2 *c) - a^2*b)*sin(3*d*x + 3*c))/(b^4*cos(8*d*x + 8*c)^2 + 16*b^4*cos(6*d*x + 6*c)^2 + 16*b^4*cos(2*d*x + 2*c)^2 + b^4*sin(8*d*x + 8*c)^2 + 16*b^4*sin (6*d*x + 6*c)^2 + 16*b^4*sin(2*d*x + 2*c)^2 - 8*b^4*cos(2*d*x + 2*c) + b^4 + 4*(64*a^2*b^2 - 48*a*b^3 + 9*b^4)*cos(4*d*x + 4*c)^2 + 4*(64*a^2*b^2 - 48*a*b^3 + 9*b^4)*sin(4*d*x + 4*c)^2 + 16*(8*a*b^3 - 3*b^4)*sin(4*d*x + 4* c)*sin(2*d*x + 2*c) - 2*(4*b^4*cos(6*d*x + 6*c) + 4*b^4*cos(2*d*x + 2*c) - b^4 + 2*(8*a*b^3 - 3*b^4)*cos(4*d*x + 4*c))*cos(8*d*x + 8*c) + 8*(4*b^4*c os(2*d*x + 2*c) - b^4 + 2*(8*a*b^3 - 3*b^4)*cos(4*d*x + 4*c))*cos(6*d*x + 6*c) - 4*(8*a*b^3 - 3*b^4 - 4*(8*a*b^3 - 3*b^4)*cos(2*d*x + 2*c))*cos(4*d* x + 4*c) - 4*(2*b^4*sin(6*d*x + 6*c) + 2*b^4*sin(2*d*x + 2*c) + (8*a*b^3 - 3*b^4)*sin(4*d*x + 4*c))*sin(8*d*x + 8*c) + 16*(2*b^4*sin(2*d*x + 2*c)...
\[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sin \left (d x + c\right )^{9}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \]
Time = 14.43 (sec) , antiderivative size = 1067, normalized size of antiderivative = 6.03 \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {{\cos \left (c+d\,x\right )}^5}{5\,b\,d}-\frac {2\,{\cos \left (c+d\,x\right )}^3}{3\,b\,d}+\frac {\cos \left (c+d\,x\right )\,\left (\frac {a-b}{b^2}+\frac {2}{b}\right )}{d}+\frac {\mathrm {atan}\left (-\frac {a^4\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^7}{a\,b^9-b^{10}}+\frac {2\,a^3\,b^2\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}+\frac {a^4\,b^9\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^{16}}{a\,b^9-b^{10}}-\frac {2\,a^7\,b^{15}}{a\,b^9-b^{10}}+\frac {2\,a^3\,b^{11}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}-\frac {2\,a^4\,b^{10}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}+\frac {a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,\sqrt {a^7\,b^9}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^{16}}{a\,b^9-b^{10}}-\frac {2\,a^7\,b^{15}}{a\,b^9-b^{10}}+\frac {2\,a^3\,b^{11}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}-\frac {2\,a^4\,b^{10}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}\right )\,\sqrt {-\frac {\sqrt {a^7\,b^9}+a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,2{}\mathrm {i}}{d}-\frac {\mathrm {atan}\left (\frac {a^4\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^7}{a\,b^9-b^{10}}-\frac {2\,a^3\,b^2\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}-\frac {a^4\,b^9\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^{16}}{a\,b^9-b^{10}}-\frac {2\,a^7\,b^{15}}{a\,b^9-b^{10}}-\frac {2\,a^3\,b^{11}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}+\frac {2\,a^4\,b^{10}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}+\frac {a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,\sqrt {a^7\,b^9}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^{16}}{a\,b^9-b^{10}}-\frac {2\,a^7\,b^{15}}{a\,b^9-b^{10}}-\frac {2\,a^3\,b^{11}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}+\frac {2\,a^4\,b^{10}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}\right )\,\sqrt {\frac {\sqrt {a^7\,b^9}-a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,2{}\mathrm {i}}{d} \]
(atan((a^4*b^9*cos(c + d*x)*(- (a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3* b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2*a^7* b^15)/(a*b^9 - b^10) + (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^10) - (2*a^ 4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) - (a^4*cos(c + d*x)*(- (a^7*b^9)^( 1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/((2*a^ 6*b^7)/(a*b^9 - b^10) + (2*a^3*b^2*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) + (a*b ^4*cos(c + d*x)*(- (a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a* b^9 - b^10)))^(1/2)*(a^7*b^9)^(1/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2* a^7*b^15)/(a*b^9 - b^10) + (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^10) - ( 2*a^4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10)))*(-((a^7*b^9)^(1/2) + a^3*b^5) /(16*(a*b^9 - b^10)))^(1/2)*2i)/d - (atan((a^4*cos(c + d*x)*((a^7*b^9)^(1/ 2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/((2*a^6* b^7)/(a*b^9 - b^10) - (2*a^3*b^2*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) - (a^4*b ^9*cos(c + d*x)*((a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^ 9 - b^10)))^(1/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2*a^7*b^15)/(a*b^9 - b^10) - (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^10) + (2*a^4*b^10*(a^7*b^ 9)^(1/2))/(a*b^9 - b^10)) + (a*b^4*cos(c + d*x)*((a^7*b^9)^(1/2)/(16*(a*b^ 9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*(a^7*b^9)^(1/2)*8i)/((2* a^6*b^16)/(a*b^9 - b^10) - (2*a^7*b^15)/(a*b^9 - b^10) - (2*a^3*b^11*(a^7* b^9)^(1/2))/(a*b^9 - b^10) + (2*a^4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10...