Integrand size = 24, antiderivative size = 178 \[ \int \frac {\csc ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {b^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{9/4} \sqrt {\sqrt {a}-\sqrt {b}} d}-\frac {b^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{9/4} \sqrt {\sqrt {a}+\sqrt {b}} d}-\frac {(a+b) \cot (c+d x)}{a^2 d}-\frac {2 \cot ^3(c+d x)}{3 a d}-\frac {\cot ^5(c+d x)}{5 a d} \]
-(a+b)*cot(d*x+c)/a^2/d-2/3*cot(d*x+c)^3/a/d-1/5*cot(d*x+c)^5/a/d+1/2*b^(3 /2)*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(9/4)/d/(a^(1/2)- b^(1/2))^(1/2)-1/2*b^(3/2)*arctan((a^(1/2)+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/ 4))/a^(9/4)/d/(a^(1/2)+b^(1/2))^(1/2)
Time = 3.62 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.98 \[ \int \frac {\csc ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\frac {15 b^{3/2} \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {15 b^{3/2} \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}+2 \cot (c+d x) \left (8 a+15 b+4 a \csc ^2(c+d x)+3 a \csc ^4(c+d x)\right )}{30 a^2 d} \]
-1/30*((15*b^(3/2)*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt [a]*Sqrt[b]]])/Sqrt[a + Sqrt[a]*Sqrt[b]] + (15*b^(3/2)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/Sqrt[-a + Sqrt[a]*Sqr t[b]] + 2*Cot[c + d*x]*(8*a + 15*b + 4*a*Csc[c + d*x]^2 + 3*a*Csc[c + d*x] ^4))/(a^2*d)
Time = 0.38 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.94, 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.
| \(\displaystyle \int \frac {\csc ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^6 \left (a-b \sin (c+d x)^4\right )}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\cot ^6(c+d x) \left (\tan ^2(c+d x)+1\right )^4}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1610 |
| \(\displaystyle \frac {\int \left (\frac {\cot ^6(c+d x)}{a}+\frac {2 \cot ^4(c+d x)}{a}+\frac {(a+b) \cot ^2(c+d x)}{a^2}+\frac {b^2 \tan ^2(c+d x)}{a^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{9/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {b^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{9/4} \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {(a+b) \cot (c+d x)}{a^2}-\frac {\cot ^5(c+d x)}{5 a}-\frac {2 \cot ^3(c+d x)}{3 a}}{d}\) |
((b^(3/2)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(9/ 4)*Sqrt[Sqrt[a] - Sqrt[b]]) - (b^(3/2)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan [c + d*x])/a^(1/4)])/(2*a^(9/4)*Sqrt[Sqrt[a] + Sqrt[b]]) - ((a + b)*Cot[c + d*x])/a^2 - (2*Cot[c + d*x]^3)/(3*a) - Cot[c + d*x]^5/(5*a))/d
3.3.10.3.1 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 = 2.02 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{2} \left (a -b \right ) \left (\frac {\left (\sqrt {a b}+a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \left (a -b \right ) \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-a \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{a^{2}}-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {a +b}{a^{2} \tan \left (d x +c \right )}-\frac {2}{3 a \tan \left (d x +c \right )^{3}}}{d}\) | \(195\) |
| default | \(\frac {\frac {b^{2} \left (a -b \right ) \left (\frac {\left (\sqrt {a b}+a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \left (a -b \right ) \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-a \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{a^{2}}-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {a +b}{a^{2} \tan \left (d x +c \right )}-\frac {2}{3 a \tan \left (d x +c \right )^{3}}}{d}\) | \(195\) |
| risch | \(-\frac {2 i \left (15 \,{\mathrm e}^{8 i \left (d x +c \right )} b -60 b \,{\mathrm e}^{6 i \left (d x +c \right )}+80 a \,{\mathrm e}^{4 i \left (d x +c \right )}+90 b \,{\mathrm e}^{4 i \left (d x +c \right )}-40 a \,{\mathrm e}^{2 i \left (d x +c \right )}-60 b \,{\mathrm e}^{2 i \left (d x +c \right )}+8 a +15 b \right )}{15 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-64 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4294967296 a^{10} d^{4}-4294967296 a^{9} b \,d^{4}\right ) \textit {\_Z}^{4}+131072 a^{5} b^{3} d^{2} \textit {\_Z}^{2}+b^{6}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {33554432 i a^{8} d^{3}}{b^{5}}-\frac {33554432 i d^{3} a^{7}}{b^{4}}\right ) \textit {\_R}^{3}+\left (-\frac {131072 d^{2} a^{6}}{b^{4}}+\frac {131072 d^{2} a^{5}}{b^{3}}\right ) \textit {\_R}^{2}+\frac {1024 i d \,a^{3} \textit {\_R}}{b^{2}}-\frac {2 a}{b}-1\right )\right )\) | \(236\) |
1/d*(b^2/a^2*(a-b)*(1/2*((a*b)^(1/2)+a)/(a-b)/(a*b)^(1/2)/(((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)-a)/(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)))-1/5/a/tan(d*x+c)^5-(a+b)/a^ 2/tan(d*x+c)-2/3/a/tan(d*x+c)^3)
Leaf count of result is larger than twice the leaf count of optimal. 1477 vs. \(2 (134) = 268\).
Time = 0.46 (sec) , antiderivative size = 1477, normalized size of antiderivative = 8.30 \[ \int \frac {\csc ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
-1/120*(8*(8*a + 15*b)*cos(d*x + c)^5 - 80*(2*a + 3*b)*cos(d*x + c)^3 - 15 *(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)*sqrt(-((a^5 - a^4 *b)*sqrt(b^7/((a^11 - 2*a^10*b + a^9*b^2)*d^4))*d^2 + b^3)/((a^5 - a^4*b)* d^2))*log(1/4*b^5*cos(d*x + c)^2 - 1/4*b^5 - 1/4*(2*(a^6*b - a^5*b^2)*d^2* cos(d*x + c)^2 - (a^6*b - a^5*b^2)*d^2)*sqrt(b^7/((a^11 - 2*a^10*b + a^9*b ^2)*d^4)) + 1/2*(a^3*b^3*d*cos(d*x + c)*sin(d*x + c) - (a^8 - a^7*b)*sqrt( b^7/((a^11 - 2*a^10*b + a^9*b^2)*d^4))*d^3*cos(d*x + c)*sin(d*x + c))*sqrt (-((a^5 - a^4*b)*sqrt(b^7/((a^11 - 2*a^10*b + a^9*b^2)*d^4))*d^2 + b^3)/(( a^5 - a^4*b)*d^2)))*sin(d*x + c) + 15*(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos( d*x + c)^2 + a^2*d)*sqrt(-((a^5 - a^4*b)*sqrt(b^7/((a^11 - 2*a^10*b + a^9* b^2)*d^4))*d^2 + b^3)/((a^5 - a^4*b)*d^2))*log(1/4*b^5*cos(d*x + c)^2 - 1/ 4*b^5 - 1/4*(2*(a^6*b - a^5*b^2)*d^2*cos(d*x + c)^2 - (a^6*b - a^5*b^2)*d^ 2)*sqrt(b^7/((a^11 - 2*a^10*b + a^9*b^2)*d^4)) - 1/2*(a^3*b^3*d*cos(d*x + c)*sin(d*x + c) - (a^8 - a^7*b)*sqrt(b^7/((a^11 - 2*a^10*b + a^9*b^2)*d^4) )*d^3*cos(d*x + c)*sin(d*x + c))*sqrt(-((a^5 - a^4*b)*sqrt(b^7/((a^11 - 2* a^10*b + a^9*b^2)*d^4))*d^2 + b^3)/((a^5 - a^4*b)*d^2)))*sin(d*x + c) + 15 *(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)*sqrt(((a^5 - a^4* b)*sqrt(b^7/((a^11 - 2*a^10*b + a^9*b^2)*d^4))*d^2 - b^3)/((a^5 - a^4*b)*d ^2))*log(-1/4*b^5*cos(d*x + c)^2 + 1/4*b^5 - 1/4*(2*(a^6*b - a^5*b^2)*d^2* cos(d*x + c)^2 - (a^6*b - a^5*b^2)*d^2)*sqrt(b^7/((a^11 - 2*a^10*b + a^...
\[ \int \frac {\csc ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int \frac {\csc ^{6}{\left (c + d x \right )}}{a - b \sin ^{4}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {\csc ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\csc \left (d x + c\right )^{6}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \]
1/15*(300*b*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) + 10*(3*b*sin(8*d*x + 8*c) - 12*b*sin(6*d*x + 6*c) + 2*(8*a + 9*b)*sin(4*d*x + 4*c) - 4*(2*a + 3*b)*si n(2*d*x + 2*c))*cos(10*d*x + 10*c) + 50*(6*b*sin(6*d*x + 6*c) - 4*(4*a + 3 *b)*sin(4*d*x + 4*c) + (8*a + 9*b)*sin(2*d*x + 2*c))*cos(8*d*x + 8*c) + 20 0*((8*a + 3*b)*sin(4*d*x + 4*c) - (4*a + 3*b)*sin(2*d*x + 2*c))*cos(6*d*x + 6*c) + 15*(a^2*d*cos(10*d*x + 10*c)^2 + 25*a^2*d*cos(8*d*x + 8*c)^2 + 10 0*a^2*d*cos(6*d*x + 6*c)^2 + 100*a^2*d*cos(4*d*x + 4*c)^2 + 25*a^2*d*cos(2 *d*x + 2*c)^2 + a^2*d*sin(10*d*x + 10*c)^2 + 25*a^2*d*sin(8*d*x + 8*c)^2 + 100*a^2*d*sin(6*d*x + 6*c)^2 + 100*a^2*d*sin(4*d*x + 4*c)^2 - 100*a^2*d*s in(4*d*x + 4*c)*sin(2*d*x + 2*c) + 25*a^2*d*sin(2*d*x + 2*c)^2 - 10*a^2*d* cos(2*d*x + 2*c) + a^2*d - 2*(5*a^2*d*cos(8*d*x + 8*c) - 10*a^2*d*cos(6*d* x + 6*c) + 10*a^2*d*cos(4*d*x + 4*c) - 5*a^2*d*cos(2*d*x + 2*c) + a^2*d)*c os(10*d*x + 10*c) - 10*(10*a^2*d*cos(6*d*x + 6*c) - 10*a^2*d*cos(4*d*x + 4 *c) + 5*a^2*d*cos(2*d*x + 2*c) - a^2*d)*cos(8*d*x + 8*c) - 20*(10*a^2*d*co s(4*d*x + 4*c) - 5*a^2*d*cos(2*d*x + 2*c) + a^2*d)*cos(6*d*x + 6*c) - 20*( 5*a^2*d*cos(2*d*x + 2*c) - a^2*d)*cos(4*d*x + 4*c) - 10*(a^2*d*sin(8*d*x + 8*c) - 2*a^2*d*sin(6*d*x + 6*c) + 2*a^2*d*sin(4*d*x + 4*c) - a^2*d*sin(2* d*x + 2*c))*sin(10*d*x + 10*c) - 50*(2*a^2*d*sin(6*d*x + 6*c) - 2*a^2*d*si n(4*d*x + 4*c) + a^2*d*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) - 100*(2*a^2*d*s in(4*d*x + 4*c) - a^2*d*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*integrate(-...
Leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (134) = 268\).
Time = 0.74 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.65 \[ \int \frac {\csc ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {15 \, {\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} b - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b^{2} - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (d x + c\right )}{\sqrt {\frac {a^{3} + \sqrt {a^{6} - {\left (a^{3} - a^{2} b\right )} a^{3}}}{a^{3} - a^{2} b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{7} - 12 \, a^{6} b + 14 \, a^{5} b^{2} - 4 \, a^{4} b^{3} - a^{3} b^{4}} - \frac {15 \, {\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} b - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b^{2} - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (d x + c\right )}{\sqrt {\frac {a^{3} - \sqrt {a^{6} - {\left (a^{3} - a^{2} b\right )} a^{3}}}{a^{3} - a^{2} b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{7} - 12 \, a^{6} b + 14 \, a^{5} b^{2} - 4 \, a^{4} b^{3} - a^{3} b^{4}} - \frac {2 \, {\left (15 \, a \tan \left (d x + c\right )^{4} + 15 \, b \tan \left (d x + c\right )^{4} + 10 \, a \tan \left (d x + c\right )^{2} + 3 \, a\right )}}{a^{2} \tan \left (d x + c\right )^{5}}}{30 \, d} \]
1/30*(15*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b - 6*sqrt(a ^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^2 - sqrt(a^2 - a*b + sqrt(a*b) *(a - b))*sqrt(a*b)*b^3)*(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) - 15*(3*sqrt(a^2 - a *b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b - 6*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^2 - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^3) *(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*(15*a*tan(d*x + c)^4 + 15*b*tan(d*x + c) ^4 + 10*a*tan(d*x + c)^2 + 3*a)/(a^2*tan(d*x + c)^5))/d
Time = 14.36 (sec) , antiderivative size = 416, normalized size of antiderivative = 2.34 \[ \int \frac {\csc ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^7\,b^6-4\,a^9\,b^4\right )-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\sqrt {a^9\,b^7}+a^5\,b^3\right )\,\left (64\,a^{14}\,b-128\,a^{13}\,b^2+64\,a^{12}\,b^3\right )}{16\,\left (a^9\,b-a^{10}\right )}\right )\,\sqrt {\frac {\sqrt {a^9\,b^7}+a^5\,b^3}{16\,\left (a^9\,b-a^{10}\right )}}}{2\,a^5\,b^7-2\,a^6\,b^6}\right )\,\sqrt {\frac {\sqrt {a^9\,b^7}+a^5\,b^3}{16\,\left (a^9\,b-a^{10}\right )}}}{d}+\frac {2\,\mathrm {atanh}\left (\frac {2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^7\,b^6-4\,a^9\,b^4\right )+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\sqrt {a^9\,b^7}-a^5\,b^3\right )\,\left (64\,a^{14}\,b-128\,a^{13}\,b^2+64\,a^{12}\,b^3\right )}{16\,\left (a^9\,b-a^{10}\right )}\right )\,\sqrt {-\frac {\sqrt {a^9\,b^7}-a^5\,b^3}{16\,\left (a^9\,b-a^{10}\right )}}}{2\,a^5\,b^7-2\,a^6\,b^6}\right )\,\sqrt {-\frac {\sqrt {a^9\,b^7}-a^5\,b^3}{16\,\left (a^9\,b-a^{10}\right )}}}{d}-\frac {\frac {1}{5\,a}+\frac {2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a+b\right )}{a^2}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \]
(2*atanh((2*(tan(c + d*x)*(4*a^7*b^6 - 4*a^9*b^4) - (tan(c + d*x)*((a^9*b^ 7)^(1/2) + a^5*b^3)*(64*a^14*b + 64*a^12*b^3 - 128*a^13*b^2))/(16*(a^9*b - a^10)))*(((a^9*b^7)^(1/2) + a^5*b^3)/(16*(a^9*b - a^10)))^(1/2))/(2*a^5*b ^7 - 2*a^6*b^6))*(((a^9*b^7)^(1/2) + a^5*b^3)/(16*(a^9*b - a^10)))^(1/2))/ d + (2*atanh((2*(tan(c + d*x)*(4*a^7*b^6 - 4*a^9*b^4) + (tan(c + d*x)*((a^ 9*b^7)^(1/2) - a^5*b^3)*(64*a^14*b + 64*a^12*b^3 - 128*a^13*b^2))/(16*(a^9 *b - a^10)))*(-((a^9*b^7)^(1/2) - a^5*b^3)/(16*(a^9*b - a^10)))^(1/2))/(2* a^5*b^7 - 2*a^6*b^6))*(-((a^9*b^7)^(1/2) - a^5*b^3)/(16*(a^9*b - a^10)))^( 1/2))/d - (1/(5*a) + (2*tan(c + d*x)^2)/(3*a) + (tan(c + d*x)^4*(a + b))/a ^2)/(d*tan(c + d*x)^5)