Integrand size = 13, antiderivative size = 178 \[ \int \frac {\tan ^5(x)}{a+b \csc (x)} \, dx=\frac {1}{16 (a+b) (1-\csc (x))^2}+\frac {5 a+7 b}{16 (a+b)^2 (1-\csc (x))}+\frac {1}{16 (a-b) (1+\csc (x))^2}+\frac {5 a-7 b}{16 (a-b)^2 (1+\csc (x))}-\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\csc (x))}{16 (a+b)^3}-\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (1+\csc (x))}{16 (a-b)^3}+\frac {b^6 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^3}-\frac {\log (\sin (x))}{a} \]
1/16/(a+b)/(1-csc(x))^2+1/16*(5*a+7*b)/(a+b)^2/(1-csc(x))+1/16/(a-b)/(1+cs c(x))^2+1/16*(5*a-7*b)/(a-b)^2/(1+csc(x))-1/16*(8*a^2+21*a*b+15*b^2)*ln(1- csc(x))/(a+b)^3-1/16*(8*a^2-21*a*b+15*b^2)*ln(1+csc(x))/(a-b)^3+b^6*ln(a+b *csc(x))/a/(a^2-b^2)^3-ln(sin(x))/a
Time = 0.87 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.97 \[ \int \frac {\tan ^5(x)}{a+b \csc (x)} \, dx=\frac {\csc (x) (b+a \sin (x)) \left (-\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (x))}{(a+b)^3}-\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (1+\sin (x))}{(a-b)^3}+\frac {16 b^6 \log (b+a \sin (x))}{a (a-b)^3 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (x))^2}+\frac {7 a+9 b}{(a+b)^2 (-1+\sin (x))}+\frac {1}{(a-b) (1+\sin (x))^2}+\frac {-7 a+9 b}{(a-b)^2 (1+\sin (x))}\right )}{16 (a+b \csc (x))} \]
(Csc[x]*(b + a*Sin[x])*(-(((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sin[x]])/(a + b)^3) - ((8*a^2 - 21*a*b + 15*b^2)*Log[1 + Sin[x]])/(a - b)^3 + (16*b^6*L og[b + a*Sin[x]])/(a*(a - b)^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[x])^2) + (7*a + 9*b)/((a + b)^2*(-1 + Sin[x])) + 1/((a - b)*(1 + Sin[x])^2) + (-7*a + 9*b)/((a - b)^2*(1 + Sin[x]))))/(16*(a + b*Csc[x]))
Time = 0.50 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4373, 615, 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.
| \(\displaystyle \int \frac {\tan ^5(x)}{a+b \csc (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cot (x)^5 (a+b \csc (x))}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle b^6 \int \frac {\sin (x)}{b (a+b \csc (x)) \left (b^2-b^2 \csc ^2(x)\right )^3}d(b \csc (x))\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle b^6 \int \left (\frac {7 b-5 a}{16 (a-b)^2 b^5 (\csc (x) b+b)^2}+\frac {\sin (x)}{a b^7}+\frac {8 a^2+21 b a+15 b^2}{16 b^6 (a+b)^3 (b-b \csc (x))}+\frac {1}{a (a-b)^3 (a+b)^3 (a+b \csc (x))}+\frac {8 a^2-21 b a+15 b^2}{16 b^6 (b-a)^3 (\csc (x) b+b)}+\frac {5 a+7 b}{16 b^5 (a+b)^2 (b-b \csc (x))^2}+\frac {1}{8 b^4 (a+b) (b-b \csc (x))^3}+\frac {1}{8 b^4 (b-a) (\csc (x) b+b)^3}\right )d(b \csc (x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b^6 \left (\frac {\log (a+b \csc (x))}{a \left (a^2-b^2\right )^3}-\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (b-b \csc (x))}{16 b^6 (a+b)^3}-\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (b \csc (x)+b)}{16 b^6 (a-b)^3}+\frac {\log (b \csc (x))}{a b^6}+\frac {5 a-7 b}{16 b^5 (a-b)^2 (b \csc (x)+b)}+\frac {5 a+7 b}{16 b^5 (a+b)^2 (b-b \csc (x))}+\frac {1}{16 b^4 (a+b) (b-b \csc (x))^2}+\frac {1}{16 b^4 (a-b) (b \csc (x)+b)^2}\right )\) |
b^6*(1/(16*b^4*(a + b)*(b - b*Csc[x])^2) + (5*a + 7*b)/(16*b^5*(a + b)^2*( b - b*Csc[x])) + 1/(16*(a - b)*b^4*(b + b*Csc[x])^2) + (5*a - 7*b)/(16*(a - b)^2*b^5*(b + b*Csc[x])) + Log[b*Csc[x]]/(a*b^6) - ((8*a^2 + 21*a*b + 15 *b^2)*Log[b - b*Csc[x]])/(16*b^6*(a + b)^3) + Log[a + b*Csc[x]]/(a*(a^2 - b^2)^3) - ((8*a^2 - 21*a*b + 15*b^2)*Log[b + b*Csc[x]])/(16*(a - b)^3*b^6) )
3.1.12.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 1.19 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.90
| method | result | size |
| default | \(\frac {b^{6} \ln \left (a \sin \left (x \right )+b \right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} a}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (x \right )-1\right )^{2}}-\frac {-7 a -9 b}{16 \left (a +b \right )^{2} \left (\sin \left (x \right )-1\right )}+\frac {\left (-8 a^{2}-21 a b -15 b^{2}\right ) \ln \left (\sin \left (x \right )-1\right )}{16 \left (a +b \right )^{3}}+\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (x \right )\right )^{2}}-\frac {7 a -9 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (x \right )\right )}+\frac {\left (-8 a^{2}+21 a b -15 b^{2}\right ) \ln \left (1+\sin \left (x \right )\right )}{16 \left (a -b \right )^{3}}\) | \(160\) |
| risch | \(-\frac {i x}{a}+\frac {15 i x \,b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i x \,a^{2}}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}-\frac {2 i x \,b^{6}}{a \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {21 i x a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {15 i x \,b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {21 i x a b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i x \,a^{2}}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {i \left (16 i a^{3} {\mathrm e}^{6 i x}-24 i a \,b^{2} {\mathrm e}^{6 i x}-5 a^{2} b \,{\mathrm e}^{7 i x}+9 b^{3} {\mathrm e}^{7 i x}+16 i a^{3} {\mathrm e}^{4 i x}-32 i a \,b^{2} {\mathrm e}^{4 i x}+3 a^{2} b \,{\mathrm e}^{5 i x}+b^{3} {\mathrm e}^{5 i x}+16 i a^{3} {\mathrm e}^{2 i x}-24 i a \,b^{2} {\mathrm e}^{2 i x}-3 a^{2} b \,{\mathrm e}^{3 i x}-b^{3} {\mathrm e}^{3 i x}+5 a^{2} b \,{\mathrm e}^{i x}-9 \,{\mathrm e}^{i x} b^{3}\right )}{4 \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i x}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i x}-i\right ) a^{2}}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {21 \ln \left ({\mathrm e}^{i x}-i\right ) a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {15 \ln \left ({\mathrm e}^{i x}-i\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {\ln \left (i+{\mathrm e}^{i x}\right ) a^{2}}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {21 \ln \left (i+{\mathrm e}^{i x}\right ) a b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {15 \ln \left (i+{\mathrm e}^{i x}\right ) b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {b^{6} \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{a \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(662\) |
b^6/(a+b)^3/(a-b)^3/a*ln(a*sin(x)+b)+1/2/(8*a+8*b)/(sin(x)-1)^2-1/16*(-7*a -9*b)/(a+b)^2/(sin(x)-1)+1/16/(a+b)^3*(-8*a^2-21*a*b-15*b^2)*ln(sin(x)-1)+ 1/2/(8*a-8*b)/(1+sin(x))^2-1/16*(7*a-9*b)/(a-b)^2/(1+sin(x))+1/16/(a-b)^3* (-8*a^2+21*a*b-15*b^2)*ln(1+sin(x))
Time = 0.37 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.46 \[ \int \frac {\tan ^5(x)}{a+b \csc (x)} \, dx=\frac {16 \, b^{6} \cos \left (x\right )^{4} \log \left (a \sin \left (x\right ) + b\right ) + 4 \, a^{6} - 8 \, a^{4} b^{2} + 4 \, a^{2} b^{4} - {\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - {\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) - 8 \, {\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2} - 2 \, {\left (2 \, a^{5} b - 4 \, a^{3} b^{3} + 2 \, a b^{5} - {\left (5 \, a^{5} b - 14 \, a^{3} b^{3} + 9 \, a b^{5}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{16 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (x\right )^{4}} \]
1/16*(16*b^6*cos(x)^4*log(a*sin(x) + b) + 4*a^6 - 8*a^4*b^2 + 4*a^2*b^4 - (8*a^6 + 3*a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4 + 15*a*b^5)*cos(x) ^4*log(sin(x) + 1) - (8*a^6 - 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*a^2*b ^4 - 15*a*b^5)*cos(x)^4*log(-sin(x) + 1) - 8*(2*a^6 - 5*a^4*b^2 + 3*a^2*b^ 4)*cos(x)^2 - 2*(2*a^5*b - 4*a^3*b^3 + 2*a*b^5 - (5*a^5*b - 14*a^3*b^3 + 9 *a*b^5)*cos(x)^2)*sin(x))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(x)^4)
\[ \int \frac {\tan ^5(x)}{a+b \csc (x)} \, dx=\int \frac {\tan ^{5}{\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.41 \[ \int \frac {\tan ^5(x)}{a+b \csc (x)} \, dx=\frac {b^{6} \log \left (a \sin \left (x\right ) + b\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac {{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) - 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (5 \, a^{2} b - 9 \, b^{3}\right )} \sin \left (x\right )^{3} + 6 \, a^{3} - 10 \, a b^{2} - 4 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (x\right )^{2} - {\left (3 \, a^{2} b - 7 \, b^{3}\right )} \sin \left (x\right )}{8 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{2}\right )}} \]
b^6*log(a*sin(x) + b)/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - 1/16*(8*a^2 - 21*a*b + 15*b^2)*log(sin(x) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 1/16* (8*a^2 + 21*a*b + 15*b^2)*log(sin(x) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/8*((5*a^2*b - 9*b^3)*sin(x)^3 + 6*a^3 - 10*a*b^2 - 4*(2*a^3 - 3*a*b^2) *sin(x)^2 - (3*a^2*b - 7*b^3)*sin(x))/((a^4 - 2*a^2*b^2 + b^4)*sin(x)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(x)^2)
Time = 0.28 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.42 \[ \int \frac {\tan ^5(x)}{a+b \csc (x)} \, dx=\frac {b^{6} \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac {{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {6 \, a^{5} - 16 \, a^{3} b^{2} + 10 \, a b^{4} + {\left (5 \, a^{4} b - 14 \, a^{2} b^{3} + 9 \, b^{5}\right )} \sin \left (x\right )^{3} - 4 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \sin \left (x\right )^{2} - {\left (3 \, a^{4} b - 10 \, a^{2} b^{3} + 7 \, b^{5}\right )} \sin \left (x\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} {\left (\sin \left (x\right ) + 1\right )}^{2} {\left (\sin \left (x\right ) - 1\right )}^{2}} \]
b^6*log(abs(a*sin(x) + b))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - 1/16*(8 *a^2 - 21*a*b + 15*b^2)*log(sin(x) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 1/16*(8*a^2 + 21*a*b + 15*b^2)*log(-sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/8*(6*a^5 - 16*a^3*b^2 + 10*a*b^4 + (5*a^4*b - 14*a^2*b^3 + 9*b^5 )*sin(x)^3 - 4*(2*a^5 - 5*a^3*b^2 + 3*a*b^4)*sin(x)^2 - (3*a^4*b - 10*a^2* b^3 + 7*b^5)*sin(x))/((a + b)^3*(a - b)^3*(sin(x) + 1)^2*(sin(x) - 1)^2)
Time = 22.46 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.50 \[ \int \frac {\tan ^5(x)}{a+b \csc (x)} \, dx=\frac {\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (3\,a\,b^2-2\,a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^7\,\left (3\,a^2\,b-7\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (11\,a^2\,b-15\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (11\,a^2\,b-15\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^2-7\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )\,\left (\frac {5\,b}{8\,{\left (a+b\right )}^2}+\frac {1}{a+b}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )-\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}-\frac {5\,b}{8\,{\left (a-b\right )}^2}+\frac {1}{a-b}\right )+\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}-\frac {b^6\,\ln \left (b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+b\right )}{-a^7+3\,a^5\,b^2-3\,a^3\,b^4+a\,b^6} \]
((2*tan(x/2)^2*(2*a*b^2 - a^3))/(a^4 + b^4 - 2*a^2*b^2) - (4*tan(x/2)^4*(3 *a*b^2 - 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) + (2*tan(x/2)^6*(2*a*b^2 - a^3))/ (a^4 + b^4 - 2*a^2*b^2) + (tan(x/2)^7*(3*a^2*b - 7*b^3))/(4*(a^4 + b^4 - 2 *a^2*b^2)) - (tan(x/2)^3*(11*a^2*b - 15*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) - (tan(x/2)^5*(11*a^2*b - 15*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (b*tan(x/ 2)*(3*a^2 - 7*b^2))/(4*(a^4 + b^4 - 2*a^2*b^2)))/(6*tan(x/2)^4 - 4*tan(x/2 )^2 - 4*tan(x/2)^6 + tan(x/2)^8 + 1) - log(tan(x/2) - 1)*((5*b)/(8*(a + b) ^2) + 1/(a + b) + b^2/(4*(a + b)^3)) - log(tan(x/2) + 1)*(b^2/(4*(a - b)^3 ) - (5*b)/(8*(a - b)^2) + 1/(a - b)) + log(tan(x/2)^2 + 1)/a - (b^6*log(b + 2*a*tan(x/2) + b*tan(x/2)^2))/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2)