Integrand size = 15, antiderivative size = 62 \[ \int \frac {\csc ^3(x)}{a+b \cos ^2(x)} \, dx=-\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \cos (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^2}-\frac {(a+3 b) \text {arctanh}(\cos (x))}{2 (a+b)^2}-\frac {\cot (x) \csc (x)}{2 (a+b)} \]
-1/2*(a+3*b)*arctanh(cos(x))/(a+b)^2-1/2*cot(x)*csc(x)/(a+b)-b^(3/2)*arcta n(cos(x)*b^(1/2)/a^(1/2))/(a+b)^2/a^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(140\) vs. \(2(62)=124\).
Time = 0.43 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.26 \[ \int \frac {\csc ^3(x)}{a+b \cos ^2(x)} \, dx=\frac {-8 b^{3/2} \arctan \left (\frac {\sqrt {b}-\sqrt {a+b} \tan \left (\frac {x}{2}\right )}{\sqrt {a}}\right )-8 b^{3/2} \arctan \left (\frac {\sqrt {b}+\sqrt {a+b} \tan \left (\frac {x}{2}\right )}{\sqrt {a}}\right )+\sqrt {a} \left (-\left ((a+b) \csc ^2\left (\frac {x}{2}\right )\right )-4 (a+3 b) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )+(a+b) \sec ^2\left (\frac {x}{2}\right )\right )}{8 \sqrt {a} (a+b)^2} \]
(-8*b^(3/2)*ArcTan[(Sqrt[b] - Sqrt[a + b]*Tan[x/2])/Sqrt[a]] - 8*b^(3/2)*A rcTan[(Sqrt[b] + Sqrt[a + b]*Tan[x/2])/Sqrt[a]] + Sqrt[a]*(-((a + b)*Csc[x /2]^2) - 4*(a + 3*b)*(Log[Cos[x/2]] - Log[Sin[x/2]]) + (a + b)*Sec[x/2]^2) )/(8*Sqrt[a]*(a + b)^2)
Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 25, 3669, 316, 397, 218, 219}
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.
| \(\displaystyle \int \frac {\csc ^3(x)}{a+b \cos ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\cos \left (x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (x+\frac {\pi }{2}\right )^2\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\cos \left (x+\frac {\pi }{2}\right )^3 \left (b \sin \left (x+\frac {\pi }{2}\right )^2+a\right )}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle -\int \frac {1}{\left (1-\cos ^2(x)\right )^2 \left (b \cos ^2(x)+a\right )}d\cos (x)\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle -\frac {\int \frac {b \cos ^2(x)+a+2 b}{\left (1-\cos ^2(x)\right ) \left (b \cos ^2(x)+a\right )}d\cos (x)}{2 (a+b)}-\frac {\cos (x)}{2 (a+b) \left (1-\cos ^2(x)\right )}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle -\frac {\frac {2 b^2 \int \frac {1}{b \cos ^2(x)+a}d\cos (x)}{a+b}+\frac {(a+3 b) \int \frac {1}{1-\cos ^2(x)}d\cos (x)}{a+b}}{2 (a+b)}-\frac {\cos (x)}{2 (a+b) \left (1-\cos ^2(x)\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {(a+3 b) \int \frac {1}{1-\cos ^2(x)}d\cos (x)}{a+b}+\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \cos (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 (a+b)}-\frac {\cos (x)}{2 (a+b) \left (1-\cos ^2(x)\right )}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \cos (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}+\frac {(a+3 b) \text {arctanh}(\cos (x))}{a+b}}{2 (a+b)}-\frac {\cos (x)}{2 (a+b) \left (1-\cos ^2(x)\right )}\) |
-1/2*((2*b^(3/2)*ArcTan[(Sqrt[b]*Cos[x])/Sqrt[a]])/(Sqrt[a]*(a + b)) + ((a + 3*b)*ArcTanh[Cos[x]])/(a + b))/(a + b) - Cos[x]/(2*(a + b)*(1 - Cos[x]^ 2))
3.1.15.3.1 Defintions of rubi rules used
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)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.41 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.53
| method | result | size |
| default | \(\frac {1}{\left (4 a +4 b \right ) \left (1+\cos \left (x \right )\right )}+\frac {\left (-a -3 b \right ) \ln \left (1+\cos \left (x \right )\right )}{4 \left (a +b \right )^{2}}-\frac {b^{2} \arctan \left (\frac {b \cos \left (x \right )}{\sqrt {a b}}\right )}{\left (a +b \right )^{2} \sqrt {a b}}+\frac {1}{\left (4 a +4 b \right ) \left (\cos \left (x \right )-1\right )}+\frac {\left (a +3 b \right ) \ln \left (\cos \left (x \right )-1\right )}{4 \left (a +b \right )^{2}}\) | \(95\) |
| risch | \(\frac {{\mathrm e}^{3 i x}+{\mathrm e}^{i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2} \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i x}-1\right ) a}{2 a^{2}+4 a b +2 b^{2}}+\frac {3 \ln \left ({\mathrm e}^{i x}-1\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i x}+1\right ) a}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {3 \ln \left ({\mathrm e}^{i x}+1\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {i \sqrt {a b}\, b \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i x}}{b}+1\right )}{2 a \left (a +b \right )^{2}}+\frac {i \sqrt {a b}\, b \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i x}}{b}+1\right )}{2 a \left (a +b \right )^{2}}\) | \(206\) |
1/(4*a+4*b)/(1+cos(x))+1/4/(a+b)^2*(-a-3*b)*ln(1+cos(x))-b^2/(a+b)^2/(a*b) ^(1/2)*arctan(b*cos(x)/(a*b)^(1/2))+1/(4*a+4*b)/(cos(x)-1)+1/4*(a+3*b)/(a+ b)^2*ln(cos(x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (50) = 100\).
Time = 0.28 (sec) , antiderivative size = 274, normalized size of antiderivative = 4.42 \[ \int \frac {\csc ^3(x)}{a+b \cos ^2(x)} \, dx=\left [\frac {2 \, {\left (b \cos \left (x\right )^{2} - b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b \cos \left (x\right )^{2} - 2 \, a \sqrt {-\frac {b}{a}} \cos \left (x\right ) - a}{b \cos \left (x\right )^{2} + a}\right ) + 2 \, {\left (a + b\right )} \cos \left (x\right ) - {\left ({\left (a + 3 \, b\right )} \cos \left (x\right )^{2} - a - 3 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left ({\left (a + 3 \, b\right )} \cos \left (x\right )^{2} - a - 3 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )}}, -\frac {4 \, {\left (b \cos \left (x\right )^{2} - b\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {\frac {b}{a}} \cos \left (x\right )\right ) - 2 \, {\left (a + b\right )} \cos \left (x\right ) + {\left ({\left (a + 3 \, b\right )} \cos \left (x\right )^{2} - a - 3 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left ({\left (a + 3 \, b\right )} \cos \left (x\right )^{2} - a - 3 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )}}\right ] \]
[1/4*(2*(b*cos(x)^2 - b)*sqrt(-b/a)*log((b*cos(x)^2 - 2*a*sqrt(-b/a)*cos(x ) - a)/(b*cos(x)^2 + a)) + 2*(a + b)*cos(x) - ((a + 3*b)*cos(x)^2 - a - 3* b)*log(1/2*cos(x) + 1/2) + ((a + 3*b)*cos(x)^2 - a - 3*b)*log(-1/2*cos(x) + 1/2))/((a^2 + 2*a*b + b^2)*cos(x)^2 - a^2 - 2*a*b - b^2), -1/4*(4*(b*cos (x)^2 - b)*sqrt(b/a)*arctan(sqrt(b/a)*cos(x)) - 2*(a + b)*cos(x) + ((a + 3 *b)*cos(x)^2 - a - 3*b)*log(1/2*cos(x) + 1/2) - ((a + 3*b)*cos(x)^2 - a - 3*b)*log(-1/2*cos(x) + 1/2))/((a^2 + 2*a*b + b^2)*cos(x)^2 - a^2 - 2*a*b - b^2)]
\[ \int \frac {\csc ^3(x)}{a+b \cos ^2(x)} \, dx=\int \frac {\csc ^{3}{\left (x \right )}}{a + b \cos ^{2}{\left (x \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (50) = 100\).
Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.69 \[ \int \frac {\csc ^3(x)}{a+b \cos ^2(x)} \, dx=-\frac {b^{2} \arctan \left (\frac {b \cos \left (x\right )}{\sqrt {a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b}} - \frac {{\left (a + 3 \, b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {{\left (a + 3 \, b\right )} \log \left (\cos \left (x\right ) - 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {\cos \left (x\right )}{2 \, {\left ({\left (a + b\right )} \cos \left (x\right )^{2} - a - b\right )}} \]
-b^2*arctan(b*cos(x)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)) - 1/4*(a + 3*b)*log(cos(x) + 1)/(a^2 + 2*a*b + b^2) + 1/4*(a + 3*b)*log(cos(x) - 1)/ (a^2 + 2*a*b + b^2) + 1/2*cos(x)/((a + b)*cos(x)^2 - a - b)
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (50) = 100\).
Time = 0.33 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.66 \[ \int \frac {\csc ^3(x)}{a+b \cos ^2(x)} \, dx=-\frac {b^{2} \arctan \left (\frac {b \cos \left (x\right )}{\sqrt {a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b}} - \frac {{\left (a + 3 \, b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {{\left (a + 3 \, b\right )} \log \left (-\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {\cos \left (x\right )}{2 \, {\left (\cos \left (x\right )^{2} - 1\right )} {\left (a + b\right )}} \]
-b^2*arctan(b*cos(x)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)) - 1/4*(a + 3*b)*log(cos(x) + 1)/(a^2 + 2*a*b + b^2) + 1/4*(a + 3*b)*log(-cos(x) + 1) /(a^2 + 2*a*b + b^2) + 1/2*cos(x)/((cos(x)^2 - 1)*(a + b))
Time = 3.02 (sec) , antiderivative size = 1138, normalized size of antiderivative = 18.35 \[ \int \frac {\csc ^3(x)}{a+b \cos ^2(x)} \, dx=\ln \left (\cos \left (x\right )-1\right )\,\left (\frac {b}{2\,{\left (a+b\right )}^2}+\frac {1}{4\,\left (a+b\right )}\right )-\frac {\cos \left (x\right )}{2\,{\sin \left (x\right )}^2\,\left (a+b\right )}-\frac {\ln \left (\cos \left (x\right )+1\right )\,\left (a+3\,b\right )}{4\,{\left (a+b\right )}^2}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {-a\,b^3}\,\left (\frac {\cos \left (x\right )\,\left (a^2\,b^3+6\,a\,b^4+13\,b^5\right )}{4\,\left (a^2+2\,a\,b+b^2\right )}+\frac {\left (\frac {2\,a^5\,b^2+12\,a^4\,b^3+28\,a^3\,b^4+32\,a^2\,b^5+18\,a\,b^6+4\,b^7}{2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {\cos \left (x\right )\,\sqrt {-a\,b^3}\,\left (-16\,a^5\,b^2-48\,a^4\,b^3-32\,a^3\,b^4+32\,a^2\,b^5+48\,a\,b^6+16\,b^7\right )}{8\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,\sqrt {-a\,b^3}}{2\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,1{}\mathrm {i}}{a^3+2\,a^2\,b+a\,b^2}+\frac {\sqrt {-a\,b^3}\,\left (\frac {\cos \left (x\right )\,\left (a^2\,b^3+6\,a\,b^4+13\,b^5\right )}{4\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\left (\frac {2\,a^5\,b^2+12\,a^4\,b^3+28\,a^3\,b^4+32\,a^2\,b^5+18\,a\,b^6+4\,b^7}{2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}+\frac {\cos \left (x\right )\,\sqrt {-a\,b^3}\,\left (-16\,a^5\,b^2-48\,a^4\,b^3-32\,a^3\,b^4+32\,a^2\,b^5+48\,a\,b^6+16\,b^7\right )}{8\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,\sqrt {-a\,b^3}}{2\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,1{}\mathrm {i}}{a^3+2\,a^2\,b+a\,b^2}}{\frac {\frac {3\,b^5}{2}+\frac {a\,b^4}{2}}{a^3+3\,a^2\,b+3\,a\,b^2+b^3}-\frac {\sqrt {-a\,b^3}\,\left (\frac {\cos \left (x\right )\,\left (a^2\,b^3+6\,a\,b^4+13\,b^5\right )}{4\,\left (a^2+2\,a\,b+b^2\right )}+\frac {\left (\frac {2\,a^5\,b^2+12\,a^4\,b^3+28\,a^3\,b^4+32\,a^2\,b^5+18\,a\,b^6+4\,b^7}{2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {\cos \left (x\right )\,\sqrt {-a\,b^3}\,\left (-16\,a^5\,b^2-48\,a^4\,b^3-32\,a^3\,b^4+32\,a^2\,b^5+48\,a\,b^6+16\,b^7\right )}{8\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,\sqrt {-a\,b^3}}{2\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )}{a^3+2\,a^2\,b+a\,b^2}+\frac {\sqrt {-a\,b^3}\,\left (\frac {\cos \left (x\right )\,\left (a^2\,b^3+6\,a\,b^4+13\,b^5\right )}{4\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\left (\frac {2\,a^5\,b^2+12\,a^4\,b^3+28\,a^3\,b^4+32\,a^2\,b^5+18\,a\,b^6+4\,b^7}{2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}+\frac {\cos \left (x\right )\,\sqrt {-a\,b^3}\,\left (-16\,a^5\,b^2-48\,a^4\,b^3-32\,a^3\,b^4+32\,a^2\,b^5+48\,a\,b^6+16\,b^7\right )}{8\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,\sqrt {-a\,b^3}}{2\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )}{a^3+2\,a^2\,b+a\,b^2}}\right )\,\sqrt {-a\,b^3}\,1{}\mathrm {i}}{a^3+2\,a^2\,b+a\,b^2} \]
log(cos(x) - 1)*(b/(2*(a + b)^2) + 1/(4*(a + b))) - cos(x)/(2*sin(x)^2*(a + b)) - (log(cos(x) + 1)*(a + 3*b))/(4*(a + b)^2) - (atan((((-a*b^3)^(1/2) *((cos(x)*(6*a*b^4 + 13*b^5 + a^2*b^3))/(4*(2*a*b + a^2 + b^2)) + (((18*a* b^6 + 4*b^7 + 32*a^2*b^5 + 28*a^3*b^4 + 12*a^4*b^3 + 2*a^5*b^2)/(2*(3*a*b^ 2 + 3*a^2*b + a^3 + b^3)) - (cos(x)*(-a*b^3)^(1/2)*(48*a*b^6 + 16*b^7 + 32 *a^2*b^5 - 32*a^3*b^4 - 48*a^4*b^3 - 16*a^5*b^2))/(8*(2*a*b + a^2 + b^2)*( a*b^2 + 2*a^2*b + a^3)))*(-a*b^3)^(1/2))/(2*(a*b^2 + 2*a^2*b + a^3)))*1i)/ (a*b^2 + 2*a^2*b + a^3) + ((-a*b^3)^(1/2)*((cos(x)*(6*a*b^4 + 13*b^5 + a^2 *b^3))/(4*(2*a*b + a^2 + b^2)) - (((18*a*b^6 + 4*b^7 + 32*a^2*b^5 + 28*a^3 *b^4 + 12*a^4*b^3 + 2*a^5*b^2)/(2*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) + (cos( x)*(-a*b^3)^(1/2)*(48*a*b^6 + 16*b^7 + 32*a^2*b^5 - 32*a^3*b^4 - 48*a^4*b^ 3 - 16*a^5*b^2))/(8*(2*a*b + a^2 + b^2)*(a*b^2 + 2*a^2*b + a^3)))*(-a*b^3) ^(1/2))/(2*(a*b^2 + 2*a^2*b + a^3)))*1i)/(a*b^2 + 2*a^2*b + a^3))/(((a*b^4 )/2 + (3*b^5)/2)/(3*a*b^2 + 3*a^2*b + a^3 + b^3) - ((-a*b^3)^(1/2)*((cos(x )*(6*a*b^4 + 13*b^5 + a^2*b^3))/(4*(2*a*b + a^2 + b^2)) + (((18*a*b^6 + 4* b^7 + 32*a^2*b^5 + 28*a^3*b^4 + 12*a^4*b^3 + 2*a^5*b^2)/(2*(3*a*b^2 + 3*a^ 2*b + a^3 + b^3)) - (cos(x)*(-a*b^3)^(1/2)*(48*a*b^6 + 16*b^7 + 32*a^2*b^5 - 32*a^3*b^4 - 48*a^4*b^3 - 16*a^5*b^2))/(8*(2*a*b + a^2 + b^2)*(a*b^2 + 2*a^2*b + a^3)))*(-a*b^3)^(1/2))/(2*(a*b^2 + 2*a^2*b + a^3))))/(a*b^2 + 2* a^2*b + a^3) + ((-a*b^3)^(1/2)*((cos(x)*(6*a*b^4 + 13*b^5 + a^2*b^3))/(...