Integrand size = 15, antiderivative size = 93 \[ \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx=\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^3}+\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {arctanh}(\sin (x))}{8 (a+b)^3}+\frac {(3 a+7 b) \sec (x) \tan (x)}{8 (a+b)^2}+\frac {\sec ^3(x) \tan (x)}{4 (a+b)} \]
1/8*(3*a^2+10*a*b+15*b^2)*arctanh(sin(x))/(a+b)^3+b^(5/2)*arctan(sin(x)*b^ (1/2)/a^(1/2))/(a+b)^3/a^(1/2)+1/8*(3*a+7*b)*sec(x)*tan(x)/(a+b)^2+1/4*sec (x)^3*tan(x)/(a+b)
Leaf count is larger than twice the leaf count of optimal. \(214\) vs. \(2(93)=186\).
Time = 1.33 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.30 \[ \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx=-\frac {\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )}{\sqrt {a}}-\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a}}+2 \left (3 a^2+10 a b+15 b^2\right ) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-2 \left (3 a^2+10 a b+15 b^2\right ) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-\frac {(a+b)^2}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^4}+\frac {(a+b)^2}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^4}+\frac {(a+b) (3 a+7 b)}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2}+\frac {(a+b) (3 a+7 b)}{-1+\sin (x)}}{16 (a+b)^3} \]
-1/16*((8*b^(5/2)*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]])/Sqrt[a] - (8*b^(5/2)*A rcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/Sqrt[a] + 2*(3*a^2 + 10*a*b + 15*b^2)*Log [Cos[x/2] - Sin[x/2]] - 2*(3*a^2 + 10*a*b + 15*b^2)*Log[Cos[x/2] + Sin[x/2 ]] - (a + b)^2/(Cos[x/2] - Sin[x/2])^4 + (a + b)^2/(Cos[x/2] + Sin[x/2])^4 + ((a + b)*(3*a + 7*b))/(Cos[x/2] + Sin[x/2])^2 + ((a + b)*(3*a + 7*b))/( -1 + Sin[x]))/(a + b)^3
Time = 0.32 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.34, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 3669, 316, 402, 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.
| \(\displaystyle \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (x)^5 \left (a+b \sin (x)^2\right )}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \int \frac {1}{\left (1-\sin ^2(x)\right )^3 \left (a+b \sin ^2(x)\right )}d\sin (x)\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\int \frac {3 b \sin ^2(x)+3 a+4 b}{\left (1-\sin ^2(x)\right )^2 \left (b \sin ^2(x)+a\right )}d\sin (x)}{4 (a+b)}+\frac {\sin (x)}{4 (a+b) \left (1-\sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2+7 b a+8 b^2+b (3 a+7 b) \sin ^2(x)}{\left (1-\sin ^2(x)\right ) \left (b \sin ^2(x)+a\right )}d\sin (x)}{2 (a+b)}+\frac {(3 a+7 b) \sin (x)}{2 (a+b) \left (1-\sin ^2(x)\right )}}{4 (a+b)}+\frac {\sin (x)}{4 (a+b) \left (1-\sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\left (3 a^2+10 a b+15 b^2\right ) \int \frac {1}{1-\sin ^2(x)}d\sin (x)}{a+b}+\frac {8 b^3 \int \frac {1}{b \sin ^2(x)+a}d\sin (x)}{a+b}}{2 (a+b)}+\frac {(3 a+7 b) \sin (x)}{2 (a+b) \left (1-\sin ^2(x)\right )}}{4 (a+b)}+\frac {\sin (x)}{4 (a+b) \left (1-\sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\left (3 a^2+10 a b+15 b^2\right ) \int \frac {1}{1-\sin ^2(x)}d\sin (x)}{a+b}+\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 (a+b)}+\frac {(3 a+7 b) \sin (x)}{2 (a+b) \left (1-\sin ^2(x)\right )}}{4 (a+b)}+\frac {\sin (x)}{4 (a+b) \left (1-\sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {arctanh}(\sin (x))}{a+b}+\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 (a+b)}+\frac {(3 a+7 b) \sin (x)}{2 (a+b) \left (1-\sin ^2(x)\right )}}{4 (a+b)}+\frac {\sin (x)}{4 (a+b) \left (1-\sin ^2(x)\right )^2}\) |
Sin[x]/(4*(a + b)*(1 - Sin[x]^2)^2) + (((8*b^(5/2)*ArcTan[(Sqrt[b]*Sin[x]) /Sqrt[a]])/(Sqrt[a]*(a + b)) + ((3*a^2 + 10*a*b + 15*b^2)*ArcTanh[Sin[x]]) /(a + b))/(2*(a + b)) + ((3*a + 7*b)*Sin[x])/(2*(a + b)*(1 - Sin[x]^2)))/( 4*(a + b))
3.4.12.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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 = 1.16 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.66
| method | result | size |
| default | \(-\frac {1}{2 \left (8 a +8 b \right ) \left (1+\sin \left (x \right )\right )^{2}}-\frac {3 a +7 b}{16 \left (a +b \right )^{2} \left (1+\sin \left (x \right )\right )}+\frac {\left (3 a^{2}+10 a b +15 b^{2}\right ) \ln \left (1+\sin \left (x \right )\right )}{16 \left (a +b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (x \right )-1\right )^{2}}-\frac {3 a +7 b}{16 \left (a +b \right )^{2} \left (\sin \left (x \right )-1\right )}+\frac {\left (-3 a^{2}-10 a b -15 b^{2}\right ) \ln \left (\sin \left (x \right )-1\right )}{16 \left (a +b \right )^{3}}+\frac {b^{3} \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{\left (a +b \right )^{3} \sqrt {a b}}\) | \(154\) |
| risch | \(-\frac {i \left (3 a \,{\mathrm e}^{7 i x}+7 b \,{\mathrm e}^{7 i x}+11 a \,{\mathrm e}^{5 i x}+15 b \,{\mathrm e}^{5 i x}-11 a \,{\mathrm e}^{3 i x}-15 b \,{\mathrm e}^{3 i x}-3 \,{\mathrm e}^{i x} a -7 \,{\mathrm e}^{i x} b \right )}{4 \left ({\mathrm e}^{2 i x}+1\right )^{4} \left (a +b \right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i x}+i\right ) a^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {5 \ln \left ({\mathrm e}^{i x}+i\right ) a b}{4 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {15 \ln \left ({\mathrm e}^{i x}+i\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 \ln \left ({\mathrm e}^{i x}-i\right ) a^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {5 \ln \left ({\mathrm e}^{i x}-i\right ) a b}{4 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {15 \ln \left ({\mathrm e}^{i x}-i\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 a \left (a +b \right )^{3}}-\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 a \left (a +b \right )^{3}}\) | \(380\) |
-1/2/(8*a+8*b)/(1+sin(x))^2-1/16*(3*a+7*b)/(a+b)^2/(1+sin(x))+1/16*(3*a^2+ 10*a*b+15*b^2)/(a+b)^3*ln(1+sin(x))+1/2/(8*a+8*b)/(sin(x)-1)^2-1/16*(3*a+7 *b)/(a+b)^2/(sin(x)-1)+1/16/(a+b)^3*(-3*a^2-10*a*b-15*b^2)*ln(sin(x)-1)+b^ 3/(a+b)^3/(a*b)^(1/2)*arctan(b*sin(x)/(a*b)^(1/2))
Time = 0.40 (sec) , antiderivative size = 327, normalized size of antiderivative = 3.52 \[ \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx=\left [\frac {8 \, b^{2} \sqrt {-\frac {b}{a}} \cos \left (x\right )^{4} \log \left (-\frac {b \cos \left (x\right )^{2} - 2 \, a \sqrt {-\frac {b}{a}} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) + {\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - {\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left ({\left (3 \, a^{2} + 10 \, a b + 7 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sin \left (x\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (x\right )^{4}}, \frac {16 \, b^{2} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {\frac {b}{a}} \sin \left (x\right )\right ) \cos \left (x\right )^{4} + {\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - {\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left ({\left (3 \, a^{2} + 10 \, a b + 7 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sin \left (x\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (x\right )^{4}}\right ] \]
[1/16*(8*b^2*sqrt(-b/a)*cos(x)^4*log(-(b*cos(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + (3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(si n(x) + 1) - (3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(-sin(x) + 1) + 2*((3*a^ 2 + 10*a*b + 7*b^2)*cos(x)^2 + 2*a^2 + 4*a*b + 2*b^2)*sin(x))/((a^3 + 3*a^ 2*b + 3*a*b^2 + b^3)*cos(x)^4), 1/16*(16*b^2*sqrt(b/a)*arctan(sqrt(b/a)*si n(x))*cos(x)^4 + (3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(sin(x) + 1) - (3*a ^2 + 10*a*b + 15*b^2)*cos(x)^4*log(-sin(x) + 1) + 2*((3*a^2 + 10*a*b + 7*b ^2)*cos(x)^2 + 2*a^2 + 4*a*b + 2*b^2)*sin(x))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(x)^4)]
\[ \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx=\int \frac {\sec ^{5}{\left (x \right )}}{a + b \sin ^{2}{\left (x \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (79) = 158\).
Time = 0.33 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.14 \[ \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx=\frac {b^{3} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b}} + \frac {{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) - 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (3 \, a + 7 \, b\right )} \sin \left (x\right )^{3} - {\left (5 \, a + 9 \, b\right )} \sin \left (x\right )}{8 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (x\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )}} \]
b^3*arctan(b*sin(x)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)) + 1/16*(3*a^2 + 10*a*b + 15*b^2)*log(sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/16*(3*a^2 + 10*a*b + 15*b^2)*log(sin(x) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/8*((3*a + 7*b)*sin(x)^3 - (5*a + 9*b)*sin(x))/((a^2 + 2 *a*b + b^2)*sin(x)^4 - 2*(a^2 + 2*a*b + b^2)*sin(x)^2 + a^2 + 2*a*b + b^2)
Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (79) = 158\).
Time = 0.29 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.90 \[ \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx=\frac {b^{3} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b}} + \frac {{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {3 \, a \sin \left (x\right )^{3} + 7 \, b \sin \left (x\right )^{3} - 5 \, a \sin \left (x\right ) - 9 \, b \sin \left (x\right )}{8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (\sin \left (x\right )^{2} - 1\right )}^{2}} \]
b^3*arctan(b*sin(x)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)) + 1/16*(3*a^2 + 10*a*b + 15*b^2)*log(sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/16*(3*a^2 + 10*a*b + 15*b^2)*log(-sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/8*(3*a*sin(x)^3 + 7*b*sin(x)^3 - 5*a*sin(x) - 9*b*sin( x))/((a^2 + 2*a*b + b^2)*(sin(x)^2 - 1)^2)
Time = 17.83 (sec) , antiderivative size = 832, normalized size of antiderivative = 8.95 \[ \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx=\frac {5\,a^3\,\sin \left (x\right )-3\,a^3\,{\sin \left (x\right )}^3+3\,a^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )+9\,a\,b^2\,\sin \left (x\right )+14\,a^2\,b\,\sin \left (x\right )-6\,a^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+3\,a^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^4-7\,a\,b^2\,{\sin \left (x\right )}^3-10\,a^2\,b\,{\sin \left (x\right )}^3+15\,a\,b^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )+10\,a^2\,b\,\mathrm {atanh}\left (\sin \left (x\right )\right )-30\,a\,b^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2-20\,a^2\,b\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+15\,a\,b^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^4+10\,a^2\,b\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^4+\mathrm {atan}\left (\frac {a\,\sin \left (x\right )\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \left (x\right )\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}+a^6\,b\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,289{}\mathrm {i}+a^3\,b^4\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,300{}\mathrm {i}+a^4\,b^3\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,190{}\mathrm {i}+a^5\,b^2\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,60{}\mathrm {i}}{9\,a^7\,b^3+60\,a^6\,b^4+190\,a^5\,b^5+300\,a^4\,b^6+225\,a^3\,b^7+64\,a^2\,b^8}\right )\,\sqrt {-a\,b^5}\,8{}\mathrm {i}-\mathrm {atan}\left (\frac {a\,\sin \left (x\right )\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \left (x\right )\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}+a^6\,b\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,289{}\mathrm {i}+a^3\,b^4\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,300{}\mathrm {i}+a^4\,b^3\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,190{}\mathrm {i}+a^5\,b^2\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,60{}\mathrm {i}}{9\,a^7\,b^3+60\,a^6\,b^4+190\,a^5\,b^5+300\,a^4\,b^6+225\,a^3\,b^7+64\,a^2\,b^8}\right )\,{\sin \left (x\right )}^2\,\sqrt {-a\,b^5}\,16{}\mathrm {i}+\mathrm {atan}\left (\frac {a\,\sin \left (x\right )\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \left (x\right )\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}+a^6\,b\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,289{}\mathrm {i}+a^3\,b^4\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,300{}\mathrm {i}+a^4\,b^3\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,190{}\mathrm {i}+a^5\,b^2\,\sin \left (x\right )\,\sqrt {-a\,b^5}\,60{}\mathrm {i}}{9\,a^7\,b^3+60\,a^6\,b^4+190\,a^5\,b^5+300\,a^4\,b^6+225\,a^3\,b^7+64\,a^2\,b^8}\right )\,{\sin \left (x\right )}^4\,\sqrt {-a\,b^5}\,8{}\mathrm {i}}{8\,a^4\,{\sin \left (x\right )}^4-16\,a^4\,{\sin \left (x\right )}^2+8\,a^4+24\,a^3\,b\,{\sin \left (x\right )}^4-48\,a^3\,b\,{\sin \left (x\right )}^2+24\,a^3\,b+24\,a^2\,b^2\,{\sin \left (x\right )}^4-48\,a^2\,b^2\,{\sin \left (x\right )}^2+24\,a^2\,b^2+8\,a\,b^3\,{\sin \left (x\right )}^4-16\,a\,b^3\,{\sin \left (x\right )}^2+8\,a\,b^3} \]
(5*a^3*sin(x) - 3*a^3*sin(x)^3 + 3*a^3*atanh(sin(x)) + atan((a*sin(x)*(-a* b^5)^(3/2)*64i - b*sin(x)*(-a*b^5)^(3/2)*64i + a^6*b*sin(x)*(-a*b^5)^(1/2) *9i + a^2*b^5*sin(x)*(-a*b^5)^(1/2)*289i + a^3*b^4*sin(x)*(-a*b^5)^(1/2)*3 00i + a^4*b^3*sin(x)*(-a*b^5)^(1/2)*190i + a^5*b^2*sin(x)*(-a*b^5)^(1/2)*6 0i)/(64*a^2*b^8 + 225*a^3*b^7 + 300*a^4*b^6 + 190*a^5*b^5 + 60*a^6*b^4 + 9 *a^7*b^3))*(-a*b^5)^(1/2)*8i + 9*a*b^2*sin(x) + 14*a^2*b*sin(x) - 6*a^3*at anh(sin(x))*sin(x)^2 + 3*a^3*atanh(sin(x))*sin(x)^4 - 7*a*b^2*sin(x)^3 - 1 0*a^2*b*sin(x)^3 + 15*a*b^2*atanh(sin(x)) + 10*a^2*b*atanh(sin(x)) - atan( (a*sin(x)*(-a*b^5)^(3/2)*64i - b*sin(x)*(-a*b^5)^(3/2)*64i + a^6*b*sin(x)* (-a*b^5)^(1/2)*9i + a^2*b^5*sin(x)*(-a*b^5)^(1/2)*289i + a^3*b^4*sin(x)*(- a*b^5)^(1/2)*300i + a^4*b^3*sin(x)*(-a*b^5)^(1/2)*190i + a^5*b^2*sin(x)*(- a*b^5)^(1/2)*60i)/(64*a^2*b^8 + 225*a^3*b^7 + 300*a^4*b^6 + 190*a^5*b^5 + 60*a^6*b^4 + 9*a^7*b^3))*sin(x)^2*(-a*b^5)^(1/2)*16i + atan((a*sin(x)*(-a* b^5)^(3/2)*64i - b*sin(x)*(-a*b^5)^(3/2)*64i + a^6*b*sin(x)*(-a*b^5)^(1/2) *9i + a^2*b^5*sin(x)*(-a*b^5)^(1/2)*289i + a^3*b^4*sin(x)*(-a*b^5)^(1/2)*3 00i + a^4*b^3*sin(x)*(-a*b^5)^(1/2)*190i + a^5*b^2*sin(x)*(-a*b^5)^(1/2)*6 0i)/(64*a^2*b^8 + 225*a^3*b^7 + 300*a^4*b^6 + 190*a^5*b^5 + 60*a^6*b^4 + 9 *a^7*b^3))*sin(x)^4*(-a*b^5)^(1/2)*8i - 30*a*b^2*atanh(sin(x))*sin(x)^2 - 20*a^2*b*atanh(sin(x))*sin(x)^2 + 15*a*b^2*atanh(sin(x))*sin(x)^4 + 10*a^2 *b*atanh(sin(x))*sin(x)^4)/(8*a^4*sin(x)^4 - 16*a^4*sin(x)^2 + 8*a*b^3 ...