Integrand size = 23, antiderivative size = 125 \[ \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\left (3 a^2-4 a b+8 b^2\right ) \text {arctanh}(\cos (c+d x))}{8 a^3 d}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b} d}-\frac {(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]
-1/8*(3*a^2-4*a*b+8*b^2)*arctanh(cos(d*x+c))/a^3/d-1/8*(3*a-4*b)*cot(d*x+c )*csc(d*x+c)/a^2/d-1/4*cot(d*x+c)*csc(d*x+c)^3/a/d+b^(5/2)*arctanh(cos(d*x +c)*b^(1/2)/(a+b)^(1/2))/a^3/d/(a+b)^(1/2)
Result contains complex when optimal does not.
Time = 6.63 (sec) , antiderivative size = 657, normalized size of antiderivative = 5.26 \[ \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {b^{5/2} \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {b} \cos \left (\frac {1}{2} (c+d x)\right )-i \sqrt {a} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-a-b}}\right ) (-2 a-b+b \cos (2 (c+d x))) \csc ^2(c+d x)}{2 a^3 \sqrt {-a-b} d \left (b+a \csc ^2(c+d x)\right )}+\frac {b^{5/2} \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {b} \cos \left (\frac {1}{2} (c+d x)\right )+i \sqrt {a} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-a-b}}\right ) (-2 a-b+b \cos (2 (c+d x))) \csc ^2(c+d x)}{2 a^3 \sqrt {-a-b} d \left (b+a \csc ^2(c+d x)\right )}+\frac {(3 a-4 b) (-2 a-b+b \cos (2 (c+d x))) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \csc ^2(c+d x)}{64 a^2 d \left (b+a \csc ^2(c+d x)\right )}+\frac {(-2 a-b+b \cos (2 (c+d x))) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^2(c+d x)}{128 a d \left (b+a \csc ^2(c+d x)\right )}+\frac {\left (3 a^2-4 a b+8 b^2\right ) (-2 a-b+b \cos (2 (c+d x))) \csc ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{16 a^3 d \left (b+a \csc ^2(c+d x)\right )}+\frac {\left (-3 a^2+4 a b-8 b^2\right ) (-2 a-b+b \cos (2 (c+d x))) \csc ^2(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{16 a^3 d \left (b+a \csc ^2(c+d x)\right )}+\frac {(-3 a+4 b) (-2 a-b+b \cos (2 (c+d x))) \csc ^2(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{64 a^2 d \left (b+a \csc ^2(c+d x)\right )}-\frac {(-2 a-b+b \cos (2 (c+d x))) \csc ^2(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right )}{128 a d \left (b+a \csc ^2(c+d x)\right )} \]
(b^(5/2)*ArcTan[(Sec[(c + d*x)/2]*(Sqrt[b]*Cos[(c + d*x)/2] - I*Sqrt[a]*Si n[(c + d*x)/2]))/Sqrt[-a - b]]*(-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x ]^2)/(2*a^3*Sqrt[-a - b]*d*(b + a*Csc[c + d*x]^2)) + (b^(5/2)*ArcTan[(Sec[ (c + d*x)/2]*(Sqrt[b]*Cos[(c + d*x)/2] + I*Sqrt[a]*Sin[(c + d*x)/2]))/Sqrt [-a - b]]*(-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2)/(2*a^3*Sqrt[-a - b]*d*(b + a*Csc[c + d*x]^2)) + ((3*a - 4*b)*(-2*a - b + b*Cos[2*(c + d*x) ])*Csc[(c + d*x)/2]^2*Csc[c + d*x]^2)/(64*a^2*d*(b + a*Csc[c + d*x]^2)) + ((-2*a - b + b*Cos[2*(c + d*x)])*Csc[(c + d*x)/2]^4*Csc[c + d*x]^2)/(128*a *d*(b + a*Csc[c + d*x]^2)) + ((3*a^2 - 4*a*b + 8*b^2)*(-2*a - b + b*Cos[2* (c + d*x)])*Csc[c + d*x]^2*Log[Cos[(c + d*x)/2]])/(16*a^3*d*(b + a*Csc[c + d*x]^2)) + ((-3*a^2 + 4*a*b - 8*b^2)*(-2*a - b + b*Cos[2*(c + d*x)])*Csc[ c + d*x]^2*Log[Sin[(c + d*x)/2]])/(16*a^3*d*(b + a*Csc[c + d*x]^2)) + ((-3 *a + 4*b)*(-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*Sec[(c + d*x)/2]^ 2)/(64*a^2*d*(b + a*Csc[c + d*x]^2)) - ((-2*a - b + b*Cos[2*(c + d*x)])*Cs c[c + d*x]^2*Sec[(c + d*x)/2]^4)/(128*a*d*(b + a*Csc[c + d*x]^2))
Time = 0.34 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3665, 316, 402, 397, 219, 221}
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.
| \(\displaystyle \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^5 \left (a+b \sin (c+d x)^2\right )}dx\) |
| \(\Big \downarrow \) 3665 |
| \(\displaystyle -\frac {\int \frac {1}{\left (1-\cos ^2(c+d x)\right )^3 \left (-b \cos ^2(c+d x)+a+b\right )}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle -\frac {\frac {\int \frac {-3 b \cos ^2(c+d x)+3 a-b}{\left (1-\cos ^2(c+d x)\right )^2 \left (-b \cos ^2(c+d x)+a+b\right )}d\cos (c+d x)}{4 a}+\frac {\cos (c+d x)}{4 a \left (1-\cos ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle -\frac {\frac {\frac {\int \frac {3 a^2-b a+4 b^2-(3 a-4 b) b \cos ^2(c+d x)}{\left (1-\cos ^2(c+d x)\right ) \left (-b \cos ^2(c+d x)+a+b\right )}d\cos (c+d x)}{2 a}+\frac {(3 a-4 b) \cos (c+d x)}{2 a \left (1-\cos ^2(c+d x)\right )}}{4 a}+\frac {\cos (c+d x)}{4 a \left (1-\cos ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle -\frac {\frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \int \frac {1}{1-\cos ^2(c+d x)}d\cos (c+d x)}{a}-\frac {8 b^3 \int \frac {1}{-b \cos ^2(c+d x)+a+b}d\cos (c+d x)}{a}}{2 a}+\frac {(3 a-4 b) \cos (c+d x)}{2 a \left (1-\cos ^2(c+d x)\right )}}{4 a}+\frac {\cos (c+d x)}{4 a \left (1-\cos ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \text {arctanh}(\cos (c+d x))}{a}-\frac {8 b^3 \int \frac {1}{-b \cos ^2(c+d x)+a+b}d\cos (c+d x)}{a}}{2 a}+\frac {(3 a-4 b) \cos (c+d x)}{2 a \left (1-\cos ^2(c+d x)\right )}}{4 a}+\frac {\cos (c+d x)}{4 a \left (1-\cos ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \text {arctanh}(\cos (c+d x))}{a}-\frac {8 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{2 a}+\frac {(3 a-4 b) \cos (c+d x)}{2 a \left (1-\cos ^2(c+d x)\right )}}{4 a}+\frac {\cos (c+d x)}{4 a \left (1-\cos ^2(c+d x)\right )^2}}{d}\) |
-((Cos[c + d*x]/(4*a*(1 - Cos[c + d*x]^2)^2) + ((((3*a^2 - 4*a*b + 8*b^2)* ArcTanh[Cos[c + d*x]])/a - (8*b^(5/2)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[ a + b]])/(a*Sqrt[a + b]))/(2*a) + ((3*a - 4*b)*Cos[c + d*x])/(2*a*(1 - Cos [c + d*x]^2)))/(4*a))/d)
3.1.84.3.1 Defintions of rubi rules used
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)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (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 - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.99 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.34
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{3} \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {1}{16 a \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {-3 a +4 b}{16 a^{2} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-3 a^{2}+4 a b -8 b^{2}\right ) \ln \left (1+\cos \left (d x +c \right )\right )}{16 a^{3}}-\frac {1}{16 a \left (\cos \left (d x +c \right )-1\right )^{2}}-\frac {-3 a +4 b}{16 a^{2} \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \ln \left (\cos \left (d x +c \right )-1\right )}{16 a^{3}}}{d}\) | \(168\) |
| default | \(\frac {\frac {b^{3} \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {1}{16 a \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {-3 a +4 b}{16 a^{2} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-3 a^{2}+4 a b -8 b^{2}\right ) \ln \left (1+\cos \left (d x +c \right )\right )}{16 a^{3}}-\frac {1}{16 a \left (\cos \left (d x +c \right )-1\right )^{2}}-\frac {-3 a +4 b}{16 a^{2} \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \ln \left (\cos \left (d x +c \right )-1\right )}{16 a^{3}}}{d}\) | \(168\) |
| risch | \(\frac {3 a \,{\mathrm e}^{7 i \left (d x +c \right )}-4 b \,{\mathrm e}^{7 i \left (d x +c \right )}-11 a \,{\mathrm e}^{5 i \left (d x +c \right )}+4 b \,{\mathrm e}^{5 i \left (d x +c \right )}-11 a \,{\mathrm e}^{3 i \left (d x +c \right )}+4 b \,{\mathrm e}^{3 i \left (d x +c \right )}+3 a \,{\mathrm e}^{i \left (d x +c \right )}-4 b \,{\mathrm e}^{i \left (d x +c \right )}}{4 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{a^{3} d}+\frac {i \sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{i \left (d x +c \right )}}{b}+1\right )}{2 \left (a +b \right ) d \,a^{3}}-\frac {i \sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{i \left (d x +c \right )}}{b}+1\right )}{2 \left (a +b \right ) d \,a^{3}}\) | \(367\) |
1/d*(b^3/a^3/((a+b)*b)^(1/2)*arctanh(b*cos(d*x+c)/((a+b)*b)^(1/2))+1/16/a/ (1+cos(d*x+c))^2-1/16*(-3*a+4*b)/a^2/(1+cos(d*x+c))+1/16/a^3*(-3*a^2+4*a*b -8*b^2)*ln(1+cos(d*x+c))-1/16/a/(cos(d*x+c)-1)^2-1/16*(-3*a+4*b)/a^2/(cos( d*x+c)-1)+1/16*(3*a^2-4*a*b+8*b^2)/a^3*ln(cos(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 288 vs. \(2 (111) = 222\).
Time = 0.34 (sec) , antiderivative size = 612, normalized size of antiderivative = 4.90 \[ \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [\frac {2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (b^{2} \cos \left (d x + c\right )^{4} - 2 \, b^{2} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{a + b}} \log \left (\frac {b \cos \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) - 2 \, {\left (5 \, a^{2} - 4 \, a b\right )} \cos \left (d x + c\right ) - {\left ({\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}}, \frac {2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (d x + c\right )^{3} - 16 \, {\left (b^{2} \cos \left (d x + c\right )^{4} - 2 \, b^{2} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \cos \left (d x + c\right )\right ) - 2 \, {\left (5 \, a^{2} - 4 \, a b\right )} \cos \left (d x + c\right ) - {\left ({\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}}\right ] \]
[1/16*(2*(3*a^2 - 4*a*b)*cos(d*x + c)^3 + 8*(b^2*cos(d*x + c)^4 - 2*b^2*co s(d*x + c)^2 + b^2)*sqrt(b/(a + b))*log((b*cos(d*x + c)^2 + 2*(a + b)*sqrt (b/(a + b))*cos(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a - b)) - 2*(5*a^2 - 4*a*b)*cos(d*x + c) - ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 + 3*a^2 - 4*a*b + 8*b^2)*log(1/2*cos(d*x + c) + 1/2) + ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 + 3*a^2 - 4*a*b + 8*b^2)*log(-1/2*cos(d*x + c) + 1/2 ))/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d), 1/16*(2*(3*a^2 - 4*a*b)*cos(d*x + c)^3 - 16*(b^2*cos(d*x + c)^4 - 2*b^2*cos(d*x + c)^2 + b^2)*sqrt(-b/(a + b))*arctan(sqrt(-b/(a + b))*cos(d*x + c)) - 2*(5*a^2 - 4*a*b)*cos(d*x + c) - ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 + 3*a^2 - 4*a*b + 8*b^2)*log(1/2*cos(d*x + c) + 1/2) + ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8 *b^2)*cos(d*x + c)^2 + 3*a^2 - 4*a*b + 8*b^2)*log(-1/2*cos(d*x + c) + 1/2) )/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)]
\[ \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\csc ^{5}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.45 \[ \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {8 \, b^{3} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3}} - \frac {2 \, {\left ({\left (3 \, a - 4 \, b\right )} \cos \left (d x + c\right )^{3} - {\left (5 \, a - 4 \, b\right )} \cos \left (d x + c\right )\right )}}{a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}} + \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{16 \, d} \]
-1/16*(8*b^3*log((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x + c) + sqrt ((a + b)*b)))/(sqrt((a + b)*b)*a^3) - 2*((3*a - 4*b)*cos(d*x + c)^3 - (5*a - 4*b)*cos(d*x + c))/(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2) + (3*a^2 - 4*a*b + 8*b^2)*log(cos(d*x + c) + 1)/a^3 - (3*a^2 - 4*a*b + 8*b^2 )*log(cos(d*x + c) - 1)/a^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (111) = 222\).
Time = 0.44 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.67 \[ \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {64 \, b^{3} \arctan \left (\frac {b \cos \left (d x + c\right ) + a + b}{\sqrt {-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{3}} + \frac {\frac {8 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {8 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} - \frac {4 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} + \frac {{\left (a^{2} - \frac {8 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {18 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {24 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {48 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{64 \, d} \]
-1/64*(64*b^3*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/(sqrt(-a*b - b^2)*a^3) + (8*a*(cos(d*x + c) - 1)/( cos(d*x + c) + 1) - 8*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/a^2 - 4*(3*a^2 - 4*a*b + 8*b^2)*log(abs (-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 + (a^2 - 8*a^2*(cos(d*x + c ) - 1)/(cos(d*x + c) + 1) + 8*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 18*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 24*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 48*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1) ^2)*(cos(d*x + c) + 1)^2/(a^3*(cos(d*x + c) - 1)^2))/d
Time = 13.56 (sec) , antiderivative size = 1105, normalized size of antiderivative = 8.84 \[ \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\text {Too large to display} \]
- ((cos(c + d*x)*(5*a - 4*b))/(8*a^2) - (cos(c + d*x)^3*(3*a - 4*b))/(8*a^ 2))/(d*(cos(c + d*x)^4 - cos(c + d*x)^2 + sin(c + d*x)^2)) - (atanh((63*b^ 4*cos(c + d*x))/(64*((63*b^4)/64 - (81*a*b^3)/256 + (27*a^2*b^2)/256 - (35 *b^5)/(32*a) + (5*b^6)/(4*a^2))) - (81*b^3*cos(c + d*x))/(256*((27*a*b^2)/ 256 - (81*b^3)/256 + (63*b^4)/(64*a) - (35*b^5)/(32*a^2) + (5*b^6)/(4*a^3) )) - (35*b^5*cos(c + d*x))/(32*((63*a*b^4)/64 - (35*b^5)/32 - (81*a^2*b^3) /256 + (27*a^3*b^2)/256 + (5*b^6)/(4*a))) + (5*b^6*cos(c + d*x))/(4*((5*b^ 6)/4 - (35*a*b^5)/32 + (63*a^2*b^4)/64 - (81*a^3*b^3)/256 + (27*a^4*b^2)/2 56)) + (27*b^2*cos(c + d*x))/(256*((27*b^2)/256 - (81*b^3)/(256*a) + (63*b ^4)/(64*a^2) - (35*b^5)/(32*a^3) + (5*b^6)/(4*a^4))))*(3*a^2 - 4*a*b + 8*b ^2))/(8*a^3*d) - (atan((((b^5*(a + b))^(1/2)*((cos(c + d*x)*(128*b^7 - 64* a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*b^3))/(64*a^4) + ((b^5*(a + b))^(1 /2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) - (cos(c + d*x)*(51 2*a^6*b^3 + 256*a^7*b^2)*(b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2 *(a^3*b + a^4)))*1i)/(a^3*b + a^4) + ((b^5*(a + b))^(1/2)*((cos(c + d*x)*( 128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*b^3))/(64*a^4) - ((b^ 5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) + (cos (c + d*x)*(512*a^6*b^3 + 256*a^7*b^2)*(b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4)))*1i)/(a^3*b + a^4))/(((5*a*b^7)/4 - b^8 - (3* a^2*b^6)/4 + (9*a^3*b^5)/32)/a^6 + ((b^5*(a + b))^(1/2)*((cos(c + d*x)*...