Integrand size = 23, antiderivative size = 140 \[ \int \frac {\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {b^4 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{9/2} \sqrt {a+b} d}-\frac {(a-b) \left (a^2+b^2\right ) \cot (c+d x)}{a^4 d}-\frac {\left (3 a^2-2 a b+b^2\right ) \cot ^3(c+d x)}{3 a^3 d}-\frac {(3 a-b) \cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^7(c+d x)}{7 a d} \]
-(a-b)*(a^2+b^2)*cot(d*x+c)/a^4/d-1/3*(3*a^2-2*a*b+b^2)*cot(d*x+c)^3/a^3/d -1/5*(3*a-b)*cot(d*x+c)^5/a^2/d-1/7*cot(d*x+c)^7/a/d+b^4*arctan((a+b)^(1/2 )*tan(d*x+c)/a^(1/2))/a^(9/2)/d/(a+b)^(1/2)
Time = 5.82 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98 \[ \int \frac {\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {b^4 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{9/2} \sqrt {a+b} d}-\frac {\cot (c+d x) \left (48 a^3-56 a^2 b+70 a b^2-105 b^3+a \left (24 a^2-28 a b+35 b^2\right ) \csc ^2(c+d x)+3 a^2 (6 a-7 b) \csc ^4(c+d x)+15 a^3 \csc ^6(c+d x)\right )}{105 a^4 d} \]
(b^4*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(9/2)*Sqrt[a + b]*d) - (Cot[c + d*x]*(48*a^3 - 56*a^2*b + 70*a*b^2 - 105*b^3 + a*(24*a^2 - 28*a* b + 35*b^2)*Csc[c + d*x]^2 + 3*a^2*(6*a - 7*b)*Csc[c + d*x]^4 + 15*a^3*Csc [c + d*x]^6))/(105*a^4*d)
Time = 0.34 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3666, 364, 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.
| \(\displaystyle \int \frac {\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^8 \left (a+b \sin (c+d x)^2\right )}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\cot ^8(c+d x) \left (\tan ^2(c+d x)+1\right )^4}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \frac {\int \left (\frac {\cot ^8(c+d x)}{a}+\frac {(3 a-b) \cot ^6(c+d x)}{a^2}+\frac {\left (3 a^2-2 b a+b^2\right ) \cot ^4(c+d x)}{a^3}+\frac {(a-b) \left (a^2+b^2\right ) \cot ^2(c+d x)}{a^4}+\frac {b^4}{a^4 \left ((a+b) \tan ^2(c+d x)+a\right )}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^4 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{9/2} \sqrt {a+b}}-\frac {(3 a-b) \cot ^5(c+d x)}{5 a^2}-\frac {(a-b) \left (a^2+b^2\right ) \cot (c+d x)}{a^4}-\frac {\left (3 a^2-2 a b+b^2\right ) \cot ^3(c+d x)}{3 a^3}-\frac {\cot ^7(c+d x)}{7 a}}{d}\) |
((b^4*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(9/2)*Sqrt[a + b]) - ((a - b)*(a^2 + b^2)*Cot[c + d*x])/a^4 - ((3*a^2 - 2*a*b + b^2)*Cot[c + d* x]^3)/(3*a^3) - ((3*a - b)*Cot[c + d*x]^5)/(5*a^2) - Cot[c + d*x]^7/(7*a)) /d
3.1.93.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1)) , x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] & & IntegerQ[p]
Time = 1.53 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.93
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{4} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{4} \sqrt {a \left (a +b \right )}}-\frac {1}{7 a \tan \left (d x +c \right )^{7}}-\frac {3 a -b}{5 a^{2} \tan \left (d x +c \right )^{5}}-\frac {3 a^{2}-2 a b +b^{2}}{3 a^{3} \tan \left (d x +c \right )^{3}}-\frac {a^{3}-a^{2} b +a \,b^{2}-b^{3}}{a^{4} \tan \left (d x +c \right )}}{d}\) | \(130\) |
| default | \(\frac {\frac {b^{4} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{4} \sqrt {a \left (a +b \right )}}-\frac {1}{7 a \tan \left (d x +c \right )^{7}}-\frac {3 a -b}{5 a^{2} \tan \left (d x +c \right )^{5}}-\frac {3 a^{2}-2 a b +b^{2}}{3 a^{3} \tan \left (d x +c \right )^{3}}-\frac {a^{3}-a^{2} b +a \,b^{2}-b^{3}}{a^{4} \tan \left (d x +c \right )}}{d}\) | \(130\) |
| risch | \(\frac {2 i \left (105 b^{3} {\mathrm e}^{12 i \left (d x +c \right )}+210 a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-630 b^{3} {\mathrm e}^{10 i \left (d x +c \right )}+560 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}-910 a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+1575 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+1680 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-1400 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+1540 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-2100 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-1008 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+1176 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-1260 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+1575 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+336 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-392 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+490 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-630 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-48 a^{3}+56 a^{2} b -70 a \,b^{2}+105 b^{3}\right )}{105 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{7}}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d \,a^{4}}+\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d \,a^{4}}\) | \(492\) |
1/d*(b^4/a^4/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-1/7/ a/tan(d*x+c)^7-1/5*(3*a-b)/a^2/tan(d*x+c)^5-1/3*(3*a^2-2*a*b+b^2)/a^3/tan( d*x+c)^3-(a^3-a^2*b+a*b^2-b^3)/a^4/tan(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (126) = 252\).
Time = 0.31 (sec) , antiderivative size = 789, normalized size of antiderivative = 5.64 \[ \int \frac {\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [-\frac {4 \, {\left (48 \, a^{5} - 8 \, a^{4} b + 14 \, a^{3} b^{2} - 35 \, a^{2} b^{3} - 105 \, a b^{4}\right )} \cos \left (d x + c\right )^{7} - 28 \, {\left (24 \, a^{5} - 4 \, a^{4} b + 7 \, a^{3} b^{2} - 10 \, a^{2} b^{3} - 45 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 140 \, {\left (6 \, a^{5} - a^{4} b + a^{3} b^{2} - a^{2} b^{3} - 9 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 105 \, {\left (b^{4} \cos \left (d x + c\right )^{6} - 3 \, b^{4} \cos \left (d x + c\right )^{4} + 3 \, b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) - 420 \, {\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right )}{420 \, {\left ({\left (a^{6} + a^{5} b\right )} d \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{6} + a^{5} b\right )} d \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{6} + a^{5} b\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} + a^{5} b\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (48 \, a^{5} - 8 \, a^{4} b + 14 \, a^{3} b^{2} - 35 \, a^{2} b^{3} - 105 \, a b^{4}\right )} \cos \left (d x + c\right )^{7} - 14 \, {\left (24 \, a^{5} - 4 \, a^{4} b + 7 \, a^{3} b^{2} - 10 \, a^{2} b^{3} - 45 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 70 \, {\left (6 \, a^{5} - a^{4} b + a^{3} b^{2} - a^{2} b^{3} - 9 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 105 \, {\left (b^{4} \cos \left (d x + c\right )^{6} - 3 \, b^{4} \cos \left (d x + c\right )^{4} + 3 \, b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 210 \, {\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right )}{210 \, {\left ({\left (a^{6} + a^{5} b\right )} d \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{6} + a^{5} b\right )} d \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{6} + a^{5} b\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} + a^{5} b\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
[-1/420*(4*(48*a^5 - 8*a^4*b + 14*a^3*b^2 - 35*a^2*b^3 - 105*a*b^4)*cos(d* x + c)^7 - 28*(24*a^5 - 4*a^4*b + 7*a^3*b^2 - 10*a^2*b^3 - 45*a*b^4)*cos(d *x + c)^5 + 140*(6*a^5 - a^4*b + a^3*b^2 - a^2*b^3 - 9*a*b^4)*cos(d*x + c) ^3 + 105*(b^4*cos(d*x + c)^6 - 3*b^4*cos(d*x + c)^4 + 3*b^4*cos(d*x + c)^2 - b^4)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4* a^2 + 5*a*b + b^2)*cos(d*x + c)^2 + 4*((2*a + b)*cos(d*x + c)^3 - (a + b)* cos(d*x + c))*sqrt(-a^2 - a*b)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos( d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2))*sin(d*x + c) - 420*(a^5 - a*b^4)*cos(d*x + c))/(((a^6 + a^5*b)*d*cos(d*x + c)^6 - 3* (a^6 + a^5*b)*d*cos(d*x + c)^4 + 3*(a^6 + a^5*b)*d*cos(d*x + c)^2 - (a^6 + a^5*b)*d)*sin(d*x + c)), -1/210*(2*(48*a^5 - 8*a^4*b + 14*a^3*b^2 - 35*a^ 2*b^3 - 105*a*b^4)*cos(d*x + c)^7 - 14*(24*a^5 - 4*a^4*b + 7*a^3*b^2 - 10* a^2*b^3 - 45*a*b^4)*cos(d*x + c)^5 + 70*(6*a^5 - a^4*b + a^3*b^2 - a^2*b^3 - 9*a*b^4)*cos(d*x + c)^3 + 105*(b^4*cos(d*x + c)^6 - 3*b^4*cos(d*x + c)^ 4 + 3*b^4*cos(d*x + c)^2 - b^4)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos( d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c)))*sin(d*x + c) - 210*(a^5 - a*b^4)*cos(d*x + c))/(((a^6 + a^5*b)*d*cos(d*x + c)^6 - 3 *(a^6 + a^5*b)*d*cos(d*x + c)^4 + 3*(a^6 + a^5*b)*d*cos(d*x + c)^2 - (a^6 + a^5*b)*d)*sin(d*x + c))]
Timed out. \[ \int \frac {\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\text {Timed out} \]
Time = 0.37 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98 \[ \int \frac {\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {105 \, b^{4} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{4}} - \frac {105 \, {\left (a^{3} - a^{2} b + a b^{2} - b^{3}\right )} \tan \left (d x + c\right )^{6} + 35 \, {\left (3 \, a^{3} - 2 \, a^{2} b + a b^{2}\right )} \tan \left (d x + c\right )^{4} + 15 \, a^{3} + 21 \, {\left (3 \, a^{3} - a^{2} b\right )} \tan \left (d x + c\right )^{2}}{a^{4} \tan \left (d x + c\right )^{7}}}{105 \, d} \]
1/105*(105*b^4*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)* a)*a^4) - (105*(a^3 - a^2*b + a*b^2 - b^3)*tan(d*x + c)^6 + 35*(3*a^3 - 2* a^2*b + a*b^2)*tan(d*x + c)^4 + 15*a^3 + 21*(3*a^3 - a^2*b)*tan(d*x + c)^2 )/(a^4*tan(d*x + c)^7))/d
Time = 0.48 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.54 \[ \int \frac {\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {105 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{4}}{\sqrt {a^{2} + a b} a^{4}} - \frac {105 \, a^{3} \tan \left (d x + c\right )^{6} - 105 \, a^{2} b \tan \left (d x + c\right )^{6} + 105 \, a b^{2} \tan \left (d x + c\right )^{6} - 105 \, b^{3} \tan \left (d x + c\right )^{6} + 105 \, a^{3} \tan \left (d x + c\right )^{4} - 70 \, a^{2} b \tan \left (d x + c\right )^{4} + 35 \, a b^{2} \tan \left (d x + c\right )^{4} + 63 \, a^{3} \tan \left (d x + c\right )^{2} - 21 \, a^{2} b \tan \left (d x + c\right )^{2} + 15 \, a^{3}}{a^{4} \tan \left (d x + c\right )^{7}}}{105 \, d} \]
1/105*(105*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d* x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)))*b^4/(sqrt(a^2 + a*b)*a^4) - (10 5*a^3*tan(d*x + c)^6 - 105*a^2*b*tan(d*x + c)^6 + 105*a*b^2*tan(d*x + c)^6 - 105*b^3*tan(d*x + c)^6 + 105*a^3*tan(d*x + c)^4 - 70*a^2*b*tan(d*x + c) ^4 + 35*a*b^2*tan(d*x + c)^4 + 63*a^3*tan(d*x + c)^2 - 21*a^2*b*tan(d*x + c)^2 + 15*a^3)/(a^4*tan(d*x + c)^7))/d
Time = 15.18 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.93 \[ \int \frac {\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {b^4\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a+b}}{\sqrt {a}}\right )}{a^{9/2}\,d\,\sqrt {a+b}}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^3-\frac {2\,a^2\,b}{3}+\frac {a\,b^2}{3}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^2\,b}{5}-\frac {3\,a^3}{5}\right )+{\mathrm {tan}\left (c+d\,x\right )}^6\,\left (a^3-a^2\,b+a\,b^2-b^3\right )+\frac {a^3}{7}}{a^4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^7} \]