Integrand size = 23, antiderivative size = 196 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=-\frac {3 b \left (8 a^2+12 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{7/2} (a+b)^{5/2} d}-\frac {(2 a+3 b) (4 a+5 b) \cot (c+d x)}{8 a^3 (a+b)^2 d}+\frac {b \csc (c+d x) \sec ^3(c+d x)}{4 a (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac {b \cot (c+d x) \left (4 a+5 b+(4 a+b) \tan ^2(c+d x)\right )}{8 a^2 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )} \]
-3/8*b*(8*a^2+12*a*b+5*b^2)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(7/2) /(a+b)^(5/2)/d-1/8*(2*a+3*b)*(4*a+5*b)*cot(d*x+c)/a^3/(a+b)^2/d+1/4*b*csc( d*x+c)*sec(d*x+c)^3/a/(a+b)/d/(a+(a+b)*tan(d*x+c)^2)^2+1/8*b*cot(d*x+c)*(4 *a+5*b+(4*a+b)*tan(d*x+c)^2)/a^2/(a+b)^2/d/(a+(a+b)*tan(d*x+c)^2)
Time = 2.82 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.09 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {(-2 a-b+b \cos (2 (c+d x))) \csc ^6(c+d x) \left (\frac {3 b \left (8 a^2+12 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right ) (2 a+b-b \cos (2 (c+d x)))^2}{(a+b)^{5/2}}+8 \sqrt {a} (2 a+b-b \cos (2 (c+d x)))^2 \cot (c+d x)+\frac {4 a^{3/2} b^2 \sin (2 (c+d x))}{a+b}+\frac {\sqrt {a} b^2 (10 a+7 b) (2 a+b-b \cos (2 (c+d x))) \sin (2 (c+d x))}{(a+b)^2}\right )}{64 a^{7/2} d \left (b+a \csc ^2(c+d x)\right )^3} \]
((-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^6*((3*b*(8*a^2 + 12*a*b + 5* b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]]*(2*a + b - b*Cos[2*(c + d* x)])^2)/(a + b)^(5/2) + 8*Sqrt[a]*(2*a + b - b*Cos[2*(c + d*x)])^2*Cot[c + d*x] + (4*a^(3/2)*b^2*Sin[2*(c + d*x)])/(a + b) + (Sqrt[a]*b^2*(10*a + 7* b)*(2*a + b - b*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(a + b)^2))/(64*a^(7/2 )*d*(b + a*Csc[c + d*x]^2)^3)
Time = 0.40 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3666, 370, 25, 439, 25, 359, 218}
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.
| \(\displaystyle \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \left (a+b \sin (c+d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\cot ^2(c+d x) \left (\tan ^2(c+d x)+1\right )^3}{\left ((a+b) \tan ^2(c+d x)+a\right )^3}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 370 |
| \(\displaystyle \frac {\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot (c+d x)}{4 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}-\frac {\int -\frac {\cot ^2(c+d x) \left (\tan ^2(c+d x)+1\right ) \left ((4 a+b) \tan ^2(c+d x)+4 a+5 b\right )}{\left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{4 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\cot ^2(c+d x) \left (\tan ^2(c+d x)+1\right ) \left ((4 a+b) \tan ^2(c+d x)+4 a+5 b\right )}{\left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{4 a (a+b)}+\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot (c+d x)}{4 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 439 |
| \(\displaystyle \frac {\frac {\frac {b \cot (c+d x) \left ((4 a+b) \tan ^2(c+d x)+4 a+5 b\right )}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\int -\frac {\cot ^2(c+d x) \left ((2 a+b) (4 a+b) \tan ^2(c+d x)+(2 a+3 b) (4 a+5 b)\right )}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{2 a (a+b)}}{4 a (a+b)}+\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot (c+d x)}{4 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\cot ^2(c+d x) \left ((2 a+b) (4 a+b) \tan ^2(c+d x)+(2 a+3 b) (4 a+5 b)\right )}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{2 a (a+b)}+\frac {b \cot (c+d x) \left ((4 a+b) \tan ^2(c+d x)+4 a+5 b\right )}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{4 a (a+b)}+\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot (c+d x)}{4 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {\frac {-\frac {3 b \left (8 a^2+12 a b+5 b^2\right ) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{a}-\frac {(2 a+3 b) (4 a+5 b) \cot (c+d x)}{a}}{2 a (a+b)}+\frac {b \cot (c+d x) \left ((4 a+b) \tan ^2(c+d x)+4 a+5 b\right )}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{4 a (a+b)}+\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot (c+d x)}{4 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {-\frac {3 b \left (8 a^2+12 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2} \sqrt {a+b}}-\frac {(2 a+3 b) (4 a+5 b) \cot (c+d x)}{a}}{2 a (a+b)}+\frac {b \cot (c+d x) \left ((4 a+b) \tan ^2(c+d x)+4 a+5 b\right )}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{4 a (a+b)}+\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot (c+d x)}{4 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}}{d}\) |
((b*Cot[c + d*x]*(1 + Tan[c + d*x]^2)^2)/(4*a*(a + b)*(a + (a + b)*Tan[c + d*x]^2)^2) + (((-3*b*(8*a^2 + 12*a*b + 5*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(3/2)*Sqrt[a + b]) - ((2*a + 3*b)*(4*a + 5*b)*Cot[c + d*x])/a)/(2*a*(a + b)) + (b*Cot[c + d*x]*(4*a + 5*b + (4*a + b)*Tan[c + d* x]^2))/(2*a*(a + b)*(a + (a + b)*Tan[c + d*x]^2)))/(4*a*(a + b)))/d
3.2.10.3.1 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(a*b*e*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1)) Int[(e*x) ^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*Simp[c*(b*c*2*(p + 1) + (b*c - a *d)*(m + 1)) + d*(b*c*2*(p + 1) + (b*c - a*d)*(m + 2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 ))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G tQ[q, 0] && !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
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 = 1.66 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\frac {\left (12 a +7 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{8 a +8 b}+\frac {3 a b \left (4 a +3 b \right ) \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}^{2}}+\frac {3 \left (8 a^{2}+12 a b +5 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{3}}-\frac {1}{a^{3} \tan \left (d x +c \right )}}{d}\) | \(160\) |
| default | \(\frac {-\frac {b \left (\frac {\frac {\left (12 a +7 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{8 a +8 b}+\frac {3 a b \left (4 a +3 b \right ) \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}^{2}}+\frac {3 \left (8 a^{2}+12 a b +5 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{3}}-\frac {1}{a^{3} \tan \left (d x +c \right )}}{d}\) | \(160\) |
| risch | \(-\frac {i \left (24 a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+36 a \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+15 b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-144 a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}-312 a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-234 a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-60 b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+128 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+464 a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}+632 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+386 a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+90 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-64 a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-224 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-214 a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-60 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+8 a^{2} b^{2}+26 a \,b^{3}+15 b^{4}\right )}{4 a^{3} \left (a +b \right )^{2} d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {3 \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 ) b}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d a}+\frac {9 \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 ) b^{2}}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{2}}+\frac {15 b^{3} \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 )}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{3}}-\frac {3 \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 ) b}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d a}-\frac {9 \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 ) b^{2}}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{2}}-\frac {15 b^{3} \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 )}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{3}}\) | \(912\) |
1/d*(-b/a^3*((1/8*(12*a+7*b)*b/(a+b)*tan(d*x+c)^3+3/8*a*b*(4*a+3*b)/(a^2+2 *a*b+b^2)*tan(d*x+c))/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2+3/8*(8*a^2+12*a* b+5*b^2)/(a^2+2*a*b+b^2)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b)) ^(1/2)))-1/a^3/tan(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 451 vs. \(2 (180) = 360\).
Time = 0.36 (sec) , antiderivative size = 1003, normalized size of antiderivative = 5.12 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
[-1/32*(4*(8*a^4*b^2 + 34*a^3*b^3 + 41*a^2*b^4 + 15*a*b^5)*cos(d*x + c)^5 - 4*(16*a^5*b + 76*a^4*b^2 + 137*a^3*b^3 + 107*a^2*b^4 + 30*a*b^5)*cos(d*x + c)^3 + 3*(8*a^4*b + 28*a^3*b^2 + 37*a^2*b^3 + 22*a*b^4 + 5*b^5 + (8*a^2 *b^3 + 12*a*b^4 + 5*b^5)*cos(d*x + c)^4 - 2*(8*a^3*b^2 + 20*a^2*b^3 + 17*a *b^4 + 5*b^5)*cos(d*x + c)^2)*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) + 4*(8*a^6 + 40*a^5*b + 92*a^4*b^2 + 111*a^3*b^3 + 66*a^2*b^4 + 15*a*b^5)*cos(d*x + c))/(((a^7*b^2 + 3*a^6*b^3 + 3*a^5*b^4 + a^4*b^5)*d*cos(d*x + c)^4 - 2*(a^8*b + 4*a^7*b^2 + 6*a^6*b^3 + 4*a^5*b^4 + a^4*b^5)*d*cos(d*x + c)^2 + (a^9 + 5*a^8*b + 10*a^7*b^2 + 10*a^6*b^3 + 5*a^5*b^4 + a^4*b^5)*d)*sin(d*x + c)), -1/16*(2*(8*a^4*b^2 + 34*a^3*b^3 + 41*a^2*b^4 + 15*a*b^5)*cos(d*x + c)^5 - 2*(16*a^5*b + 76*a^4*b^2 + 137*a^3 *b^3 + 107*a^2*b^4 + 30*a*b^5)*cos(d*x + c)^3 - 3*(8*a^4*b + 28*a^3*b^2 + 37*a^2*b^3 + 22*a*b^4 + 5*b^5 + (8*a^2*b^3 + 12*a*b^4 + 5*b^5)*cos(d*x + c )^4 - 2*(8*a^3*b^2 + 20*a^2*b^3 + 17*a*b^4 + 5*b^5)*cos(d*x + c)^2)*sqrt(a ^2 + a*b)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*c os(d*x + c)*sin(d*x + c)))*sin(d*x + c) + 2*(8*a^6 + 40*a^5*b + 92*a^4*b^2 + 111*a^3*b^3 + 66*a^2*b^4 + 15*a*b^5)*cos(d*x + c))/(((a^7*b^2 + 3*a^...
Timed out. \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.48 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.38 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=-\frac {\frac {3 \, {\left (8 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {{\left (8 \, a^{4} + 32 \, a^{3} b + 60 \, a^{2} b^{2} + 51 \, a b^{3} + 15 \, b^{4}\right )} \tan \left (d x + c\right )^{4} + 8 \, a^{4} + 16 \, a^{3} b + 8 \, a^{2} b^{2} + {\left (16 \, a^{4} + 48 \, a^{3} b + 60 \, a^{2} b^{2} + 25 \, a b^{3}\right )} \tan \left (d x + c\right )^{2}}{{\left (a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} \tan \left (d x + c\right )^{5} + 2 \, {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \tan \left (d x + c\right )}}{8 \, d} \]
-1/8*(3*(8*a^2*b + 12*a*b^2 + 5*b^3)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a^5 + 2*a^4*b + a^3*b^2)*sqrt((a + b)*a)) + ((8*a^4 + 32*a^3*b + 60*a^2*b^2 + 51*a*b^3 + 15*b^4)*tan(d*x + c)^4 + 8*a^4 + 16*a^3*b + 8*a^2 *b^2 + (16*a^4 + 48*a^3*b + 60*a^2*b^2 + 25*a*b^3)*tan(d*x + c)^2)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*tan(d*x + c)^5 + 2*(a^7 + 3*a^ 6*b + 3*a^5*b^2 + a^4*b^3)*tan(d*x + c)^3 + (a^7 + 2*a^6*b + a^5*b^2)*tan( d*x + c)))/d
Time = 0.40 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.18 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=-\frac {\frac {3 \, {\left (8 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {a^{2} + a b}} + \frac {12 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 19 \, a b^{3} \tan \left (d x + c\right )^{3} + 7 \, b^{4} \tan \left (d x + c\right )^{3} + 12 \, a^{2} b^{2} \tan \left (d x + c\right ) + 9 \, a b^{3} \tan \left (d x + c\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}} + \frac {8}{a^{3} \tan \left (d x + c\right )}}{8 \, d} \]
-1/8*(3*(8*a^2*b + 12*a*b^2 + 5*b^3)*(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)))/((a^5 + 2*a^4*b + a^3*b^2)*sqrt(a^2 + a*b)) + (12*a^2*b^2*tan(d*x + c)^3 + 19*a *b^3*tan(d*x + c)^3 + 7*b^4*tan(d*x + c)^3 + 12*a^2*b^2*tan(d*x + c) + 9*a *b^3*tan(d*x + c))/((a^5 + 2*a^4*b + a^3*b^2)*(a*tan(d*x + c)^2 + b*tan(d* x + c)^2 + a)^2) + 8/(a^3*tan(d*x + c)))/d
Time = 15.80 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.28 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=-\frac {\frac {1}{a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (8\,a^3+24\,a^2\,b+36\,a\,b^2+15\,b^3\right )}{8\,a^3\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (16\,a^3+48\,a^2\,b+60\,a\,b^2+25\,b^3\right )}{8\,a^2\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^5\,\left (a^2+2\,a\,b+b^2\right )+a^2\,\mathrm {tan}\left (c+d\,x\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (2\,a^2+2\,b\,a\right )\right )}-\frac {3\,b\,\mathrm {atan}\left (\frac {3\,b\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^5+2\,a^4\,b+a^3\,b^2\right )\,\left (8\,a^2+12\,a\,b+5\,b^2\right )}{a^{7/2}\,{\left (a+b\right )}^{3/2}\,\left (24\,a^2\,b+36\,a\,b^2+15\,b^3\right )}\right )\,\left (8\,a^2+12\,a\,b+5\,b^2\right )}{8\,a^{7/2}\,d\,{\left (a+b\right )}^{5/2}} \]
- (1/a + (tan(c + d*x)^4*(36*a*b^2 + 24*a^2*b + 8*a^3 + 15*b^3))/(8*a^3*(a + b)) + (tan(c + d*x)^2*(60*a*b^2 + 48*a^2*b + 16*a^3 + 25*b^3))/(8*a^2*( 2*a*b + a^2 + b^2)))/(d*(tan(c + d*x)^5*(2*a*b + a^2 + b^2) + a^2*tan(c + d*x) + tan(c + d*x)^3*(2*a*b + 2*a^2))) - (3*b*atan((3*b*tan(c + d*x)*(2*a ^4*b + a^5 + a^3*b^2)*(12*a*b + 8*a^2 + 5*b^2))/(a^(7/2)*(a + b)^(3/2)*(36 *a*b^2 + 24*a^2*b + 15*b^3)))*(12*a*b + 8*a^2 + 5*b^2))/(8*a^(7/2)*d*(a + b)^(5/2))