Integrand size = 23, antiderivative size = 162 \[ \int \frac {\csc ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {b^2 (6 a+5 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{7/2} (a+b)^{3/2} d}-\frac {\left (2 a^2-a b-5 b^2\right ) \cot (c+d x)}{2 a^3 (a+b) d}-\frac {(2 a+5 b) \cot ^3(c+d x)}{6 a^2 (a+b) d}+\frac {b \csc ^3(c+d x) \sec (c+d x)}{2 a (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )} \]
1/2*b^2*(6*a+5*b)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(7/2)/(a+b)^(3/ 2)/d-1/2*(2*a^2-a*b-5*b^2)*cot(d*x+c)/a^3/(a+b)/d-1/6*(2*a+5*b)*cot(d*x+c) ^3/a^2/(a+b)/d+1/2*b*csc(d*x+c)^3*sec(d*x+c)/a/(a+b)/d/(a+(a+b)*tan(d*x+c) ^2)
Time = 2.33 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.25 \[ \int \frac {\csc ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {(-2 a-b+b \cos (2 (c+d x))) \csc ^4(c+d x) \left (\frac {3 b^2 (6 a+5 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right ) (-2 a-b+b \cos (2 (c+d x)))}{(a+b)^{3/2}}+4 \sqrt {a} (a-3 b) (2 a+b-b \cos (2 (c+d x))) \cot (c+d x)+2 a^{3/2} (2 a+b-b \cos (2 (c+d x))) \cot (c+d x) \csc ^2(c+d x)-\frac {3 \sqrt {a} b^3 \sin (2 (c+d x))}{a+b}\right )}{24 a^{7/2} d \left (b+a \csc ^2(c+d x)\right )^2} \]
((-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^4*((3*b^2*(6*a + 5*b)*ArcTan [(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]]*(-2*a - b + b*Cos[2*(c + d*x)]))/(a + b)^(3/2) + 4*Sqrt[a]*(a - 3*b)*(2*a + b - b*Cos[2*(c + d*x)])*Cot[c + d*x ] + 2*a^(3/2)*(2*a + b - b*Cos[2*(c + d*x)])*Cot[c + d*x]*Csc[c + d*x]^2 - (3*Sqrt[a]*b^3*Sin[2*(c + d*x)])/(a + b)))/(24*a^(7/2)*d*(b + a*Csc[c + d *x]^2)^2)
Time = 0.37 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3666, 370, 25, 437, 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.
| \(\displaystyle \int \frac {\csc ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^4 \left (a+b \sin (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\cot ^4(c+d x) \left (\tan ^2(c+d x)+1\right )^3}{\left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 370 |
| \(\displaystyle \frac {\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot ^3(c+d x)}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\int -\frac {\cot ^4(c+d x) \left (\tan ^2(c+d x)+1\right ) \left ((2 a+b) \tan ^2(c+d x)+2 a+5 b\right )}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{2 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\cot ^4(c+d x) \left (\tan ^2(c+d x)+1\right ) \left ((2 a+b) \tan ^2(c+d x)+2 a+5 b\right )}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{2 a (a+b)}+\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot ^3(c+d x)}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
| \(\Big \downarrow \) 437 |
| \(\displaystyle \frac {\frac {\int \left (\frac {(2 a+5 b) \cot ^4(c+d x)}{a}+\frac {\left (2 a^2-b a-5 b^2\right ) \cot ^2(c+d x)}{a^2}+\frac {b^2 (6 a+5 b)}{a^2 \left ((a+b) \tan ^2(c+d x)+a\right )}\right )d\tan (c+d x)}{2 a (a+b)}+\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot ^3(c+d x)}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {b^2 (6 a+5 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a+b}}-\frac {\left (2 a^2-a b-5 b^2\right ) \cot (c+d x)}{a^2}-\frac {(2 a+5 b) \cot ^3(c+d x)}{3 a}}{2 a (a+b)}+\frac {b \left (\tan ^2(c+d x)+1\right )^2 \cot ^3(c+d x)}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
(((b^2*(6*a + 5*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(5/2)*Sq rt[a + b]) - ((2*a^2 - a*b - 5*b^2)*Cot[c + d*x])/a^2 - ((2*a + 5*b)*Cot[c + d*x]^3)/(3*a))/(2*a*(a + b)) + (b*Cot[c + d*x]^3*(1 + Tan[c + d*x]^2)^2 )/(2*a*(a + b)*(a + (a + b)*Tan[c + d*x]^2)))/d
3.2.5.3.1 Defintions of rubi rules used
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)^(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*( a + b*x^2)^p*(c + d*x^2)^q*(e + f*x^2)^r, x], x] /; FreeQ[{a, b, c, d, e, f , g, m}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
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 = 1.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.75
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{2} \left (\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\left (6 a +5 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{a^{3}}-\frac {1}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {a -2 b}{a^{3} \tan \left (d x +c \right )}}{d}\) | \(122\) |
| default | \(\frac {\frac {b^{2} \left (\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\left (6 a +5 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{a^{3}}-\frac {1}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {a -2 b}{a^{3} \tan \left (d x +c \right )}}{d}\) | \(122\) |
| risch | \(\frac {i \left (-18 a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-15 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+36 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+102 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+60 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+48 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-20 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-158 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-90 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-16 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+12 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+82 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+60 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+4 a^{2} b -8 a \,b^{2}-15 b^{3}\right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left (a +b \right ) \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}-\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{2}}-\frac {5 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{3}}+\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{2}}+\frac {5 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{3}}\) | \(666\) |
1/d*(b^2/a^3*(1/2*b/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)+1/2 *(6*a+5*b)/(a+b)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))) -1/3/a^2/tan(d*x+c)^3-(a-2*b)/a^3/tan(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (146) = 292\).
Time = 0.30 (sec) , antiderivative size = 843, normalized size of antiderivative = 5.20 \[ \int \frac {\csc ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [-\frac {4 \, {\left (4 \, a^{4} b - 4 \, a^{3} b^{2} - 23 \, a^{2} b^{3} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} - 8 \, {\left (2 \, a^{5} + 3 \, a^{4} b - 12 \, a^{3} b^{2} - 28 \, a^{2} b^{3} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left ({\left (6 \, a b^{3} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b^{2} + 11 \, a b^{3} + 5 \, b^{4} - {\left (6 \, a^{2} b^{2} + 17 \, a b^{3} + 10 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\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 ) + 12 \, {\left (2 \, a^{5} + 2 \, a^{4} b - 6 \, a^{3} b^{2} - 11 \, a^{2} b^{3} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )}{24 \, {\left ({\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} d \cos \left (d x + c\right )^{4} - {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} + 2 \, a^{4} b^{3}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (4 \, a^{4} b - 4 \, a^{3} b^{2} - 23 \, a^{2} b^{3} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} - 4 \, {\left (2 \, a^{5} + 3 \, a^{4} b - 12 \, a^{3} b^{2} - 28 \, a^{2} b^{3} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left ({\left (6 \, a b^{3} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b^{2} + 11 \, a b^{3} + 5 \, b^{4} - {\left (6 \, a^{2} b^{2} + 17 \, a b^{3} + 10 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\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 ) + 6 \, {\left (2 \, a^{5} + 2 \, a^{4} b - 6 \, a^{3} b^{2} - 11 \, a^{2} b^{3} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )}{12 \, {\left ({\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} d \cos \left (d x + c\right )^{4} - {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} + 2 \, a^{4} b^{3}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
[-1/24*(4*(4*a^4*b - 4*a^3*b^2 - 23*a^2*b^3 - 15*a*b^4)*cos(d*x + c)^5 - 8 *(2*a^5 + 3*a^4*b - 12*a^3*b^2 - 28*a^2*b^3 - 15*a*b^4)*cos(d*x + c)^3 + 3 *((6*a*b^3 + 5*b^4)*cos(d*x + c)^4 + 6*a^2*b^2 + 11*a*b^3 + 5*b^4 - (6*a^2 *b^2 + 17*a*b^3 + 10*b^4)*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) + 12*(2*a^5 + 2*a^4*b - 6*a^3*b^2 - 1 1*a^2*b^3 - 5*a*b^4)*cos(d*x + c))/(((a^6*b + 2*a^5*b^2 + a^4*b^3)*d*cos(d *x + c)^4 - (a^7 + 4*a^6*b + 5*a^5*b^2 + 2*a^4*b^3)*d*cos(d*x + c)^2 + (a^ 7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*d)*sin(d*x + c)), -1/12*(2*(4*a^4*b - 4 *a^3*b^2 - 23*a^2*b^3 - 15*a*b^4)*cos(d*x + c)^5 - 4*(2*a^5 + 3*a^4*b - 12 *a^3*b^2 - 28*a^2*b^3 - 15*a*b^4)*cos(d*x + c)^3 + 3*((6*a*b^3 + 5*b^4)*co s(d*x + c)^4 + 6*a^2*b^2 + 11*a*b^3 + 5*b^4 - (6*a^2*b^2 + 17*a*b^3 + 10*b ^4)*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)*cos(d*x + c)*sin(d*x + c)))*sin(d*x + c) + 6*(2*a ^5 + 2*a^4*b - 6*a^3*b^2 - 11*a^2*b^3 - 5*a*b^4)*cos(d*x + c))/(((a^6*b + 2*a^5*b^2 + a^4*b^3)*d*cos(d*x + c)^4 - (a^7 + 4*a^6*b + 5*a^5*b^2 + 2*a^4 *b^3)*d*cos(d*x + c)^2 + (a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*d)*sin(d*x + c))]
\[ \int \frac {\csc ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{\left (a + b \sin ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.38 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06 \[ \int \frac {\csc ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {3 \, {\left (6 \, a b^{2} + 5 \, b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} a}} - \frac {3 \, {\left (2 \, a^{3} - 6 \, a b^{2} - 5 \, b^{3}\right )} \tan \left (d x + c\right )^{4} + 2 \, a^{3} + 2 \, a^{2} b + 2 \, {\left (4 \, a^{3} - a^{2} b - 5 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \tan \left (d x + c\right )^{5} + {\left (a^{5} + a^{4} b\right )} \tan \left (d x + c\right )^{3}}}{6 \, d} \]
1/6*(3*(6*a*b^2 + 5*b^3)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a^ 4 + a^3*b)*sqrt((a + b)*a)) - (3*(2*a^3 - 6*a*b^2 - 5*b^3)*tan(d*x + c)^4 + 2*a^3 + 2*a^2*b + 2*(4*a^3 - a^2*b - 5*a*b^2)*tan(d*x + c)^2)/((a^5 + 2* a^4*b + a^3*b^2)*tan(d*x + c)^5 + (a^5 + a^4*b)*tan(d*x + c)^3))/d
Time = 0.36 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {3 \, b^{3} \tan \left (d x + c\right )}{{\left (a^{4} + a^{3} b\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}} + \frac {3 \, {\left (6 \, a b^{2} + 5 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{{\left (a^{4} + a^{3} b\right )} \sqrt {a^{2} + a b}} - \frac {2 \, {\left (3 \, a \tan \left (d x + c\right )^{2} - 6 \, b \tan \left (d x + c\right )^{2} + a\right )}}{a^{3} \tan \left (d x + c\right )^{3}}}{6 \, d} \]
1/6*(3*b^3*tan(d*x + c)/((a^4 + a^3*b)*(a*tan(d*x + c)^2 + b*tan(d*x + c)^ 2 + a)) + 3*(6*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^4 + a^3* b)*sqrt(a^2 + a*b)) - 2*(3*a*tan(d*x + c)^2 - 6*b*tan(d*x + c)^2 + a)/(a^3 *tan(d*x + c)^3))/d
Time = 14.87 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.01 \[ \int \frac {\csc ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {b^2\,\mathrm {atan}\left (\frac {b^2\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^4+b\,a^3\right )\,\left (6\,a+5\,b\right )}{a^{7/2}\,\left (5\,b^3+6\,a\,b^2\right )\,\sqrt {a+b}}\right )\,\left (6\,a+5\,b\right )}{2\,a^{7/2}\,d\,{\left (a+b\right )}^{3/2}}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (4\,a-5\,b\right )}{3\,a^2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (-2\,a^3+6\,a\,b^2+5\,b^3\right )}{2\,a^3\,\left (a+b\right )}}{d\,\left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+a\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )} \]
(b^2*atan((b^2*tan(c + d*x)*(a^3*b + a^4)*(6*a + 5*b))/(a^(7/2)*(6*a*b^2 + 5*b^3)*(a + b)^(1/2)))*(6*a + 5*b))/(2*a^(7/2)*d*(a + b)^(3/2)) - (1/(3*a ) + (tan(c + d*x)^2*(4*a - 5*b))/(3*a^2) - (tan(c + d*x)^4*(6*a*b^2 - 2*a^ 3 + 5*b^3))/(2*a^3*(a + b)))/(d*(tan(c + d*x)^5*(a + b) + a*tan(c + d*x)^3 ))