Integrand size = 14, antiderivative size = 144 \[ \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{5/2} d}+\frac {b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )} \]
1/8*(8*a^2+8*a*b+3*b^2)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(5/2)/(a+ b)^(5/2)/d+1/4*b*cos(d*x+c)*sin(d*x+c)/a/(a+b)/d/(a+b*sin(d*x+c)^2)^2+3/8* b*(2*a+b)*cos(d*x+c)*sin(d*x+c)/a^2/(a+b)^2/d/(a+b*sin(d*x+c)^2)
Time = 11.72 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{5/2}}+\frac {\sqrt {a} b \left (16 a^2+16 a b+3 b^2-3 b (2 a+b) \cos (2 (c+d x))\right ) \sin (2 (c+d x))}{(a+b)^2 (2 a+b-b \cos (2 (c+d x)))^2}}{8 a^{5/2} d} \]
(((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a + b)^(5/2) + (Sqrt[a]*b*(16*a^2 + 16*a*b + 3*b^2 - 3*b*(2*a + b)*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/((a + b)^2*(2*a + b - b*Cos[2*(c + d*x)])^2))/(8 *a^(5/2)*d)
Time = 0.53 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 3663, 25, 3042, 3652, 27, 3042, 3660, 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.
| \(\displaystyle \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \sin (c+d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 3663 |
| \(\displaystyle \frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {-2 b \sin ^2(c+d x)+4 a+3 b}{\left (b \sin ^2(c+d x)+a\right )^2}dx}{4 a (a+b)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-2 b \sin ^2(c+d x)+4 a+3 b}{\left (b \sin ^2(c+d x)+a\right )^2}dx}{4 a (a+b)}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-2 b \sin (c+d x)^2+4 a+3 b}{\left (b \sin (c+d x)^2+a\right )^2}dx}{4 a (a+b)}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 3652 |
| \(\displaystyle \frac {\frac {\int \frac {8 a^2+8 b a+3 b^2}{b \sin ^2(c+d x)+a}dx}{2 a (a+b)}+\frac {3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{2 a d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{b \sin ^2(c+d x)+a}dx}{2 a (a+b)}+\frac {3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{2 a d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{b \sin (c+d x)^2+a}dx}{2 a (a+b)}+\frac {3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{2 a d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 3660 |
| \(\displaystyle \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{2 a d (a+b)}+\frac {3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{2 a d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a+b)^{3/2}}+\frac {3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{2 a d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
(b*Cos[c + d*x]*Sin[c + d*x])/(4*a*(a + b)*d*(a + b*Sin[c + d*x]^2)^2) + ( ((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*a^ (3/2)*(a + b)^(3/2)*d) + (3*b*(2*a + b)*Cos[c + d*x]*Sin[c + d*x])/(2*a*(a + b)*d*(a + b*Sin[c + d*x]^2)))/(4*a*(a + b))
3.2.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x ]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Simp[1/(2* a*(a + b)*(p + 1)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*( p + 1) + b*(2*p + 3)) + 2*(A*b - a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[1/(a + (a + b)*ff^2*x^ 2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Simp[1/(2*a*(p + 1)*(a + b)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && LtQ[p, -1]
Time = 1.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.03
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (8 a +3 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{8 a^{2} \left (a +b \right )}+\frac {b \left (8 a +5 b \right ) \tan \left (d x +c \right )}{8 a \left (a^{2}+2 a b +b^{2}\right )}}{{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}^{2}}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 a^{2} \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}}{d}\) | \(148\) |
| default | \(\frac {\frac {\frac {\left (8 a +3 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{8 a^{2} \left (a +b \right )}+\frac {b \left (8 a +5 b \right ) \tan \left (d x +c \right )}{8 a \left (a^{2}+2 a b +b^{2}\right )}}{{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}^{2}}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 a^{2} \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}}{d}\) | \(148\) |
| risch | \(-\frac {i \left (-8 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-8 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+48 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+72 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+42 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-40 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-40 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+6 a \,b^{2}+3 b^{3}\right )}{4 a^{2} \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}}-\frac {\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 )^{2} d}-\frac {\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 {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}}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{2}}+\frac {\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 )^{2} d}+\frac {\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 {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}}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{2}}\) | \(782\) |
1/d*((1/8*(8*a+3*b)/a^2*b/(a+b)*tan(d*x+c)^3+1/8*b*(8*a+5*b)/a/(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+1/8*(8*a^2+8*a*b+3*b^ 2)/a^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)))
Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (130) = 260\).
Time = 0.30 (sec) , antiderivative size = 843, normalized size of antiderivative = 5.85 \[ \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\left [-\frac {{\left ({\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 24 \, a^{3} b + 27 \, a^{2} b^{2} + 14 \, a b^{3} + 3 \, b^{4} - 2 \, {\left (8 \, a^{3} b + 16 \, a^{2} b^{2} + 11 \, a b^{3} + 3 \, 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 ) + 4 \, {\left (3 \, {\left (2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (8 \, a^{4} b + 19 \, a^{3} b^{2} + 14 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{32 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b + 4 \, a^{6} b^{2} + 6 \, a^{5} b^{3} + 4 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}\right )} d\right )}}, -\frac {{\left ({\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 24 \, a^{3} b + 27 \, a^{2} b^{2} + 14 \, a b^{3} + 3 \, b^{4} - 2 \, {\left (8 \, a^{3} b + 16 \, a^{2} b^{2} + 11 \, a b^{3} + 3 \, 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 ) + 2 \, {\left (3 \, {\left (2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (8 \, a^{4} b + 19 \, a^{3} b^{2} + 14 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b + 4 \, a^{6} b^{2} + 6 \, a^{5} b^{3} + 4 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}\right )} d\right )}}\right ] \]
[-1/32*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(d*x + c)^4 + 8*a^4 + 24*a^3*b + 27*a^2*b^2 + 14*a*b^3 + 3*b^4 - 2*(8*a^3*b + 16*a^2*b^2 + 11*a*b^3 + 3*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)) + 4*(3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(d*x + c)^3 - (8*a^4*b + 19*a^3 *b^2 + 14*a^2*b^3 + 3*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^6*b^2 + 3*a^5 *b^3 + 3*a^4*b^4 + a^3*b^5)*d*cos(d*x + c)^4 - 2*(a^7*b + 4*a^6*b^2 + 6*a^ 5*b^3 + 4*a^4*b^4 + a^3*b^5)*d*cos(d*x + c)^2 + (a^8 + 5*a^7*b + 10*a^6*b^ 2 + 10*a^5*b^3 + 5*a^4*b^4 + a^3*b^5)*d), -1/16*(((8*a^2*b^2 + 8*a*b^3 + 3 *b^4)*cos(d*x + c)^4 + 8*a^4 + 24*a^3*b + 27*a^2*b^2 + 14*a*b^3 + 3*b^4 - 2*(8*a^3*b + 16*a^2*b^2 + 11*a*b^3 + 3*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))) + 2*(3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(d*x + c)^3 - (8*a^4*b + 19*a^3*b^2 + 14*a^2*b^3 + 3*a*b^4)*cos(d*x + c))*sin(d*x + c)) /((a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*d*cos(d*x + c)^4 - 2*(a^7*b + 4*a^6*b^2 + 6*a^5*b^3 + 4*a^4*b^4 + a^3*b^5)*d*cos(d*x + c)^2 + (a^8 + 5 *a^7*b + 10*a^6*b^2 + 10*a^5*b^3 + 5*a^4*b^4 + a^3*b^5)*d)]
Timed out. \[ \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.39 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {{\left (8 \, a^{2} b + 11 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (d x + c\right )^{3} + {\left (8 \, a^{2} b + 5 \, a b^{2}\right )} \tan \left (d x + c\right )}{a^{6} + 2 \, a^{5} b + a^{4} b^{2} + {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (d x + c\right )^{2}}}{8 \, d} \]
1/8*((8*a^2 + 8*a*b + 3*b^2)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/ ((a^4 + 2*a^3*b + a^2*b^2)*sqrt((a + b)*a)) + ((8*a^2*b + 11*a*b^2 + 3*b^3 )*tan(d*x + c)^3 + (8*a^2*b + 5*a*b^2)*tan(d*x + c))/(a^6 + 2*a^5*b + a^4* b^2 + (a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*tan(d*x + c)^4 + 2 *(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*tan(d*x + c)^2))/d
Time = 0.33 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {a^{2} + a b}} + \frac {8 \, a^{2} b \tan \left (d x + c\right )^{3} + 11 \, a b^{2} \tan \left (d x + c\right )^{3} + 3 \, b^{3} \tan \left (d x + c\right )^{3} + 8 \, a^{2} b \tan \left (d x + c\right ) + 5 \, a b^{2} \tan \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}}}{8 \, d} \]
1/8*((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)))*(8*a^2 + 8*a*b + 3*b^2)/((a^4 + 2*a^3 *b + a^2*b^2)*sqrt(a^2 + a*b)) + (8*a^2*b*tan(d*x + c)^3 + 11*a*b^2*tan(d* x + c)^3 + 3*b^3*tan(d*x + c)^3 + 8*a^2*b*tan(d*x + c) + 5*a*b^2*tan(d*x + c))/((a^4 + 2*a^3*b + a^2*b^2)*(a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)^ 2))/d
Time = 14.12 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,b^2+8\,a\,b\right )}{8\,a^2\,\left (a+b\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (5\,b^2+8\,a\,b\right )}{8\,a\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2+2\,a\,b+b^2\right )+a^2+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )\right )}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )\,\left (8\,a^2+8\,a\,b+3\,b^2\right )}{8\,a^{5/2}\,d\,{\left (a+b\right )}^{5/2}} \]
((tan(c + d*x)^3*(8*a*b + 3*b^2))/(8*a^2*(a + b)) + (tan(c + d*x)*(8*a*b + 5*b^2))/(8*a*(2*a*b + a^2 + b^2)))/(d*(tan(c + d*x)^4*(2*a*b + a^2 + b^2) + a^2 + tan(c + d*x)^2*(2*a*b + 2*a^2))) + (atan((tan(c + d*x)*(2*a + 2*b )*(2*a*b + a^2 + b^2))/(2*a^(1/2)*(a + b)^(5/2)))*(8*a*b + 8*a^2 + 3*b^2)) /(8*a^(5/2)*d*(a + b)^(5/2))