Integrand size = 23, antiderivative size = 148 \[ \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {(4 a-b) x}{2 b^3}+\frac {a^{3/2} (4 a+5 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 b^3 (a+b)^{3/2} d}-\frac {a (2 a+b) \tan (c+d x)}{2 b^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )} \]
-1/2*(4*a-b)*x/b^3+1/2*a^(3/2)*(4*a+5*b)*arctan((a+b)^(1/2)*tan(d*x+c)/a^( 1/2))/b^3/(a+b)^(3/2)/d-1/2*a*(2*a+b)*tan(d*x+c)/b^2/(a+b)/d/(a+(a+b)*tan( d*x+c)^2)-1/2*sin(d*x+c)^2*tan(d*x+c)/b/d/(a+(a+b)*tan(d*x+c)^2)
Time = 1.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {-2 (4 a-b) (c+d x)+\frac {2 a^{3/2} (4 a+5 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{3/2}}+b \left (-1-\frac {2 a^2}{(a+b) (2 a+b-b \cos (2 (c+d x)))}\right ) \sin (2 (c+d x))}{4 b^3 d} \]
(-2*(4*a - b)*(c + d*x) + (2*a^(3/2)*(4*a + 5*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a + b)^(3/2) + b*(-1 - (2*a^2)/((a + b)*(2*a + b - b*C os[2*(c + d*x)])))*Sin[2*(c + d*x)])/(4*b^3*d)
Time = 0.40 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3666, 372, 440, 27, 397, 216, 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.
| \(\displaystyle \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{\left (a+b \sin (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\tan ^6(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2 \left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {\tan ^2(c+d x) \left (3 a-(a-b) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{2 b}-\frac {\tan ^3(c+d x)}{2 b \left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 \left (a (2 a+b)-\left (2 a^2+2 b a-b^2\right ) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 b (a+b)}-\frac {a (2 a+b) \tan (c+d x)}{b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{2 b}-\frac {\tan ^3(c+d x)}{2 b \left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a (2 a+b)-\left (2 a^2+2 b a-b^2\right ) \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{b (a+b)}-\frac {a (2 a+b) \tan (c+d x)}{b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{2 b}-\frac {\tan ^3(c+d x)}{2 b \left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {a^2 (4 a+5 b) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{b}-\frac {(4 a-b) (a+b) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{b}}{b (a+b)}-\frac {a (2 a+b) \tan (c+d x)}{b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{2 b}-\frac {\tan ^3(c+d x)}{2 b \left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {a^2 (4 a+5 b) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{b}-\frac {(4 a-b) (a+b) \arctan (\tan (c+d x))}{b}}{b (a+b)}-\frac {a (2 a+b) \tan (c+d x)}{b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{2 b}-\frac {\tan ^3(c+d x)}{2 b \left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {a^{3/2} (4 a+5 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b \sqrt {a+b}}-\frac {(4 a-b) (a+b) \arctan (\tan (c+d x))}{b}}{b (a+b)}-\frac {a (2 a+b) \tan (c+d x)}{b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{2 b}-\frac {\tan ^3(c+d x)}{2 b \left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\) |
(-1/2*Tan[c + d*x]^3/(b*(1 + Tan[c + d*x]^2)*(a + (a + b)*Tan[c + d*x]^2)) + ((-(((4*a - b)*(a + b)*ArcTan[Tan[c + d*x]])/b) + (a^(3/2)*(4*a + 5*b)* ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(b*Sqrt[a + b]))/(b*(a + b)) - (a*(2*a + b)*Tan[c + d*x])/(b*(a + b)*(a + (a + b)*Tan[c + d*x]^2)))/(2*b ))/d
3.1.100.3.1 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 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[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 1]
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 = 1.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{2} \left (-\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\left (4 a +5 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{b^{3}}-\frac {\frac {\tan \left (d x +c \right ) b}{2+2 \left (\tan ^{2}\left (d x +c \right )\right )}+\frac {\left (4 a -b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{b^{3}}}{d}\) | \(134\) |
| default | \(\frac {\frac {a^{2} \left (-\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\left (4 a +5 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{b^{3}}-\frac {\frac {\tan \left (d x +c \right ) b}{2+2 \left (\tan ^{2}\left (d x +c \right )\right )}+\frac {\left (4 a -b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{b^{3}}}{d}\) | \(134\) |
| risch | \(-\frac {2 a x}{b^{3}}+\frac {x}{2 b^{2}}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {i a^{2} \left (2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}{b^{3} \left (a +b \right ) d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}-\frac {\sqrt {-a \left (a +b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{\left (a +b \right )^{2} d \,b^{3}}-\frac {5 \sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{4 \left (a +b \right )^{2} d \,b^{2}}+\frac {\sqrt {-a \left (a +b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{\left (a +b \right )^{2} d \,b^{3}}+\frac {5 \sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{4 \left (a +b \right )^{2} d \,b^{2}}\) | \(358\) |
1/d*(1/b^3*a^2*(-1/2*b/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)+ 1/2*(4*a+5*b)/(a+b)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2 )))-1/b^3*(1/2*b*tan(d*x+c)/(1+tan(d*x+c)^2)+1/2*(4*a-b)*arctan(tan(d*x+c) )))
Time = 0.32 (sec) , antiderivative size = 623, normalized size of antiderivative = 4.21 \[ \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [-\frac {4 \, {\left (4 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x \cos \left (d x + c\right )^{2} - 4 \, {\left (4 \, a^{3} + 7 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} d x + {\left (4 \, a^{3} + 9 \, a^{2} b + 5 \, a b^{2} - {\left (4 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {a}{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^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{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 ({\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a b^{4} + b^{5}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} d\right )}}, -\frac {2 \, {\left (4 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (4 \, a^{3} + 7 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} d x - {\left (4 \, a^{3} + 9 \, a^{2} b + 5 \, a b^{2} - {\left (4 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a b^{4} + b^{5}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} d\right )}}\right ] \]
[-1/8*(4*(4*a^2*b + 3*a*b^2 - b^3)*d*x*cos(d*x + c)^2 - 4*(4*a^3 + 7*a^2*b + 2*a*b^2 - b^3)*d*x + (4*a^3 + 9*a^2*b + 5*a*b^2 - (4*a^2*b + 5*a*b^2)*c os(d*x + c)^2)*sqrt(-a/(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^2 + 3*a*b + b^2)*cos(d* x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(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*((a*b^2 + b^3)*cos(d*x + c)^3 - (2*a^2*b + 2*a*b^ 2 + b^3)*cos(d*x + c))*sin(d*x + c))/((a*b^4 + b^5)*d*cos(d*x + c)^2 - (a^ 2*b^3 + 2*a*b^4 + b^5)*d), -1/4*(2*(4*a^2*b + 3*a*b^2 - b^3)*d*x*cos(d*x + c)^2 - 2*(4*a^3 + 7*a^2*b + 2*a*b^2 - b^3)*d*x - (4*a^3 + 9*a^2*b + 5*a*b ^2 - (4*a^2*b + 5*a*b^2)*cos(d*x + c)^2)*sqrt(a/(a + b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c)*sin(d*x + c)) ) + 2*((a*b^2 + b^3)*cos(d*x + c)^3 - (2*a^2*b + 2*a*b^2 + b^3)*cos(d*x + c))*sin(d*x + c))/((a*b^4 + b^5)*d*cos(d*x + c)^2 - (a^2*b^3 + 2*a*b^4 + b ^5)*d)]
Timed out. \[ \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.22 \[ \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\left (4 \, a^{3} + 5 \, a^{2} b\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {{\left (2 \, a^{2} + 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (2 \, a^{2} + a b\right )} \tan \left (d x + c\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{4} + a^{2} b^{2} + a b^{3} + {\left (2 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{2}} - \frac {{\left (d x + c\right )} {\left (4 \, a - b\right )}}{b^{3}}}{2 \, d} \]
1/2*((4*a^3 + 5*a^2*b)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a*b^ 3 + b^4)*sqrt((a + b)*a)) - ((2*a^2 + 2*a*b + b^2)*tan(d*x + c)^3 + (2*a^2 + a*b)*tan(d*x + c))/((a^2*b^2 + 2*a*b^3 + b^4)*tan(d*x + c)^4 + a^2*b^2 + a*b^3 + (2*a^2*b^2 + 3*a*b^3 + b^4)*tan(d*x + c)^2) - (d*x + c)*(4*a - b )/b^3)/d
Time = 0.40 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\left (4 \, a^{3} + 5 \, a^{2} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{{\left (a b^{3} + b^{4}\right )} \sqrt {a^{2} + a b}} - \frac {2 \, a^{2} \tan \left (d x + c\right )^{3} + 2 \, a b \tan \left (d x + c\right )^{3} + b^{2} \tan \left (d x + c\right )^{3} + 2 \, a^{2} \tan \left (d x + c\right ) + a b \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{4} + b \tan \left (d x + c\right )^{4} + 2 \, a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a b^{2} + b^{3}\right )}} - \frac {{\left (d x + c\right )} {\left (4 \, a - b\right )}}{b^{3}}}{2 \, d} \]
1/2*((4*a^3 + 5*a^2*b)*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arct an((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)))/((a*b^3 + b^4)*sqrt (a^2 + a*b)) - (2*a^2*tan(d*x + c)^3 + 2*a*b*tan(d*x + c)^3 + b^2*tan(d*x + c)^3 + 2*a^2*tan(d*x + c) + a*b*tan(d*x + c))/((a*tan(d*x + c)^4 + b*tan (d*x + c)^4 + 2*a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)*(a*b^2 + b^3)) - (d*x + c)*(4*a - b)/b^3)/d
Time = 16.77 (sec) , antiderivative size = 2295, normalized size of antiderivative = 15.51 \[ \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
(atan((((a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*((tan(c + d*x)*(96*a^5*b - 4* a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5)) + (((2*a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) + ( tan(c + d*x)*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*(80*a*b^9 + 16*b^10 + 14 4*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))/(2*(a*b^4 + b^5)*(3*a*b^5 + b^6 + 3 *a^2*b^4 + a^3*b^3)))*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2))/(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3))*1i)/(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3) + ((a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*((tan(c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a ^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5)) - (((2 *a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) - (tan(c + d*x) *(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*(80*a*b^9 + 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))/(2*(a*b^4 + b^5)*(3*a*b^5 + b^6 + 3*a^2*b^4 + a ^3*b^3)))*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2))/(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3))*1i)/(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3))/((16*a^5*b + 8*a^6 + (5*a^2*b^4)/4 - (13*a^3*b^3)/2 + (3*a^4*b^2)/2)/(a*b^6 + b^7) + ((a + ( 5*b)/4)*(-a^3*(a + b)^3)^(1/2)*((tan(c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5)) + (((2*a *b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) + (tan(c + d*x)*( a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*(80*a*b^9 + 16*b^10 + 144*a^2*b^8 + 11 2*a^3*b^7 + 32*a^4*b^6))/(2*(a*b^4 + b^5)*(3*a*b^5 + b^6 + 3*a^2*b^4 + ...