Integrand size = 24, antiderivative size = 343 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=-\frac {\left (4 a-10 \sqrt {a} \sqrt {b}+3 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{5/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} b^{3/2} d}+\frac {\left (4 a+10 \sqrt {a} \sqrt {b}+3 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{5/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} b^{3/2} d}-\frac {\tan (c+d x) \left (a (a+3 b)+\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)\right )}{8 (a-b)^3 d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )^2}-\frac {\tan (c+d x) \left (\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}+\frac {\left (2 a^2+15 a b+3 b^2\right ) \tan ^2(c+d x)}{(a-b)^2}\right )}{32 a b d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \]
-1/64*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*(4*a+3*b-10*a^(1/ 2)*b^(1/2))/a^(5/4)/b^(3/2)/d/(a^(1/2)-b^(1/2))^(5/2)+1/64*arctan((a^(1/2) +b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*(4*a+3*b+10*a^(1/2)*b^(1/2))/a^(5/4)/b ^(3/2)/d/(a^(1/2)+b^(1/2))^(5/2)-1/8*tan(d*x+c)*(a*(a+3*b)+(a^2+6*a*b+b^2) *tan(d*x+c)^2)/(a-b)^3/d/(a+2*a*tan(d*x+c)^2+(a-b)*tan(d*x+c)^4)^2-1/32*ta n(d*x+c)*(2*a*(a^2-a*b-8*b^2)/(a-b)^3+(2*a^2+15*a*b+3*b^2)*tan(d*x+c)^2/(a -b)^2)/a/b/d/(a+2*a*tan(d*x+c)^2+(a-b)*tan(d*x+c)^4)
Time = 6.23 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \sqrt {b} \left (4 a+10 \sqrt {a} \sqrt {b}+3 b\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{a \sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \sqrt {b} \left (4 a-10 \sqrt {a} \sqrt {b}+3 b\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{a \sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {4 b \left (4 a^2-19 a b-3 b^2+3 b (a+b) \cos (2 (c+d x))\right ) \sin (2 (c+d x))}{a (8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x)))}-\frac {128 (a-b) b (2 a+b-b \cos (2 (c+d x))) \sin (2 (c+d x))}{(-8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))^2}}{64 (a-b)^2 b^2 d} \]
(((Sqrt[a] - Sqrt[b])^2*Sqrt[b]*(4*a + 10*Sqrt[a]*Sqrt[b] + 3*b)*ArcTan[(( Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(a*Sqrt[a + S qrt[a]*Sqrt[b]]) + ((Sqrt[a] + Sqrt[b])^2*Sqrt[b]*(4*a - 10*Sqrt[a]*Sqrt[b ] + 3*b)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt [b]]])/(a*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + (4*b*(4*a^2 - 19*a*b - 3*b^2 + 3*b *(a + b)*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(a*(8*a - 3*b + 4*b*Cos[2*(c + d*x)] - b*Cos[4*(c + d*x)])) - (128*(a - b)*b*(2*a + b - b*Cos[2*(c + d* x)])*Sin[2*(c + d*x)])/(-8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d *x)])^2)/(64*(a - b)^2*b^2*d)
Time = 0.83 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3696, 1672, 27, 2206, 27, 1480, 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.
| \(\displaystyle \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{\left (a-b \sin (c+d x)^4\right )^3}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\tan ^6(c+d x) \left (\tan ^2(c+d x)+1\right )^2}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^3}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1672 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 \left (\frac {8 a^2 b \tan ^6(c+d x)}{a-b}-\frac {16 a^2 b^2 \tan ^4(c+d x)}{(a-b)^2}-\frac {a^2 b \left (5 a^2+6 b a-3 b^2\right ) \tan ^2(c+d x)}{(a-b)^3}+\frac {a^3 b (a+3 b)}{(a-b)^3}\right )}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{16 a^2 b}-\frac {\tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\frac {8 a^2 b \tan ^6(c+d x)}{a-b}-\frac {16 a^2 b^2 \tan ^4(c+d x)}{(a-b)^2}-\frac {a^2 b \left (5 a^2+6 b a-3 b^2\right ) \tan ^2(c+d x)}{(a-b)^3}+\frac {a^3 b (a+3 b)}{(a-b)^3}}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{8 a^2 b}-\frac {\tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 2206 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {2 a^3 b \left (2 a (a+2 b)-\left (2 a^2-17 b a+3 b^2\right ) \tan ^2(c+d x)\right )}{(a-b)^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{8 a^2 b}-\frac {a \tan (c+d x) \left (\frac {\left (2 a^2+15 a b+3 b^2\right ) \tan ^2(c+d x)}{(a-b)^2}+\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {\tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {a \int \frac {2 a (a+2 b)-\left (2 a^2-17 b a+3 b^2\right ) \tan ^2(c+d x)}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{4 (a-b)^2}-\frac {a \tan (c+d x) \left (\frac {\left (2 a^2+15 a b+3 b^2\right ) \tan ^2(c+d x)}{(a-b)^2}+\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {\tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\frac {a \left (\frac {\left (\sqrt {a}-\sqrt {b}\right )^3 \left (10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {b}}-\frac {\left (\sqrt {a}+\sqrt {b}\right )^3 \left (-10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {b}}\right )}{4 (a-b)^2}-\frac {a \tan (c+d x) \left (\frac {\left (2 a^2+15 a b+3 b^2\right ) \tan ^2(c+d x)}{(a-b)^2}+\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {\tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {a \left (\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \left (10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {b} \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \left (-10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {b} \sqrt {\sqrt {a}-\sqrt {b}}}\right )}{4 (a-b)^2}-\frac {a \tan (c+d x) \left (\frac {\left (2 a^2+15 a b+3 b^2\right ) \tan ^2(c+d x)}{(a-b)^2}+\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {\tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
(-1/8*(Tan[c + d*x]*(a*(a + 3*b) + (a^2 + 6*a*b + b^2)*Tan[c + d*x]^2))/(( a - b)^3*(a + 2*a*Tan[c + d*x]^2 + (a - b)*Tan[c + d*x]^4)^2) + ((a*(-1/2* ((Sqrt[a] + Sqrt[b])^2*(4*a - 10*Sqrt[a]*Sqrt[b] + 3*b)*ArcTan[(Sqrt[Sqrt[ a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(a^(1/4)*Sqrt[Sqrt[a] - Sqrt[b]]*Sqr t[b]) + ((Sqrt[a] - Sqrt[b])^2*(4*a + 10*Sqrt[a]*Sqrt[b] + 3*b)*ArcTan[(Sq rt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(1/4)*Sqrt[Sqrt[a] + Sq rt[b]]*Sqrt[b])))/(4*(a - b)^2) - (a*Tan[c + d*x]*((2*a*(a^2 - a*b - 8*b^2 ))/(a - b)^3 + ((2*a^2 + 15*a*b + 3*b^2)*Tan[c + d*x]^2)/(a - b)^2))/(4*(a + 2*a*Tan[c + d*x]^2 + (a - b)*Tan[c + d*x]^4)))/(8*a^2*b))/d
3.3.31.3.1 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[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_), x_Symbol] :> With[{f = Coeff[PolynomialRemainder[x^m*(d + e*x^2)^q , a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[x^m*(d + e*x^ 2)^q, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a *b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c*x^4)^(p + 1)* Simp[ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[x^m*(d + e*x^ 2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], x], x], x]] /; FreeQ[{a, b, c, d, e}, x ] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[q, 1] && IGtQ[m/2, 0]
Int[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[Poly nomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^ 4)^(p + 1)*((a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)/(2*a*(p + 1)*(b ^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c *x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[Px, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4* p + 7)*(b*d - 2*a*e)*x^2, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Px, x ^2] && Expon[Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2) ^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] & & IntegerQ[m/2] && IntegerQ[p]
Time = 4.79 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.26
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\left (2 a^{2}+15 a b +3 b^{2}\right ) \left (\tan ^{7}\left (d x +c \right )\right )}{32 \left (a -b \right ) a b}-\frac {\left (3 a^{2}+14 a b -5 b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{16 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 a^{2}+19 a b -b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{32 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {a \left (a +2 b \right ) \tan \left (d x +c \right )}{16 b \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (-2 a^{2} \sqrt {a b}+17 a b \sqrt {a b}-3 b^{2} \sqrt {a b}+4 a^{3}-15 a^{2} b -a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (-2 a^{2} \sqrt {a b}+17 a b \sqrt {a b}-3 b^{2} \sqrt {a b}-4 a^{3}+15 a^{2} b +a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 a b \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(432\) |
| default | \(\frac {\frac {-\frac {\left (2 a^{2}+15 a b +3 b^{2}\right ) \left (\tan ^{7}\left (d x +c \right )\right )}{32 \left (a -b \right ) a b}-\frac {\left (3 a^{2}+14 a b -5 b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{16 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 a^{2}+19 a b -b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{32 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {a \left (a +2 b \right ) \tan \left (d x +c \right )}{16 b \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (-2 a^{2} \sqrt {a b}+17 a b \sqrt {a b}-3 b^{2} \sqrt {a b}+4 a^{3}-15 a^{2} b -a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (-2 a^{2} \sqrt {a b}+17 a b \sqrt {a b}-3 b^{2} \sqrt {a b}-4 a^{3}+15 a^{2} b +a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 a b \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(432\) |
| risch | \(\text {Expression too large to display}\) | \(2377\) |
1/d*((-1/32*(2*a^2+15*a*b+3*b^2)/(a-b)/a/b*tan(d*x+c)^7-1/16*(3*a^2+14*a*b -5*b^2)/b/(a^2-2*a*b+b^2)*tan(d*x+c)^5-1/32*(6*a^2+19*a*b-b^2)/b/(a^2-2*a* b+b^2)*tan(d*x+c)^3-1/16*a*(a+2*b)/b/(a^2-2*a*b+b^2)*tan(d*x+c))/(tan(d*x+ c)^4*a-b*tan(d*x+c)^4+2*a*tan(d*x+c)^2+a)^2+1/32/a/b/(a^2-2*a*b+b^2)*(a-b) *(1/2*(-2*a^2*(a*b)^(1/2)+17*a*b*(a*b)^(1/2)-3*b^2*(a*b)^(1/2)+4*a^3-15*a^ 2*b-a*b^2)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)* tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))+1/2*(-2*a^2*(a*b)^(1/2)+17*a*b*( a*b)^(1/2)-3*b^2*(a*b)^(1/2)-4*a^3+15*a^2*b+a*b^2)/(a*b)^(1/2)/(a-b)/(((a* b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^( 1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 5961 vs. \(2 (291) = 582\).
Time = 2.03 (sec) , antiderivative size = 5961, normalized size of antiderivative = 17.38 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sin \left (d x + c\right )^{6}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \]
-1/16*(4*(32*a^3*b^2 - 84*a^2*b^3 - 83*a*b^4 + 21*b^5)*cos(4*d*x + 4*c)*si n(2*d*x + 2*c) + ((4*a^2*b^3 - 13*a*b^4 + 3*b^5)*sin(14*d*x + 14*c) - 3*(8 *a^2*b^3 - 33*a*b^4 + 7*b^5)*sin(12*d*x + 12*c) + (64*a^3*b^2 + 68*a^2*b^3 - 225*a*b^4 + 63*b^5)*sin(10*d*x + 10*c) - 3*(128*a^3*b^2 + 32*a^2*b^3 - 61*a*b^4 + 35*b^5)*sin(8*d*x + 8*c) - (64*a^3*b^2 + 452*a^2*b^3 - 9*a*b^4 - 105*b^5)*sin(6*d*x + 6*c) + 3*(40*a^2*b^3 - 29*a*b^4 - 21*b^5)*sin(4*d*x + 4*c) - (4*a^2*b^3 - 37*a*b^4 - 21*b^5)*sin(2*d*x + 2*c))*cos(16*d*x + 1 6*c) + 2*(2*(32*a^3*b^2 - 84*a^2*b^3 - 83*a*b^4 + 21*b^5)*sin(12*d*x + 12* c) - 8*(64*a^3*b^2 - 84*a^2*b^3 - 43*a*b^4 + 21*b^5)*sin(10*d*x + 10*c) - (512*a^4*b - 3584*a^3*b^2 + 1388*a^2*b^3 - 11*a*b^4 - 315*b^5)*sin(8*d*x + 8*c) + 16*(172*a^2*b^3 - 37*a*b^4 - 21*b^5)*sin(6*d*x + 6*c) + 2*(32*a^3* b^2 - 372*a^2*b^3 + 289*a*b^4 + 105*b^5)*sin(4*d*x + 4*c) + 8*(4*a^2*b^3 - 25*a*b^4 - 9*b^5)*sin(2*d*x + 2*c))*cos(14*d*x + 14*c) - 2*(2*(512*a^4*b - 672*a^3*b^2 + 1228*a^2*b^3 + 21*a*b^4 - 147*b^5)*sin(10*d*x + 10*c) - 3* (3072*a^4*b - 6272*a^3*b^2 + 2920*a^2*b^3 - 413*a*b^4 - 245*b^5)*sin(8*d*x + 8*c) - 2*(512*a^4*b + 3936*a^3*b^2 - 6740*a^2*b^3 + 1281*a*b^4 + 441*b^ 5)*sin(6*d*x + 6*c) + 12*(192*a^3*b^2 - 416*a^2*b^3 + 161*a*b^4 + 49*b^5)* sin(4*d*x + 4*c) - 2*(32*a^3*b^2 - 372*a^2*b^3 + 289*a*b^4 + 105*b^5)*sin( 2*d*x + 2*c))*cos(12*d*x + 12*c) - 2*((8192*a^5 + 27136*a^4*b - 37696*a^3* b^2 + 17644*a^2*b^3 - 2079*a*b^4 - 735*b^5)*sin(8*d*x + 8*c) + 8*(1024*...
Leaf count of result is larger than twice the leaf count of optimal. 2231 vs. \(2 (291) = 582\).
Time = 1.40 (sec) , antiderivative size = 2231, normalized size of antiderivative = 6.50 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
1/64*(((6*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^4 - 63*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b + 109*sqrt(a^2 - a*b + sqrt(a*b )*(a - b))*sqrt(a*b)*a^2*b^2 - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a* b)*a*b^3 - 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*b^4)*(a^3*b - 2 *a^2*b^2 + a*b^3)^2*abs(-a + b) + 2*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b)) *a^8*b - 9*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^7*b^2 - 4*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^6*b^3 + 34*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^5* b^4 - 33*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^4*b^5 + 7*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^3*b^6 + 2*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^7 )*abs(a^3*b - 2*a^2*b^2 + a*b^3)*abs(-a + b) - (12*sqrt(a^2 - a*b + sqrt(a *b)*(a - b))*sqrt(a*b)*a^11*b - 117*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sq rt(a*b)*a^10*b^2 + 431*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^9*b ^3 - 773*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^8*b^4 + 703*sqrt( a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^7*b^5 - 279*sqrt(a^2 - a*b + sq rt(a*b)*(a - b))*sqrt(a*b)*a^6*b^6 + 5*sqrt(a^2 - a*b + sqrt(a*b)*(a - b)) *sqrt(a*b)*a^5*b^7 + 17*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^4* b^8 + sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b^9)*abs(-a + b))* (pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/sqrt((a^4*b - 2*a^3*b^ 2 + a^2*b^3 + sqrt((a^4*b - 2*a^3*b^2 + a^2*b^3)^2 - (a^4*b - 2*a^3*b^2 + a^2*b^3)*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)))/(a^4*b - 3*a^3*b^2 +...
Time = 20.09 (sec) , antiderivative size = 6391, normalized size of antiderivative = 18.63 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
- ((tan(c + d*x)^3*(19*a*b + 6*a^2 - b^2))/(32*(a^2*b - 2*a*b^2 + b^3)) + (a*tan(c + d*x)*(a + 2*b))/(16*(a^2*b - 2*a*b^2 + b^3)) + (tan(c + d*x)^7* (15*a*b + 2*a^2 + 3*b^2))/(32*a*(a*b - b^2)) + (tan(c + d*x)^5*(14*a*b + 3 *a^2 - 5*b^2))/(16*(a - b)*(a*b - b^2)))/(d*(tan(c + d*x)^8*(a^2 - 2*a*b + b^2) + a^2 - tan(c + d*x)^4*(2*a*b - 6*a^2) - tan(c + d*x)^6*(4*a*b - 4*a ^2) + 4*a^2*tan(c + d*x)^2)) - (atan(((((65536*a^3*b^7 - 163840*a^4*b^6 + 98304*a^5*b^5 + 32768*a^6*b^4 - 32768*a^7*b^3)/(32768*(a^2*b^6 - 3*a^3*b^5 + 3*a^4*b^4 - a^5*b^3)) - (tan(c + d*x)*((9*b^3*(a^5*b^9)^(1/2) - 80*a^3* (a^5*b^9)^(1/2) - 15*a^3*b^7 + 30*a^4*b^6 + 229*a^5*b^5 - 116*a^6*b^4 + 16 *a^7*b^3 - 86*a*b^2*(a^5*b^9)^(1/2) + 301*a^2*b*(a^5*b^9)^(1/2))/(16384*(a ^5*b^11 - 5*a^6*b^10 + 10*a^7*b^9 - 10*a^8*b^8 + 5*a^9*b^7 - a^10*b^6)))^( 1/2)*(16384*a^3*b^8 - 81920*a^4*b^7 + 163840*a^5*b^6 - 163840*a^6*b^5 + 81 920*a^7*b^4 - 16384*a^8*b^3))/(256*(a*b^5 - 3*a^2*b^4 + 3*a^3*b^3 - a^4*b^ 2)))*((9*b^3*(a^5*b^9)^(1/2) - 80*a^3*(a^5*b^9)^(1/2) - 15*a^3*b^7 + 30*a^ 4*b^6 + 229*a^5*b^5 - 116*a^6*b^4 + 16*a^7*b^3 - 86*a*b^2*(a^5*b^9)^(1/2) + 301*a^2*b*(a^5*b^9)^(1/2))/(16384*(a^5*b^11 - 5*a^6*b^10 + 10*a^7*b^9 - 10*a^8*b^8 + 5*a^9*b^7 - a^10*b^6)))^(1/2) + (tan(c + d*x)*(16*a^5 - 116*a ^4*b - 101*a*b^4 + 9*b^5 + 331*a^2*b^3 + 149*a^3*b^2))/(256*(a*b^5 - 3*a^2 *b^4 + 3*a^3*b^3 - a^4*b^2)))*((9*b^3*(a^5*b^9)^(1/2) - 80*a^3*(a^5*b^9)^( 1/2) - 15*a^3*b^7 + 30*a^4*b^6 + 229*a^5*b^5 - 116*a^6*b^4 + 16*a^7*b^3...