Integrand size = 24, antiderivative size = 217 \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {\left (\sqrt {a}-2 \sqrt {b}\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{8 \sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )^{3/2} b^{5/4} d}+\frac {\left (\sqrt {a}+2 \sqrt {b}\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{8 \sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )^{3/2} b^{5/4} d}-\frac {\cos (c+d x) \left (a+b-b \cos ^2(c+d x)\right )}{4 (a-b) b d \left (a-b+2 b \cos ^2(c+d x)-b \cos ^4(c+d x)\right )} \]
-1/4*cos(d*x+c)*(a+b-b*cos(d*x+c)^2)/(a-b)/b/d/(a-b+2*b*cos(d*x+c)^2-b*cos (d*x+c)^4)+1/8*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))*(a^(1/2) -2*b^(1/2))/b^(5/4)/d/a^(1/2)/(a^(1/2)-b^(1/2))^(3/2)+1/8*arctanh(b^(1/4)* cos(d*x+c)/(a^(1/2)+b^(1/2))^(1/2))*(a^(1/2)+2*b^(1/2))/b^(5/4)/d/a^(1/2)/ (a^(1/2)+b^(1/2))^(3/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 2.58 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.16 \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=-\frac {\frac {32 \cos (c+d x) (2 a+b-b \cos (2 (c+d x)))}{8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))}+i \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-8 a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+22 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+4 i a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-11 i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+8 a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-22 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-4 i a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+11 i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6-i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{32 (a-b) b d} \]
-1/32*((32*Cos[c + d*x]*(2*a + b - b*Cos[2*(c + d*x)]))/(8*a - 3*b + 4*b*C os[2*(c + d*x)] - b*Cos[4*(c + d*x)]) + I*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8 & , (-2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x ] - #1)] + I*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] - 8*a*ArcTan[Sin[c + d*x] /(Cos[c + d*x] - #1)]*#1^2 + 22*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] *#1^2 + (4*I)*a*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (11*I)*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 8*a*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - # 1)]*#1^4 - 22*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - (4*I)*a*Lo g[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 + (11*I)*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 + 2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^6 - I*b*Log [1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^6)/(-(b*#1) - 8*a*#1^3 + 3*b*#1^3 - 3*b* #1^5 + b*#1^7) & ])/((a - b)*b*d)
Time = 0.41 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3694, 1517, 27, 1480, 218, 221}
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.
| \(\displaystyle \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^5}{\left (a-b \sin (c+d x)^4\right )^2}dx\) |
| \(\Big \downarrow \) 3694 |
| \(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(c+d x)\right )^2}{\left (-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b\right )^2}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 1517 |
| \(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\int \frac {2 a \left (b \cos ^2(c+d x)+a-3 b\right )}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{8 a b (a-b)}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\int \frac {b \cos ^2(c+d x)+a-3 b}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{4 b (a-b)}}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\frac {1}{2} \sqrt {b} \left (\frac {a-2 b}{\sqrt {a}}+\sqrt {b}\right ) \int \frac {1}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}-b \cos ^2(c+d x)}d\cos (c+d x)-\frac {1}{2} \sqrt {b} \left (\frac {a-2 b}{\sqrt {a}}-\sqrt {b}\right ) \int \frac {1}{-b \cos ^2(c+d x)-\left (\sqrt {a}-\sqrt {b}\right ) \sqrt {b}}d\cos (c+d x)}{4 b (a-b)}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\frac {1}{2} \sqrt {b} \left (\frac {a-2 b}{\sqrt {a}}+\sqrt {b}\right ) \int \frac {1}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}-b \cos ^2(c+d x)}d\cos (c+d x)+\frac {\left (\frac {a-2 b}{\sqrt {a}}-\sqrt {b}\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt [4]{b} \sqrt {\sqrt {a}-\sqrt {b}}}}{4 b (a-b)}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\frac {\left (\frac {a-2 b}{\sqrt {a}}-\sqrt {b}\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt [4]{b} \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {\left (\frac {a-2 b}{\sqrt {a}}+\sqrt {b}\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt [4]{b} \sqrt {\sqrt {a}+\sqrt {b}}}}{4 b (a-b)}}{d}\) |
-((-1/4*((((a - 2*b)/Sqrt[a] - Sqrt[b])*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt [Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(1/4)) + (((a - 2*b)/Sq rt[a] + Sqrt[b])*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/ (2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(1/4)))/((a - b)*b) + (Cos[c + d*x]*(a + b - b*Cos[c + d*x]^2))/(4*(a - b)*b*(a - b + 2*b*Cos[c + d*x]^2 - b*Cos[c + d* x]^4)))/d)
3.3.14.3.1 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_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x _Symbol] :> With[{f = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(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)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(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]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ [c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 2.08 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.87
| method | result | size |
| derivativedivides | \(\frac {-\frac {-\frac {\cos ^{3}\left (d x +c \right )}{4 \left (a -b \right )}+\frac {\left (a +b \right ) \cos \left (d x +c \right )}{4 b \left (a -b \right )}}{a -b +2 b \left (\cos ^{2}\left (d x +c \right )\right )-b \left (\cos ^{4}\left (d x +c \right )\right )}-\frac {\frac {\left (\sqrt {a b}+2 b -a \right ) \arctan \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\left (\sqrt {a b}-2 b +a \right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}}{4 \left (a -b \right )}}{d}\) | \(188\) |
| default | \(\frac {-\frac {-\frac {\cos ^{3}\left (d x +c \right )}{4 \left (a -b \right )}+\frac {\left (a +b \right ) \cos \left (d x +c \right )}{4 b \left (a -b \right )}}{a -b +2 b \left (\cos ^{2}\left (d x +c \right )\right )-b \left (\cos ^{4}\left (d x +c \right )\right )}-\frac {\frac {\left (\sqrt {a b}+2 b -a \right ) \arctan \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\left (\sqrt {a b}-2 b +a \right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}}{4 \left (a -b \right )}}{d}\) | \(188\) |
| risch | \(-\frac {b \,{\mathrm e}^{7 i \left (d x +c \right )}-4 a \,{\mathrm e}^{5 i \left (d x +c \right )}-b \,{\mathrm e}^{5 i \left (d x +c \right )}-4 a \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{3 i \left (d x +c \right )}+b \,{\mathrm e}^{i \left (d x +c \right )}}{2 b \left (a -b \right ) d \left ({\mathrm e}^{8 i \left (d x +c \right )} b -4 b \,{\mathrm e}^{6 i \left (d x +c \right )}-16 a \,{\mathrm e}^{4 i \left (d x +c \right )}+6 b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}-\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{5} b^{5} d^{4}-3 a^{4} b^{6} d^{4}+3 a^{3} b^{7} d^{4}-a^{2} b^{8} d^{4}\right ) \textit {\_Z}^{4}+\left (8 a^{3} b^{3} d^{2}-8 a^{2} b^{4} d^{2}-32 a \,b^{5} d^{2}\right ) \textit {\_Z}^{2}-16 a^{2}+128 a b -256 b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {i a^{4} b^{5} d^{3}}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}+\frac {3 i a^{3} b^{6} d^{3}}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}-\frac {3 i a^{2} b^{7} d^{3}}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}+\frac {i a \,b^{8} d^{3}}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}\right ) \textit {\_R}^{3}+\left (-\frac {2 i a^{4} b d}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}+\frac {16 i a^{3} b^{2} d}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}-\frac {50 i a^{2} b^{3} d}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}+\frac {52 i a \,b^{4} d}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}+\frac {16 i b^{5} d}{2 a^{3}-18 a^{2} b +56 a \,b^{2}-64 b^{3}}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{32}\) | \(604\) |
1/d*(-(-1/4/(a-b)*cos(d*x+c)^3+1/4*(a+b)/b/(a-b)*cos(d*x+c))/(a-b+2*b*cos( d*x+c)^2-b*cos(d*x+c)^4)-1/4/(a-b)*(1/2*((a*b)^(1/2)+2*b-a)/(a*b)^(1/2)/(( (a*b)^(1/2)-b)*b)^(1/2)*arctan(cos(d*x+c)*b/(((a*b)^(1/2)-b)*b)^(1/2))-1/2 *((a*b)^(1/2)-2*b+a)/(a*b)^(1/2)/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh(cos(d*x +c)*b/(((a*b)^(1/2)+b)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 2507 vs. \(2 (169) = 338\).
Time = 0.60 (sec) , antiderivative size = 2507, normalized size of antiderivative = 11.55 \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \]
-1/16*(4*b*cos(d*x + c)^3 - ((a*b^2 - b^3)*d*cos(d*x + c)^4 - 2*(a*b^2 - b ^3)*d*cos(d*x + c)^2 - (a^2*b - 2*a*b^2 + b^3)*d)*sqrt(((a^4*b^2 - 3*a^3*b ^3 + 3*a^2*b^4 - a*b^5)*d^2*sqrt((a^4 - 10*a^3*b + 41*a^2*b^2 - 80*a*b^3 + 64*b^4)/((a^7*b^5 - 6*a^6*b^6 + 15*a^5*b^7 - 20*a^4*b^8 + 15*a^3*b^9 - 6* a^2*b^10 + a*b^11)*d^4)) + a^2 - a*b - 4*b^2)/((a^4*b^2 - 3*a^3*b^3 + 3*a^ 2*b^4 - a*b^5)*d^2))*log((a^3 - 9*a^2*b + 28*a*b^2 - 32*b^3)*cos(d*x + c) - (2*(a^4*b^5 - 3*a^3*b^6 + 3*a^2*b^7 - a*b^8)*d^3*sqrt((a^4 - 10*a^3*b + 41*a^2*b^2 - 80*a*b^3 + 64*b^4)/((a^7*b^5 - 6*a^6*b^6 + 15*a^5*b^7 - 20*a^ 4*b^8 + 15*a^3*b^9 - 6*a^2*b^10 + a*b^11)*d^4)) - (a^4*b - 8*a^3*b^2 + 23* a^2*b^3 - 24*a*b^4)*d)*sqrt(((a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 - a*b^5)*d^2 *sqrt((a^4 - 10*a^3*b + 41*a^2*b^2 - 80*a*b^3 + 64*b^4)/((a^7*b^5 - 6*a^6* b^6 + 15*a^5*b^7 - 20*a^4*b^8 + 15*a^3*b^9 - 6*a^2*b^10 + a*b^11)*d^4)) + a^2 - a*b - 4*b^2)/((a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 - a*b^5)*d^2))) + ((a *b^2 - b^3)*d*cos(d*x + c)^4 - 2*(a*b^2 - b^3)*d*cos(d*x + c)^2 - (a^2*b - 2*a*b^2 + b^3)*d)*sqrt(-((a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 - a*b^5)*d^2*sq rt((a^4 - 10*a^3*b + 41*a^2*b^2 - 80*a*b^3 + 64*b^4)/((a^7*b^5 - 6*a^6*b^6 + 15*a^5*b^7 - 20*a^4*b^8 + 15*a^3*b^9 - 6*a^2*b^10 + a*b^11)*d^4)) - a^2 + a*b + 4*b^2)/((a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 - a*b^5)*d^2))*log((a^3 - 9*a^2*b + 28*a*b^2 - 32*b^3)*cos(d*x + c) - (2*(a^4*b^5 - 3*a^3*b^6 + 3* a^2*b^7 - a*b^8)*d^3*sqrt((a^4 - 10*a^3*b + 41*a^2*b^2 - 80*a*b^3 + 64*...
Timed out. \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{2}} \,d x } \]
1/2*(4*b^2*cos(2*d*x + 2*c)*cos(d*x + c) + 4*b^2*sin(2*d*x + 2*c)*sin(d*x + c) - b^2*cos(d*x + c) - 4*(4*a*b + b^2)*sin(3*d*x + 3*c)*sin(2*d*x + 2*c ) - (b^2*cos(7*d*x + 7*c) + b^2*cos(d*x + c) - (4*a*b + b^2)*cos(5*d*x + 5 *c) - (4*a*b + b^2)*cos(3*d*x + 3*c))*cos(8*d*x + 8*c) + (4*b^2*cos(6*d*x + 6*c) + 4*b^2*cos(2*d*x + 2*c) - b^2 + 2*(8*a*b - 3*b^2)*cos(4*d*x + 4*c) )*cos(7*d*x + 7*c) + 4*(b^2*cos(d*x + c) - (4*a*b + b^2)*cos(5*d*x + 5*c) - (4*a*b + b^2)*cos(3*d*x + 3*c))*cos(6*d*x + 6*c) + (4*a*b + b^2 - 2*(32* a^2 - 4*a*b - 3*b^2)*cos(4*d*x + 4*c) - 4*(4*a*b + b^2)*cos(2*d*x + 2*c))* cos(5*d*x + 5*c) - 2*((32*a^2 - 4*a*b - 3*b^2)*cos(3*d*x + 3*c) - (8*a*b - 3*b^2)*cos(d*x + c))*cos(4*d*x + 4*c) + (4*a*b + b^2 - 4*(4*a*b + b^2)*co s(2*d*x + 2*c))*cos(3*d*x + 3*c) + 2*((a*b^3 - b^4)*d*cos(8*d*x + 8*c)^2 + 16*(a*b^3 - b^4)*d*cos(6*d*x + 6*c)^2 + 4*(64*a^3*b - 112*a^2*b^2 + 57*a* b^3 - 9*b^4)*d*cos(4*d*x + 4*c)^2 + 16*(a*b^3 - b^4)*d*cos(2*d*x + 2*c)^2 + (a*b^3 - b^4)*d*sin(8*d*x + 8*c)^2 + 16*(a*b^3 - b^4)*d*sin(6*d*x + 6*c) ^2 + 4*(64*a^3*b - 112*a^2*b^2 + 57*a*b^3 - 9*b^4)*d*sin(4*d*x + 4*c)^2 + 16*(8*a^2*b^2 - 11*a*b^3 + 3*b^4)*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16 *(a*b^3 - b^4)*d*sin(2*d*x + 2*c)^2 - 8*(a*b^3 - b^4)*d*cos(2*d*x + 2*c) + (a*b^3 - b^4)*d - 2*(4*(a*b^3 - b^4)*d*cos(6*d*x + 6*c) + 2*(8*a^2*b^2 - 11*a*b^3 + 3*b^4)*d*cos(4*d*x + 4*c) + 4*(a*b^3 - b^4)*d*cos(2*d*x + 2*c) - (a*b^3 - b^4)*d)*cos(8*d*x + 8*c) + 8*(2*(8*a^2*b^2 - 11*a*b^3 + 3*b^...
\[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{2}} \,d x } \]
Time = 16.69 (sec) , antiderivative size = 3839, normalized size of antiderivative = 17.69 \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \]
(cos(c + d*x)^3/(4*(a - b)) - (cos(c + d*x)*(a + b))/(4*b*(a - b)))/(d*(a - b + 2*b*cos(c + d*x)^2 - b*cos(c + d*x)^4)) - (atan(((((768*a*b^4 - 1024 *a^2*b^3 + 256*a^3*b^2)/(64*(a^2 - 2*a*b + b^2)) - (cos(c + d*x)*(-(a^2*(a ^3*b^5)^(1/2) + 8*b^2*(a^3*b^5)^(1/2) - 4*a*b^5 - a^2*b^4 + a^3*b^3 - 5*a* b*(a^3*b^5)^(1/2))/(256*(a^2*b^8 - 3*a^3*b^7 + 3*a^4*b^6 - a^5*b^5)))^(1/2 )*(256*a*b^6 - 512*a^2*b^5 + 256*a^3*b^4))/(4*(a^2 - 2*a*b + b^2)))*(-(a^2 *(a^3*b^5)^(1/2) + 8*b^2*(a^3*b^5)^(1/2) - 4*a*b^5 - a^2*b^4 + a^3*b^3 - 5 *a*b*(a^3*b^5)^(1/2))/(256*(a^2*b^8 - 3*a^3*b^7 + 3*a^4*b^6 - a^5*b^5)))^( 1/2) + (cos(c + d*x)*(a^2*b - 3*a*b^2 + 4*b^3))/(4*(a^2 - 2*a*b + b^2)))*( -(a^2*(a^3*b^5)^(1/2) + 8*b^2*(a^3*b^5)^(1/2) - 4*a*b^5 - a^2*b^4 + a^3*b^ 3 - 5*a*b*(a^3*b^5)^(1/2))/(256*(a^2*b^8 - 3*a^3*b^7 + 3*a^4*b^6 - a^5*b^5 )))^(1/2)*1i - (((768*a*b^4 - 1024*a^2*b^3 + 256*a^3*b^2)/(64*(a^2 - 2*a*b + b^2)) + (cos(c + d*x)*(-(a^2*(a^3*b^5)^(1/2) + 8*b^2*(a^3*b^5)^(1/2) - 4*a*b^5 - a^2*b^4 + a^3*b^3 - 5*a*b*(a^3*b^5)^(1/2))/(256*(a^2*b^8 - 3*a^3 *b^7 + 3*a^4*b^6 - a^5*b^5)))^(1/2)*(256*a*b^6 - 512*a^2*b^5 + 256*a^3*b^4 ))/(4*(a^2 - 2*a*b + b^2)))*(-(a^2*(a^3*b^5)^(1/2) + 8*b^2*(a^3*b^5)^(1/2) - 4*a*b^5 - a^2*b^4 + a^3*b^3 - 5*a*b*(a^3*b^5)^(1/2))/(256*(a^2*b^8 - 3* a^3*b^7 + 3*a^4*b^6 - a^5*b^5)))^(1/2) - (cos(c + d*x)*(a^2*b - 3*a*b^2 + 4*b^3))/(4*(a^2 - 2*a*b + b^2)))*(-(a^2*(a^3*b^5)^(1/2) + 8*b^2*(a^3*b^5)^ (1/2) - 4*a*b^5 - a^2*b^4 + a^3*b^3 - 5*a*b*(a^3*b^5)^(1/2))/(256*(a^2*...