Integrand size = 24, antiderivative size = 113 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}+\frac {\left (a-2 \sqrt {a} \sqrt {b}+b\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}-\frac {\sin (c+d x)}{b d} \]
-sin(d*x+c)/b/d+1/2*arctan(b^(1/4)*sin(d*x+c)/a^(1/4))*(a^(1/2)+b^(1/2))^2 /a^(3/4)/b^(5/4)/d+1/2*arctanh(b^(1/4)*sin(d*x+c)/a^(1/4))*(a+b-2*a^(1/2)* b^(1/2))/a^(3/4)/b^(5/4)/d
Result contains complex when optimal does not.
Time = 0.22 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {-\left (\sqrt {a}-\sqrt {b}\right )^2 \log \left (\sqrt [4]{a}-\sqrt [4]{b} \sin (c+d x)\right )+i \left (\left (\sqrt {a}+\sqrt {b}\right )^2 \log \left (\sqrt [4]{a}-i \sqrt [4]{b} \sin (c+d x)\right )-\left (\sqrt {a}+\sqrt {b}\right )^2 \log \left (\sqrt [4]{a}+i \sqrt [4]{b} \sin (c+d x)\right )-i \left (\sqrt {a}-\sqrt {b}\right )^2 \log \left (\sqrt [4]{a}+\sqrt [4]{b} \sin (c+d x)\right )\right )-4 a^{3/4} \sqrt [4]{b} \sin (c+d x)}{4 a^{3/4} b^{5/4} d} \]
(-((Sqrt[a] - Sqrt[b])^2*Log[a^(1/4) - b^(1/4)*Sin[c + d*x]]) + I*((Sqrt[a ] + Sqrt[b])^2*Log[a^(1/4) - I*b^(1/4)*Sin[c + d*x]] - (Sqrt[a] + Sqrt[b]) ^2*Log[a^(1/4) + I*b^(1/4)*Sin[c + d*x]] - I*(Sqrt[a] - Sqrt[b])^2*Log[a^( 1/4) + b^(1/4)*Sin[c + d*x]]) - 4*a^(3/4)*b^(1/4)*Sin[c + d*x])/(4*a^(3/4) *b^(5/4)*d)
Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3702, 1485, 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.
| \(\displaystyle \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{a-b \sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3702 |
| \(\displaystyle \frac {\int \frac {\left (1-\sin ^2(c+d x)\right )^2}{a-b \sin ^4(c+d x)}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 1485 |
| \(\displaystyle \frac {\int \left (\frac {-2 b \sin ^2(c+d x)+a+b}{b \left (a-b \sin ^4(c+d x)\right )}-\frac {1}{b}\right )d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4}}+\frac {\left (-2 \sqrt {a} \sqrt {b}+a+b\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4}}-\frac {\sin (c+d x)}{b}}{d}\) |
(((Sqrt[a] + Sqrt[b])^2*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4) *b^(5/4)) + ((a - 2*Sqrt[a]*Sqrt[b] + b)*ArcTanh[(b^(1/4)*Sin[c + d*x])/a^ (1/4)])/(2*a^(3/4)*b^(5/4)) - Sin[c + d*x]/b)/d
3.5.5.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> Int[Expa ndIntegrand[(d + e*x^2)^q/(a + c*x^4), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[q]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x _)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
Time = 1.24 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.35
| method | result | size |
| derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {\frac {\left (a +b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}+\frac {2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{b}}{d}\) | \(152\) |
| default | \(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {\frac {\left (a +b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}+\frac {2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{b}}{d}\) | \(152\) |
| risch | \(\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (256 a^{3} b^{5} d^{4} \textit {\_Z}^{4}+\left (128 a^{3} b^{3} d^{2}+128 a^{2} b^{4} d^{2}\right ) \textit {\_Z}^{2}-a^{4}+4 a^{3} b -6 a^{2} b^{2}+4 a \,b^{3}-b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {256 i a^{3} b^{4} d^{3} \textit {\_R}^{3}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\left (\frac {8 i a^{4} b d}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {120 i a^{3} b^{2} d}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {120 i a^{2} b^{3} d}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {8 i a \,b^{4} d}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}-\frac {a^{4}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {4 a^{3} b}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {10 a^{2} b^{2}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {4 a \,b^{3}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {b^{4}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}\right )\right )\) | \(512\) |
1/d*(-sin(d*x+c)/b+1/b*(1/4*(a+b)*(1/b*a)^(1/4)/a*(ln((sin(d*x+c)+(1/b*a)^ (1/4))/(sin(d*x+c)-(1/b*a)^(1/4)))+2*arctan(sin(d*x+c)/(1/b*a)^(1/4)))+1/2 /(1/b*a)^(1/4)*(2*arctan(sin(d*x+c)/(1/b*a)^(1/4))-ln((sin(d*x+c)+(1/b*a)^ (1/4))/(sin(d*x+c)-(1/b*a)^(1/4))))))
Leaf count of result is larger than twice the leaf count of optimal. 1041 vs. \(2 (85) = 170\).
Time = 0.44 (sec) , antiderivative size = 1041, normalized size of antiderivative = 9.21 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {b d \sqrt {-\frac {a b^{2} d^{2} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} + 4 \, a + 4 \, b}{a b^{2} d^{2}}} \log \left (\frac {1}{2} \, {\left (a^{4} + 4 \, a^{3} b - 10 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sin \left (d x + c\right ) + \frac {1}{2} \, {\left (2 \, a^{3} b^{4} d^{3} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} - {\left (a^{4} b + 7 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + a b^{4}\right )} d\right )} \sqrt {-\frac {a b^{2} d^{2} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} + 4 \, a + 4 \, b}{a b^{2} d^{2}}}\right ) - b d \sqrt {\frac {a b^{2} d^{2} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} - 4 \, a - 4 \, b}{a b^{2} d^{2}}} \log \left (\frac {1}{2} \, {\left (a^{4} + 4 \, a^{3} b - 10 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sin \left (d x + c\right ) + \frac {1}{2} \, {\left (2 \, a^{3} b^{4} d^{3} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} + {\left (a^{4} b + 7 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + a b^{4}\right )} d\right )} \sqrt {\frac {a b^{2} d^{2} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} - 4 \, a - 4 \, b}{a b^{2} d^{2}}}\right ) - b d \sqrt {-\frac {a b^{2} d^{2} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} + 4 \, a + 4 \, b}{a b^{2} d^{2}}} \log \left (-\frac {1}{2} \, {\left (a^{4} + 4 \, a^{3} b - 10 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sin \left (d x + c\right ) + \frac {1}{2} \, {\left (2 \, a^{3} b^{4} d^{3} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} - {\left (a^{4} b + 7 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + a b^{4}\right )} d\right )} \sqrt {-\frac {a b^{2} d^{2} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} + 4 \, a + 4 \, b}{a b^{2} d^{2}}}\right ) + b d \sqrt {\frac {a b^{2} d^{2} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} - 4 \, a - 4 \, b}{a b^{2} d^{2}}} \log \left (-\frac {1}{2} \, {\left (a^{4} + 4 \, a^{3} b - 10 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sin \left (d x + c\right ) + \frac {1}{2} \, {\left (2 \, a^{3} b^{4} d^{3} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} + {\left (a^{4} b + 7 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + a b^{4}\right )} d\right )} \sqrt {\frac {a b^{2} d^{2} \sqrt {\frac {a^{4} + 12 \, a^{3} b + 38 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}}{a^{3} b^{5} d^{4}}} - 4 \, a - 4 \, b}{a b^{2} d^{2}}}\right ) + 4 \, \sin \left (d x + c\right )}{4 \, b d} \]
-1/4*(b*d*sqrt(-(a*b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + 4*a + 4*b)/(a*b^2*d^2))*log(1/2*(a^4 + 4*a^3*b - 10* a^2*b^2 + 4*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*a^3*b^4*d^3*sqrt((a^4 + 12* a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) - (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt(-(a*b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + 4*a + 4*b)/(a*b^2*d^2))) - b*d*sqrt((a* b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) - 4*a - 4*b)/(a*b^2*d^2))*log(1/2*(a^4 + 4*a^3*b - 10*a^2*b^2 + 4*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*a^3*b^4*d^3*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4) *d)*sqrt((a*b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a ^3*b^5*d^4)) - 4*a - 4*b)/(a*b^2*d^2))) - b*d*sqrt(-(a*b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + 4*a + 4*b)/(a*b^ 2*d^2))*log(-1/2*(a^4 + 4*a^3*b - 10*a^2*b^2 + 4*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*a^3*b^4*d^3*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/ (a^3*b^5*d^4)) - (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt(-(a*b^2*d ^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + 4* a + 4*b)/(a*b^2*d^2))) + b*d*sqrt((a*b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2 *b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) - 4*a - 4*b)/(a*b^2*d^2))*log(-1/2*( a^4 + 4*a^3*b - 10*a^2*b^2 + 4*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*a^3*b...
Timed out. \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {\frac {2 \, {\left (b {\left (2 \, \sqrt {a} + \sqrt {b}\right )} + a \sqrt {b}\right )} \arctan \left (\frac {\sqrt {b} \sin \left (d x + c\right )}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {{\left (b {\left (2 \, \sqrt {a} - \sqrt {b}\right )} - a \sqrt {b}\right )} \log \left (\frac {\sqrt {b} \sin \left (d x + c\right ) - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} \sin \left (d x + c\right ) + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}}}{b} - \frac {4 \, \sin \left (d x + c\right )}{b}}{4 \, d} \]
1/4*((2*(b*(2*sqrt(a) + sqrt(b)) + a*sqrt(b))*arctan(sqrt(b)*sin(d*x + c)/ sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + (b*(2*sqr t(a) - sqrt(b)) - a*sqrt(b))*log((sqrt(b)*sin(d*x + c) - sqrt(sqrt(a)*sqrt (b)))/(sqrt(b)*sin(d*x + c) + sqrt(sqrt(a)*sqrt(b))))/(sqrt(a)*sqrt(sqrt(a )*sqrt(b))*sqrt(b)))/b - 4*sin(d*x + c)/b)/d
Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (85) = 170\).
Time = 0.64 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.75 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\frac {8 \, \sin \left (d x + c\right )}{b} - \frac {2 \, \sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} - 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{a b^{3}} - \frac {2 \, \sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} - 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{a b^{3}} - \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} + 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{a b^{3}} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} + 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{a b^{3}}}{8 \, d} \]
-1/8*(8*sin(d*x + c)/b - 2*sqrt(2)*((-a*b^3)^(1/4)*(a*b + b^2) - 2*(-a*b^3 )^(3/4))*arctan(1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) + 2*sin(d*x + c))/(-a/b) ^(1/4))/(a*b^3) - 2*sqrt(2)*((-a*b^3)^(1/4)*(a*b + b^2) - 2*(-a*b^3)^(3/4) )*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) - 2*sin(d*x + c))/(-a/b)^(1/4) )/(a*b^3) - sqrt(2)*((-a*b^3)^(1/4)*(a*b + b^2) + 2*(-a*b^3)^(3/4))*log(si n(d*x + c)^2 + sqrt(2)*(-a/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/(a*b^3) + s qrt(2)*((-a*b^3)^(1/4)*(a*b + b^2) + 2*(-a*b^3)^(3/4))*log(sin(d*x + c)^2 - sqrt(2)*(-a/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/(a*b^3))/d
Time = 14.40 (sec) , antiderivative size = 1097, normalized size of antiderivative = 9.71 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {8\,b^3\,\sin \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^3\,b^5}}{16\,a\,b^5}-\frac {1}{4\,a\,b}-\frac {1}{4\,b^2}+\frac {3\,\sqrt {a^3\,b^5}}{8\,a^2\,b^4}+\frac {\sqrt {a^3\,b^5}}{16\,a^3\,b^3}}}{\frac {2\,\sqrt {a^3\,b^5}}{a^2}-24\,a\,b+\frac {14\,\sqrt {a^3\,b^5}}{b^2}-4\,a^2-4\,b^2+\frac {14\,\sqrt {a^3\,b^5}}{a\,b}+\frac {2\,a\,\sqrt {a^3\,b^5}}{b^3}}+\frac {48\,a\,b^2\,\sin \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^3\,b^5}}{16\,a\,b^5}-\frac {1}{4\,a\,b}-\frac {1}{4\,b^2}+\frac {3\,\sqrt {a^3\,b^5}}{8\,a^2\,b^4}+\frac {\sqrt {a^3\,b^5}}{16\,a^3\,b^3}}}{\frac {2\,\sqrt {a^3\,b^5}}{a^2}-24\,a\,b+\frac {14\,\sqrt {a^3\,b^5}}{b^2}-4\,a^2-4\,b^2+\frac {14\,\sqrt {a^3\,b^5}}{a\,b}+\frac {2\,a\,\sqrt {a^3\,b^5}}{b^3}}+\frac {8\,a^2\,b\,\sin \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^3\,b^5}}{16\,a\,b^5}-\frac {1}{4\,a\,b}-\frac {1}{4\,b^2}+\frac {3\,\sqrt {a^3\,b^5}}{8\,a^2\,b^4}+\frac {\sqrt {a^3\,b^5}}{16\,a^3\,b^3}}}{\frac {2\,\sqrt {a^3\,b^5}}{a^2}-24\,a\,b+\frac {14\,\sqrt {a^3\,b^5}}{b^2}-4\,a^2-4\,b^2+\frac {14\,\sqrt {a^3\,b^5}}{a\,b}+\frac {2\,a\,\sqrt {a^3\,b^5}}{b^3}}\right )\,\sqrt {\frac {a^2\,\sqrt {a^3\,b^5}+b^2\,\sqrt {a^3\,b^5}-4\,a^2\,b^4-4\,a^3\,b^3+6\,a\,b\,\sqrt {a^3\,b^5}}{16\,a^3\,b^5}}}{d}-\frac {2\,\mathrm {atanh}\left (\frac {8\,b^3\,\sin \left (c+d\,x\right )\,\sqrt {-\frac {1}{4\,b^2}-\frac {1}{4\,a\,b}-\frac {\sqrt {a^3\,b^5}}{16\,a\,b^5}-\frac {3\,\sqrt {a^3\,b^5}}{8\,a^2\,b^4}-\frac {\sqrt {a^3\,b^5}}{16\,a^3\,b^3}}}{24\,a\,b+\frac {2\,\sqrt {a^3\,b^5}}{a^2}+\frac {14\,\sqrt {a^3\,b^5}}{b^2}+4\,a^2+4\,b^2+\frac {14\,\sqrt {a^3\,b^5}}{a\,b}+\frac {2\,a\,\sqrt {a^3\,b^5}}{b^3}}+\frac {48\,a\,b^2\,\sin \left (c+d\,x\right )\,\sqrt {-\frac {1}{4\,b^2}-\frac {1}{4\,a\,b}-\frac {\sqrt {a^3\,b^5}}{16\,a\,b^5}-\frac {3\,\sqrt {a^3\,b^5}}{8\,a^2\,b^4}-\frac {\sqrt {a^3\,b^5}}{16\,a^3\,b^3}}}{24\,a\,b+\frac {2\,\sqrt {a^3\,b^5}}{a^2}+\frac {14\,\sqrt {a^3\,b^5}}{b^2}+4\,a^2+4\,b^2+\frac {14\,\sqrt {a^3\,b^5}}{a\,b}+\frac {2\,a\,\sqrt {a^3\,b^5}}{b^3}}+\frac {8\,a^2\,b\,\sin \left (c+d\,x\right )\,\sqrt {-\frac {1}{4\,b^2}-\frac {1}{4\,a\,b}-\frac {\sqrt {a^3\,b^5}}{16\,a\,b^5}-\frac {3\,\sqrt {a^3\,b^5}}{8\,a^2\,b^4}-\frac {\sqrt {a^3\,b^5}}{16\,a^3\,b^3}}}{24\,a\,b+\frac {2\,\sqrt {a^3\,b^5}}{a^2}+\frac {14\,\sqrt {a^3\,b^5}}{b^2}+4\,a^2+4\,b^2+\frac {14\,\sqrt {a^3\,b^5}}{a\,b}+\frac {2\,a\,\sqrt {a^3\,b^5}}{b^3}}\right )\,\sqrt {-\frac {a^2\,\sqrt {a^3\,b^5}+b^2\,\sqrt {a^3\,b^5}+4\,a^2\,b^4+4\,a^3\,b^3+6\,a\,b\,\sqrt {a^3\,b^5}}{16\,a^3\,b^5}}}{d}-\frac {\sin \left (c+d\,x\right )}{b\,d} \]
(2*atanh((8*b^3*sin(c + d*x)*((a^3*b^5)^(1/2)/(16*a*b^5) - 1/(4*a*b) - 1/( 4*b^2) + (3*(a^3*b^5)^(1/2))/(8*a^2*b^4) + (a^3*b^5)^(1/2)/(16*a^3*b^3))^( 1/2))/((2*(a^3*b^5)^(1/2))/a^2 - 24*a*b + (14*(a^3*b^5)^(1/2))/b^2 - 4*a^2 - 4*b^2 + (14*(a^3*b^5)^(1/2))/(a*b) + (2*a*(a^3*b^5)^(1/2))/b^3) + (48*a *b^2*sin(c + d*x)*((a^3*b^5)^(1/2)/(16*a*b^5) - 1/(4*a*b) - 1/(4*b^2) + (3 *(a^3*b^5)^(1/2))/(8*a^2*b^4) + (a^3*b^5)^(1/2)/(16*a^3*b^3))^(1/2))/((2*( a^3*b^5)^(1/2))/a^2 - 24*a*b + (14*(a^3*b^5)^(1/2))/b^2 - 4*a^2 - 4*b^2 + (14*(a^3*b^5)^(1/2))/(a*b) + (2*a*(a^3*b^5)^(1/2))/b^3) + (8*a^2*b*sin(c + d*x)*((a^3*b^5)^(1/2)/(16*a*b^5) - 1/(4*a*b) - 1/(4*b^2) + (3*(a^3*b^5)^( 1/2))/(8*a^2*b^4) + (a^3*b^5)^(1/2)/(16*a^3*b^3))^(1/2))/((2*(a^3*b^5)^(1/ 2))/a^2 - 24*a*b + (14*(a^3*b^5)^(1/2))/b^2 - 4*a^2 - 4*b^2 + (14*(a^3*b^5 )^(1/2))/(a*b) + (2*a*(a^3*b^5)^(1/2))/b^3))*((a^2*(a^3*b^5)^(1/2) + b^2*( a^3*b^5)^(1/2) - 4*a^2*b^4 - 4*a^3*b^3 + 6*a*b*(a^3*b^5)^(1/2))/(16*a^3*b^ 5))^(1/2))/d - (2*atanh((8*b^3*sin(c + d*x)*(- 1/(4*b^2) - 1/(4*a*b) - (a^ 3*b^5)^(1/2)/(16*a*b^5) - (3*(a^3*b^5)^(1/2))/(8*a^2*b^4) - (a^3*b^5)^(1/2 )/(16*a^3*b^3))^(1/2))/(24*a*b + (2*(a^3*b^5)^(1/2))/a^2 + (14*(a^3*b^5)^( 1/2))/b^2 + 4*a^2 + 4*b^2 + (14*(a^3*b^5)^(1/2))/(a*b) + (2*a*(a^3*b^5)^(1 /2))/b^3) + (48*a*b^2*sin(c + d*x)*(- 1/(4*b^2) - 1/(4*a*b) - (a^3*b^5)^(1 /2)/(16*a*b^5) - (3*(a^3*b^5)^(1/2))/(8*a^2*b^4) - (a^3*b^5)^(1/2)/(16*a^3 *b^3))^(1/2))/(24*a*b + (2*(a^3*b^5)^(1/2))/a^2 + (14*(a^3*b^5)^(1/2))/...