\(\int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [338]

Optimal result

Integrand size = 24, antiderivative size = 113 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}+\frac {\left (a-2 \sqrt {a} \sqrt {b}+b\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}-\frac {\sin (c+d x)}{b d} \] Output:

1/2*(a^(1/2)+b^(1/2))^2*arctan(b^(1/4)*sin(d*x+c)/a^(1/4))/a^(3/4)/b^(5/4) 
/d+1/2*(a-2*a^(1/2)*b^(1/2)+b)*arctanh(b^(1/4)*sin(d*x+c)/a^(1/4))/a^(3/4) 
/b^(5/4)/d-sin(d*x+c)/b/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {-\left (\sqrt {a}-\sqrt {b}\right )^2 \log \left (\sqrt [4]{a}-\sqrt [4]{b} \sin (c+d x)\right )+i \left (\left (\sqrt {a}+\sqrt {b}\right )^2 \log \left (\sqrt [4]{a}-i \sqrt [4]{b} \sin (c+d x)\right )-\left (\sqrt {a}+\sqrt {b}\right )^2 \log \left (\sqrt [4]{a}+i \sqrt [4]{b} \sin (c+d x)\right )-i \left (\sqrt {a}-\sqrt {b}\right )^2 \log \left (\sqrt [4]{a}+\sqrt [4]{b} \sin (c+d x)\right )\right )-4 a^{3/4} \sqrt [4]{b} \sin (c+d x)}{4 a^{3/4} b^{5/4} d} \] Input:

Integrate[Cos[c + d*x]^5/(a - b*Sin[c + d*x]^4),x]
 

Output:

(-((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)
 

Rubi [A] (verified)

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

\(\displaystyle \frac {\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4}}+\frac {\left (-2 \sqrt {a} \sqrt {b}+a+b\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4}}-\frac {\sin (c+d x)}{b}}{d}\)

Input:

Int[Cos[c + d*x]^5/(a - b*Sin[c + d*x]^4),x]
 

Output:

(((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
 

Defintions of rubi rules used

rule 1485

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3702

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])
 

Maple [A] (verified)

Time = 1.62 (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	\(\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (256 a^{3} b^{5} d^{4} \textit {\_Z}^{4}+\left (128 a^{3} b^{3} d^{2}+128 a^{2} b^{4} d^{2}\right ) \textit {\_Z}^{2}-a^{4}+4 a^{3} b -6 a^{2} b^{2}+4 a \,b^{3}-b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {256 i a^{3} b^{4} d^{3} \textit {\_R}^{3}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\left (\frac {8 i a^{4} b d}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {120 i a^{3} b^{2} d}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {120 i a^{2} b^{3} d}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {8 i a \,b^{4} d}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}-\frac {a^{4}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {4 a^{3} b}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {10 a^{2} b^{2}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {4 a \,b^{3}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {b^{4}}{a^{4}+4 a^{3} b -10 a^{2} b^{2}+4 a \,b^{3}+b^{4}}\right )\right )\)	\(512\)

Input:

int(cos(d*x+c)^5/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/b*sin(d*x+c)+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))))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1041 vs. \(2 (85) = 170\).

Time = 0.20 (sec) , antiderivative size = 1041, normalized size of antiderivative = 9.21 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
 

Output:

-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...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5/(a-b*sin(d*x+c)**4),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {\frac {2 \, {\left (b {\left (2 \, \sqrt {a} + \sqrt {b}\right )} + a \sqrt {b}\right )} \arctan \left (\frac {\sqrt {b} \sin \left (d x + c\right )}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {{\left (b {\left (2 \, \sqrt {a} - \sqrt {b}\right )} - a \sqrt {b}\right )} \log \left (\frac {\sqrt {b} \sin \left (d x + c\right ) - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} \sin \left (d x + c\right ) + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}}}{b} - \frac {4 \, \sin \left (d x + c\right )}{b}}{4 \, d} \] Input:

integrate(cos(d*x+c)^5/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
 

Output:

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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (85) = 170\).

Time = 0.42 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.85 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\sin \left (d x + c\right )}{b d} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} - 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3} d} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} - 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3} d} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} + 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{8 \, a b^{3} d} - \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} + 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{8 \, a b^{3} d} \] Input:

integrate(cos(d*x+c)^5/(a-b*sin(d*x+c)^4),x, algorithm="giac")
 

Output:

-sin(d*x + c)/(b*d) + 1/4*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*d) + 1/4*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*d) + 1/8*sqrt(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) - 1/8*sqrt(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)
 

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 1097, normalized size of antiderivative = 9.71 \[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:

int(cos(c + d*x)^5/(a - b*sin(c + d*x)^4),x)
 

Output:

(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))/...
 

Reduce [F]

\[ \int \frac {\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\left (\int \frac {\cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )^{4} b -a}d x \right ) \] Input:

int(cos(d*x+c)^5/(a-b*sin(d*x+c)^4),x)
 

Output:

 - int(cos(c + d*x)**5/(sin(c + d*x)**4*b - a),x)