\(\int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [344]

Optimal result

Integrand size = 24, antiderivative size = 155 \[ \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {5 x}{2 b}-\frac {\left (\sqrt {a}-\sqrt {b}\right )^{5/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/2} d}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^{5/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/2} d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \] Output:

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

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {-10 b (c+d x)+\frac {2 \left (\sqrt {a}+\sqrt {b}\right )^3 \sqrt {b} \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {2 \left (\sqrt {a}-\sqrt {b}\right )^3 \sqrt {b} \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}-b \sin (2 (c+d x))}{4 b^2 d} \] Input:

Integrate[Cos[c + d*x]^6/(a - b*Sin[c + d*x]^4),x]
 

Output:

(-10*b*(c + d*x) + (2*(Sqrt[a] + Sqrt[b])^3*Sqrt[b]*ArcTan[((Sqrt[a] + Sqr 
t[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt[a]* 
Sqrt[b]]) + (2*(Sqrt[a] - Sqrt[b])^3*Sqrt[b]*ArcTanh[((Sqrt[a] - Sqrt[b])* 
Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt 
[b]]) - b*Sin[2*(c + d*x)])/(4*b^2*d)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3703, 1484, 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.

\(\displaystyle \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^6}{a-b \sin (c+d x)^4}dx\)

\(\Big \downarrow \) 3703

\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right )^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 1484

\(\displaystyle \frac {\int \left (\frac {2 (a-b) \tan ^2(c+d x)+3 a+b}{b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {2}{b \left (\tan ^2(c+d x)+1\right )}-\frac {1}{b \left (\tan ^2(c+d x)+1\right )^2}\right )d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {\left (\sqrt {a}-\sqrt {b}\right )^{5/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/2}}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^{5/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/2}}-\frac {5 \arctan (\tan (c+d x))}{2 b}-\frac {\tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}}{d}\)

Input:

Int[Cos[c + d*x]^6/(a - b*Sin[c + d*x]^4),x]
 

Output:

((-5*ArcTan[Tan[c + d*x]])/(2*b) - ((Sqrt[a] - Sqrt[b])^(5/2)*ArcTan[(Sqrt 
[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(3/2)) + ((Sqrt[a 
] + Sqrt[b])^(5/2)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)]) 
/(2*a^(3/4)*b^(3/2)) - Tan[c + d*x]/(2*b*(1 + Tan[c + d*x]^2)))/d
 

Defintions of rubi rules used

rule 1484

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb 
ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + 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 3703

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Su 
bst[Int[(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]
 

Maple [A] (verified)

Time = 2.17 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {\left (a -b \right ) \left (\frac {\left (a +b +2 \sqrt {a b}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (-a -b +2 \sqrt {a b}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{b}-\frac {\frac {\tan \left (d x +c \right )}{2 \tan \left (d x +c \right )^{2}+2}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{2}}{b}}{d}\)	\(179\)
default	\(\frac {\frac {\left (a -b \right ) \left (\frac {\left (a +b +2 \sqrt {a b}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (-a -b +2 \sqrt {a b}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{b}-\frac {\frac {\tan \left (d x +c \right )}{2 \tan \left (d x +c \right )^{2}+2}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{2}}{b}}{d}\)	\(179\)
risch	\(-\frac {5 x}{2 b}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (256 a^{3} b^{6} d^{4} \textit {\_Z}^{4}+\left (32 a^{4} b^{3} d^{2}+320 a^{3} b^{4} d^{2}+160 a^{2} b^{5} d^{2}\right ) \textit {\_Z}^{2}+a^{5}-5 a^{4} b +10 a^{3} b^{2}-10 a^{2} b^{3}+5 a \,b^{4}-b^{5}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {256 i a^{3} b^{5} d^{3} \textit {\_R}^{3}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}+\left (\frac {32 a^{4} b^{3} d^{2}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}-\frac {64 a^{3} b^{4} d^{2}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}+\frac {32 a^{2} b^{5} d^{2}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}\right ) \textit {\_R}^{2}+\left (\frac {56 i a^{4} b^{2} d}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}+\frac {280 i a^{3} b^{3} d}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}+\frac {168 i a^{2} b^{4} d}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}+\frac {8 i d \,b^{5} a}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}\right ) \textit {\_R} +\frac {2 a^{5}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}+\frac {11 a^{4} b}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}-\frac {28 a^{3} b^{2}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}+\frac {14 a^{2} b^{3}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}+\frac {2 a \,b^{4}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}-\frac {b^{5}}{5 a^{4} b -14 a^{2} b^{3}+8 a \,b^{4}+b^{5}}\right )\right )\)	\(646\)

Input:

int(cos(d*x+c)^6/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/b*(a-b)*(1/2*(a+b+2*(a*b)^(1/2))/(a*b)^(1/2)/(((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*(-a-b 
+2*(a*b)^(1/2))/(a*b)^(1/2)/(((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)))-1/b*(1/2*tan(d*x+c)/(tan(d*x+c)^2+ 
1)+5/2*arctan(tan(d*x+c))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1751 vs. \(2 (111) = 222\).

Time = 0.44 (sec) , antiderivative size = 1751, normalized size of antiderivative = 11.30 \[ \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^6/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
 

Output:

1/8*(b*d*sqrt((a*b^3*d^2*sqrt((25*a^4 + 100*a^3*b + 110*a^2*b^2 + 20*a*b^3 
 + b^4)/(a^3*b^5*d^4)) - a^2 - 10*a*b - 5*b^2)/(a*b^3*d^2))*log(5/4*a^4 - 
7/2*a^2*b^2 + 2*a*b^3 + 1/4*b^4 - 1/4*(5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4) 
*cos(d*x + c)^2 + 1/2*(2*a^3*b^4*d^3*sqrt((25*a^4 + 100*a^3*b + 110*a^2*b^ 
2 + 20*a*b^3 + b^4)/(a^3*b^5*d^4))*cos(d*x + c)*sin(d*x + c) + (5*a^4*b + 
15*a^3*b^2 + 11*a^2*b^3 + a*b^4)*d*cos(d*x + c)*sin(d*x + c))*sqrt((a*b^3* 
d^2*sqrt((25*a^4 + 100*a^3*b + 110*a^2*b^2 + 20*a*b^3 + b^4)/(a^3*b^5*d^4) 
) - a^2 - 10*a*b - 5*b^2)/(a*b^3*d^2)) + 1/4*(2*(a^4*b^2 - 2*a^3*b^3 + a^2 
*b^4)*d^2*cos(d*x + c)^2 - (a^4*b^2 - 2*a^3*b^3 + a^2*b^4)*d^2)*sqrt((25*a 
^4 + 100*a^3*b + 110*a^2*b^2 + 20*a*b^3 + b^4)/(a^3*b^5*d^4))) - b*d*sqrt( 
(a*b^3*d^2*sqrt((25*a^4 + 100*a^3*b + 110*a^2*b^2 + 20*a*b^3 + b^4)/(a^3*b 
^5*d^4)) - a^2 - 10*a*b - 5*b^2)/(a*b^3*d^2))*log(5/4*a^4 - 7/2*a^2*b^2 + 
2*a*b^3 + 1/4*b^4 - 1/4*(5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4)*cos(d*x + c)^ 
2 - 1/2*(2*a^3*b^4*d^3*sqrt((25*a^4 + 100*a^3*b + 110*a^2*b^2 + 20*a*b^3 + 
 b^4)/(a^3*b^5*d^4))*cos(d*x + c)*sin(d*x + c) + (5*a^4*b + 15*a^3*b^2 + 1 
1*a^2*b^3 + a*b^4)*d*cos(d*x + c)*sin(d*x + c))*sqrt((a*b^3*d^2*sqrt((25*a 
^4 + 100*a^3*b + 110*a^2*b^2 + 20*a*b^3 + b^4)/(a^3*b^5*d^4)) - a^2 - 10*a 
*b - 5*b^2)/(a*b^3*d^2)) + 1/4*(2*(a^4*b^2 - 2*a^3*b^3 + a^2*b^4)*d^2*cos( 
d*x + c)^2 - (a^4*b^2 - 2*a^3*b^3 + a^2*b^4)*d^2)*sqrt((25*a^4 + 100*a^3*b 
 + 110*a^2*b^2 + 20*a*b^3 + b^4)/(a^3*b^5*d^4))) + b*d*sqrt(-(a*b^3*d^2...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**6/(a-b*sin(d*x+c)**4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\cos \left (d x + c\right )^{6}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:

integrate(cos(d*x+c)^6/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
 

Output:

-1/4*(4*b*d*integrate(-4*(4*(a*b + 3*b^2)*cos(6*d*x + 6*c)^2 + 4*(40*a^2 - 
 23*a*b + 3*b^2)*cos(4*d*x + 4*c)^2 + 4*(a*b + 3*b^2)*cos(2*d*x + 2*c)^2 + 
 4*(a*b + 3*b^2)*sin(6*d*x + 6*c)^2 + 4*(40*a^2 - 23*a*b + 3*b^2)*sin(4*d* 
x + 4*c)^2 + 2*(8*a^2 + 41*a*b - 13*b^2)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) 
 + 4*(a*b + 3*b^2)*sin(2*d*x + 2*c)^2 - ((a*b + 3*b^2)*cos(6*d*x + 6*c) + 
2*(5*a*b - b^2)*cos(4*d*x + 4*c) + (a*b + 3*b^2)*cos(2*d*x + 2*c))*cos(8*d 
*x + 8*c) - (a*b + 3*b^2 - 2*(8*a^2 + 41*a*b - 13*b^2)*cos(4*d*x + 4*c) - 
8*(a*b + 3*b^2)*cos(2*d*x + 2*c))*cos(6*d*x + 6*c) - 2*(5*a*b - b^2 - (8*a 
^2 + 41*a*b - 13*b^2)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - (a*b + 3*b^2)*c 
os(2*d*x + 2*c) - ((a*b + 3*b^2)*sin(6*d*x + 6*c) + 2*(5*a*b - b^2)*sin(4* 
d*x + 4*c) + (a*b + 3*b^2)*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 2*((8*a^2 
+ 41*a*b - 13*b^2)*sin(4*d*x + 4*c) + 4*(a*b + 3*b^2)*sin(2*d*x + 2*c))*si 
n(6*d*x + 6*c))/(b^3*cos(8*d*x + 8*c)^2 + 16*b^3*cos(6*d*x + 6*c)^2 + 16*b 
^3*cos(2*d*x + 2*c)^2 + b^3*sin(8*d*x + 8*c)^2 + 16*b^3*sin(6*d*x + 6*c)^2 
 + 16*b^3*sin(2*d*x + 2*c)^2 - 8*b^3*cos(2*d*x + 2*c) + b^3 + 4*(64*a^2*b 
- 48*a*b^2 + 9*b^3)*cos(4*d*x + 4*c)^2 + 4*(64*a^2*b - 48*a*b^2 + 9*b^3)*s 
in(4*d*x + 4*c)^2 + 16*(8*a*b^2 - 3*b^3)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) 
 - 2*(4*b^3*cos(6*d*x + 6*c) + 4*b^3*cos(2*d*x + 2*c) - b^3 + 2*(8*a*b^2 - 
 3*b^3)*cos(4*d*x + 4*c))*cos(8*d*x + 8*c) + 8*(4*b^3*cos(2*d*x + 2*c) - b 
^3 + 2*(8*a*b^2 - 3*b^3)*cos(4*d*x + 4*c))*cos(6*d*x + 6*c) - 4*(8*a*b^...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 995 vs. \(2 (111) = 222\).

Time = 0.75 (sec) , antiderivative size = 995, normalized size of antiderivative = 6.42 \[ \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^6/(a-b*sin(d*x+c)^4),x, algorithm="giac")
 

Output:

-1/2*(5*(d*x + c)/b + (2*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)* 
a^2 - 6*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b - sqrt(a^2 - a*b 
 + sqrt(a*b)*(a - b))*sqrt(a*b)*b^2)*b^2*abs(-a + b) - (9*sqrt(a^2 - a*b + 
 sqrt(a*b)*(a - b))*a^3*b - 15*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^2 
 - 9*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^3 - sqrt(a^2 - a*b + sqrt(a*b 
)*(a - b))*b^4)*abs(-a + b)*abs(b) + (3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b) 
)*sqrt(a*b)*a^3*b - 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^ 
2 - 7*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^3 - sqrt(a^2 - a*b 
 + sqrt(a*b)*(a - b))*sqrt(a*b)*b^4)*abs(-a + b))*(pi*floor((d*x + c)/pi + 
 1/2) + arctan(tan(d*x + c)/sqrt((a*b + sqrt(a^2*b^2 - (a*b - b^2)*a*b))/( 
a*b - b^2))))/((3*a^5*b^2 - 12*a^4*b^3 + 14*a^3*b^4 - 4*a^2*b^5 - a*b^6)*a 
bs(b)) - (2*(3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2 - 6*sqrt( 
a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a*b - sqrt(a^2 - a*b - sqrt(a*b)* 
(a - b))*sqrt(a*b)*b^2)*b^2*abs(-a + b) + (9*sqrt(a^2 - a*b - sqrt(a*b)*(a 
 - b))*a^3*b - 15*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b^2 - 9*sqrt(a^2 
 - a*b - sqrt(a*b)*(a - b))*a*b^3 - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*b^ 
4)*abs(-a + b)*abs(b) + (3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a 
^3*b - 3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 - 7*sqrt(a^ 
2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^3 - sqrt(a^2 - a*b - sqrt(a*b)* 
(a - b))*sqrt(a*b)*b^4)*abs(-a + b))*(pi*floor((d*x + c)/pi + 1/2) + ar...
 

Mupad [B] (verification not implemented)

Time = 37.85 (sec) , antiderivative size = 3088, normalized size of antiderivative = 19.92 \[ \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^6/(a - b*sin(c + d*x)^4),x)
 

Output:

(atan((a^4*b^8*sin(c + d*x)*(-(5*a^2*(a^3*b^7)^(1/2) + b^2*(a^3*b^7)^(1/2) 
 + 5*a^2*b^5 + 10*a^3*b^4 + a^4*b^3 + 10*a*b*(a^3*b^7)^(1/2))/(16*a^3*b^6) 
)^(1/2)*240i - a^3*b^9*sin(c + d*x)*(-(5*a^2*(a^3*b^7)^(1/2) + b^2*(a^3*b^ 
7)^(1/2) + 5*a^2*b^5 + 10*a^3*b^4 + a^4*b^3 + 10*a*b*(a^3*b^7)^(1/2))/(16* 
a^3*b^6))^(1/2)*108i - a^5*b^7*sin(c + d*x)*(-(5*a^2*(a^3*b^7)^(1/2) + b^2 
*(a^3*b^7)^(1/2) + 5*a^2*b^5 + 10*a^3*b^4 + a^4*b^3 + 10*a*b*(a^3*b^7)^(1/ 
2))/(16*a^3*b^6))^(1/2)*80i - a^6*b^6*sin(c + d*x)*(-(5*a^2*(a^3*b^7)^(1/2 
) + b^2*(a^3*b^7)^(1/2) + 5*a^2*b^5 + 10*a^3*b^4 + a^4*b^3 + 10*a*b*(a^3*b 
^7)^(1/2))/(16*a^3*b^6))^(1/2)*120i + a^7*b^5*sin(c + d*x)*(-(5*a^2*(a^3*b 
^7)^(1/2) + b^2*(a^3*b^7)^(1/2) + 5*a^2*b^5 + 10*a^3*b^4 + a^4*b^3 + 10*a* 
b*(a^3*b^7)^(1/2))/(16*a^3*b^6))^(1/2)*60i + a^8*b^4*sin(c + d*x)*(-(5*a^2 
*(a^3*b^7)^(1/2) + b^2*(a^3*b^7)^(1/2) + 5*a^2*b^5 + 10*a^3*b^4 + a^4*b^3 
+ 10*a*b*(a^3*b^7)^(1/2))/(16*a^3*b^6))^(1/2)*8i - a^3*b^11*sin(c + d*x)*( 
-(5*a^2*(a^3*b^7)^(1/2) + b^2*(a^3*b^7)^(1/2) + 5*a^2*b^5 + 10*a^3*b^4 + a 
^4*b^3 + 10*a*b*(a^3*b^7)^(1/2))/(16*a^3*b^6))^(3/2)*64i + a^4*b^10*sin(c 
+ d*x)*(-(5*a^2*(a^3*b^7)^(1/2) + b^2*(a^3*b^7)^(1/2) + 5*a^2*b^5 + 10*a^3 
*b^4 + a^4*b^3 + 10*a*b*(a^3*b^7)^(1/2))/(16*a^3*b^6))^(3/2)*128i + a^5*b^ 
9*sin(c + d*x)*(-(5*a^2*(a^3*b^7)^(1/2) + b^2*(a^3*b^7)^(1/2) + 5*a^2*b^5 
+ 10*a^3*b^4 + a^4*b^3 + 10*a*b*(a^3*b^7)^(1/2))/(16*a^3*b^6))^(3/2)*6080i 
 + a^6*b^8*sin(c + d*x)*(-(5*a^2*(a^3*b^7)^(1/2) + b^2*(a^3*b^7)^(1/2) ...
 

Reduce [F]

\[ \int \frac {\cos ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\left (\int \frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )^{4} b -a}d x \right ) \] Input:

int(cos(d*x+c)^6/(a-b*sin(d*x+c)^4),x)
 

Output:

 - int(cos(c + d*x)**6/(sin(c + d*x)**4*b - a),x)