\(\int \frac {\cos ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [345]

Optimal result

Integrand size = 24, antiderivative size = 127 \[ \int \frac {\cos ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {x}{b}+\frac {\left (\sqrt {a}-\sqrt {b}\right )^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b d}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {-2 (c+d x)+\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}}{2 b d} \] Input:

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

Output:

(-2*(c + d*x) + ((Sqrt[a] + Sqrt[b])^2*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + 
 d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]]) - ( 
(Sqrt[a] - Sqrt[b])^2*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + 
 Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]))/(2*b*d)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.03, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3703

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

\(\Big \downarrow \) 1484

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-(ArcTan[Tan[c + d*x]]/b) + ((Sqrt[a] - Sqrt[b])^(3/2)*ArcTan[(Sqrt[Sqrt[ 
a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*b) + ((Sqrt[a] + Sqrt[b]) 
^(3/2)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)* 
b))/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 = 1.00 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{b}+\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{b}}{d}\)	\(148\)
default	\(\frac {-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{b}+\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{b}}{d}\)	\(148\)
risch	\(-\frac {x}{b}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (256 a^{3} b^{4} d^{4} \textit {\_Z}^{4}+\left (32 a^{3} b^{2} d^{2}+96 d^{2} b^{3} a^{2}\right ) \textit {\_Z}^{2}+a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {128 i a^{3} b^{3} d^{3} \textit {\_R}^{3}}{3 a^{2} b -2 b^{2} a -b^{3}}+\left (\frac {32 a^{3} b^{2} d^{2}}{3 a^{2} b -2 b^{2} a -b^{3}}-\frac {32 a^{2} b^{3} d^{2}}{3 a^{2} b -2 b^{2} a -b^{3}}\right ) \textit {\_R}^{2}+\left (-\frac {8 i a^{3} b d}{3 a^{2} b -2 b^{2} a -b^{3}}-\frac {48 i a^{2} b^{2} d}{3 a^{2} b -2 b^{2} a -b^{3}}-\frac {8 i d \,b^{3} a}{3 a^{2} b -2 b^{2} a -b^{3}}\right ) \textit {\_R} +\frac {2 a^{3}}{3 a^{2} b -2 b^{2} a -b^{3}}+\frac {a^{2} b}{3 a^{2} b -2 b^{2} a -b^{3}}-\frac {4 b^{2} a}{3 a^{2} b -2 b^{2} a -b^{3}}+\frac {b^{3}}{3 a^{2} b -2 b^{2} a -b^{3}}\right )\right )\)	\(378\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1197 vs. \(2 (91) = 182\).

Time = 0.28 (sec) , antiderivative size = 1197, normalized size of antiderivative = 9.43 \[ \int \frac {\cos ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*(b*sqrt((a*b^2*d^2*sqrt((9*a^2 + 6*a*b + b^2)/(a^3*b^3*d^4)) - a - 3*b 
)/(a*b^2*d^2))*log(1/4*(3*a^2 - 2*a*b - b^2)*cos(d*x + c)^2 - 3/4*a^2 + 1/ 
2*a*b + 1/4*b^2 + 1/2*(a^3*b^2*d^3*sqrt((9*a^2 + 6*a*b + b^2)/(a^3*b^3*d^4 
))*cos(d*x + c)*sin(d*x + c) + (3*a^2*b + a*b^2)*d*cos(d*x + c)*sin(d*x + 
c))*sqrt((a*b^2*d^2*sqrt((9*a^2 + 6*a*b + b^2)/(a^3*b^3*d^4)) - a - 3*b)/( 
a*b^2*d^2)) - 1/4*(2*(a^3*b - a^2*b^2)*d^2*cos(d*x + c)^2 - (a^3*b - a^2*b 
^2)*d^2)*sqrt((9*a^2 + 6*a*b + b^2)/(a^3*b^3*d^4))) - b*sqrt((a*b^2*d^2*sq 
rt((9*a^2 + 6*a*b + b^2)/(a^3*b^3*d^4)) - a - 3*b)/(a*b^2*d^2))*log(1/4*(3 
*a^2 - 2*a*b - b^2)*cos(d*x + c)^2 - 3/4*a^2 + 1/2*a*b + 1/4*b^2 - 1/2*(a^ 
3*b^2*d^3*sqrt((9*a^2 + 6*a*b + b^2)/(a^3*b^3*d^4))*cos(d*x + c)*sin(d*x + 
 c) + (3*a^2*b + a*b^2)*d*cos(d*x + c)*sin(d*x + c))*sqrt((a*b^2*d^2*sqrt( 
(9*a^2 + 6*a*b + b^2)/(a^3*b^3*d^4)) - a - 3*b)/(a*b^2*d^2)) - 1/4*(2*(a^3 
*b - a^2*b^2)*d^2*cos(d*x + c)^2 - (a^3*b - a^2*b^2)*d^2)*sqrt((9*a^2 + 6* 
a*b + b^2)/(a^3*b^3*d^4))) + b*sqrt(-(a*b^2*d^2*sqrt((9*a^2 + 6*a*b + b^2) 
/(a^3*b^3*d^4)) + a + 3*b)/(a*b^2*d^2))*log(-1/4*(3*a^2 - 2*a*b - b^2)*cos 
(d*x + c)^2 + 3/4*a^2 - 1/2*a*b - 1/4*b^2 + 1/2*(a^3*b^2*d^3*sqrt((9*a^2 + 
 6*a*b + b^2)/(a^3*b^3*d^4))*cos(d*x + c)*sin(d*x + c) - (3*a^2*b + a*b^2) 
*d*cos(d*x + c)*sin(d*x + c))*sqrt(-(a*b^2*d^2*sqrt((9*a^2 + 6*a*b + b^2)/ 
(a^3*b^3*d^4)) + a + 3*b)/(a*b^2*d^2)) - 1/4*(2*(a^3*b - a^2*b^2)*d^2*cos( 
d*x + c)^2 - (a^3*b - a^2*b^2)*d^2)*sqrt((9*a^2 + 6*a*b + b^2)/(a^3*b^3...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-(b*integrate(-8*(4*b^2*cos(6*d*x + 6*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + 4* 
b^2*sin(6*d*x + 6*c)^2 + 4*b^2*sin(2*d*x + 2*c)^2 + 4*(8*a^2 - 3*a*b)*cos( 
4*d*x + 4*c)^2 - b^2*cos(2*d*x + 2*c) + 4*(8*a^2 - 3*a*b)*sin(4*d*x + 4*c) 
^2 + 6*(4*a*b - b^2)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - (b^2*cos(6*d*x + 
6*c) + 2*a*b*cos(4*d*x + 4*c) + b^2*cos(2*d*x + 2*c))*cos(8*d*x + 8*c) + ( 
8*b^2*cos(2*d*x + 2*c) - b^2 + 6*(4*a*b - b^2)*cos(4*d*x + 4*c))*cos(6*d*x 
 + 6*c) - 2*(a*b - 3*(4*a*b - b^2)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - (b 
^2*sin(6*d*x + 6*c) + 2*a*b*sin(4*d*x + 4*c) + b^2*sin(2*d*x + 2*c))*sin(8 
*d*x + 8*c) + 2*(4*b^2*sin(2*d*x + 2*c) + 3*(4*a*b - b^2)*sin(4*d*x + 4*c) 
)*sin(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)*sin(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 
^2 - 3*b^3 - 4*(8*a*b^2 - 3*b^3)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - 4*(2 
*b^3*sin(6*d*x + 6*c) + 2*b^3*sin(2*d*x + 2*c) + (8*a*b^2 - 3*b^3)*sin(4*d 
*x + 4*c))*sin(8*d*x + 8*c) + 16*(2*b^3*sin(2*d*x + 2*c) + (8*a*b^2 - 3...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 906 vs. \(2 (91) = 182\).

Time = 0.76 (sec) , antiderivative size = 906, normalized size of antiderivative = 7.13 \[ \int \frac {\cos ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(2*(d*x + c)/b + ((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) - (3*sqrt(a^2 - a*b + s 
qrt(a*b)*(a - b))*a^3*b - 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^2 - 
7*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))*s 
qrt(a*b)*a^2*b^2 - 6*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)*abs(b)) + ((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) - (3*sqrt(a^2 - a*b - 
 sqrt(a*b)*(a - b))*a^3*b - 3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b^2 
- 7*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^2*b^2 - 6*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*flo 
or((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 - ...
 

Mupad [B] (verification not implemented)

Time = 37.08 (sec) , antiderivative size = 4299, normalized size of antiderivative = 33.85 \[ \int \frac {\cos ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

atan((90*a^4*tan(c + d*x))/(10*a*b^3 + 132*a^3*b - 90*a^4 - 2*b^4 - 68*a^2 
*b^2 + (18*a^5)/b) - (18*a^5*tan(c + d*x))/(10*a*b^4 - 90*a^4*b + 18*a^5 - 
 2*b^5 - 68*a^2*b^3 + 132*a^3*b^2) + (2*b^4*tan(c + d*x))/(10*a*b^3 + 132* 
a^3*b - 90*a^4 - 2*b^4 - 68*a^2*b^2 + (18*a^5)/b) + (68*a^2*b^2*tan(c + d* 
x))/(10*a*b^3 + 132*a^3*b - 90*a^4 - 2*b^4 - 68*a^2*b^2 + (18*a^5)/b) - (1 
0*a*b^3*tan(c + d*x))/(10*a*b^3 + 132*a^3*b - 90*a^4 - 2*b^4 - 68*a^2*b^2 
+ (18*a^5)/b) - (132*a^3*b*tan(c + d*x))/(10*a*b^3 + 132*a^3*b - 90*a^4 - 
2*b^4 - 68*a^2*b^2 + (18*a^5)/b))/(b*d) + (atan(((tan(c + d*x)*(30*a*b^4 - 
 30*a^4*b + 6*a^5 - 6*b^5 - 60*a^2*b^3 + 60*a^3*b^2) + (-(3*a*(a^3*b^5)^(1 
/2) + b*(a^3*b^5)^(1/2) + 3*a^2*b^3 + a^3*b^2)/(16*a^3*b^4))^(1/2)*(36*a*b 
^5 - 12*a^5*b - 4*b^6 + ((-(3*a*(a^3*b^5)^(1/2) + b*(a^3*b^5)^(1/2) + 3*a^ 
2*b^3 + a^3*b^2)/(16*a^3*b^4))^(1/2)*(64*a*b^7 + 256*a^2*b^6 - 896*a^3*b^5 
 + 768*a^4*b^4 - 192*a^5*b^3 + tan(c + d*x)*(-(3*a*(a^3*b^5)^(1/2) + b*(a^ 
3*b^5)^(1/2) + 3*a^2*b^3 + a^3*b^2)/(16*a^3*b^4))^(1/2)*(768*a^2*b^7 - 768 
*a^3*b^6 - 768*a^4*b^5 + 768*a^5*b^4)) + tan(c + d*x)*(80*a*b^6 - 16*b^7 + 
 224*a^2*b^5 - 480*a^3*b^4 + 48*a^4*b^3 + 144*a^5*b^2))*(-(3*a*(a^3*b^5)^( 
1/2) + b*(a^3*b^5)^(1/2) + 3*a^2*b^3 + a^3*b^2)/(16*a^3*b^4))^(1/2) - 72*a 
^2*b^4 + 40*a^3*b^3 + 12*a^4*b^2))*(-(3*a*(a^3*b^5)^(1/2) + b*(a^3*b^5)^(1 
/2) + 3*a^2*b^3 + a^3*b^2)/(16*a^3*b^4))^(1/2)*1i + (tan(c + d*x)*(30*a*b^ 
4 - 30*a^4*b + 6*a^5 - 6*b^5 - 60*a^2*b^3 + 60*a^3*b^2) - (-(3*a*(a^3*b...
 

Reduce [F]

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

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

Output:

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