\(\int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [343]

Optimal result

Integrand size = 24, antiderivative size = 182 \[ \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {(8 a+35 b) x}{8 b^2}+\frac {\left (\sqrt {a}-\sqrt {b}\right )^{7/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^2 d}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^{7/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^2 d}-\frac {11 \cos (c+d x) \sin (c+d x)}{8 b d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d} \] Output:

-1/8*(8*a+35*b)*x/b^2+1/2*(a^(1/2)-b^(1/2))^(7/2)*arctan((a^(1/2)-b^(1/2)) 
^(1/2)*tan(d*x+c)/a^(1/4))/a^(3/4)/b^2/d+1/2*(a^(1/2)+b^(1/2))^(7/2)*arcta 
n((a^(1/2)+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(3/4)/b^2/d-11/8*cos(d*x+c 
)*sin(d*x+c)/b/d-1/4*cos(d*x+c)^3*sin(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {4 (8 a+35 b) (c+d x)-\frac {16 \left (\sqrt {a}+\sqrt {b}\right )^4 \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {16 \left (\sqrt {a}-\sqrt {b}\right )^4 \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}+24 b \sin (2 (c+d x))+b \sin (4 (c+d x))}{32 b^2 d} \] Input:

Integrate[Cos[c + d*x]^8/(a - b*Sin[c + d*x]^4),x]
 

Output:

-1/32*(4*(8*a + 35*b)*(c + d*x) - (16*(Sqrt[a] + Sqrt[b])^4*ArcTan[((Sqrt[ 
a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + 
Sqrt[a]*Sqrt[b]]) + (16*(Sqrt[a] - Sqrt[b])^4*ArcTanh[((Sqrt[a] - Sqrt[b]) 
*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqr 
t[b]]) + 24*b*Sin[2*(c + d*x)] + b*Sin[4*(c + d*x)])/(b^2*d)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.10, 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.

\(\displaystyle \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^8}{a-b \sin (c+d x)^4}dx\)

\(\Big \downarrow \) 3703

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

\(\Big \downarrow \) 1484

\(\displaystyle \frac {\int \left (\frac {-a-3 b}{b^2 \left (\tan ^2(c+d x)+1\right )}+\frac {a^2+6 b a+b^2+(a-b) (a+3 b) \tan ^2(c+d x)}{b^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {2}{b \left (\tan ^2(c+d x)+1\right )^2}-\frac {1}{b \left (\tan ^2(c+d x)+1\right )^3}\right )d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\left (\sqrt {a}-\sqrt {b}\right )^{7/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^2}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^{7/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^2}-\frac {(a+3 b) \arctan (\tan (c+d x))}{b^2}-\frac {11 \arctan (\tan (c+d x))}{8 b}-\frac {11 \tan (c+d x)}{8 b \left (\tan ^2(c+d x)+1\right )}-\frac {\tan (c+d x)}{4 b \left (\tan ^2(c+d x)+1\right )^2}}{d}\)

Input:

Int[Cos[c + d*x]^8/(a - b*Sin[c + d*x]^4),x]
 

Output:

((-11*ArcTan[Tan[c + d*x]])/(8*b) - ((a + 3*b)*ArcTan[Tan[c + d*x]])/b^2 + 
 ((Sqrt[a] - Sqrt[b])^(7/2)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/ 
a^(1/4)])/(2*a^(3/4)*b^2) + ((Sqrt[a] + Sqrt[b])^(7/2)*ArcTan[(Sqrt[Sqrt[a 
] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^2) - Tan[c + d*x]/(4*b*( 
1 + Tan[c + d*x]^2)^2) - (11*Tan[c + d*x])/(8*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 = 4.11 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\frac {11 \tan \left (d x +c \right )^{3} b}{8}+\frac {13 \tan \left (d x +c \right ) b}{8}}{\left (\tan \left (d x +c \right )^{2}+1\right )^{2}}+\frac {\left (8 a +35 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{b^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (a \sqrt {a b}+3 \sqrt {a b}\, b +3 a b +b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (a \sqrt {a b}+3 \sqrt {a b}\, b -3 a b -b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{b^{2}}}{d}\)	\(224\)
default	\(\frac {-\frac {\frac {\frac {11 \tan \left (d x +c \right )^{3} b}{8}+\frac {13 \tan \left (d x +c \right ) b}{8}}{\left (\tan \left (d x +c \right )^{2}+1\right )^{2}}+\frac {\left (8 a +35 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{b^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (a \sqrt {a b}+3 \sqrt {a b}\, b +3 a b +b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (a \sqrt {a b}+3 \sqrt {a b}\, b -3 a b -b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{b^{2}}}{d}\)	\(224\)
risch	\(\text {Expression too large to display}\)	\(1433\)

Input:

int(cos(d*x+c)^8/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/b^2*((11/8*tan(d*x+c)^3*b+13/8*tan(d*x+c)*b)/(tan(d*x+c)^2+1)^2+1/ 
8*(8*a+35*b)*arctan(tan(d*x+c)))+1/b^2*(a-b)*(1/2*(a*(a*b)^(1/2)+3*(a*b)^( 
1/2)*b+3*a*b+b^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*(a*b)^(1/2)+3*(a*b)^(1/2 
)*b-3*a*b-b^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))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2433 vs. \(2 (140) = 280\).

Time = 0.73 (sec) , antiderivative size = 2433, normalized size of antiderivative = 13.37 \[ \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^8/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
 

Output:

1/8*(b^2*d*sqrt(-(a*b^4*d^2*sqrt((49*a^6 + 490*a^5*b + 1519*a^4*b^2 + 1484 
*a^3*b^3 + 511*a^2*b^4 + 42*a*b^5 + b^6)/(a^3*b^7*d^4)) + a^3 + 21*a^2*b + 
 35*a*b^2 + 7*b^3)/(a*b^4*d^2))*log(7/4*a^6 + 7/2*a^5*b - 63/4*a^4*b^2 + 9 
*a^3*b^3 + 25/4*a^2*b^4 - 9/2*a*b^5 - 1/4*b^6 - 1/4*(7*a^6 + 14*a^5*b - 63 
*a^4*b^2 + 36*a^3*b^3 + 25*a^2*b^4 - 18*a*b^5 - b^6)*cos(d*x + c)^2 + 1/2* 
((a^4*b^5 + 3*a^3*b^6)*d^3*sqrt((49*a^6 + 490*a^5*b + 1519*a^4*b^2 + 1484* 
a^3*b^3 + 511*a^2*b^4 + 42*a*b^5 + b^6)/(a^3*b^7*d^4))*cos(d*x + c)*sin(d* 
x + c) - (21*a^5*b^2 + 112*a^4*b^3 + 98*a^3*b^4 + 24*a^2*b^5 + a*b^6)*d*co 
s(d*x + c)*sin(d*x + c))*sqrt(-(a*b^4*d^2*sqrt((49*a^6 + 490*a^5*b + 1519* 
a^4*b^2 + 1484*a^3*b^3 + 511*a^2*b^4 + 42*a*b^5 + b^6)/(a^3*b^7*d^4)) + a^ 
3 + 21*a^2*b + 35*a*b^2 + 7*b^3)/(a*b^4*d^2)) - 1/4*(2*(a^5*b^3 - 3*a^4*b^ 
4 + 3*a^3*b^5 - a^2*b^6)*d^2*cos(d*x + c)^2 - (a^5*b^3 - 3*a^4*b^4 + 3*a^3 
*b^5 - a^2*b^6)*d^2)*sqrt((49*a^6 + 490*a^5*b + 1519*a^4*b^2 + 1484*a^3*b^ 
3 + 511*a^2*b^4 + 42*a*b^5 + b^6)/(a^3*b^7*d^4))) - b^2*d*sqrt(-(a*b^4*d^2 
*sqrt((49*a^6 + 490*a^5*b + 1519*a^4*b^2 + 1484*a^3*b^3 + 511*a^2*b^4 + 42 
*a*b^5 + b^6)/(a^3*b^7*d^4)) + a^3 + 21*a^2*b + 35*a*b^2 + 7*b^3)/(a*b^4*d 
^2))*log(7/4*a^6 + 7/2*a^5*b - 63/4*a^4*b^2 + 9*a^3*b^3 + 25/4*a^2*b^4 - 9 
/2*a*b^5 - 1/4*b^6 - 1/4*(7*a^6 + 14*a^5*b - 63*a^4*b^2 + 36*a^3*b^3 + 25* 
a^2*b^4 - 18*a*b^5 - b^6)*cos(d*x + c)^2 - 1/2*((a^4*b^5 + 3*a^3*b^6)*d^3* 
sqrt((49*a^6 + 490*a^5*b + 1519*a^4*b^2 + 1484*a^3*b^3 + 511*a^2*b^4 + ...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**8/(a-b*sin(d*x+c)**4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\cos \left (d x + c\right )^{8}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:

integrate(cos(d*x+c)^8/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
 

Output:

-1/32*(32*b^2*d*integrate(-16*(4*(a*b^2 + b^3)*cos(6*d*x + 6*c)^2 + 2*(8*a 
^3 + 29*a^2*b - 20*a*b^2 + 3*b^3)*cos(4*d*x + 4*c)^2 + 4*(a*b^2 + b^3)*cos 
(2*d*x + 2*c)^2 + 4*(a*b^2 + b^3)*sin(6*d*x + 6*c)^2 + 2*(8*a^3 + 29*a^2*b 
 - 20*a*b^2 + 3*b^3)*sin(4*d*x + 4*c)^2 + 2*(10*a^2*b + 13*a*b^2 - 5*b^3)* 
sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*(a*b^2 + b^3)*sin(2*d*x + 2*c)^2 - ( 
(a*b^2 + b^3)*cos(6*d*x + 6*c) + (a^2*b + 4*a*b^2 - b^3)*cos(4*d*x + 4*c) 
+ (a*b^2 + b^3)*cos(2*d*x + 2*c))*cos(8*d*x + 8*c) - (a*b^2 + b^3 - 2*(10* 
a^2*b + 13*a*b^2 - 5*b^3)*cos(4*d*x + 4*c) - 8*(a*b^2 + b^3)*cos(2*d*x + 2 
*c))*cos(6*d*x + 6*c) - (a^2*b + 4*a*b^2 - b^3 - 2*(10*a^2*b + 13*a*b^2 - 
5*b^3)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - (a*b^2 + b^3)*cos(2*d*x + 2*c) 
 - ((a*b^2 + b^3)*sin(6*d*x + 6*c) + (a^2*b + 4*a*b^2 - b^3)*sin(4*d*x + 4 
*c) + (a*b^2 + b^3)*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 2*((10*a^2*b + 13 
*a*b^2 - 5*b^3)*sin(4*d*x + 4*c) + 4*(a*b^2 + b^3)*sin(2*d*x + 2*c))*sin(6 
*d*x + 6*c))/(b^4*cos(8*d*x + 8*c)^2 + 16*b^4*cos(6*d*x + 6*c)^2 + 16*b^4* 
cos(2*d*x + 2*c)^2 + b^4*sin(8*d*x + 8*c)^2 + 16*b^4*sin(6*d*x + 6*c)^2 + 
16*b^4*sin(2*d*x + 2*c)^2 - 8*b^4*cos(2*d*x + 2*c) + b^4 + 4*(64*a^2*b^2 - 
 48*a*b^3 + 9*b^4)*cos(4*d*x + 4*c)^2 + 4*(64*a^2*b^2 - 48*a*b^3 + 9*b^4)* 
sin(4*d*x + 4*c)^2 + 16*(8*a*b^3 - 3*b^4)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c 
) - 2*(4*b^4*cos(6*d*x + 6*c) + 4*b^4*cos(2*d*x + 2*c) - b^4 + 2*(8*a*b^3 
- 3*b^4)*cos(4*d*x + 4*c))*cos(8*d*x + 8*c) + 8*(4*b^4*cos(2*d*x + 2*c)...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 836 vs. \(2 (140) = 280\).

Time = 0.89 (sec) , antiderivative size = 836, normalized size of antiderivative = 4.59 \[ \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^8/(a-b*sin(d*x+c)^4),x, algorithm="giac")
 

Output:

1/8*(4*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^4 + 12*sqrt(a^2 - a*b + sq 
rt(a*b)*(a - b))*a^3*b - 34*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^2 - 
12*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 - 12*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3 + 12*s 
qrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b + 28*sqrt(a^2 - a*b + s 
qrt(a*b)*(a - b))*sqrt(a*b)*a*b^2 + 4*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))* 
sqrt(a*b)*b^3)*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/sqrt((a 
*b^2 + sqrt(a^2*b^4 - (a*b^2 - b^3)*a*b^2))/(a*b^2 - b^3))))*abs(-a + b)/( 
3*a^5*b^2 - 12*a^4*b^3 + 14*a^3*b^4 - 4*a^2*b^5 - a*b^6) + 4*(3*sqrt(a^2 - 
 a*b - sqrt(a*b)*(a - b))*a^4 + 12*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^3 
*b - 34*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b^2 - 12*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 + 12*sq 
rt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^3 - 12*sqrt(a^2 - a*b - sqrt 
(a*b)*(a - b))*sqrt(a*b)*a^2*b - 28*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sq 
rt(a*b)*a*b^2 - 4*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^3)*(pi*f 
loor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/sqrt((a*b^2 - sqrt(a^2*b^4 
- (a*b^2 - b^3)*a*b^2))/(a*b^2 - b^3))))*abs(-a + b)/(3*a^5*b^2 - 12*a^4*b 
^3 + 14*a^3*b^4 - 4*a^2*b^5 - a*b^6) - (d*x + c)*(8*a + 35*b)/b^2 - (11*ta 
n(d*x + c)^3 + 13*tan(d*x + c))/((tan(d*x + c)^2 + 1)^2*b))/d
 

Mupad [B] (verification not implemented)

Time = 37.91 (sec) , antiderivative size = 8773, normalized size of antiderivative = 48.20 \[ \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^8/(a - b*sin(c + d*x)^4),x)
 

Output:

- (atan(((((373728*a^3*b^8 - 256*b^11 - 208208*a^2*b^9 - 17552*a*b^10 + 35 
296*a^4*b^7 - 240464*a^5*b^6 + 29040*a^6*b^5 + 27648*a^7*b^4 + 768*a^8*b^3 
)/(64*b^5) - (((4096*a*b^12 + 53248*a^2*b^11 - 129024*a^3*b^10 + 69632*a^4 
*b^9 + 14336*a^5*b^8 - 12288*a^6*b^7)/(64*b^5) - (tan(c + d*x)*(-(7*a^3*(a 
^3*b^9)^(1/2) + b^3*(a^3*b^9)^(1/2) + 7*a^2*b^7 + 35*a^3*b^6 + 21*a^4*b^5 
+ a^5*b^4 + 21*a*b^2*(a^3*b^9)^(1/2) + 35*a^2*b*(a^3*b^9)^(1/2))/(16*a^3*b 
^8))^(1/2)*(12288*a^2*b^11 - 12288*a^3*b^10 - 12288*a^4*b^9 + 12288*a^5*b^ 
8))/(16*b^4))*(-(7*a^3*(a^3*b^9)^(1/2) + b^3*(a^3*b^9)^(1/2) + 7*a^2*b^7 + 
 35*a^3*b^6 + 21*a^4*b^5 + a^5*b^4 + 21*a*b^2*(a^3*b^9)^(1/2) + 35*a^2*b*( 
a^3*b^9)^(1/2))/(16*a^3*b^8))^(1/2) + (tan(c + d*x)*(256*a*b^10 + 256*b^11 
 - 70832*a^2*b^9 + 61136*a^3*b^8 + 53616*a^4*b^7 - 12432*a^5*b^6 - 29696*a 
^6*b^5 - 2304*a^7*b^4))/(16*b^4))*(-(7*a^3*(a^3*b^9)^(1/2) + b^3*(a^3*b^9) 
^(1/2) + 7*a^2*b^7 + 35*a^3*b^6 + 21*a^4*b^5 + a^5*b^4 + 21*a*b^2*(a^3*b^9 
)^(1/2) + 35*a^2*b*(a^3*b^9)^(1/2))/(16*a^3*b^8))^(1/2))*(-(7*a^3*(a^3*b^9 
)^(1/2) + b^3*(a^3*b^9)^(1/2) + 7*a^2*b^7 + 35*a^3*b^6 + 21*a^4*b^5 + a^5* 
b^4 + 21*a*b^2*(a^3*b^9)^(1/2) + 35*a^2*b*(a^3*b^9)^(1/2))/(16*a^3*b^8))^( 
1/2) + (tan(c + d*x)*(336*a^8*b - 1497*a*b^8 + 96*a^9 - 1257*b^9 + 21499*a 
^2*b^7 - 41861*a^3*b^6 + 27109*a^4*b^5 + 3077*a^5*b^4 - 9223*a^6*b^3 + 172 
1*a^7*b^2))/(16*b^4))*(-(7*a^3*(a^3*b^9)^(1/2) + b^3*(a^3*b^9)^(1/2) + 7*a 
^2*b^7 + 35*a^3*b^6 + 21*a^4*b^5 + a^5*b^4 + 21*a*b^2*(a^3*b^9)^(1/2) +...
 

Reduce [F]

\[ \int \frac {\cos ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\left (\int \frac {\cos \left (d x +c \right )^{8}}{\sin \left (d x +c \right )^{4} b -a}d x \right ) \] Input:

int(cos(d*x+c)^8/(a-b*sin(d*x+c)^4),x)
 

Output:

 - int(cos(c + d*x)**8/(sin(c + d*x)**4*b - a),x)