\(\int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [479]

Optimal result

Integrand size = 21, antiderivative size = 250 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {x}{b^4}-\frac {a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} b^4 (a+b)^{7/2} d}-\frac {a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{6 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \sin (c+d x)}{6 b^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \] Output:

x/b^4-a*(2*a^6-7*a^4*b^2+8*a^2*b^4-8*b^6)*arctan((a-b)^(1/2)*tan(1/2*d*x+1 
/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/b^4/(a+b)^(7/2)/d-1/3*a^2*cos(d*x+c)^2*sin( 
d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*a^3*(3*a^2-8*b^2)*sin(d*x+c)/b 
^3/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2-1/6*a^2*(9*a^4-28*a^2*b^2+34*b^4)*sin( 
d*x+c)/b^3/(a^2-b^2)^3/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 3.49 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {6 (c+d x)-\frac {6 a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}-\frac {2 a^4 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^3}+\frac {a^3 b \left (7 a^2-12 b^2\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}+\frac {a^2 b \left (-11 a^4+32 a^2 b^2-36 b^4\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))}}{6 b^4 d} \] Input:

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

Output:

(6*(c + d*x) - (6*a*(2*a^6 - 7*a^4*b^2 + 8*a^2*b^4 - 8*b^6)*ArcTanh[((a - 
b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(7/2) - (2*a^4*b*Sin[ 
c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^3) + (a^3*b*(7*a^2 - 12*b^ 
2)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])^2) + (a^2*b*(-1 
1*a^4 + 32*a^2*b^2 - 36*b^4)*Sin[c + d*x])/((a - b)^3*(a + b)^3*(a + b*Cos 
[c + d*x])))/(6*b^4*d)
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.22, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3271, 3042, 3510, 3042, 3500, 27, 3042, 3214, 3042, 3138, 218}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 3271

\(\displaystyle -\frac {\int \frac {\cos (c+d x) \left (2 a^2-3 b \cos (c+d x) a-3 \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3}dx}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 a^2-3 b \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3510

\(\displaystyle -\frac {\frac {\int \frac {2 b \left (3 a^2-8 b^2\right ) a^2+\left (3 a^4-10 b^2 a^2+12 b^4\right ) \cos (c+d x) a-6 b \left (a^2-b^2\right )^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^2}dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {2 b \left (3 a^2-8 b^2\right ) a^2+\left (3 a^4-10 b^2 a^2+12 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-6 b \left (a^2-b^2\right )^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3500

\(\displaystyle -\frac {\frac {\frac {a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {3 \left (2 b \cos (c+d x) \left (a^2-b^2\right )^3+a b^2 \left (a^4-2 b^2 a^2+6 b^4\right )\right )}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 \int \frac {2 b \cos (c+d x) \left (a^2-b^2\right )^3+a b^2 \left (a^4-2 b^2 a^2+6 b^4\right )}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 \int \frac {2 b \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-b^2\right )^3+a b^2 \left (a^4-2 b^2 a^2+6 b^4\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {\frac {\frac {a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 \left (2 x \left (a^2-b^2\right )^3-a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right ) \int \frac {1}{a+b \cos (c+d x)}dx\right )}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 \left (2 x \left (a^2-b^2\right )^3-a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {\frac {\frac {a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 \left (2 x \left (a^2-b^2\right )^3-\frac {2 a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d}\right )}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 \left (2 x \left (a^2-b^2\right )^3-\frac {2 a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}\right )}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^3 \left (3 a^2-8 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

Input:

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

Output:

-1/3*(a^2*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x 
])^3) - (-1/2*(a^3*(3*a^2 - 8*b^2)*Sin[c + d*x])/(b^2*(a^2 - b^2)*d*(a + b 
*Cos[c + d*x])^2) + ((-3*(2*(a^2 - b^2)^3*x - (2*a*(2*a^6 - 7*a^4*b^2 + 8* 
a^2*b^4 - 8*b^6)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt 
[a - b]*Sqrt[a + b]*d)))/(b*(a^2 - b^2)) + (a^2*(9*a^4 - 28*a^2*b^2 + 34*b 
^4)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(2*b^2*(a^2 - b^2) 
))/(3*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* 
(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin 
[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x] 
, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
IntegersQ[2*m, 2*n])
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 3510

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 1.70 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.44

method	result	size
derivativedivides	\(\frac {-\frac {2 a \left (\frac {\frac {\left (2 a^{4}-a^{3} b -6 a^{2} b^{2}+4 a \,b^{3}+12 b^{4}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {2 \left (3 a^{4}-11 a^{2} b^{2}+18 b^{4}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (2 a^{4}+a^{3} b -6 a^{2} b^{2}-4 a \,b^{3}+12 b^{4}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {\left (2 a^{6}-7 a^{4} b^{2}+8 b^{4} a^{2}-8 b^{6}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\)	\(361\)
default	\(\frac {-\frac {2 a \left (\frac {\frac {\left (2 a^{4}-a^{3} b -6 a^{2} b^{2}+4 a \,b^{3}+12 b^{4}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {2 \left (3 a^{4}-11 a^{2} b^{2}+18 b^{4}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (2 a^{4}+a^{3} b -6 a^{2} b^{2}-4 a \,b^{3}+12 b^{4}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {\left (2 a^{6}-7 a^{4} b^{2}+8 b^{4} a^{2}-8 b^{6}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\)	\(361\)
risch	\(\text {Expression too large to display}\)	\(1051\)

Input:

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

Output:

1/d*(-2*a/b^4*((1/2*(2*a^4-a^3*b-6*a^2*b^2+4*a*b^3+12*b^4)*a*b/(a-b)/(a^3+ 
3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5+2/3*(3*a^4-11*a^2*b^2+18*b^4)*a* 
b/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*(2*a^4+a^3*b-6* 
a^2*b^2-4*a*b^3+12*b^4)*a*b/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/ 
2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3+1/2*(2*a^6-7*a 
^4*b^2+8*a^2*b^4-8*b^6)/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)* 
arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))+2/b^4*arctan(tan(1/2 
*d*x+1/2*c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 688 vs. \(2 (235) = 470\).

Time = 0.20 (sec) , antiderivative size = 1445, normalized size of antiderivative = 5.78 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/12*(12*(a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d*x*cos(d*x 
 + c)^3 + 36*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*x*co 
s(d*x + c)^2 + 36*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d 
*x*cos(d*x + c) + 12*(a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)* 
d*x - 3*(2*a^10 - 7*a^8*b^2 + 8*a^6*b^4 - 8*a^4*b^6 + (2*a^7*b^3 - 7*a^5*b 
^5 + 8*a^3*b^7 - 8*a*b^9)*cos(d*x + c)^3 + 3*(2*a^8*b^2 - 7*a^6*b^4 + 8*a^ 
4*b^6 - 8*a^2*b^8)*cos(d*x + c)^2 + 3*(2*a^9*b - 7*a^7*b^3 + 8*a^5*b^5 - 8 
*a^3*b^7)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 
- b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + 
c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 2*(6* 
a^10*b - 23*a^8*b^3 + 43*a^6*b^5 - 26*a^4*b^7 + (11*a^8*b^3 - 43*a^6*b^5 + 
 68*a^4*b^7 - 36*a^2*b^9)*cos(d*x + c)^2 + 15*(a^9*b^2 - 4*a^7*b^4 + 7*a^5 
*b^6 - 4*a^3*b^8)*cos(d*x + c))*sin(d*x + c))/((a^8*b^7 - 4*a^6*b^9 + 6*a^ 
4*b^11 - 4*a^2*b^13 + b^15)*d*cos(d*x + c)^3 + 3*(a^9*b^6 - 4*a^7*b^8 + 6* 
a^5*b^10 - 4*a^3*b^12 + a*b^14)*d*cos(d*x + c)^2 + 3*(a^10*b^5 - 4*a^8*b^7 
 + 6*a^6*b^9 - 4*a^4*b^11 + a^2*b^13)*d*cos(d*x + c) + (a^11*b^4 - 4*a^9*b 
^6 + 6*a^7*b^8 - 4*a^5*b^10 + a^3*b^12)*d), 1/6*(6*(a^8*b^3 - 4*a^6*b^5 + 
6*a^4*b^7 - 4*a^2*b^9 + b^11)*d*x*cos(d*x + c)^3 + 18*(a^9*b^2 - 4*a^7*b^4 
 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*x*cos(d*x + c)^2 + 18*(a^10*b - 4*a^8 
*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*x*cos(d*x + c) + 6*(a^11 - 4*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 531 vs. \(2 (235) = 470\).

Time = 0.60 (sec) , antiderivative size = 531, normalized size of antiderivative = 2.12 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

1/3*(3*(2*a^7 - 7*a^5*b^2 + 8*a^3*b^4 - 8*a*b^6)*(pi*floor(1/2*(d*x + c)/p 
i + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x 
 + 1/2*c))/sqrt(a^2 - b^2)))/((a^6*b^4 - 3*a^4*b^6 + 3*a^2*b^8 - b^10)*sqr 
t(a^2 - b^2)) - (6*a^8*tan(1/2*d*x + 1/2*c)^5 - 15*a^7*b*tan(1/2*d*x + 1/2 
*c)^5 - 6*a^6*b^2*tan(1/2*d*x + 1/2*c)^5 + 45*a^5*b^3*tan(1/2*d*x + 1/2*c) 
^5 - 6*a^4*b^4*tan(1/2*d*x + 1/2*c)^5 - 60*a^3*b^5*tan(1/2*d*x + 1/2*c)^5 
+ 36*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 + 12*a^8*tan(1/2*d*x + 1/2*c)^3 - 56*a 
^6*b^2*tan(1/2*d*x + 1/2*c)^3 + 116*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 - 72*a^ 
2*b^6*tan(1/2*d*x + 1/2*c)^3 + 6*a^8*tan(1/2*d*x + 1/2*c) + 15*a^7*b*tan(1 
/2*d*x + 1/2*c) - 6*a^6*b^2*tan(1/2*d*x + 1/2*c) - 45*a^5*b^3*tan(1/2*d*x 
+ 1/2*c) - 6*a^4*b^4*tan(1/2*d*x + 1/2*c) + 60*a^3*b^5*tan(1/2*d*x + 1/2*c 
) + 36*a^2*b^6*tan(1/2*d*x + 1/2*c))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b 
^9)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^3) + 3*( 
d*x + c)/b^4)/d
 

Mupad [B] (verification not implemented)

Time = 59.96 (sec) , antiderivative size = 7247, normalized size of antiderivative = 28.99 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*atan((((((8*(16*a*b^20 - 4*b^21 + 12*a^2*b^19 - 64*a^3*b^18 - 20*a^4*b^ 
17 + 110*a^5*b^16 + 30*a^6*b^15 - 110*a^7*b^14 - 30*a^8*b^13 + 70*a^9*b^12 
 + 14*a^10*b^11 - 26*a^11*b^10 - 2*a^12*b^9 + 4*a^13*b^8))/(a*b^19 + b^20 
- 5*a^2*b^18 - 5*a^3*b^17 + 10*a^4*b^16 + 10*a^5*b^15 - 10*a^6*b^14 - 10*a 
^7*b^13 + 5*a^8*b^12 + 5*a^9*b^11 - a^10*b^10 - a^11*b^9) - (tan(c/2 + (d* 
x)/2)*(8*a*b^21 - 8*a^2*b^20 - 48*a^3*b^19 + 48*a^4*b^18 + 120*a^5*b^17 - 
120*a^6*b^16 - 160*a^7*b^15 + 160*a^8*b^14 + 120*a^9*b^13 - 120*a^10*b^12 
- 48*a^11*b^11 + 48*a^12*b^10 + 8*a^13*b^9 - 8*a^14*b^8)*8i)/(b^4*(a*b^16 
+ b^17 - 5*a^2*b^15 - 5*a^3*b^14 + 10*a^4*b^13 + 10*a^5*b^12 - 10*a^6*b^11 
 - 10*a^7*b^10 + 5*a^8*b^9 + 5*a^9*b^8 - a^10*b^7 - a^11*b^6)))*1i)/b^4 + 
(8*tan(c/2 + (d*x)/2)*(8*a^14 - 8*a^13*b - 8*a*b^13 + 4*b^14 + 44*a^2*b^12 
 + 48*a^3*b^11 - 92*a^4*b^10 - 120*a^5*b^9 + 156*a^6*b^8 + 160*a^7*b^7 - 1 
64*a^8*b^6 - 120*a^9*b^5 + 117*a^10*b^4 + 48*a^11*b^3 - 48*a^12*b^2))/(a*b 
^16 + b^17 - 5*a^2*b^15 - 5*a^3*b^14 + 10*a^4*b^13 + 10*a^5*b^12 - 10*a^6* 
b^11 - 10*a^7*b^10 + 5*a^8*b^9 + 5*a^9*b^8 - a^10*b^7 - a^11*b^6))/b^4 - ( 
(((8*(16*a*b^20 - 4*b^21 + 12*a^2*b^19 - 64*a^3*b^18 - 20*a^4*b^17 + 110*a 
^5*b^16 + 30*a^6*b^15 - 110*a^7*b^14 - 30*a^8*b^13 + 70*a^9*b^12 + 14*a^10 
*b^11 - 26*a^11*b^10 - 2*a^12*b^9 + 4*a^13*b^8))/(a*b^19 + b^20 - 5*a^2*b^ 
18 - 5*a^3*b^17 + 10*a^4*b^16 + 10*a^5*b^15 - 10*a^6*b^14 - 10*a^7*b^13 + 
5*a^8*b^12 + 5*a^9*b^11 - a^10*b^10 - a^11*b^9) + (tan(c/2 + (d*x)/2)*(...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 1981, normalized size of antiderivative = 7.92 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq 
rt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**7*b**3 + 42*sqrt(a**2 - b 
**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos 
(c + d*x)*sin(c + d*x)**2*a**5*b**5 - 48*sqrt(a**2 - b**2)*atan((tan((c + 
d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d* 
x)**2*a**3*b**7 + 48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + 
 d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a*b**9 + 36*sq 
rt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - 
 b**2))*cos(c + d*x)*a**9*b - 114*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2) 
*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**7*b**3 + 102*s 
qrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 
- b**2))*cos(c + d*x)*a**5*b**5 - 96*sqrt(a**2 - b**2)*atan((tan((c + d*x) 
/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**3*b**7 - 48 
*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a** 
2 - b**2))*cos(c + d*x)*a*b**9 - 36*sqrt(a**2 - b**2)*atan((tan((c + d*x)/ 
2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**8*b**2 + 
126*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( 
a**2 - b**2))*sin(c + d*x)**2*a**6*b**4 - 144*sqrt(a**2 - b**2)*atan((tan( 
(c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a* 
*4*b**6 + 144*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x...