3.5.82 \(\int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [482]

3.5.82.1 Optimal result
3.5.82.2 Mathematica [A] (verified)
3.5.82.3 Rubi [A] (verified)
3.5.82.4 Maple [A] (verified)
3.5.82.5 Fricas [B] (verification not implemented)
3.5.82.6 Sympy [F(-1)]
3.5.82.7 Maxima [F(-2)]
3.5.82.8 Giac [B] (verification not implemented)
3.5.82.9 Mupad [B] (verification not implemented)

3.5.82.1 Optimal result

Integrand size = 19, antiderivative size = 192 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {b \left (4 a^2+b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2+3 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \]

output
-b*(4*a^2+b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7 
/2)/(a+b)^(7/2)/d+1/3*a*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*(2*a 
^2+3*b^2)*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2+1/6*a*(2*a^2+13*b^2) 
*sin(d*x+c)/(a^2-b^2)^3/d/(a+b*cos(d*x+c))
 
3.5.82.2 Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {-\frac {6 b \left (4 a^2+b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}+\frac {\left (6 a^5+10 a^3 b^2-a b^4-3 b \left (-2 a^4-9 a^2 b^2+b^4\right ) \cos (c+d x)+a b^2 \left (2 a^2+13 b^2\right ) \cos ^2(c+d x)\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}}{6 d} \]

input
Integrate[Cos[c + d*x]/(a + b*Cos[c + d*x])^4,x]
 
output
((-6*b*(4*a^2 + b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]) 
/(-a^2 + b^2)^(7/2) + ((6*a^5 + 10*a^3*b^2 - a*b^4 - 3*b*(-2*a^4 - 9*a^2*b 
^2 + b^4)*Cos[c + d*x] + a*b^2*(2*a^2 + 13*b^2)*Cos[c + d*x]^2)*Sin[c + d* 
x])/((a - b)^3*(a + b)^3*(a + b*Cos[c + d*x])^3))/(6*d)
 
3.5.82.3 Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {3042, 3233, 3042, 3233, 25, 3042, 3233, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

\(\displaystyle \frac {a \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {-\frac {\int -\frac {3 b \left (4 a^2+b^2\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (2 a^2+3 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {3 b \left (4 a^2+b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (2 a^2+3 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {3 b \left (4 a^2+b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (2 a^2+3 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {a \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {6 b \left (4 a^2+b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (2 a^2+3 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {a \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {6 b \left (4 a^2+b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (2 a^2+3 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}\)

input
Int[Cos[c + d*x]/(a + b*Cos[c + d*x])^4,x]
 
output
(a*Sin[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) - (-1/2*((2*a^2 
+ 3*b^2)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((6*b*(4*a 
^2 + b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b] 
*Sqrt[a + b]*(a^2 - b^2)*d) - (a*(2*a^2 + 13*b^2)*Sin[c + d*x])/((a^2 - b^ 
2)*d*(a + b*Cos[c + d*x])))/(2*(a^2 - b^2)))/(3*(a^2 - b^2))
 

3.5.82.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3233
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 
3.5.82.4 Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.53

method result size
derivativedivides \(\frac {-\frac {2 \left (-\frac {\left (2 a^{3}+2 a^{2} b +6 a \,b^{2}+b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 \left (3 a^{2}+7 b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{3}-2 a^{2} b +6 a \,b^{2}-b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{3}}-\frac {b \left (4 a^{2}+b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) \(293\)
default \(\frac {-\frac {2 \left (-\frac {\left (2 a^{3}+2 a^{2} b +6 a \,b^{2}+b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 \left (3 a^{2}+7 b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{3}-2 a^{2} b +6 a \,b^{2}-b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{3}}-\frac {b \left (4 a^{2}+b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) \(293\)
risch \(\frac {i \left (12 a^{2} b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+3 b^{6} {\mathrm e}^{5 i \left (d x +c \right )}+60 a^{3} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+15 a \,b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+8 a^{6} {\mathrm e}^{3 i \left (d x +c \right )}+64 a^{4} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+78 a^{2} b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+24 a^{5} b \,{\mathrm e}^{2 i \left (d x +c \right )}+102 a^{3} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+24 a \,b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+12 a^{4} b^{2} {\mathrm e}^{i \left (d x +c \right )}+66 a^{2} b^{4} {\mathrm e}^{i \left (d x +c \right )}-3 b^{6} {\mathrm e}^{i \left (d x +c \right )}+2 a^{3} b^{3}+13 a \,b^{5}\right )}{3 b \left (a^{2}-b^{2}\right )^{3} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{3}}-\frac {2 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {2 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) \(608\)

input
int(cos(d*x+c)/(a+cos(d*x+c)*b)^4,x,method=_RETURNVERBOSE)
 
output
1/d*(-2*(-1/2*(2*a^3+2*a^2*b+6*a*b^2+b^3)/(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^2+7*b^2)*a/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*t 
an(1/2*d*x+1/2*c)^3-1/2*(2*a^3-2*a^2*b+6*a*b^2-b^3)/(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-b*tan(1/2*d*x+1/2*c) 
^2+a+b)^3-b*(4*a^2+b^2)/(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)))
 
3.5.82.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (177) = 354\).

Time = 0.31 (sec) , antiderivative size = 891, normalized size of antiderivative = 4.64 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\left [\frac {3 \, {\left (4 \, a^{5} b + a^{3} b^{3} + {\left (4 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (6 \, a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + a b^{6} + {\left (2 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 13 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, a^{6} b + 7 \, a^{4} b^{3} - 10 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}}, -\frac {3 \, {\left (4 \, a^{5} b + a^{3} b^{3} + {\left (4 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (6 \, a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + a b^{6} + {\left (2 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 13 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, a^{6} b + 7 \, a^{4} b^{3} - 10 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}}\right ] \]

input
integrate(cos(d*x+c)/(a+b*cos(d*x+c))^4,x, algorithm="fricas")
 
output
[1/12*(3*(4*a^5*b + a^3*b^3 + (4*a^2*b^4 + b^6)*cos(d*x + c)^3 + 3*(4*a^3* 
b^3 + a*b^5)*cos(d*x + c)^2 + 3*(4*a^4*b^2 + a^2*b^4)*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^7 + 4*a^5*b^2 - 11*a^3*b^4 + 
 a*b^6 + (2*a^5*b^2 + 11*a^3*b^4 - 13*a*b^6)*cos(d*x + c)^2 + 3*(2*a^6*b + 
 7*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a 
^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c)^3 + 3*(a^9*b^2 - 4*a 
^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*(a^10*b - 4* 
a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9* 
b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d), -1/6*(3*(4*a^5*b + a^3*b^3 + (4 
*a^2*b^4 + b^6)*cos(d*x + c)^3 + 3*(4*a^3*b^3 + a*b^5)*cos(d*x + c)^2 + 3* 
(4*a^4*b^2 + a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c 
) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (6*a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 
 a*b^6 + (2*a^5*b^2 + 11*a^3*b^4 - 13*a*b^6)*cos(d*x + c)^2 + 3*(2*a^6*b + 
 7*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a 
^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c)^3 + 3*(a^9*b^2 - 4*a 
^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*(a^10*b - 4* 
a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9* 
b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d)]
 
3.5.82.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)/(a+b*cos(d*x+c))**4,x)
 
output
Timed out
 
3.5.82.7 Maxima [F(-2)]

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

input
integrate(cos(d*x+c)/(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
 
3.5.82.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (177) = 354\).

Time = 0.34 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.22 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{3}}}{3 \, d} \]

input
integrate(cos(d*x+c)/(a+b*cos(d*x+c))^4,x, algorithm="giac")
 
output
1/3*(3*(4*a^2*b + b^3)*(pi*floor(1/2*(d*x + c)/pi + 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 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + (6*a^5*tan(1/2* 
d*x + 1/2*c)^5 - 6*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 12*a^3*b^2*tan(1/2*d*x + 
 1/2*c)^5 - 27*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 12*a*b^4*tan(1/2*d*x + 1/2 
*c)^5 + 3*b^5*tan(1/2*d*x + 1/2*c)^5 + 12*a^5*tan(1/2*d*x + 1/2*c)^3 + 16* 
a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 28*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 6*a^5*t 
an(1/2*d*x + 1/2*c) + 6*a^4*b*tan(1/2*d*x + 1/2*c) + 12*a^3*b^2*tan(1/2*d* 
x + 1/2*c) + 27*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*a*b^4*tan(1/2*d*x + 1/2* 
c) - 3*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*t 
an(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^3))/d
 
3.5.82.9 Mupad [B] (verification not implemented)

Time = 17.62 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.99 \[ \int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^3+2\,a^2\,b+6\,a\,b^2+b^3\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^3+7\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3-2\,a^2\,b+6\,a\,b^2-b^3\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2+b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}\,\left (4\,a^2\,b+b^3\right )}\right )\,\left (4\,a^2+b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

input
int(cos(c + d*x)/(a + b*cos(c + d*x))^4,x)
 
output
((tan(c/2 + (d*x)/2)^5*(6*a*b^2 + 2*a^2*b + 2*a^3 + b^3))/((a + b)^3*(a - 
b)) + (4*tan(c/2 + (d*x)/2)^3*(7*a*b^2 + 3*a^3))/(3*(a + b)^2*(a^2 - 2*a*b 
 + b^2)) + (tan(c/2 + (d*x)/2)*(6*a*b^2 - 2*a^2*b + 2*a^3 - b^3))/((a + b) 
*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/(d*(3*a*b^2 - tan(c/2 + (d*x)/2)^4*(3*a 
*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) - tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b 
- 3*a^3 + 3*b^3) + 3*a^2*b + a^3 + b^3 + tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3 
*a^2*b + a^3 - b^3))) - (b*atan((b*tan(c/2 + (d*x)/2)*(4*a^2 + b^2)*(2*a - 
 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(2*(a + b)^(1/2)*(a - b)^(7/2)*(4*a 
^2*b + b^3)))*(4*a^2 + b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))