\(\int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [460]

Optimal result

Integrand size = 21, antiderivative size = 166 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (6 a^2+b^2\right ) x}{2 b^4}-\frac {2 a^3 \left (3 a^2-4 b^2\right ) \text {arctanh}\left (\frac {(a-b) \sin (c+d x)}{\sqrt {-a^2+b^2} (1+\cos (c+d x))}\right )}{b^4 \left (-a^2+b^2\right )^{3/2} d}-\frac {2 a \sin (c+d x)}{b^3 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {a^4 \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

1/2*(6*a^2+b^2)*x/b^4-2*a^3*(3*a^2-4*b^2)*arctanh((a-b)*sin(d*x+c)/(-a^2+b 
^2)^(1/2)/(1+cos(d*x+c)))/b^4/(-a^2+b^2)^(3/2)/d-2*a*sin(d*x+c)/b^3/d+1/2* 
cos(d*x+c)*sin(d*x+c)/b^2/d-a^4*sin(d*x+c)/b^3/(a^2-b^2)/d/(a+b*cos(d*x+c) 
)
 

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \left (6 a^2+b^2\right ) (c+d x)-\frac {8 a^3 \left (3 a^2-4 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}-8 a b \sin (c+d x)-\frac {4 a^4 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+b^2 \sin (2 (c+d x))}{4 b^4 d} \] Input:

Integrate[Cos[c + d*x]^4/(a + b*Cos[c + d*x])^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.36, 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, 3528, 25, 3042, 3502, 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.

\(\displaystyle \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

Int[Cos[c + d*x]^4/(a + b*Cos[c + d*x])^2,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 3502

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

rule 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

Maple [A] (verified)

Time = 1.57 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (-2 a b -\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-2 a b +\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (6 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}-\frac {2 a^{3} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (3 a^{2}-4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}}{d}\)	\(218\)
default	\(\frac {\frac {\frac {2 \left (\left (-2 a b -\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-2 a b +\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (6 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}-\frac {2 a^{3} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (3 a^{2}-4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}}{d}\)	\(218\)
risch	\(\frac {3 x \,a^{2}}{b^{4}}+\frac {x}{2 b^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{d \,b^{3}}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{d \,b^{3}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {2 i a^{4} \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{4} \left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {3 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {3 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}\)	\(502\)

Input:

int(cos(d*x+c)^4/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2/b^4*(((-2*a*b-1/2*b^2)*tan(1/2*d*x+1/2*c)^3+(-2*a*b+1/2*b^2)*tan(1/ 
2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(6*a^2+b^2)*arctan(tan(1/2*d* 
x+1/2*c)))-2*a^3/b^4*(a*b/(a^2-b^2)*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*a^2-4*b^2)/(a-b)/(a+b)/((a-b)*(a+b))^( 
1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 651, normalized size of antiderivative = 3.92 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [\frac {{\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} d x \cos \left (d x + c\right ) + {\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} d x - {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + {\left (3 \, a^{5} b - 4 \, a^{3} b^{3}\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 ) - {\left (6 \, a^{6} b - 10 \, a^{4} b^{3} + 4 \, a^{2} b^{5} - {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d\right )}}, \frac {{\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} d x \cos \left (d x + c\right ) + {\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} d x - 2 \, {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + {\left (3 \, a^{5} b - 4 \, a^{3} b^{3}\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^{6} b - 10 \, a^{4} b^{3} + 4 \, a^{2} b^{5} - {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
 

Output:

[1/2*((6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*d*x*cos(d*x + c) + (6*a^7 - 
 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*d*x - (3*a^6 - 4*a^4*b^2 + (3*a^5*b - 4*a 
^3*b^3)*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)) - (6*a^6* 
b - 10*a^4*b^3 + 4*a^2*b^5 - (a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^2 + 
3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2* 
a^2*b^7 + b^9)*d*cos(d*x + c) + (a^5*b^4 - 2*a^3*b^6 + a*b^8)*d), 1/2*((6* 
a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*d*x*cos(d*x + c) + (6*a^7 - 11*a^5*b 
^2 + 4*a^3*b^4 + a*b^6)*d*x - 2*(3*a^6 - 4*a^4*b^2 + (3*a^5*b - 4*a^3*b^3) 
*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^6*b - 10*a^4*b^3 + 4*a^2*b^5 - (a^4*b^3 - 2*a^2*b 
^5 + b^7)*cos(d*x + c)^2 + 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c))*s 
in(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c) + (a^5*b^4 - 2*a^ 
3*b^6 + a*b^8)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**4/(a+b*cos(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^2,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 [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.58 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} - \frac {4 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (6 \, a^{2} + b^{2}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {2 \, {\left (4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \] Input:

integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/2*(4*a^4*tan(1/2*d*x + 1/2*c)/((a^2*b^3 - b^5)*(a*tan(1/2*d*x + 1/2*c)^ 
2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) - 4*(3*a^5 - 4*a^3*b^2)*(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^2*b^4 - b^6)*sqrt(a^2 - b^2 
)) - (6*a^2 + b^2)*(d*x + c)/b^4 + 2*(4*a*tan(1/2*d*x + 1/2*c)^3 + b*tan(1 
/2*d*x + 1/2*c)^3 + 4*a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/((t 
an(1/2*d*x + 1/2*c)^2 + 1)^2*b^3))/d
 

Mupad [B] (verification not implemented)

Time = 48.49 (sec) , antiderivative size = 3751, normalized size of antiderivative = 22.60 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^4/(a + b*cos(c + d*x))^2,x)
 

Output:

(atan(((((8*tan(c/2 + (d*x)/2)*(72*a^10 - 72*a^9*b - 2*a*b^9 + b^10 + 11*a 
^2*b^8 - 20*a^3*b^7 + 23*a^4*b^6 - 26*a^5*b^5 + 17*a^6*b^4 + 120*a^7*b^3 - 
 120*a^8*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + ((a^2*6i + b^2*1i)*((8* 
(2*b^15 + 6*a^2*b^13 - 16*a^3*b^12 - 14*a^4*b^11 + 28*a^5*b^10 + 6*a^6*b^9 
 - 12*a^7*b^8))/(a*b^11 + b^12 - a^2*b^10 - a^3*b^9) - (4*tan(c/2 + (d*x)/ 
2)*(a^2*6i + b^2*1i)*(8*a*b^13 - 8*a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 
8*a^5*b^9 - 8*a^6*b^8))/(b^4*(a*b^8 + b^9 - a^2*b^7 - a^3*b^6))))/(2*b^4)) 
*(a^2*6i + b^2*1i)*1i)/(2*b^4) + (((8*tan(c/2 + (d*x)/2)*(72*a^10 - 72*a^9 
*b - 2*a*b^9 + b^10 + 11*a^2*b^8 - 20*a^3*b^7 + 23*a^4*b^6 - 26*a^5*b^5 + 
17*a^6*b^4 + 120*a^7*b^3 - 120*a^8*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) 
 - ((a^2*6i + b^2*1i)*((8*(2*b^15 + 6*a^2*b^13 - 16*a^3*b^12 - 14*a^4*b^11 
 + 28*a^5*b^10 + 6*a^6*b^9 - 12*a^7*b^8))/(a*b^11 + b^12 - a^2*b^10 - a^3* 
b^9) + (4*tan(c/2 + (d*x)/2)*(a^2*6i + b^2*1i)*(8*a*b^13 - 8*a^2*b^12 - 16 
*a^3*b^11 + 16*a^4*b^10 + 8*a^5*b^9 - 8*a^6*b^8))/(b^4*(a*b^8 + b^9 - a^2* 
b^7 - a^3*b^6))))/(2*b^4))*(a^2*6i + b^2*1i)*1i)/(2*b^4))/((16*(108*a^11 - 
 54*a^10*b + 4*a^3*b^8 - 4*a^4*b^7 + 41*a^5*b^6 - 9*a^6*b^5 + 63*a^7*b^4 + 
 81*a^8*b^3 - 216*a^9*b^2))/(a*b^11 + b^12 - a^2*b^10 - a^3*b^9) - (((8*ta 
n(c/2 + (d*x)/2)*(72*a^10 - 72*a^9*b - 2*a*b^9 + b^10 + 11*a^2*b^8 - 20*a^ 
3*b^7 + 23*a^4*b^6 - 26*a^5*b^5 + 17*a^6*b^4 + 120*a^7*b^3 - 120*a^8*b^2)) 
/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + ((a^2*6i + b^2*1i)*((8*(2*b^15 + 6...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 623, normalized size of antiderivative = 3.75 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {-\sin \left (d x +c \right )^{3} b^{7}+\sin \left (d x +c \right ) b^{7}+6 a^{7} c -\sin \left (d x +c \right )^{3} a^{4} b^{3}+a \,b^{6} c +a \,b^{6} d x +16 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4} b^{2}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{5} b^{2}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{4}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{6}+6 \cos \left (d x +c \right ) a^{6} b c -11 \cos \left (d x +c \right ) a^{4} b^{3} c +4 \cos \left (d x +c \right ) a^{2} b^{5} c -11 a^{5} b^{2} d x +4 a^{3} b^{4} d x +\cos \left (d x +c \right ) b^{7} c +\cos \left (d x +c \right ) b^{7} d x -12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{5} b +16 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{3} b^{3}+6 \cos \left (d x +c \right ) a^{6} b d x -11 \cos \left (d x +c \right ) a^{4} b^{3} d x +4 \cos \left (d x +c \right ) a^{2} b^{5} d x -12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{6}+2 \sin \left (d x +c \right )^{3} a^{2} b^{5}-6 \sin \left (d x +c \right ) a^{6} b +11 \sin \left (d x +c \right ) a^{4} b^{3}-6 \sin \left (d x +c \right ) a^{2} b^{5}+6 a^{7} d x -11 a^{5} b^{2} c +4 a^{3} b^{4} c}{2 b^{4} d \left (\cos \left (d x +c \right ) a^{4} b -2 \cos \left (d x +c \right ) a^{2} b^{3}+\cos \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

int(cos(d*x+c)^4/(a+b*cos(d*x+c))^2,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)*a**5*b + 16*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**3 
 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*a**6 + 16*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan 
((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**4*b**2 - 3*cos(c + d*x)*sin(c + d*x 
)*a**5*b**2 + 6*cos(c + d*x)*sin(c + d*x)*a**3*b**4 - 3*cos(c + d*x)*sin(c 
 + d*x)*a*b**6 + 6*cos(c + d*x)*a**6*b*c + 6*cos(c + d*x)*a**6*b*d*x - 11* 
cos(c + d*x)*a**4*b**3*c - 11*cos(c + d*x)*a**4*b**3*d*x + 4*cos(c + d*x)* 
a**2*b**5*c + 4*cos(c + d*x)*a**2*b**5*d*x + cos(c + d*x)*b**7*c + cos(c + 
 d*x)*b**7*d*x - sin(c + d*x)**3*a**4*b**3 + 2*sin(c + d*x)**3*a**2*b**5 - 
 sin(c + d*x)**3*b**7 - 6*sin(c + d*x)*a**6*b + 11*sin(c + d*x)*a**4*b**3 
- 6*sin(c + d*x)*a**2*b**5 + sin(c + d*x)*b**7 + 6*a**7*c + 6*a**7*d*x - 1 
1*a**5*b**2*c - 11*a**5*b**2*d*x + 4*a**3*b**4*c + 4*a**3*b**4*d*x + a*b** 
6*c + a*b**6*d*x)/(2*b**4*d*(cos(c + d*x)*a**4*b - 2*cos(c + d*x)*a**2*b** 
3 + cos(c + d*x)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))