\(\int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [480]

Optimal result

Integrand size = 21, antiderivative size = 222 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {b \left (3 a^2+2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \] Output:

-b*(3*a^2+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^2*cos(d*x+c)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d 
*x+c))^3-1/6*a^2*(2*a^2-7*b^2)*sin(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c 
))^2+1/6*a*(2*a^4-5*a^2*b^2+18*b^4)*sin(d*x+c)/b^2/(a^2-b^2)^3/d/(a+b*cos( 
d*x+c))
 

Mathematica [A] (verified)

Time = 1.80 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {-\frac {6 b \left (3 a^2+2 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}+\frac {a \left (4 a^4+11 a^2 b^2+3 a b \left (a^2+9 b^2\right ) \cos (c+d x)+\left (2 a^4-5 a^2 b^2+18 b^4\right ) \cos ^2(c+d x)\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}}{6 d} \] Input:

Integrate[Cos[c + d*x]^3/(a + b*Cos[c + d*x])^4,x]
 

Output:

((-6*b*(3*a^2 + 2*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2] 
])/(-a^2 + b^2)^(7/2) + (a*(4*a^4 + 11*a^2*b^2 + 3*a*b*(a^2 + 9*b^2)*Cos[c 
 + d*x] + (2*a^4 - 5*a^2*b^2 + 18*b^4)*Cos[c + d*x]^2)*Sin[c + d*x])/((a - 
 b)^3*(a + b)^3*(a + b*Cos[c + d*x])^3))/(6*d)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.20, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3271, 3042, 3500, 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.

\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

\(\displaystyle -\frac {\frac {a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {3 b^3 \left (3 a^2+2 b^2\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}}{2 b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 b^3 \left (3 a^2+2 b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}}{2 b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 b^3 \left (3 a^2+2 b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}}{2 b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {\frac {a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {6 b^3 \left (3 a^2+2 b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {6 b^3 \left (3 a^2+2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}\)

Input:

Int[Cos[c + d*x]^3/(a + b*Cos[c + d*x])^4,x]
 

Output:

-1/3*(a^2*Cos[c + d*x]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]) 
^3) - ((a^2*(2*a^2 - 7*b^2)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c 
+ d*x])^2) - ((-6*b^3*(3*a^2 + 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^4 - 5*a^ 
2*b^2 + 18*b^4)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(2*b*( 
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 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]
 

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]
 

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.27

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (2 a^{2}+3 a b +6 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {2 \left (a^{2}+9 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{2}-3 a b +6 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(281\)
default	\(\frac {-\frac {2 \left (-\frac {\left (2 a^{2}+3 a b +6 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {2 \left (a^{2}+9 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{2}-3 a b +6 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(281\)
risch	\(\frac {i a \left (6 a^{5} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-18 a^{3} b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+27 a \,b^{6} {\mathrm e}^{5 i \left (d x +c \right )}+12 a^{6} b \,{\mathrm e}^{4 i \left (d x +c \right )}-36 a^{4} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+81 a^{2} b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+18 b^{7} {\mathrm e}^{4 i \left (d x +c \right )}+8 a^{7} {\mathrm e}^{3 i \left (d x +c \right )}-8 a^{5} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+42 a^{3} b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+108 a \,b^{6} {\mathrm e}^{3 i \left (d x +c \right )}+12 a^{6} b \,{\mathrm e}^{2 i \left (d x +c \right )}-18 a^{4} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+120 a^{2} b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+36 b^{7} {\mathrm e}^{2 i \left (d x +c \right )}+6 a^{5} b^{2} {\mathrm e}^{i \left (d x +c \right )}-12 a^{3} b^{4} {\mathrm e}^{i \left (d x +c \right )}+81 a \,b^{6} {\mathrm e}^{i \left (d x +c \right )}+2 a^{4} b^{3}-5 a^{2} b^{5}+18 b^{7}\right )}{3 b^{3} \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 {3 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 )}{2 \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 )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 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 )}{2 \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 )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\)	\(696\)

Input:

int(cos(d*x+c)^3/(a+cos(d*x+c)*b)^4,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2*(-1/2*(2*a^2+3*a*b+6*b^2)*a/(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*(a^2+9*b^2)*a/(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^2-3*a*b+6*b^2)*a/(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-b*(3 
*a^2+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)))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 893, normalized size of antiderivative = 4.02 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^4,x, algorithm="fricas")
 

Output:

[1/12*(3*(3*a^5*b + 2*a^3*b^3 + (3*a^2*b^4 + 2*b^6)*cos(d*x + c)^3 + 3*(3* 
a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 + 3*(3*a^4*b^2 + 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*(4*a^7 + 7*a^5*b^2 - 11*a 
^3*b^4 + (2*a^7 - 7*a^5*b^2 + 23*a^3*b^4 - 18*a*b^6)*cos(d*x + c)^2 + 3*(a 
^6*b + 8*a^4*b^3 - 9*a^2*b^5)*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*(3*a^5*b + 2*a^3*b^3 + ( 
3*a^2*b^4 + 2*b^6)*cos(d*x + c)^3 + 3*(3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 
 + 3*(3*a^4*b^2 + 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))) - (4*a^7 + 7*a^5*b^2 - 11*a^ 
3*b^4 + (2*a^7 - 7*a^5*b^2 + 23*a^3*b^4 - 18*a*b^6)*cos(d*x + c)^2 + 3*(a^ 
6*b + 8*a^4*b^3 - 9*a^2*b^5)*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)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**3/(a+b*cos(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^3/(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 [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{3}}}{3 \, d} \] Input:

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^4,x, algorithm="giac")
 

Output:

1/3*(3*(3*a^2*b + 2*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 - 3*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 6*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 + 18*a*b^4*tan(1/2*d*x + 1/ 
2*c)^5 + 4*a^5*tan(1/2*d*x + 1/2*c)^3 + 32*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 
- 36*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 6*a^5*tan(1/2*d*x + 1/2*c) + 3*a^4*b*t 
an(1/2*d*x + 1/2*c) + 6*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) + 18*a*b^4*tan(1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^ 
4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^3)) 
/d
 

Mupad [B] (verification not implemented)

Time = 45.64 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.70 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^3+9\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^3+3\,a^2\,b+6\,a\,b^2\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3-3\,a^2\,b+6\,a\,b^2\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+2\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\left (3\,a^2\,b+2\,b^3\right )\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (3\,a^2+2\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \] Input:

int(cos(c + d*x)^3/(a + b*cos(c + d*x))^4,x)
 

Output:

((4*tan(c/2 + (d*x)/2)^3*(9*a*b^2 + a^3))/(3*(a + b)^2*(a^2 - 2*a*b + b^2) 
) + (tan(c/2 + (d*x)/2)^5*(6*a*b^2 + 3*a^2*b + 2*a^3))/((a + b)^3*(a - b)) 
 + (tan(c/2 + (d*x)/2)*(6*a*b^2 - 3*a^2*b + 2*a^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)*(3*a^2 + 2*b^2)*(2*a - 2*b)*(3*a*b 
^2 - 3*a^2*b + a^3 - b^3))/(2*(3*a^2*b + 2*b^3)*(a + b)^(1/2)*(a - b)^(7/2 
)))*(3*a^2 + 2*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 1076, normalized size of antiderivative = 4.85 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^3/(a+b*cos(d*x+c))^4,x)
 

Output:

( - 18*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**2*b**4 - 12*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*b**6 + 54*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**4*b**2 + 54* 
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**2*b**4 + 12*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)*b**6 - 54*sqr 
t(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**3*b**3 - 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*b**5 + 1 
8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a* 
*2 - b**2))*a**5*b + 66*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan(( 
c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**3*b**3 + 36*sqrt(a**2 - b**2)*atan((t 
an((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a*b**5 - 3*cos( 
c + d*x)*sin(c + d*x)*a**6*b - 24*cos(c + d*x)*sin(c + d*x)*a**4*b**3 + 27 
*cos(c + d*x)*sin(c + d*x)*a**2*b**5 + 2*sin(c + d*x)**3*a**7 - 7*sin(c + 
d*x)**3*a**5*b**2 + 23*sin(c + d*x)**3*a**3*b**4 - 18*sin(c + d*x)**3*a*b* 
*6 - 6*sin(c + d*x)*a**7 - 12*sin(c + d*x)*a**3*b**4 + 18*sin(c + d*x)*a*b 
**6)/(6*d*(cos(c + d*x)*sin(c + d*x)**2*a**8*b**3 - 4*cos(c + d*x)*sin(...