\(\int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx\) [450]

Optimal result

Integrand size = 21, antiderivative size = 148 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {2 a^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}+\frac {\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2872, 3128, 3102, 2814, 2738, 211} \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 a^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {a x \left (2 a^2+b^2\right )}{2 b^4}+\frac {\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac {a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

[In]

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

[Out]

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

Rule 211

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

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[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 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(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 2872

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

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1)/(b*f*(m + 2))), x] + Dist[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 3128

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] + Dist[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])))

Rubi steps \begin{align*} \text {integral}& = \frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {\cos (c+d x) \left (2 a+2 b \cos (c+d x)-3 a \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b} \\ & = -\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {-3 a^2+a b \cos (c+d x)+2 \left (3 a^2+2 b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^2} \\ & = \frac {\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {-3 a^2 b-3 a \left (2 a^2+b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3} \\ & = -\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {a^4 \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^4} \\ & = -\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\left (2 a^4\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d} \\ & = -\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {2 a^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}+\frac {\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-6 a \left (2 a^2+b^2\right ) (c+d x)-\frac {24 a^4 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+3 b \left (4 a^2+3 b^2\right ) \sin (c+d x)-3 a b^2 \sin (2 (c+d x))+b^3 \sin (3 (c+d x))}{12 b^4 d} \]

[In]

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

[Out]

(-6*a*(2*a^2 + b^2)*(c + d*x) - (24*a^4*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2]
+ 3*b*(4*a^2 + 3*b^2)*Sin[c + d*x] - 3*a*b^2*Sin[2*(c + d*x)] + b^3*Sin[3*(c + d*x)])/(12*b^4*d)

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {\frac {2 a^{4} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-a^{2} b -\frac {1}{2} a \,b^{2}-b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 a^{2} b -\frac {2}{3} b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{2} b -b^{3}+\frac {1}{2} a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\)	\(179\)
default	\(\frac {\frac {2 a^{4} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-a^{2} b -\frac {1}{2} a \,b^{2}-b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 a^{2} b -\frac {2}{3} b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{2} b -b^{3}+\frac {1}{2} a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\)	\(179\)
risch	\(-\frac {a^{3} x}{b^{4}}-\frac {a x}{2 b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{8 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{8 b d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {\sin \left (3 d x +3 c \right )}{12 d b}-\frac {a \sin \left (2 d x +2 c \right )}{4 b^{2} d}\)	\(280\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.70 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{4} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} d x - {\left (6 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}, \frac {6 \, \sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} d x + {\left (6 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \]

[In]

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

[Out]

[-1/6*(3*sqrt(-a^2 + b^2)*a^4*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*c
os(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)) + 3*(2*a^5 - a^3
*b^2 - a*b^4)*d*x - (6*a^4*b - 2*a^2*b^3 - 4*b^5 + 2*(a^2*b^3 - b^5)*cos(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)*cos(
d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d), 1/6*(6*sqrt(a^2 - b^2)*a^4*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2
- b^2)*sin(d*x + c))) - 3*(2*a^5 - a^3*b^2 - a*b^4)*d*x + (6*a^4*b - 2*a^2*b^3 - 4*b^5 + 2*(a^2*b^3 - b^5)*co
s(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.68 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{4}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {3 \, {\left (2 \, a^{3} + a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {2 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(12*(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^4/(sqrt(a^2 - b^2)*b^4) + 3*(2*a^3 + a*b^2)*(d*x + c)/b^4 - 2*(6*a^2*tan(1/2*d*x
+ 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*a^2*tan(1/2*d*x + 1/2*c)^3 + 4*b
^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*b^2*tan(1/2*d*x + 1/2*
c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d

Mupad [B] (verification not implemented)

Time = 15.34 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {3\,\sin \left (c+d\,x\right )}{4\,b\,d}+\frac {\sin \left (3\,c+3\,d\,x\right )}{12\,b\,d}-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^2\,d}+\frac {a^2\,\sin \left (c+d\,x\right )}{b^3\,d}-\frac {2\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}-\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d}+\frac {a^4\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,2{}\mathrm {i}}{b^4\,d\,\sqrt {b^2-a^2}} \]

[In]

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

[Out]

(3*sin(c + d*x))/(4*b*d) + sin(3*c + 3*d*x)/(12*b*d) - (a*sin(2*c + 2*d*x))/(4*b^2*d) + (a^2*sin(c + d*x))/(b^
3*d) - (2*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^4*d) - (a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x
)/2)))/(b^2*d) + (a^4*atan(((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2))*1i)/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^
(1/2)))*2i)/(b^4*d*(b^2 - a^2)^(1/2))