\(\int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [459]

Optimal result

Integrand size = 21, antiderivative size = 266 \[ \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {a \left (4 a^2+b^2\right ) x}{b^5}+\frac {2 a^4 \left (4 a^2-5 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^5 (a+b)^{3/2} d}+\frac {\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac {a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

-a*(4*a^2+b^2)*x/b^5+2*a^4*(4*a^2-5*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/ 
2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^5/(a+b)^(3/2)/d+1/3*(12*a^4-7*a^2*b^2-2*b^ 
4)*sin(d*x+c)/b^4/(a^2-b^2)/d-a*(2*a^2-b^2)*cos(d*x+c)*sin(d*x+c)/b^3/(a^2 
-b^2)/d+1/3*(4*a^2-b^2)*cos(d*x+c)^2*sin(d*x+c)/b^2/(a^2-b^2)/d-a^2*cos(d* 
x+c)^3*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.56 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.66 \[ \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {-12 a (2 a-i b) (2 a+i b) (c+d x)+\frac {24 a^4 \left (4 a^2-5 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+9 b \left (4 a^2+b^2\right ) \sin (c+d x)+\frac {12 a^5 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-6 a b^2 \sin (2 (c+d x))+b^3 \sin (3 (c+d x))}{12 b^5 d} \] Input:

Integrate[Cos[c + d*x]^5/(a + b*Cos[c + d*x])^2,x]
 

Output:

(-12*a*(2*a - I*b)*(2*a + I*b)*(c + d*x) + (24*a^4*(4*a^2 - 5*b^2)*ArcTanh 
[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 9*b*(4 
*a^2 + b^2)*Sin[c + d*x] + (12*a^5*b*Sin[c + d*x])/((a - b)*(a + b)*(a + b 
*Cos[c + d*x])) - 6*a*b^2*Sin[2*(c + d*x)] + b^3*Sin[3*(c + d*x)])/(12*b^5 
*d)
 

Rubi [A] (verified)

Time = 1.57 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 3271, 3042, 3528, 25, 3042, 3528, 27, 3042, 3502, 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.

\(\displaystyle \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {-\frac {\frac {\int -\frac {2 \left (3 \left (2 a^2-b^2\right ) a^2-b \left (2 a^2+b^2\right ) \cos (c+d x) a-\left (12 a^4-7 b^2 a^2-2 b^4\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {3 \left (2 a^2-b^2\right ) a^2-b \left (2 a^2+b^2\right ) \cos (c+d x) a-\left (12 a^4-7 b^2 a^2-2 b^4\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)}dx}{b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {3 \left (2 a^2-b^2\right ) a^2-b \left (2 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+\left (-12 a^4+7 b^2 a^2+2 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {-\frac {-\frac {\frac {\int \frac {3 \left (b \left (2 a^2-b^2\right ) a^2+\left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \cos (c+d x) a\right )}{a+b \cos (c+d x)}dx}{b}-\frac {\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \int \frac {b \left (2 a^2-b^2\right ) a^2+\left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \cos (c+d x) a}{a+b \cos (c+d x)}dx}{b}-\frac {\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \int \frac {b \left (2 a^2-b^2\right ) a^2+\left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2+b^2\right )}{b}-\frac {a^4 \left (4 a^2-5 b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}-\frac {\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2+b^2\right )}{b}-\frac {a^4 \left (4 a^2-5 b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2+b^2\right )}{b}-\frac {2 a^4 \left (4 a^2-5 b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}-\frac {-\frac {3 a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2+b^2\right )}{b}-\frac {2 a^4 \left (4 a^2-5 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}\)

Input:

Int[Cos[c + d*x]^5/(a + b*Cos[c + d*x])^2,x]
 

Output:

-((a^2*Cos[c + d*x]^3*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) 
) - (-1/3*((4*a^2 - b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(b*d) - ((-3*a*(2*a^ 
2 - b^2)*Cos[c + d*x]*Sin[c + d*x])/(b*d) - ((3*((a*(a^2 - b^2)*(4*a^2 + b 
^2)*x)/b - (2*a^4*(4*a^2 - 5*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sq 
rt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)))/b - ((12*a^4 - 7*a^2*b^2 - 2*b 
^4)*Sin[c + d*x])/(b*d))/b)/(3*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 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 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 = 1.86 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {\frac {2 a^{4} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (4 a^{2}-5 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5}}-\frac {2 \left (\frac {\left (-3 a^{2} b -b^{2} a -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-6 a^{2} b -\frac {2}{3} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-3 a^{2} b +b^{2} a -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+a \left (4 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{5}}}{d}\)	\(257\)
default	\(\frac {\frac {2 a^{4} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (4 a^{2}-5 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5}}-\frac {2 \left (\frac {\left (-3 a^{2} b -b^{2} a -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-6 a^{2} b -\frac {2}{3} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-3 a^{2} b +b^{2} a -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+a \left (4 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{5}}}{d}\)	\(257\)
risch	\(-\frac {4 a^{3} x}{b^{5}}-\frac {a x}{b^{3}}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d \,b^{3}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d \,b^{4}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{8 d \,b^{2}}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d \,b^{4}}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d \,b^{2}}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 d \,b^{3}}+\frac {2 i a^{5} \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{5} \left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {4 a^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{5}}+\frac {5 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {4 a^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{5}}-\frac {5 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {\sin \left (3 d x +3 c \right )}{12 b^{2} d}\)	\(562\)

Input:

int(cos(d*x+c)^5/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 747, normalized size of antiderivative = 2.81 \[ \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
 

Output:

[-1/6*(6*(4*a^7*b - 7*a^5*b^3 + 2*a^3*b^5 + a*b^7)*d*x*cos(d*x + c) + 6*(4 
*a^8 - 7*a^6*b^2 + 2*a^4*b^4 + a^2*b^6)*d*x + 3*(4*a^7 - 5*a^5*b^2 + (4*a^ 
6*b - 5*a^4*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)) 
- 2*(12*a^7*b - 19*a^5*b^3 + 5*a^3*b^5 + 2*a*b^7 + (a^4*b^4 - 2*a^2*b^6 + 
b^8)*cos(d*x + c)^3 - 2*(a^5*b^3 - 2*a^3*b^5 + a*b^7)*cos(d*x + c)^2 + 2*( 
3*a^6*b^2 - 5*a^4*b^4 + a^2*b^6 + b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b 
^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c) + (a^5*b^5 - 2*a^3*b^7 + a*b^9)*d), 
-1/3*(3*(4*a^7*b - 7*a^5*b^3 + 2*a^3*b^5 + a*b^7)*d*x*cos(d*x + c) + 3*(4* 
a^8 - 7*a^6*b^2 + 2*a^4*b^4 + a^2*b^6)*d*x - 3*(4*a^7 - 5*a^5*b^2 + (4*a^6 
*b - 5*a^4*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))) - (12*a^7*b - 19*a^5*b^3 + 5*a^3*b^5 + 2* 
a*b^7 + (a^4*b^4 - 2*a^2*b^6 + b^8)*cos(d*x + c)^3 - 2*(a^5*b^3 - 2*a^3*b^ 
5 + a*b^7)*cos(d*x + c)^2 + 2*(3*a^6*b^2 - 5*a^4*b^4 + a^2*b^6 + b^8)*cos( 
d*x + c))*sin(d*x + c))/((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c) + (a^ 
5*b^5 - 2*a^3*b^7 + a*b^9)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5/(a+b*cos(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^5/(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 = 0.51 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} - \frac {6 \, {\left (4 \, a^{6} - 5 \, a^{4} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{5} - b^{7}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, {\left (4 \, a^{3} + a b^{2}\right )} {\left (d x + c\right )}}{b^{5}} + \frac {2 \, {\left (9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}}}{3 \, d} \] Input:

integrate(cos(d*x+c)^5/(a+b*cos(d*x+c))^2,x, algorithm="giac")
 

Output:

1/3*(6*a^5*tan(1/2*d*x + 1/2*c)/((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)) - 6*(4*a^6 - 5*a^4*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^5 - b^7)*sqrt(a^2 - b^2) 
) - 3*(4*a^3 + a*b^2)*(d*x + c)/b^5 + 2*(9*a^2*tan(1/2*d*x + 1/2*c)^5 + 3* 
a*b*tan(1/2*d*x + 1/2*c)^5 + 3*b^2*tan(1/2*d*x + 1/2*c)^5 + 18*a^2*tan(1/2 
*d*x + 1/2*c)^3 + 2*b^2*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*tan(1/2*d*x + 1/2*c 
) - 3*a*b*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x 
 + 1/2*c)^2 + 1)^3*b^4))/d
 

Mupad [B] (verification not implemented)

Time = 48.23 (sec) , antiderivative size = 3852, normalized size of antiderivative = 14.48 \[ \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^5/(a + b*cos(c + d*x))^2,x)
 

Output:

- ((2*tan(c/2 + (d*x)/2)^3*(8*a*b^4 - 6*a^4*b - 36*a^5 - b^5 + 7*a^2*b^3 + 
 19*a^3*b^2))/(3*b^4*(a + b)*(a - b)) - (2*tan(c/2 + (d*x)/2)^7*(4*a^5 - 2 
*a^4*b + b^5 + a^2*b^3 - 3*a^3*b^2))/(b^4*(a + b)*(a - b)) + (2*tan(c/2 + 
(d*x)/2)^5*(8*a*b^4 + 6*a^4*b - 36*a^5 + b^5 - 7*a^2*b^3 + 19*a^3*b^2))/(3 
*b^4*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2)*(b^5 - 4*a^5 - 2*a^4*b + a^2 
*b^3 + 3*a^3*b^2))/(b^4*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2)^8 
*(a - b) + tan(c/2 + (d*x)/2)^2*(4*a + 2*b) + tan(c/2 + (d*x)/2)^6*(4*a - 
2*b) + 6*a*tan(c/2 + (d*x)/2)^4)) - (2*a*atan(((a*(4*a^2 + b^2)*((32*tan(c 
/2 + (d*x)/2)*(32*a^12 - 32*a^11*b + a^2*b^10 - 2*a^3*b^9 + 7*a^4*b^8 - 12 
*a^5*b^7 + 7*a^6*b^6 - 2*a^7*b^5 + 2*a^8*b^4 + 48*a^9*b^3 - 48*a^10*b^2))/ 
(a*b^10 + b^11 - a^2*b^9 - a^3*b^8) + (a*(4*a^2 + b^2)*((32*(a*b^17 + a^3* 
b^15 - 5*a^4*b^14 - 4*a^5*b^13 + 9*a^6*b^12 + 2*a^7*b^11 - 4*a^8*b^10))/(a 
*b^14 + b^15 - a^2*b^13 - a^3*b^12) - (a*tan(c/2 + (d*x)/2)*(4*a^2 + b^2)* 
(2*a*b^15 - 2*a^2*b^14 - 4*a^3*b^13 + 4*a^4*b^12 + 2*a^5*b^11 - 2*a^6*b^10 
)*32i)/(b^5*(a*b^10 + b^11 - a^2*b^9 - a^3*b^8)))*1i)/b^5))/b^5 + (a*(4*a^ 
2 + b^2)*((32*tan(c/2 + (d*x)/2)*(32*a^12 - 32*a^11*b + a^2*b^10 - 2*a^3*b 
^9 + 7*a^4*b^8 - 12*a^5*b^7 + 7*a^6*b^6 - 2*a^7*b^5 + 2*a^8*b^4 + 48*a^9*b 
^3 - 48*a^10*b^2))/(a*b^10 + b^11 - a^2*b^9 - a^3*b^8) - (a*(4*a^2 + b^2)* 
((32*(a*b^17 + a^3*b^15 - 5*a^4*b^14 - 4*a^5*b^13 + 9*a^6*b^12 + 2*a^7*b^1 
1 - 4*a^8*b^10))/(a*b^14 + b^15 - a^2*b^13 - a^3*b^12) + (a*tan(c/2 + (...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 686, normalized size of antiderivative = 2.58 \[ \int \frac {\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^5/(a+b*cos(d*x+c))^2,x)
 

Output:

(24*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**6*b - 30*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**3 + 
24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a 
**2 - b**2))*a**7 - 30*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**2 - cos(c + d*x)*sin(c + d*x)**3* 
a**4*b**4 + 2*cos(c + d*x)*sin(c + d*x)**3*a**2*b**6 - cos(c + d*x)*sin(c 
+ d*x)**3*b**8 + 6*cos(c + d*x)*sin(c + d*x)*a**6*b**2 - 9*cos(c + d*x)*si 
n(c + d*x)*a**4*b**4 + 3*cos(c + d*x)*sin(c + d*x)*b**8 - 12*cos(c + d*x)* 
a**7*b*c - 12*cos(c + d*x)*a**7*b*d*x + 21*cos(c + d*x)*a**5*b**3*c + 21*c 
os(c + d*x)*a**5*b**3*d*x - 6*cos(c + d*x)*a**3*b**5*c - 6*cos(c + d*x)*a* 
*3*b**5*d*x - 3*cos(c + d*x)*a*b**7*c - 3*cos(c + d*x)*a*b**7*d*x + 2*sin( 
c + d*x)**3*a**5*b**3 - 4*sin(c + d*x)**3*a**3*b**5 + 2*sin(c + d*x)**3*a* 
b**7 + 12*sin(c + d*x)*a**7*b - 21*sin(c + d*x)*a**5*b**3 + 9*sin(c + d*x) 
*a**3*b**5 - 12*a**8*c - 12*a**8*d*x + 21*a**6*b**2*c + 21*a**6*b**2*d*x - 
 6*a**4*b**4*c - 6*a**4*b**4*d*x - 3*a**2*b**6*c - 3*a**2*b**6*d*x)/(3*b** 
5*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))