\(\int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [470]

Optimal result

Integrand size = 21, antiderivative size = 221 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {3 a x}{b^4}+\frac {3 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}+\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac {a^2 \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {3 a^3 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.00 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {-6 a (c+d x)-\frac {6 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+2 b \sin (c+d x)-\frac {a^4 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac {a^3 b \left (5 a^2-8 b^2\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}}{2 b^4 d} \] Input:

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

Output:

(-6*a*(c + d*x) - (6*a^2*(2*a^4 - 5*a^2*b^2 + 4*b^4)*ArcTanh[((a - b)*Tan[ 
(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 2*b*Sin[c + d*x] - ( 
a^4*b*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (a^3*b*(5*a 
^2 - 8*b^2)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*b 
^4*d)
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.17, 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, 3510, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3510

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

-1/2*(a^2*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x 
])^2) - ((-3*a^3*(a^2 - 2*b^2)*Sin[c + d*x])/(b^2*(a^2 - b^2)*d*(a + b*Cos 
[c + d*x])) + ((3*(2*a*(a^2 - b^2)^2*x - (2*a^2*(2*a^4 - 5*a^2*b^2 + 4*b^4 
)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a 
+ b]*d)))/b - ((3*a^2 - 2*b^2)*(a^2 - b^2)*Sin[c + d*x])/d)/(b^2*(a^2 - b^ 
2)))/(2*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 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 3510

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+3 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{4}}+\frac {2 a^{2} \left (\frac {\frac {\left (4 a^{2}-a b -8 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a^{2}+a b -8 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {3 \left (2 a^{4}-5 a^{2} b^{2}+4 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}}{d}\)	\(266\)
default	\(\frac {-\frac {2 \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+3 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{4}}+\frac {2 a^{2} \left (\frac {\frac {\left (4 a^{2}-a b -8 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a^{2}+a b -8 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {3 \left (2 a^{4}-5 a^{2} b^{2}+4 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}}{d}\)	\(266\)
risch	\(-\frac {3 a x}{b^{4}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,b^{3}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,b^{3}}+\frac {i a^{3} \left (6 a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-9 a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+10 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-11 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-8 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+14 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-23 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}+5 a^{2} b^{2}-8 b^{4}\right )}{b^{4} \left (a^{2}-b^{2}\right )^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {3 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 )^{2} \left (a -b \right )^{2} d \,b^{4}}+\frac {15 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 )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{2}}-\frac {6 a^{2} \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 )^{2} \left (a -b \right )^{2} d}+\frac {3 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}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{4}}-\frac {15 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}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{2}}+\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(724\)

Input:

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

Output:

1/d*(-2/b^4*(-b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+3*a*arctan(tan 
(1/2*d*x+1/2*c)))+2*a^2/b^4*((1/2*(4*a^2-a*b-8*b^2)*a*b/(a-b)/(a^2+2*a*b+b 
^2)*tan(1/2*d*x+1/2*c)^3+1/2*(4*a^2+a*b-8*b^2)*a*b/(a+b)/(a^2-2*a*b+b^2)*t 
an(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)^2+3 
/2*(2*a^4-5*a^2*b^2+4*b^4)/(a^4-2*a^2*b^2+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))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 480 vs. \(2 (206) = 412\).

Time = 0.14 (sec) , antiderivative size = 1029, normalized size of antiderivative = 4.66 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/4*(12*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*x*cos(d*x + c)^2 + 2 
4*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d*x*cos(d*x + c) + 12*(a^9 - 3 
*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*x + 3*(2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 + ( 
2*a^6*b^2 - 5*a^4*b^4 + 4*a^2*b^6)*cos(d*x + c)^2 + 2*(2*a^7*b - 5*a^5*b^3 
 + 4*a^3*b^5)*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^8*b - 17*a^6*b^3 + 13*a^4*b^5 - 2*a^2*b^7 + 2*(a^6*b^3 - 3*a^4*b^5 + 
 3*a^2*b^7 - b^9)*cos(d*x + c)^2 + (9*a^7*b^2 - 25*a^5*b^4 + 20*a^3*b^6 - 
4*a*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - 
b^12)*d*cos(d*x + c)^2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3*b^9 - a*b^11)*d*co 
s(d*x + c) + (a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d), -1/2*(6*(a^7 
*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*x*cos(d*x + c)^2 + 12*(a^8*b - 3*a 
^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d*x*cos(d*x + c) + 6*(a^9 - 3*a^7*b^2 + 3*a^ 
5*b^4 - a^3*b^6)*d*x - 3*(2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 + (2*a^6*b^2 - 5*a 
^4*b^4 + 4*a^2*b^6)*cos(d*x + c)^2 + 2*(2*a^7*b - 5*a^5*b^3 + 4*a^3*b^5)*c 
os(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^8*b - 17*a^6*b^3 + 13*a^4*b^5 - 2*a^2*b^7 + 2*(a^6* 
b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(d*x + c)^2 + (9*a^7*b^2 - 25*a^5*b^ 
4 + 20*a^3*b^6 - 4*a*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^6 - 3*a^4...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^3,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 = 0.60 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {3 \, {\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a^{2} - b^{2}}} - \frac {4 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} + \frac {3 \, {\left (d x + c\right )} a}{b^{4}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b^{3}}}{d} \] Input:

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

Output:

-(3*(2*a^6 - 5*a^4*b^2 + 4*a^2*b^4)*(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))/sq 
rt(a^2 - b^2)))/((a^4*b^4 - 2*a^2*b^6 + b^8)*sqrt(a^2 - b^2)) - (4*a^6*tan 
(1/2*d*x + 1/2*c)^3 - 5*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 7*a^4*b^2*tan(1/2*d 
*x + 1/2*c)^3 + 8*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*a^6*tan(1/2*d*x + 1/2 
*c) + 5*a^5*b*tan(1/2*d*x + 1/2*c) - 7*a^4*b^2*tan(1/2*d*x + 1/2*c) - 8*a^ 
3*b^3*tan(1/2*d*x + 1/2*c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*d*x + 
1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) + 3*(d*x + c)*a/b^4 - 2*ta 
n(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*b^3))/d
 

Mupad [B] (verification not implemented)

Time = 50.34 (sec) , antiderivative size = 5350, normalized size of antiderivative = 24.21 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(c/2 + (d*x)/2)^5*(2*a*b^4 - 3*a^4*b + 6*a^5 - 2*b^5 + 4*a^2*b^3 - 12 
*a^3*b^2))/((a*b^3 - b^4)*(a + b)^2) + (tan(c/2 + (d*x)/2)*(2*a*b^4 + 3*a^ 
4*b + 6*a^5 + 2*b^5 - 4*a^2*b^3 - 12*a^3*b^2))/((a + b)*(b^5 - 2*a*b^4 + a 
^2*b^3)) + (2*tan(c/2 + (d*x)/2)^3*(6*a^6 - 2*b^6 + 6*a^2*b^4 - 13*a^4*b^2 
))/(b*(a*b^2 - b^3)*(a + b)^2*(a - b)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*( 
2*a*b + 3*a^2 - b^2) + tan(c/2 + (d*x)/2)^6*(a^2 - 2*a*b + b^2) + a^2 + b^ 
2 - tan(c/2 + (d*x)/2)^4*(2*a*b - 3*a^2 + b^2))) - (6*a*atan(((3*a*((8*tan 
(c/2 + (d*x)/2)*(72*a^12 - 72*a^11*b + 36*a^2*b^10 - 72*a^3*b^9 + 36*a^4*b 
^8 + 288*a^5*b^7 - 288*a^6*b^6 - 432*a^7*b^5 + 441*a^8*b^4 + 288*a^9*b^3 - 
 288*a^10*b^2))/(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a 
^5*b^8 - a^6*b^7 - a^7*b^6) + (a*((24*(4*a*b^17 - 8*a^2*b^16 - 12*a^3*b^15 
 + 26*a^4*b^14 + 14*a^5*b^13 - 32*a^6*b^12 - 8*a^7*b^11 + 18*a^8*b^10 + 2* 
a^9*b^9 - 4*a^10*b^8))/(a*b^15 + b^16 - 3*a^2*b^14 - 3*a^3*b^13 + 3*a^4*b^ 
12 + 3*a^5*b^11 - a^6*b^10 - a^7*b^9) - (a*tan(c/2 + (d*x)/2)*(8*a*b^17 - 
8*a^2*b^16 - 32*a^3*b^15 + 32*a^4*b^14 + 48*a^5*b^13 - 48*a^6*b^12 - 32*a^ 
7*b^11 + 32*a^8*b^10 + 8*a^9*b^9 - 8*a^10*b^8)*24i)/(b^4*(a*b^12 + b^13 - 
3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6)))*3i) 
/b^4))/b^4 + (3*a*((8*tan(c/2 + (d*x)/2)*(72*a^12 - 72*a^11*b + 36*a^2*b^1 
0 - 72*a^3*b^9 + 36*a^4*b^8 + 288*a^5*b^7 - 288*a^6*b^6 - 432*a^7*b^5 + 44 
1*a^8*b^4 + 288*a^9*b^3 - 288*a^10*b^2))/(a*b^12 + b^13 - 3*a^2*b^11 - ...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1271, normalized size of antiderivative = 5.75 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^4/(a+b*cos(d*x+c))^3,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**7*b - 60*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**5*b**3 + 
48*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**5 - 12*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**6*b* 
*2 + 30*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/s 
qrt(a**2 - b**2))*sin(c + d*x)**2*a**4*b**4 - 24*sqrt(a**2 - b**2)*atan((t 
an((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2 
*a**2*b**6 + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x) 
/2)*b)/sqrt(a**2 - b**2))*a**8 - 18*sqrt(a**2 - b**2)*atan((tan((c + d*x)/ 
2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**6*b**2 - 6*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**4 + 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**2*b**6 + 9*cos(c + d*x)*sin(c + d*x)*a**7*b**2 - 2 
5*cos(c + d*x)*sin(c + d*x)*a**5*b**4 + 20*cos(c + d*x)*sin(c + d*x)*a**3* 
b**6 - 4*cos(c + d*x)*sin(c + d*x)*a*b**8 - 12*cos(c + d*x)*a**8*b*c - 12* 
cos(c + d*x)*a**8*b*d*x + 36*cos(c + d*x)*a**6*b**3*c + 36*cos(c + d*x)*a* 
*6*b**3*d*x - 36*cos(c + d*x)*a**4*b**5*c - 36*cos(c + d*x)*a**4*b**5*d*x 
+ 12*cos(c + d*x)*a**2*b**7*c + 12*cos(c + d*x)*a**2*b**7*d*x - 2*sin(c...