\(\int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx\) [75]

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {18 \sin (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}+\frac {12 \sin (c+d x)}{35 a^4 d (1+\cos (c+d x))}-\frac {\cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {8 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3} \] Output:

-18/35*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2+12/35*sin(d*x+c)/a^4/d/(1+cos(d*x 
+c))-1/7*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+8/35*sin(d*x+c)/a/d/ 
(a+a*cos(d*x+c))^3
 

Mathematica [A] (verified)

Time = 2.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.49 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\left (2+8 \cos (c+d x)+13 \cos ^2(c+d x)+12 \cos ^3(c+d x)\right ) \sin (c+d x)}{35 a^4 d (1+\cos (c+d x))^4} \] Input:

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

Output:

((2 + 8*Cos[c + d*x] + 13*Cos[c + d*x]^2 + 12*Cos[c + d*x]^3)*Sin[c + d*x] 
)/(35*a^4*d*(1 + Cos[c + d*x])^4)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3244, 27, 3042, 3447, 3042, 3498, 27, 3042, 3229, 3042, 3127}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3244

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3498

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3229

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3127

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

Input:

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

Output:

-1/7*(Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^4) - (2*((-4*a* 
Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + (3*((3*Sin[c + d*x])/(d*(1 + 
Cos[c + d*x])^2) - (2*a*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))))/(5*a^2))) 
/(7*a^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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3127

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + 
 d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b 
^2, 0]
 

rule 3229

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/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1))   I 
nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && 
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

rule 3244

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

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3498

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 - a* 
B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 
/(a^2*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b 
*B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
 B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.50

method	result	size
parallelrisch	\(-\frac {\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a^{4} d}\)	\(57\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(58\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(58\)
risch	\(\frac {2 i \left (35 \,{\mathrm e}^{6 i \left (d x +c \right )}+105 \,{\mathrm e}^{5 i \left (d x +c \right )}+210 \,{\mathrm e}^{4 i \left (d x +c \right )}+210 \,{\mathrm e}^{3 i \left (d x +c \right )}+147 \,{\mathrm e}^{2 i \left (d x +c \right )}+49 \,{\mathrm e}^{i \left (d x +c \right )}+12\right )}{35 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(91\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{40 a d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{70 a d}+\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{280 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{140 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{56 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{3}}\)	\(152\)

Input:

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

Output:

-1/56*(tan(1/2*d*x+1/2*c)^6-21/5*tan(1/2*d*x+1/2*c)^4+7*tan(1/2*d*x+1/2*c) 
^2-7)*tan(1/2*d*x+1/2*c)/a^4/d
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left (12 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{35 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:

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

Output:

1/35*(12*cos(d*x + c)^3 + 13*cos(d*x + c)^2 + 8*cos(d*x + c) + 2)*sin(d*x 
+ c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c) 
^2 + 4*a^4*d*cos(d*x + c) + a^4*d)
 

Sympy [A] (verification not implemented)

Time = 2.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} - \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} + \frac {3 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-tan(c/2 + d*x/2)**7/(56*a**4*d) + 3*tan(c/2 + d*x/2)**5/(40*a* 
*4*d) - tan(c/2 + d*x/2)**3/(8*a**4*d) + tan(c/2 + d*x/2)/(8*a**4*d), Ne(d 
, 0)), (x*cos(c)**3/(a*cos(c) + a)**4, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{280 \, a^{4} d} \] Input:

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

Output:

1/280*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c 
) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos( 
d*x + c) + 1)^7)/(a^4*d)
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.52 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{280 \, a^{4} d} \] Input:

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

Output:

-1/280*(5*tan(1/2*d*x + 1/2*c)^7 - 21*tan(1/2*d*x + 1/2*c)^5 + 35*tan(1/2* 
d*x + 1/2*c)^3 - 35*tan(1/2*d*x + 1/2*c))/(a^4*d)
 

Mupad [B] (verification not implemented)

Time = 40.92 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.51 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-35\right )}{280\,a^4\,d} \] Input:

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

Output:

-(tan(c/2 + (d*x)/2)*(35*tan(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^4 + 
5*tan(c/2 + (d*x)/2)^6 - 35))/(280*a^4*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.51 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+35\right )}{280 a^{4} d} \] Input:

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

Output:

(tan((c + d*x)/2)*( - 5*tan((c + d*x)/2)**6 + 21*tan((c + d*x)/2)**4 - 35* 
tan((c + d*x)/2)**2 + 35))/(280*a**4*d)