3.1.68 \(\int \frac {\cos (c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^4} \, dx\) [68]

3.1.68.1 Optimal result

Integrand size = 31, antiderivative size = 138 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {(23 A-54 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac {4 (2 A+9 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (3 A-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3} \]

output

1/105*(23*A-54*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2+4/105*(2*A+9*C)*sin(d* 
x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A+C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x 
+c))^4-2/35*(3*A-4*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3
 

3.1.68.2 Mathematica [A] (verified)

Time = 2.23 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.30 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (70 (2 A+9 C) \sin \left (\frac {d x}{2}\right )-70 (2 A+9 C) \sin \left (c+\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )+441 C \sin \left (c+\frac {3 d x}{2}\right )-315 C \sin \left (2 c+\frac {3 d x}{2}\right )+56 A \sin \left (2 c+\frac {5 d x}{2}\right )+147 C \sin \left (2 c+\frac {5 d x}{2}\right )-105 C \sin \left (3 c+\frac {5 d x}{2}\right )+8 A \sin \left (3 c+\frac {7 d x}{2}\right )+36 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{420 a^4 d (1+\cos (c+d x))^4} \]

input

Integrate[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^4,x]
 

output

(Cos[(c + d*x)/2]*Sec[c/2]*(70*(2*A + 9*C)*Sin[(d*x)/2] - 70*(2*A + 9*C)*S 
in[c + (d*x)/2] + 168*A*Sin[c + (3*d*x)/2] + 441*C*Sin[c + (3*d*x)/2] - 31 
5*C*Sin[2*c + (3*d*x)/2] + 56*A*Sin[2*c + (5*d*x)/2] + 147*C*Sin[2*c + (5* 
d*x)/2] - 105*C*Sin[3*c + (5*d*x)/2] + 8*A*Sin[3*c + (7*d*x)/2] + 36*C*Sin 
[3*c + (7*d*x)/2]))/(420*a^4*d*(1 + Cos[c + d*x])^4)
 

3.1.68.3 Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3521, 3042, 3447, 3042, 3498, 25, 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.

\(\displaystyle \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int \frac {\cos (c+d x) (a (5 A-2 C)-a (A-6 C) \cos (c+d x))}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a (5 A-2 C)-a (A-6 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\int \frac {a (5 A-2 C) \cos (c+d x)-a (A-6 C) \cos ^2(c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (5 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )-a (A-6 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3498

\(\displaystyle \frac {-\frac {\int -\frac {6 a^2 (3 A-4 C)-5 a^2 (A-6 C) \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a (3 A-4 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {6 a^2 (3 A-4 C)-5 a^2 (A-6 C) \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a (3 A-4 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {6 a^2 (3 A-4 C)-5 a^2 (A-6 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {2 a (3 A-4 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3229

\(\displaystyle \frac {\frac {\frac {4}{3} a (2 A+9 C) \int \frac {1}{\cos (c+d x) a+a}dx+\frac {(23 A-54 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (3 A-4 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {4}{3} a (2 A+9 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {(23 A-54 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (3 A-4 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3127

\(\displaystyle \frac {\frac {\frac {4 a (2 A+9 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)}+\frac {(23 A-54 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (3 A-4 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}\)

input

Int[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^4,x]
 

output

-1/7*((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^4) + (( 
-2*a*(3*A - 4*C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + (((23*A - 54 
*C)*Sin[c + d*x])/(3*d*(1 + Cos[c + d*x])^2) + (4*a*(2*A + 9*C)*Sin[c + d* 
x])/(3*d*(a + a*Cos[c + d*x])))/(5*a^2))/(7*a^2)
 

3.1.68.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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]
 

rule 3521

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

3.1.68.4 Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.55

method	result	size
parallelrisch	\(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 \left (A -3 C \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+7 \left (-\frac {A}{3}+C \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 A -7 C \right )}{56 a^{4} d}\)	\(76\)
derivativedivides	\(\frac {\frac {\left (-A -C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A +3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (A -3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(90\)
default	\(\frac {\frac {\left (-A -C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A +3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (A -3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(90\)
risch	\(\frac {2 i \left (105 C \,{\mathrm e}^{6 i \left (d x +c \right )}+315 C \,{\mathrm e}^{5 i \left (d x +c \right )}+140 A \,{\mathrm e}^{4 i \left (d x +c \right )}+630 C \,{\mathrm e}^{4 i \left (d x +c \right )}+140 A \,{\mathrm e}^{3 i \left (d x +c \right )}+630 C \,{\mathrm e}^{3 i \left (d x +c \right )}+168 A \,{\mathrm e}^{2 i \left (d x +c \right )}+441 C \,{\mathrm e}^{2 i \left (d x +c \right )}+56 A \,{\mathrm e}^{i \left (d x +c \right )}+147 C \,{\mathrm e}^{i \left (d x +c \right )}+8 A +36 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(150\)
norman	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (A +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}+\frac {\left (5 A +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (11 A -3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{70 a d}-\frac {\left (11 A -3 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{140 a d}+\frac {\left (19 A +3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 a d}-\frac {\left (73 A -39 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{840 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a^{3}}\)	\(193\)

input

int(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+cos(d*x+c)*a)^4,x,method=_RETURNVERBO 
SE)
 

output

-1/56*tan(1/2*d*x+1/2*c)*((A+C)*tan(1/2*d*x+1/2*c)^6+7/5*(A-3*C)*tan(1/2*d 
*x+1/2*c)^4+7*(-1/3*A+C)*tan(1/2*d*x+1/2*c)^2-7*A-7*C)/a^4/d
 

3.1.68.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left (4 \, {\left (2 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (32 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (13 \, A + 6 \, C\right )} \cos \left (d x + c\right ) + 13 \, A + 6 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

input

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="f 
ricas")
 

output

1/105*(4*(2*A + 9*C)*cos(d*x + c)^3 + (32*A + 39*C)*cos(d*x + c)^2 + 4*(13 
*A + 6*C)*cos(d*x + c) + 13*A + 6*C)*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)
 

3.1.68.6 Sympy [A] (verification not implemented)

Time = 2.10 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.32 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} - \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} - \frac {C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} + \frac {3 C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

input

integrate(cos(d*x+c)*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**4,x)
 

output

Piecewise((-A*tan(c/2 + d*x/2)**7/(56*a**4*d) - A*tan(c/2 + d*x/2)**5/(40* 
a**4*d) + A*tan(c/2 + d*x/2)**3/(24*a**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d 
) - C*tan(c/2 + d*x/2)**7/(56*a**4*d) + 3*C*tan(c/2 + d*x/2)**5/(40*a**4*d 
) - C*tan(c/2 + d*x/2)**3/(8*a**4*d) + C*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d 
, 0)), (x*(A + C*cos(c)**2)*cos(c)/(a*cos(c) + a)**4, True))
 

3.1.68.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.27 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

input

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="m 
axima")
 

output

1/840*(A*(105*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 - 15*sin(d*x + c)^7/ 
(cos(d*x + c) + 1)^7)/a^4 + 3*C*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35*s 
in(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
 

3.1.68.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=-\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 63 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

input

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="g 
iac")
 

output

-1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 + 15*C*tan(1/2*d*x + 1/2*c)^7 + 21*A*t 
an(1/2*d*x + 1/2*c)^5 - 63*C*tan(1/2*d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1 
/2*c)^3 + 105*C*tan(1/2*d*x + 1/2*c)^3 - 105*A*tan(1/2*d*x + 1/2*c) - 105* 
C*tan(1/2*d*x + 1/2*c))/(a^4*d)
 

3.1.68.9 Mupad [B] (verification not implemented)

Time = 1.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.61 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-3\,C\right )}{24\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-3\,C\right )}{40\,a^4}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{8\,a^4}}{d} \]

input

int((cos(c + d*x)*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^4,x)
 

output

-((tan(c/2 + (d*x)/2)^7*(A + C))/(56*a^4) - (tan(c/2 + (d*x)/2)^3*(A - 3*C 
))/(24*a^4) + (tan(c/2 + (d*x)/2)^5*(A - 3*C))/(40*a^4) - (tan(c/2 + (d*x) 
/2)*(A + C))/(8*a^4))/d