\(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [258]

Optimal result

Integrand size = 31, antiderivative size = 263 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=-\frac {\left (4 a A b-6 a^2 B-b^2 B\right ) x}{2 b^4}+\frac {2 a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^4 (a+b)^{3/2} d}+\frac {\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

-1/2*(4*A*a*b-6*B*a^2-B*b^2)*x/b^4+2*a^2*(2*A*a^2*b-3*A*b^3-3*B*a^3+4*B*a* 
b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^4/(a 
+b)^(3/2)/d+(2*A*a^2*b-A*b^3-3*B*a^3+2*B*a*b^2)*sin(d*x+c)/b^3/(a^2-b^2)/d 
-1/2*(2*A*a*b-3*B*a^2+B*b^2)*cos(d*x+c)*sin(d*x+c)/b^2/(a^2-b^2)/d+a*(A*b- 
B*a)*cos(d*x+c)^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 2.73 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \left (-4 a A b+6 a^2 B+b^2 B\right ) (c+d x)-\frac {8 a^2 \left (-2 a^2 A b+3 A b^3+3 a^3 B-4 a b^2 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+4 b (A b-2 a B) \sin (c+d x)+\frac {4 a^3 b (A b-a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+b^2 B \sin (2 (c+d x))}{4 b^4 d} \] Input:

Integrate[(Cos[c + d*x]^3*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3468, 25, 3042, 3528, 25, 3042, 3502, 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.

\(\displaystyle \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

\(\displaystyle \frac {-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right )}{b}-\frac {2 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \sin (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right )}{b}-\frac {2 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \sin (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right )}{b}-\frac {4 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \sin (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {\frac {x \left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right )}{b}-\frac {4 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \sin (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}\)

Input:

Int[(Cos[c + d*x]^3*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^2,x]
 

Output:

(a*(A*b - a*B)*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c 
+ d*x])) + (-1/2*((2*a*A*b - 3*a^2*B + b^2*B)*Cos[c + d*x]*Sin[c + d*x])/( 
b*d) - ((((a^2 - b^2)*(4*a*A*b - 6*a^2*B - b^2*B)*x)/b - (4*a^2*(2*a^2*A*b 
 - 3*A*b^3 - 3*a^3*B + 4*a*b^2*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sq 
rt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b - (2*(2*a^2*A*b - A*b^3 - 3*a 
^3*B + 2*a*b^2*B)*Sin[c + d*x])/(b*d))/(2*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 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 3468

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

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 = 1.98 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {\frac {2 a^{2} \left (\frac {a \left (A b -B a \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 A \,a^{2} b -3 A \,b^{3}-3 a^{3} B +4 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}-\frac {2 \left (\frac {\left (-A \,b^{2}+2 B a b +\frac {1}{2} B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A \,b^{2}+2 B a b -\frac {1}{2} B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (4 A a b -6 a^{2} B -B \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\)	\(267\)
default	\(\frac {\frac {2 a^{2} \left (\frac {a \left (A b -B a \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 A \,a^{2} b -3 A \,b^{3}-3 a^{3} B +4 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4}}-\frac {2 \left (\frac {\left (-A \,b^{2}+2 B a b +\frac {1}{2} B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A \,b^{2}+2 B a b -\frac {1}{2} B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (4 A a b -6 a^{2} B -B \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\)	\(267\)
risch	\(-\frac {2 x A a}{b^{3}}+\frac {3 x \,a^{2} B}{b^{4}}+\frac {B x}{2 b^{2}}-\frac {2 i a^{3} \left (-A b +B a \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{4} \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 {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 d \,b^{2}}+\frac {i B \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B a}{d \,b^{3}}-\frac {i B \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B a}{d \,b^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 d \,b^{2}}-\frac {2 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 ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 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 ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}+\frac {3 a^{5} \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 ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}-\frac {4 a^{3} \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 ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 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 ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 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 ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {3 a^{5} \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 ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}+\frac {4 a^{3} \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 ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}\)	\(918\)

Input:

int(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(2*a^2/b^4*(a*(A*b-B*a)*b/(a^2-b^2)*tan(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*A*a^2*b-3*A*b^3-3*B*a^3+4*B*a*b^2) 
/(a-b)/(a+b)/((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*b^2+2*B*a*b+1/2*B*b^2)*tan(1/2*d*x+1/2*c)^3+(-A*b 
^2+2*B*a*b-1/2*B*b^2)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*( 
4*A*a*b-6*B*a^2-B*b^2)*arctan(tan(1/2*d*x+1/2*c))))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 964, normalized size of antiderivative = 3.67 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="f 
ricas")
 

Output:

[1/2*((6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4*b^3 + 8*A*a^3*b^4 + 4*B*a^2*b^5 
- 4*A*a*b^6 + B*b^7)*d*x*cos(d*x + c) + (6*B*a^7 - 4*A*a^6*b - 11*B*a^5*b^ 
2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*d*x + (3*B*a^6 - 2* 
A*a^5*b - 4*B*a^4*b^2 + 3*A*a^3*b^3 + (3*B*a^5*b - 2*A*a^4*b^2 - 4*B*a^3*b 
^3 + 3*A*a^2*b^4)*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)*si 
n(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) 
 - (6*B*a^6*b - 4*A*a^5*b^2 - 10*B*a^4*b^3 + 6*A*a^3*b^4 + 4*B*a^2*b^5 - 2 
*A*a*b^6 - (B*a^4*b^3 - 2*B*a^2*b^5 + B*b^7)*cos(d*x + c)^2 + (3*B*a^5*b^2 
 - 2*A*a^4*b^3 - 6*B*a^3*b^4 + 4*A*a^2*b^5 + 3*B*a*b^6 - 2*A*b^7)*cos(d*x 
+ c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c) + (a^5*b^4 
 - 2*a^3*b^6 + a*b^8)*d), 1/2*((6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4*b^3 + 8 
*A*a^3*b^4 + 4*B*a^2*b^5 - 4*A*a*b^6 + B*b^7)*d*x*cos(d*x + c) + (6*B*a^7 
- 4*A*a^6*b - 11*B*a^5*b^2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a 
*b^6)*d*x - 2*(3*B*a^6 - 2*A*a^5*b - 4*B*a^4*b^2 + 3*A*a^3*b^3 + (3*B*a^5* 
b - 2*A*a^4*b^2 - 4*B*a^3*b^3 + 3*A*a^2*b^4)*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))) - (6*B*a^6*b 
 - 4*A*a^5*b^2 - 10*B*a^4*b^3 + 6*A*a^3*b^4 + 4*B*a^2*b^5 - 2*A*a*b^6 - (B 
*a^4*b^3 - 2*B*a^2*b^5 + B*b^7)*cos(d*x + c)^2 + (3*B*a^5*b^2 - 2*A*a^4*b^ 
3 - 6*B*a^3*b^4 + 4*A*a^2*b^5 + 3*B*a*b^6 - 2*A*b^7)*cos(d*x + c))*sin(...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="m 
axima")
 

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 = 0.17 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (3 \, B a^{5} - 2 \, A a^{4} b - 4 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {4 \, {\left (B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {{\left (6 \, B a^{2} - 4 \, A a b + B b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {2 \, {\left (4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \] Input:

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 38.01 (sec) , antiderivative size = 6744, normalized size of antiderivative = 25.64 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^3*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^2,x)
 

Output:

(a^2*atan(((a^2*(-(a + b)^3*(a - b)^3)^(1/2)*((8*tan(c/2 + (d*x)/2)*(72*B^ 
2*a^10 + B^2*b^10 - 2*B^2*a*b^9 - 72*B^2*a^9*b + 16*A^2*a^2*b^8 - 32*A^2*a 
^3*b^7 + 20*A^2*a^4*b^6 + 64*A^2*a^5*b^5 - 64*A^2*a^6*b^4 - 32*A^2*a^7*b^3 
 + 32*A^2*a^8*b^2 + 11*B^2*a^2*b^8 - 20*B^2*a^3*b^7 + 23*B^2*a^4*b^6 - 26* 
B^2*a^5*b^5 + 17*B^2*a^6*b^4 + 120*B^2*a^7*b^3 - 120*B^2*a^8*b^2 - 8*A*B*a 
*b^9 - 96*A*B*a^9*b + 16*A*B*a^2*b^8 - 40*A*B*a^3*b^7 + 64*A*B*a^4*b^6 - 4 
0*A*B*a^5*b^5 - 176*A*B*a^6*b^4 + 176*A*B*a^7*b^3 + 96*A*B*a^8*b^2))/(a*b^ 
8 + b^9 - a^2*b^7 - a^3*b^6) + (a^2*((8*(2*B*b^15 + 12*A*a^2*b^13 + 12*A*a 
^3*b^12 - 20*A*a^4*b^11 - 4*A*a^5*b^10 + 8*A*a^6*b^9 + 6*B*a^2*b^13 - 16*B 
*a^3*b^12 - 14*B*a^4*b^11 + 28*B*a^5*b^10 + 6*B*a^6*b^9 - 12*B*a^7*b^8 - 8 
*A*a*b^14))/(a*b^11 + b^12 - a^2*b^10 - a^3*b^9) - (8*a^2*tan(c/2 + (d*x)/ 
2)*(-(a + b)^3*(a - b)^3)^(1/2)*(3*A*b^3 + 3*B*a^3 - 2*A*a^2*b - 4*B*a*b^2 
)*(8*a*b^13 - 8*a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 8*a^5*b^9 - 8*a^6*b 
^8))/((a*b^8 + b^9 - a^2*b^7 - a^3*b^6)*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^ 
6*b^4)))*(-(a + b)^3*(a - b)^3)^(1/2)*(3*A*b^3 + 3*B*a^3 - 2*A*a^2*b - 4*B 
*a*b^2))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))*(3*A*b^3 + 3*B*a^3 - 2* 
A*a^2*b - 4*B*a*b^2)*1i)/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4) + (a^2*( 
-(a + b)^3*(a - b)^3)^(1/2)*((8*tan(c/2 + (d*x)/2)*(72*B^2*a^10 + B^2*b^10 
 - 2*B^2*a*b^9 - 72*B^2*a^9*b + 16*A^2*a^2*b^8 - 32*A^2*a^3*b^7 + 20*A^2*a 
^4*b^6 + 64*A^2*a^5*b^5 - 64*A^2*a^6*b^4 - 32*A^2*a^7*b^3 + 32*A^2*a^8*...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3}+\cos \left (d x +c \right )^{2} a^{2} b^{2} d x -\cos \left (d x +c \right )^{2} b^{4} d x +\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}+\sin \left (d x +c \right )^{2} a^{2} b^{2} d x -\sin \left (d x +c \right )^{2} b^{4} d x -2 \sin \left (d x +c \right ) a^{3} b +2 \sin \left (d x +c \right ) a \,b^{3}+2 a^{4} d x -2 a^{2} b^{2} d x}{2 b^{3} d \left (a^{2}-b^{2}\right )} \] Input:

int(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x)
 

Output:

( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*a**3 + cos(c + d*x)**2*a**2*b**2*d*x - cos(c + d*x)**2*b** 
4*d*x + cos(c + d*x)*sin(c + d*x)*a**2*b**2 - cos(c + d*x)*sin(c + d*x)*b* 
*4 + sin(c + d*x)**2*a**2*b**2*d*x - sin(c + d*x)**2*b**4*d*x - 2*sin(c + 
d*x)*a**3*b + 2*sin(c + d*x)*a*b**3 + 2*a**4*d*x - 2*a**2*b**2*d*x)/(2*b** 
3*d*(a**2 - b**2))