\(\int \frac {(B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [266]

Optimal result

Integrand size = 40, antiderivative size = 79 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{a^2 d}-\frac {(4 B-C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(B-C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:

B*arctanh(sin(d*x+c))/a^2/d-1/3*(4*B-C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/ 
3*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(170\) vs. \(2(79)=158\).

Time = 1.04 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.15 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (6 B \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+(B-C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (4 B-C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+(B-C) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \] Input:

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

Output:

(-2*Cos[(c + d*x)/2]*(6*B*Cos[(c + d*x)/2]^3*(Log[Cos[(c + d*x)/2] - Sin[( 
c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (B - C)*Sec[c/2 
]*Sin[(d*x)/2] + 2*(4*B - C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + (B 
 - C)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {3042, 3508, 3042, 3457, 3042, 3457, 27, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

\(\displaystyle \int \frac {\sec (c+d x) (B+C \cos (c+d x))}{(a \cos (c+d x)+a)^2}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int 3 a^2 B \sec (c+d x)dx}{a^2}-\frac {(4 B-C) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 B \int \sec (c+d x)dx-\frac {(4 B-C) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {3 B \text {arctanh}(\sin (c+d x))}{d}-\frac {(4 B-C) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((B - C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((3*B*ArcTanh[Sin 
[c + d*x]])/d - ((4*B - C)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])))/(3*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 3457

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

rule 3508

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

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\frac {-6 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (B -C \right )+9 B -3 C \right )}{6 a^{2} d}\)	\(75\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(91\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(91\)
risch	\(-\frac {2 i \left (3 B \,{\mathrm e}^{2 i \left (d x +c \right )}+9 B \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}+4 B -C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}\)	\(110\)
norman	\(\frac {-\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (3 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (4 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}+\frac {\left (5 B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}-\frac {\left (5 B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\)	\(204\)

Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x,method 
=_RETURNVERBOSE)
 

Output:

1/6*(-6*B*ln(tan(1/2*d*x+1/2*c)-1)+6*B*ln(tan(1/2*d*x+1/2*c)+1)-tan(1/2*d* 
x+1/2*c)*(tan(1/2*d*x+1/2*c)^2*(B-C)+9*B-3*C))/a^2/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.66 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (4 \, B - C\right )} \cos \left (d x + c\right ) + 5 \, B - 2 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x, 
 algorithm="fricas")
 

Output:

1/6*(3*(B*cos(d*x + c)^2 + 2*B*cos(d*x + c) + B)*log(sin(d*x + c) + 1) - 3 
*(B*cos(d*x + c)^2 + 2*B*cos(d*x + c) + B)*log(-sin(d*x + c) + 1) - 2*((4* 
B - C)*cos(d*x + c) + 5*B - 2*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a 
^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F]

\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+a*cos(d*x+c))**2 
,x)
 

Output:

(Integral(B*cos(c + d*x)*sec(c + d*x)**2/(cos(c + d*x)**2 + 2*cos(c + d*x) 
 + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2/(cos(c + d*x)**2 + 
2*cos(c + d*x) + 1), x))/a**2
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.84 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x, 
 algorithm="maxima")
 

Output:

-1/6*(B*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) 
 + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin( 
d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) - C*(3*sin(d*x + c)/(cos(d*x + c) + 
1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.43 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {6 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x, 
 algorithm="giac")
 

Output:

1/6*(6*B*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*B*log(abs(tan(1/2*d*x 
+ 1/2*c) - 1))/a^2 - (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1 
/2*c)^3 + 9*B*a^4*tan(1/2*d*x + 1/2*c) - 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6 
)/d
 

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B}{a^2}+\frac {B-C}{2\,a^2}\right )}{d} \] Input:

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

Output:

(2*B*atanh(tan(c/2 + (d*x)/2)))/(a^2*d) - (tan(c/2 + (d*x)/2)^3*(B - C))/( 
6*a^2*d) - (tan(c/2 + (d*x)/2)*(B/a^2 + (B - C)/(2*a^2)))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{6 a^{2} d} \] Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x)
 

Output:

( - 6*log(tan((c + d*x)/2) - 1)*b + 6*log(tan((c + d*x)/2) + 1)*b - tan((c 
 + d*x)/2)**3*b + tan((c + d*x)/2)**3*c - 9*tan((c + d*x)/2)*b + 3*tan((c 
+ d*x)/2)*c)/(6*a**2*d)