\(\int \frac {(a B+b B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [294]

Optimal result

Integrand size = 34, antiderivative size = 88 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 b^2 B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}-\frac {b B \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {B \tan (c+d x)}{a d} \] Output:

2*b^2*B*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^2/(a-b)^(1/2) 
/(a+b)^(1/2)/d-b*B*arctanh(sin(d*x+c))/a^2/d+B*tan(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.32 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {B \left (-\frac {2 b^2 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+a \tan (c+d x)\right )}{a^2 d} \] Input:

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

Output:

(B*((-2*b^2*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^ 
2 + b^2] + b*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x) 
/2] + Sin[(c + d*x)/2]]) + a*Tan[c + d*x]))/(a^2*d)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {2011, 3042, 3281, 25, 27, 3042, 3226, 3042, 3138, 218, 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.

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

\(\displaystyle B \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3281

\(\displaystyle B \left (\frac {\int -\frac {b \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\tan (c+d x)}{a d}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle B \left (\frac {\tan (c+d x)}{a d}-\frac {\int \frac {b \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle B \left (\frac {\tan (c+d x)}{a d}-\frac {b \int \frac {\sec (c+d x)}{a+b \cos (c+d x)}dx}{a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle B \left (\frac {\tan (c+d x)}{a d}-\frac {b \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\right )\)

\(\Big \downarrow \) 3226

\(\displaystyle B \left (\frac {\tan (c+d x)}{a d}-\frac {b \left (\frac {\int \sec (c+d x)dx}{a}-\frac {b \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle B \left (\frac {\tan (c+d x)}{a d}-\frac {b \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}\right )\)

\(\Big \downarrow \) 3138

\(\displaystyle B \left (\frac {\tan (c+d x)}{a d}-\frac {b \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle B \left (\frac {\tan (c+d x)}{a d}-\frac {b \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}\right )\)

\(\Big \downarrow \) 4257

\(\displaystyle B \left (\frac {\tan (c+d x)}{a d}-\frac {b \left (\frac {\text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}\right )\)

Input:

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

Output:

B*(-((b*((-2*b*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt 
[a - b]*Sqrt[a + b]*d) + ArcTanh[Sin[c + d*x]]/(a*d)))/a) + Tan[c + d*x]/( 
a*d))
 

Defintions of rubi rules used

rule 25

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

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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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 3226

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), 
 x], x] - Simp[d/(b*c - a*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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 = 2.86 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {2 B \left (-\frac {1}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2}}-\frac {1}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2}}+\frac {b^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d}\)	\(125\)
default	\(\frac {2 B \left (-\frac {1}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2}}-\frac {1}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2}}+\frac {b^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d}\)	\(125\)
risch	\(\frac {2 i B}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {b^{2} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b^{2} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {B b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {B b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}\)	\(221\)

Input:

int((B*a+b*B*cos(d*x+c))*sec(d*x+c)^2/(a+cos(d*x+c)*b)^2,x,method=_RETURNV 
ERBOSE)
 

Output:

2/d*B*(-1/2/a/(tan(1/2*d*x+1/2*c)-1)+1/2*b/a^2*ln(tan(1/2*d*x+1/2*c)-1)-1/ 
2/a/(tan(1/2*d*x+1/2*c)+1)-1/2*b/a^2*ln(tan(1/2*d*x+1/2*c)+1)+b^2/a^2/((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. 165 vs. \(2 (79) = 158\).

Time = 0.13 (sec) , antiderivative size = 398, normalized size of antiderivative = 4.52 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} B b^{2} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (B a^{2} b - B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{2} b - B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{3} - B a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, \frac {2 \, \sqrt {a^{2} - b^{2}} B b^{2} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (B a^{2} b - B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{2} b - B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{3} - B a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \] Input:

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorith 
m="fricas")
 

Output:

[-1/2*(sqrt(-a^2 + b^2)*B*b^2*cos(d*x + c)*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)) + (B* 
a^2*b - B*b^3)*cos(d*x + c)*log(sin(d*x + c) + 1) - (B*a^2*b - B*b^3)*cos( 
d*x + c)*log(-sin(d*x + c) + 1) - 2*(B*a^3 - B*a*b^2)*sin(d*x + c))/((a^4 
- a^2*b^2)*d*cos(d*x + c)), 1/2*(2*sqrt(a^2 - b^2)*B*b^2*arctan(-(a*cos(d* 
x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c) - (B*a^2*b - B*b^ 
3)*cos(d*x + c)*log(sin(d*x + c) + 1) + (B*a^2*b - B*b^3)*cos(d*x + c)*log 
(-sin(d*x + c) + 1) + 2*(B*a^3 - B*a*b^2)*sin(d*x + c))/((a^4 - a^2*b^2)*d 
*cos(d*x + c))]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorith 
m="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 = 0.19 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.76 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} B b^{2}}{\sqrt {a^{2} - b^{2}} a^{2}} - \frac {B b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac {B b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a}}{d} \] Input:

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorith 
m="giac")
 

Output:

(2*(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)))*B*b^2/(sqrt(a^2 - b 
^2)*a^2) - B*b*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 + B*b*log(abs(tan(1/ 
2*d*x + 1/2*c) - 1))/a^2 - 2*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c) 
^2 - 1)*a))/d
 

Mupad [B] (verification not implemented)

Time = 25.18 (sec) , antiderivative size = 326, normalized size of antiderivative = 3.70 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2\,B\,\left (\frac {a^3\,\sin \left (c+d\,x\right )}{2}-\frac {a\,b^2\,\sin \left (c+d\,x\right )}{2}\right )}{a^2\,d\,\cos \left (c+d\,x\right )\,\left (a^2-b^2\right )}-\frac {2\,B\,\left (a^2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+b^2\,\mathrm {atanh}\left (\frac {a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (b^2-a^2\right )}^{3/2}-2\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+3\,a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a\,b^2-a^3\right )}^2}\right )\,\sqrt {b^2-a^2}\right )}{a^2\,d\,\left (a^2-b^2\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.28 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {b \left (2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}+\sin \left (d x +c \right ) a^{3}-\sin \left (d x +c \right ) a \,b^{2}\right )}{\cos \left (d x +c \right ) a^{2} d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

(b*(2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*cos(c + d*x)*b**2 + cos(c + d*x)*log(tan((c + d*x)/2) - 1) 
*a**2*b - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**3 - cos(c + d*x)*log(t 
an((c + d*x)/2) + 1)*a**2*b + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**3 
+ sin(c + d*x)*a**3 - sin(c + d*x)*a*b**2))/(cos(c + d*x)*a**2*d*(a**2 - b 
**2))