\(\int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\) [1552]

Optimal result

Integrand size = 29, antiderivative size = 135 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {(A+B) \log (1-\sin (c+d x))}{2 (a+b)^2 d}+\frac {(A-B) \log (1+\sin (c+d x))}{2 (a-b)^2 d}-\frac {\left (2 a A b-a^2 B-b^2 B\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {A b-a B}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \] Output:

-1/2*(A+B)*ln(1-sin(d*x+c))/(a+b)^2/d+1/2*(A-B)*ln(1+sin(d*x+c))/(a-b)^2/d 
-(2*A*a*b-B*a^2-B*b^2)*ln(a+b*sin(d*x+c))/(a^2-b^2)^2/d+(A*b-B*a)/(a^2-b^2 
)/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.32 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {-\frac {B ((-a+b) \log (1-\sin (c+d x))+(a+b) \log (1+\sin (c+d x))-2 b \log (a+b \sin (c+d x)))}{2 b (-a+b) (a+b)}+b \left (A-\frac {a B}{b}\right ) \left (-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 b}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{d} \] Input:

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

Output:

(-1/2*(B*((-a + b)*Log[1 - Sin[c + d*x]] + (a + b)*Log[1 + Sin[c + d*x]] - 
 2*b*Log[a + b*Sin[c + d*x]]))/(b*(-a + b)*(a + b)) + b*(A - (a*B)/b)*(-1/ 
2*Log[1 - Sin[c + d*x]]/(b*(a + b)^2) + Log[1 + Sin[c + d*x]]/(2*(a - b)^2 
*b) - (2*a*Log[a + b*Sin[c + d*x]])/((a - b)^2*(a + b)^2) + 1/((a^2 - b^2) 
*(a + b*Sin[c + d*x]))))/d
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 27, 657, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-1/2*((A + B)*Log[b - b*Sin[c + d*x]])/(a + b)^2 - ((2*a*A*b - a^2*B - b^ 
2*B)*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 + ((A - B)*Log[b + b*Sin[c + d 
*x]])/(2*(a - b)^2) + (A*b - a*B)/((a^2 - b^2)*(a + b*Sin[c + d*x])))/d
 

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 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3316

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

Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {\left (A -B \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}-\frac {\left (2 A a b -a^{2} B -B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {A b -B a}{\left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )}+\frac {\left (-A -B \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(128\)
default	\(\frac {\frac {\left (A -B \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}-\frac {\left (2 A a b -a^{2} B -B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {A b -B a}{\left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )}+\frac {\left (-A -B \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(128\)
parallelrisch	\(\frac {-2 \left (a +b \sin \left (d x +c \right )\right ) a \left (A a b -\frac {1}{2} a^{2} B -\frac {1}{2} B \,b^{2}\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-a \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a +b \right ) \left (a \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-b \sin \left (d x +c \right ) \left (a -b \right ) \left (A b -B a \right )\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a d \left (a +b \sin \left (d x +c \right )\right )}\)	\(184\)
norman	\(\frac {-\frac {2 b \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (a^{2}-b^{2}\right )}-\frac {2 b \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a \left (a^{2}-b^{2}\right )}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {\left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\left (2 A a b -a^{2} B -B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\)	\(259\)
risch	\(-\frac {2 i B \,b^{2} c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i a^{2} B x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {4 i A a b c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i B \,b^{2} x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {i A c}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {i A x}{a^{2}+2 a b +b^{2}}+\frac {i B x}{a^{2}+2 a b +b^{2}}+\frac {i B x}{a^{2}-2 a b +b^{2}}+\frac {i B c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i B c}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {i A x}{a^{2}-2 a b +b^{2}}-\frac {i A c}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 i a^{2} B c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {4 i A a b x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i \left (A b -B a \right ) {\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A a b}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) a^{2} B}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B \,b^{2}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\)	\(665\)

Input:

int(sec(d*x+c)*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE 
)
 

Output:

1/d*(1/2*(A-B)/(a-b)^2*ln(1+sin(d*x+c))-(2*A*a*b-B*a^2-B*b^2)/(a+b)^2/(a-b 
)^2*ln(a+b*sin(d*x+c))+(A*b-B*a)/(a+b)/(a-b)/(a+b*sin(d*x+c))+1/2*(-A-B)/( 
a+b)^2*ln(sin(d*x+c)-1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (129) = 258\).

Time = 0.22 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.10 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \, B a^{3} - 2 \, A a^{2} b - 2 \, B a b^{2} + 2 \, A b^{3} - 2 \, {\left (B a^{3} - 2 \, A a^{2} b + B a b^{2} + {\left (B a^{2} b - 2 \, A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left ({\left (A - B\right )} a^{3} + 2 \, {\left (A - B\right )} a^{2} b + {\left (A - B\right )} a b^{2} + {\left ({\left (A - B\right )} a^{2} b + 2 \, {\left (A - B\right )} a b^{2} + {\left (A - B\right )} b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + B\right )} a^{3} - 2 \, {\left (A + B\right )} a^{2} b + {\left (A + B\right )} a b^{2} + {\left ({\left (A + B\right )} a^{2} b - 2 \, {\left (A + B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \] Input:

integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="fri 
cas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (B a - A b\right )}}{a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}}{2 \, d} \] Input:

integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="max 
ima")
 

Output:

1/2*(2*(B*a^2 - 2*A*a*b + B*b^2)*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 
+ b^4) + (A - B)*log(sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) - (A + B)*log(s 
in(d*x + c) - 1)/(a^2 + 2*a*b + b^2) - 2*(B*a - A*b)/(a^3 - a*b^2 + (a^2*b 
 - b^3)*sin(d*x + c)))/d
 

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.35 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {{\left (B a^{2} b - 2 \, A a b^{2} + B b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b d - 2 \, a^{2} b^{3} d + b^{5} d} - \frac {{\left (A + B\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} + \frac {{\left (A - B\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} - \frac {B a^{3} - A a^{2} b - B a b^{2} + A b^{3}}{{\left (b \sin \left (d x + c\right ) + a\right )} {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} d} \] Input:

integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="gia 
c")
 

Output:

(B*a^2*b - 2*A*a*b^2 + B*b^3)*log(abs(b*sin(d*x + c) + a))/(a^4*b*d - 2*a^ 
2*b^3*d + b^5*d) - 1/2*(A + B)*log(abs(-sin(d*x + c) + 1))/(a^2*d + 2*a*b* 
d + b^2*d) + 1/2*(A - B)*log(abs(-sin(d*x + c) - 1))/(a^2*d - 2*a*b*d + b^ 
2*d) - (B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)/((b*sin(d*x + c) + a)*(a + b)^2 
*(a - b)^2*d)
 

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {A\,b-B\,a}{d\,\left (a^2-b^2\right )\,\left (a+b\,\sin \left (c+d\,x\right )\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {A}{2}+\frac {B}{2}\right )}{d\,{\left (a+b\right )}^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (B\,a^2-2\,A\,a\,b+B\,b^2\right )}{d\,{\left (a^2-b^2\right )}^2}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {A}{2}-\frac {B}{2}\right )}{d\,{\left (a-b\right )}^2} \] Input:

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

Output:

(A*b - B*a)/(d*(a^2 - b^2)*(a + b*sin(c + d*x))) - (log(sin(c + d*x) - 1)* 
(A/2 + B/2))/(d*(a + b)^2) + (log(a + b*sin(c + d*x))*(B*a^2 + B*b^2 - 2*A 
*a*b))/(d*(a^2 - b^2)^2) + (log(sin(c + d*x) + 1)*(A/2 - B/2))/(d*(a - b)^ 
2)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.77 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b}{d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

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