\(\int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [438]

Optimal result

Integrand size = 19, antiderivative size = 104 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b)^2 d}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 d}-\frac {2 a b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {b}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {b \left (-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 b}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{d} \] Input:

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

Output:

(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 = 0.33 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3147, 477, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (c+d x) (a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3147

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

\(\Big \downarrow \) 477

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 477

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
a^p   Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 
]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & 
& NiceSqrtQ[-b/a] &&  !FractionalPowerFactorQ[Rt[-b/a, 2]]
 

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 3147

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {b}{\left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 a b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\)	\(93\)
default	\(\frac {\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {b}{\left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 a b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\)	\(93\)
parallelrisch	\(\frac {\left (-2 b^{2} a^{2} \sin \left (d x +c \right )-2 a^{3} b \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-a \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a +b \right ) \left (\left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-b^{2} \sin \left (d x +c \right ) \left (a -b \right )\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a d \left (a +b \sin \left (d x +c \right )\right )}\)	\(162\)
norman	\(-\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}\)	\(173\)
risch	\(-\frac {i x}{a^{2}-2 a b +b^{2}}-\frac {i c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i x}{a^{2}+2 a b +b^{2}}+\frac {i c}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {4 i a b x}{a^{4}-2 b^{2} a^{2}+b^{4}}+\frac {4 i a b c}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )} b -b +2 i {\mathrm e}^{i \left (d x +c \right )} a \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}\)	\(295\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.81 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, a^{2} b - 2 \, b^{3} - 4 \, {\left (a b^{2} \sin \left (d x + c\right ) + a^{2} b\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{3} + 2 \, a^{2} b + a b^{2} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \] Input:

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

Output:

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

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.13 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, a b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, b}{a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.39 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \, a b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b d - 2 \, a^{2} b^{3} d + b^{5} d} - \frac {\log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} + \frac {\log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} + \frac {a^{2} b - b^{3}}{{\left (b \sin \left (d x + c\right ) + a\right )} {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 16.47 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.94 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,{\left (a-b\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,{\left (a+b\right )}^2}+\frac {b}{d\,\left (a^2-b^2\right )\,\left (a+b\,\sin \left (c+d\,x\right )\right )}-\frac {2\,a\,b\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,{\left (a^2-b^2\right )}^2} \] Input:

int(1/(cos(c + d*x)*(a + b*sin(c + d*x))^2),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 387, normalized size of antiderivative = 3.72 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a \,b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a \,b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right ) a \,b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{2} b +a^{2} b -b^{3}}{d \left (\sin \left (d x +c \right ) a^{4} b -2 \sin \left (d x +c \right ) a^{2} b^{3}+\sin \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

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

Output:

( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**2*b + 2*log(tan((c + d*x)/2) 
 - 1)*sin(c + d*x)*a*b**2 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b**3 - 
log(tan((c + d*x)/2) - 1)*a**3 + 2*log(tan((c + d*x)/2) - 1)*a**2*b - log( 
tan((c + d*x)/2) - 1)*a*b**2 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**2 
*b + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a*b**2 + log(tan((c + d*x)/2 
) + 1)*sin(c + d*x)*b**3 + log(tan((c + d*x)/2) + 1)*a**3 + 2*log(tan((c + 
 d*x)/2) + 1)*a**2*b + log(tan((c + d*x)/2) + 1)*a*b**2 - 2*log(tan((c + d 
*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a*b**2 - 2*log(tan((c 
 + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**2*b + a**2*b - b**3)/(d*(si 
n(c + d*x)*a**4*b - 2*sin(c + d*x)*a**2*b**3 + sin(c + d*x)*b**5 + a**5 - 
2*a**3*b**2 + a*b**4))