Integrand size = 19, antiderivative size = 75 \[ \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right ) d} \] Output:
-1/2*ln(1-sin(d*x+c))/(a+b)/d+1/2*ln(1+sin(d*x+c))/(a-b)/d-b*ln(a+b*sin(d* x+c))/(a^2-b^2)/d
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85 \[ \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {(-a+b) \log (1-\sin (c+d x))+(a+b) \log (1+\sin (c+d x))-2 b \log (a+b \sin (c+d x))}{2 (a-b) (a+b) d} \] Input:
Integrate[Sec[c + d*x]/(a + b*Sin[c + d*x]),x]
Output:
((-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*(a - b)*(a + b)*d)
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x) (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b \int \frac {1}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {\int \left (-\frac {b^2}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {b}{2 (a+b) (b-b \sin (c+d x))}+\frac {b}{2 (a-b) (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{b d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {b^2 \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {b \log (b-b \sin (c+d x))}{2 (a+b)}+\frac {b \log (b \sin (c+d x)+b)}{2 (a-b)}}{b d}\) |
Input:
Int[Sec[c + d*x]/(a + b*Sin[c + d*x]),x]
Output:
(-1/2*(b*Log[b - b*Sin[c + d*x]])/(a + b) - (b^2*Log[a + b*Sin[c + d*x]])/ (a^2 - b^2) + (b*Log[b + b*Sin[c + d*x]])/(2*(a - b)))/(b*d)
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]]
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]
Time = 0.62 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a -b \right ) \left (a +b \right )}}{d}\) | \(71\) |
default | \(\frac {\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a -b \right ) \left (a +b \right )}}{d}\) | \(71\) |
parallelrisch | \(\frac {-b \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \right )+\left (-a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (a +b \right )}{d \left (a^{2}-b^{2}\right )}\) | \(81\) |
norman | \(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \left (a +b \right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}{d \left (a^{2}-b^{2}\right )}\) | \(92\) |
risch | \(-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {2 i b x}{a^{2}-b^{2}}+\frac {2 i b c}{d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{2}-b^{2}\right )}\) | \(175\) |
Input:
int(sec(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/(2*a-2*b)*ln(1+sin(d*x+c))-1/(2*a+2*b)*ln(sin(d*x+c)-1)-b/(a-b)/(a+ b)*ln(a+b*sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83 \[ \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, b \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a - b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \] Input:
integrate(sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/2*(2*b*log(b*sin(d*x + c) + a) - (a + b)*log(sin(d*x + c) + 1) + (a - b )*log(-sin(d*x + c) + 1))/((a^2 - b^2)*d)
\[ \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Integral(sec(c + d*x)/(a + b*sin(c + d*x)), x)
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85 \[ \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \] Input:
integrate(sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/2*(2*b*log(b*sin(d*x + c) + a)/(a^2 - b^2) - log(sin(d*x + c) + 1)/(a - b) + log(sin(d*x + c) - 1)/(a + b))/d
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b d - b^{3} d} + \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a d - b d\right )}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a d + b d\right )}} \] Input:
integrate(sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-b^2*log(abs(b*sin(d*x + c) + a))/(a^2*b*d - b^3*d) + 1/2*log(abs(sin(d*x + c) + 1))/(a*d - b*d) - 1/2*log(abs(sin(d*x + c) - 1))/(a*d + b*d)
Time = 0.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,\left (a-b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,\left (a+b\right )}-\frac {b\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^2-b^2\right )} \] Input:
int(1/(cos(c + d*x)*(a + b*sin(c + d*x))),x)
Output:
log(sin(c + d*x) + 1)/(2*d*(a - b)) - log(sin(c + d*x) - 1)/(2*d*(a + b)) - (b*log(a + b*sin(c + d*x)))/(d*(a^2 - b^2))
Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.39 \[ \int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b}{d \left (a^{2}-b^{2}\right )} \] Input:
int(sec(d*x+c)/(a+b*sin(d*x+c)),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))