Integrand size = 19, antiderivative size = 37 \[ \int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{2 a d}+\frac {1}{2 d (a+a \sin (c+d x))} \] Output:
1/2*arctanh(sin(d*x+c))/a/d+1/2/d/(a+a*sin(d*x+c))
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))+\frac {1}{1+\sin (c+d x)}}{2 a d} \] Input:
Integrate[Tan[c + d*x]/(a + a*Sin[c + d*x]),x]
Output:
(ArcTanh[Sin[c + d*x]] + (1 + Sin[c + d*x])^(-1))/(2*a*d)
Time = 0.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.57, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3185, 3042, 3086, 15, 3091, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3185 |
\(\displaystyle \frac {\int \sec ^2(c+d x) \tan (c+d x)dx}{a}-\frac {\int \sec (c+d x) \tan ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)dx}{a}-\frac {\int \sec (c+d x) \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \sec (c+d x)d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x) \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\sec ^2(c+d x)}{2 a d}-\frac {\int \sec (c+d x) \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {\sec ^2(c+d x)}{2 a d}-\frac {\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \sec (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec ^2(c+d x)}{2 a d}-\frac {\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sec ^2(c+d x)}{2 a d}-\frac {\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d}}{a}\) |
Input:
Int[Tan[c + d*x]/(a + a*Sin[c + d*x]),x]
Output:
Sec[c + d*x]^2/(2*a*d) - (-1/2*ArcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan [c + d*x])/(2*d))/a
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{4}+\frac {1}{2+2 \sin \left (d x +c \right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{4}}{d a}\) | \(43\) |
default | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{4}+\frac {1}{2+2 \sin \left (d x +c \right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{4}}{d a}\) | \(43\) |
risch | \(\frac {i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}\) | \(76\) |
Input:
int(tan(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/4*ln(sin(d*x+c)-1)+1/2/(1+sin(d*x+c))+1/4*ln(1+sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.57 \[ \int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2}{4 \, {\left (a d \sin \left (d x + c\right ) + a d\right )}} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/4*((sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - (sin(d*x + c) + 1)*log(-si n(d*x + c) + 1) + 2)/(a*d*sin(d*x + c) + a*d)
\[ \int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\tan {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a} + \frac {2}{a \sin \left (d x + c\right ) + a}}{4 \, d} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/4*(log(sin(d*x + c) + 1)/a - log(sin(d*x + c) - 1)/a + 2/(a*sin(d*x + c) + a))/d
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.49 \[ \int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, a d} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, a d} + \frac {1}{2 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/4*log(abs(sin(d*x + c) + 1))/(a*d) - 1/4*log(abs(sin(d*x + c) - 1))/(a*d ) + 1/2/(a*d*(sin(d*x + c) + 1))
Time = 17.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.65 \[ \int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \] Input:
int(tan(c + d*x)/(a + a*sin(c + d*x)),x)
Output:
atanh(tan(c/2 + (d*x)/2))/(a*d) - tan(c/2 + (d*x)/2)/(d*(a + 2*a*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))
Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.30 \[ \int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+1}{2 a d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(tan(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2) - 1)*sin(c + d*x) - log(tan((c + d*x)/2) - 1) + l og(tan((c + d*x)/2) + 1)*sin(c + d*x) + log(tan((c + d*x)/2) + 1) + 1)/(2* a*d*(sin(c + d*x) + 1))