Integrand size = 19, antiderivative size = 60 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{4 a^2 d}+\frac {1}{4 d (a+a \sin (c+d x))^2}-\frac {1}{4 d \left (a^2+a^2 \sin (c+d x)\right )} \] Output:
1/4*arctanh(sin(d*x+c))/a^2/d+1/4/d/(a+a*sin(d*x+c))^2-1/4/d/(a^2+a^2*sin( d*x+c))
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.60 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))-\frac {\sin (c+d x)}{(1+\sin (c+d x))^2}}{4 a^2 d} \] Input:
Integrate[Tan[c + d*x]/(a + a*Sin[c + d*x])^2,x]
Output:
(ArcTanh[Sin[c + d*x]] - Sin[c + d*x]/(1 + Sin[c + d*x])^2)/(4*a^2*d)
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3186, 86, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {a \sin (c+d x)}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (-\frac {1}{2 (\sin (c+d x) a+a)^3}+\frac {1}{4 \left (a^2-a^2 \sin ^2(c+d x)\right ) a}+\frac {1}{4 (\sin (c+d x) a+a)^2 a}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\text {arctanh}(\sin (c+d x))}{4 a^2}-\frac {1}{4 a (a \sin (c+d x)+a)}+\frac {1}{4 (a \sin (c+d x)+a)^2}}{d}\) |
Input:
Int[Tan[c + d*x]/(a + a*Sin[c + d*x])^2,x]
Output:
(ArcTanh[Sin[c + d*x]]/(4*a^2) + 1/(4*(a + a*Sin[c + d*x])^2) - 1/(4*a*(a + a*Sin[c + d*x])))/d
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.67 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8}}{d \,a^{2}}\) | \(55\) |
default | \(\frac {\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8}}{d \,a^{2}}\) | \(55\) |
risch | \(-\frac {i \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d \,a^{2}}\) | \(88\) |
Input:
int(tan(d*x+c)/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(1/4/(1+sin(d*x+c))^2-1/4/(1+sin(d*x+c))+1/8*ln(1+sin(d*x+c))-1/8* ln(sin(d*x+c)-1))
Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.73 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \sin \left (d x + c\right ) - 2 \, a^{2} d\right )}} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/8*((cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(sin(d*x + c) + 1) - (cos(d* x + c)^2 - 2*sin(d*x + c) - 2)*log(-sin(d*x + c) + 1) + 2*sin(d*x + c))/(a ^2*d*cos(d*x + c)^2 - 2*a^2*d*sin(d*x + c) - 2*a^2*d)
\[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\tan {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c))**2,x)
Output:
Integral(tan(c + d*x)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, \sin \left (d x + c\right )}{a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, d} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-1/8*(2*sin(d*x + c)/(a^2*sin(d*x + c)^2 + 2*a^2*sin(d*x + c) + a^2) - log (sin(d*x + c) + 1)/a^2 + log(sin(d*x + c) - 1)/a^2)/d
Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.58 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2 \right |}\right )}{16 \, a^{2} d} - \frac {\log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2 \right |}\right )}{16 \, a^{2} d} - \frac {\frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 6}{16 \, a^{2} d {\left (\frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2\right )}} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/16*log(abs(1/sin(d*x + c) + sin(d*x + c) + 2))/(a^2*d) - 1/16*log(abs(1/ sin(d*x + c) + sin(d*x + c) - 2))/(a^2*d) - 1/16*(1/sin(d*x + c) + sin(d*x + c) + 6)/(a^2*d*(1/sin(d*x + c) + sin(d*x + c) + 2))
Time = 18.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.93 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\right )} \] Input:
int(tan(c + d*x)/(a + a*sin(c + d*x))^2,x)
Output:
atanh(tan(c/2 + (d*x)/2))/(2*a^2*d) - (tan(c/2 + (d*x)/2)/2 + tan(c/2 + (d *x)/2)^3/2)/(d*(6*a^2*tan(c/2 + (d*x)/2)^2 + 4*a^2*tan(c/2 + (d*x)/2)^3 + a^2*tan(c/2 + (d*x)/2)^4 + a^2 + 4*a^2*tan(c/2 + (d*x)/2)))
Time = 0.18 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.50 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right )^{2}+1}{8 a^{2} d \left (\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1\right )} \] Input:
int(tan(d*x+c)/(a+a*sin(d*x+c))^2,x)
Output:
( - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 2*log(tan((c + d*x)/2) - 1) + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 2*log(tan( (c + d*x)/2) + 1) + sin(c + d*x)**2 + 1)/(8*a**2*d*(sin(c + d*x)**2 + 2*si n(c + d*x) + 1))