Integrand size = 27, antiderivative size = 110 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {b^3 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right ) d} \] Output:
-csc(d*x+c)/a/d-1/2*ln(1-sin(d*x+c))/(a+b)/d-b*ln(sin(d*x+c))/a^2/d+1/2*ln (1+sin(d*x+c))/(a-b)/d-b^3*ln(a+b*sin(d*x+c))/a^2/(a^2-b^2)/d
Time = 0.42 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.98 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \left (-\frac {\csc (c+d x)}{a b}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)}-\frac {\log (\sin (c+d x))}{a^2}+\frac {\log (1+\sin (c+d x))}{2 (a-b) b}-\frac {b^2 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )}\right )}{d} \] Input:
Integrate[(Csc[c + d*x]^2*Sec[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(b*(-(Csc[c + d*x]/(a*b)) - Log[1 - Sin[c + d*x]]/(2*b*(a + b)) - Log[Sin[ c + d*x]]/a^2 + Log[1 + Sin[c + d*x]]/(2*(a - b)*b) - (b^2*Log[a + b*Sin[c + d*x]])/(a^2*(a^2 - b^2))))/d
Time = 0.39 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3316, 27, 615, 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.
| \(\displaystyle \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x) (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b \int \frac {\csc ^2(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \int \frac {\csc ^2(c+d x)}{b^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^3 \int \left (\frac {\csc ^2(c+d x)}{a b^4}-\frac {\csc (c+d x)}{a^2 b^3}+\frac {1}{2 b^3 (a+b) (b-b \sin (c+d x))}-\frac {1}{a^2 (a-b) (a+b) (a+b \sin (c+d x))}-\frac {1}{2 b^3 (b-a) (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^3 \left (-\frac {\log (b \sin (c+d x))}{a^2 b^2}-\frac {\log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )}-\frac {\csc (c+d x)}{a b^3}-\frac {\log (b-b \sin (c+d x))}{2 b^3 (a+b)}+\frac {\log (b \sin (c+d x)+b)}{2 b^3 (a-b)}\right )}{d}\) |
Input:
Int[(Csc[c + d*x]^2*Sec[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(b^3*(-(Csc[c + d*x]/(a*b^3)) - Log[b*Sin[c + d*x]]/(a^2*b^2) - Log[b - b* Sin[c + d*x]]/(2*b^3*(a + b)) - Log[a + b*Sin[c + d*x]]/(a^2*(a^2 - b^2)) + Log[b + b*Sin[c + d*x]]/(2*(a - b)*b^3)))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
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]
Time = 0.54 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.93
| method | result | size |
| derivativedivides | \(\frac {-\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\) | \(102\) |
| default | \(\frac {-\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\) | \(102\) |
| parallelrisch | \(\frac {-\ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) b^{3}-a^{2} \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a +b \right ) \left (a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\left (a -b \right ) \left (a \csc \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{2}\right )}{d \left (a^{4}-a^{2} b^{2}\right )}\) | \(136\) |
| norman | \(\frac {-\frac {1}{2 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \left (a +b \right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}-\frac {b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{2} d \left (a^{2}-b^{2}\right )}\) | \(156\) |
| risch | \(-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {2 i b^{3} x}{a^{2} \left (a^{2}-b^{2}\right )}+\frac {2 i b^{3} c}{a^{2} d \left (a^{2}-b^{2}\right )}+\frac {2 i b x}{a^{2}}+\frac {2 i b c}{a^{2} d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{2} d \left (a^{2}-b^{2}\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(261\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-b^3/(a+b)/(a-b)/a^2*ln(a+b*sin(d*x+c))-1/a/sin(d*x+c)-1/a^2*b*ln(sin (d*x+c))+1/(2*a-2*b)*ln(1+sin(d*x+c))-1/(2*a+2*b)*ln(sin(d*x+c)-1))
Time = 0.18 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.30 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + 2 \, a^{3} - 2 \, a b^{2} + 2 \, {\left (a^{2} b - b^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - {\left (a^{3} + a^{2} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + {\left (a^{3} - a^{2} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/2*(2*b^3*log(b*sin(d*x + c) + a)*sin(d*x + c) + 2*a^3 - 2*a*b^2 + 2*(a^ 2*b - b^3)*log(1/2*sin(d*x + c))*sin(d*x + c) - (a^3 + a^2*b)*log(sin(d*x + c) + 1)*sin(d*x + c) + (a^3 - a^2*b)*log(-sin(d*x + c) + 1)*sin(d*x + c) )/((a^4 - a^2*b^2)*d*sin(d*x + c))
\[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)**2*sec(c + d*x)/(a + b*sin(c + d*x)), x)
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - a^{2} b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac {2 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}} + \frac {2}{a \sin \left (d x + c\right )}}{2 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/2*(2*b^3*log(b*sin(d*x + c) + a)/(a^4 - a^2*b^2) - log(sin(d*x + c) + 1 )/(a - b) + log(sin(d*x + c) - 1)/(a + b) + 2*b*log(sin(d*x + c))/a^2 + 2/ (a*sin(d*x + c)))/d
Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b d - a^{2} b^{3} d} + \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a d - b d\right )}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a d + b d\right )}} - \frac {b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2} d} - \frac {1}{a d \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-b^4*log(abs(b*sin(d*x + c) + a))/(a^4*b*d - a^2*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) - b*log(abs(sin(d*x + c)))/(a^2*d) - 1/(a*d*sin(d*x + c))
Time = 19.64 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,\left (a+b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,\left (a-b\right )}+\frac {1}{a\,\sin \left (c+d\,x\right )}+\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{a^2}+\frac {b^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a^4-a^2\,b^2}}{d} \] Input:
int(1/(cos(c + d*x)*sin(c + d*x)^2*(a + b*sin(c + d*x))),x)
Output:
-(log(sin(c + d*x) - 1)/(2*(a + b)) - log(sin(c + d*x) + 1)/(2*(a - b)) + 1/(a*sin(c + d*x)) + (b*log(sin(c + d*x)))/a^2 + (b^3*log(a + b*sin(c + d* x)))/(a^4 - a^2*b^2))/d
Time = 0.19 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.90 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) b^{3}-a^{3}+a \,b^{2}}{\sin \left (d x +c \right ) a^{2} d \left (a^{2}-b^{2}\right )} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**3 + log(tan((c + d*x)/2) - 1 )*sin(c + d*x)*a**2*b + log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**3 + log( tan((c + d*x)/2) + 1)*sin(c + d*x)*a**2*b - log(tan((c + d*x)/2)**2*a + 2* tan((c + d*x)/2)*b + a)*sin(c + d*x)*b**3 - log(tan((c + d*x)/2))*sin(c + d*x)*a**2*b + log(tan((c + d*x)/2))*sin(c + d*x)*b**3 - a**3 + a*b**2)/(si n(c + d*x)*a**2*d*(a**2 - b**2))