Integrand size = 27, antiderivative size = 132 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {b^4 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d} \]
b*csc(d*x+c)/a^2/d-1/2*csc(d*x+c)^2/a/d-1/2*ln(1-sin(d*x+c))/(a+b)/d+(a^2+ b^2)*ln(sin(d*x+c))/a^3/d-1/2*ln(1+sin(d*x+c))/(a-b)/d+b^4*ln(a+b*sin(d*x+ c))/a^3/(a^2-b^2)/d
Time = 0.38 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \left (\frac {\csc (c+d x)}{a^2}-\frac {\csc ^2(c+d x)}{2 a b}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)}+\frac {\log (\sin (c+d x))}{a b}+\frac {b \log (\sin (c+d x))}{a^3}-\frac {\log (1+\sin (c+d x))}{2 (a-b) b}+\frac {b^3 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )}\right )}{d} \]
(b*(Csc[c + d*x]/a^2 - Csc[c + d*x]^2/(2*a*b) - Log[1 - Sin[c + d*x]]/(2*b *(a + b)) + Log[Sin[c + d*x]]/(a*b) + (b*Log[Sin[c + d*x]])/a^3 - Log[1 + Sin[c + d*x]]/(2*(a - b)*b) + (b^3*Log[a + b*Sin[c + d*x]])/(a^3*(a^2 - b^ 2))))/d
Time = 0.38 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.04, 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.
| \(\displaystyle \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^3 \cos (c+d x) (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b \int \frac {\csc ^3(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^4 \int \frac {\csc ^3(c+d x)}{b^3 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^4 \int \left (\frac {\csc ^3(c+d x)}{a b^5}-\frac {\csc ^2(c+d x)}{a^2 b^4}+\frac {\left (a^2+b^2\right ) \csc (c+d x)}{a^3 b^5}+\frac {1}{2 b^4 (a+b) (b-b \sin (c+d x))}+\frac {1}{a^3 (a-b) (a+b) (a+b \sin (c+d x))}+\frac {1}{2 b^4 (b-a) (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^4 \left (\frac {\csc (c+d x)}{a^2 b^3}+\frac {\log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )}+\frac {\left (a^2+b^2\right ) \log (b \sin (c+d x))}{a^3 b^4}-\frac {\csc ^2(c+d x)}{2 a b^4}-\frac {\log (b-b \sin (c+d x))}{2 b^4 (a+b)}-\frac {\log (b \sin (c+d x)+b)}{2 b^4 (a-b)}\right )}{d}\) |
(b^4*(Csc[c + d*x]/(a^2*b^3) - Csc[c + d*x]^2/(2*a*b^4) + ((a^2 + b^2)*Log [b*Sin[c + d*x]])/(a^3*b^4) - Log[b - b*Sin[c + d*x]]/(2*b^4*(a + b)) + Lo g[a + b*Sin[c + d*x]]/(a^3*(a^2 - b^2)) - Log[b + b*Sin[c + d*x]]/(2*(a - b)*b^4)))/d
3.14.37.3.1 Defintions of rubi rules used
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 = 0.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{3}}-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\) | \(120\) |
| default | \(\frac {\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{3}}-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\) | \(120\) |
| parallelrisch | \(\frac {8 \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) b^{4}-8 a^{3} \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-8 \left (\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}+\frac {\left (\left (-8 a^{2}-8 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 b \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 b \right )\right ) a \right ) \left (a -b \right )}{8}\right ) \left (a +b \right )}{8 a^{5} d -8 a^{3} b^{2} d}\) | \(180\) |
| norman | \(\frac {-\frac {1}{8 a d}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2} d}+\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}+\frac {b^{4} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{3} d \left (a^{2}-b^{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a +b \right ) d}\) | \(199\) |
| risch | \(\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {i c}{d \left (a -b \right )}+\frac {2 i \left (-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {i x}{a -b}-\frac {2 i c}{d a}-\frac {2 i b^{4} c}{a^{3} d \left (a^{2}-b^{2}\right )}-\frac {2 i b^{2} c}{a^{3} d}-\frac {2 i b^{4} x}{a^{3} \left (a^{2}-b^{2}\right )}-\frac {2 i x}{a}-\frac {2 i b^{2} x}{a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}+\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{3} d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}\) | \(330\) |
1/d*(b^4/(a+b)/(a-b)/a^3*ln(a+b*sin(d*x+c))-1/2/a/sin(d*x+c)^2+(a^2+b^2)/a ^3*ln(sin(d*x+c))+1/a^2*b/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 = 0.62 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.70 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^{4} - a^{2} b^{2} + 2 \, {\left (b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (a^{4} - b^{4} - {\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (a^{4} + a^{3} b - {\left (a^{4} + a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} - a^{3} b - {\left (a^{4} - a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{5} - a^{3} b^{2}\right )} d\right )}} \]
1/2*(a^4 - a^2*b^2 + 2*(b^4*cos(d*x + c)^2 - b^4)*log(b*sin(d*x + c) + a) - 2*(a^4 - b^4 - (a^4 - b^4)*cos(d*x + c)^2)*log(-1/2*sin(d*x + c)) + (a^4 + a^3*b - (a^4 + a^3*b)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + (a^4 - a^ 3*b - (a^4 - a^3*b)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a^3*b - a* b^3)*sin(d*x + c))/((a^5 - a^3*b^2)*d*cos(d*x + c)^2 - (a^5 - a^3*b^2)*d)
\[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5} - a^{3} b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} + \frac {2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]
1/2*(2*b^4*log(b*sin(d*x + c) + a)/(a^5 - a^3*b^2) - log(sin(d*x + c) + 1) /(a - b) - log(sin(d*x + c) - 1)/(a + b) + 2*(a^2 + b^2)*log(sin(d*x + c)) /a^3 + (2*b*sin(d*x + c) - a)/(a^2*sin(d*x + c)^2))/d
Time = 0.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b - a^{3} b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2} + 3 \, b^{2} \sin \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) + a^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \]
1/2*(2*b^5*log(abs(b*sin(d*x + c) + a))/(a^5*b - a^3*b^3) - log(abs(sin(d* x + c) + 1))/(a - b) - log(abs(sin(d*x + c) - 1))/(a + b) + 2*(a^2 + b^2)* log(abs(sin(d*x + c)))/a^3 - (3*a^2*sin(d*x + c)^2 + 3*b^2*sin(d*x + c)^2 - 2*a*b*sin(d*x + c) + a^2)/(a^3*sin(d*x + c)^2))/d
Time = 11.79 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )\right )\,\left (a^2+b^2\right )}{a^3\,d}-\frac {\frac {1}{2\,a}-\frac {b\,\sin \left (c+d\,x\right )}{a^2}}{d\,{\sin \left (c+d\,x\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,\left (a+b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,\left (a-b\right )}+\frac {b^4\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^5-a^3\,b^2\right )} \]