Integrand size = 27, antiderivative size = 156 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {\log (\sin (c+d x))}{a d}-\frac {(2 a-3 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac {b^4 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sin (c+d x))}+\frac {1}{4 (a-b) d (1+\sin (c+d x))} \]
-1/4*(2*a+3*b)*ln(1-sin(d*x+c))/(a+b)^2/d+ln(sin(d*x+c))/a/d-1/4*(2*a-3*b) *ln(1+sin(d*x+c))/(a-b)^2/d-b^4*ln(a+b*sin(d*x+c))/a/(a^2-b^2)^2/d+1/4/(a+ b)/d/(1-sin(d*x+c))+1/4/(a-b)/d/(1+sin(d*x+c))
Time = 0.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^4 \left (-\frac {(2 a+3 b) \log (1-\sin (c+d x))}{b^4 (a+b)^2}+\frac {4 \log (\sin (c+d x))}{a b^4}-\frac {(2 a-3 b) \log (1+\sin (c+d x))}{(a-b)^2 b^4}-\frac {4 \log (a+b \sin (c+d x))}{a (a-b)^2 (a+b)^2}-\frac {1}{b^4 (a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) b^4 (1+\sin (c+d x))}\right )}{4 d} \]
(b^4*(-(((2*a + 3*b)*Log[1 - Sin[c + d*x]])/(b^4*(a + b)^2)) + (4*Log[Sin[ c + d*x]])/(a*b^4) - ((2*a - 3*b)*Log[1 + Sin[c + d*x]])/((a - b)^2*b^4) - (4*Log[a + b*Sin[c + d*x]])/(a*(a - b)^2*(a + b)^2) - 1/(b^4*(a + b)*(-1 + Sin[c + d*x])) + 1/((a - b)*b^4*(1 + Sin[c + d*x]))))/(4*d)
Time = 0.41 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x) \cos (c+d x)^3 (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b^3 \int \frac {\csc (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^4 \int \frac {\csc (c+d x)}{b (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \frac {b^4 \int \left (\frac {3 b-2 a}{4 (a-b)^2 b^4 (\sin (c+d x) b+b)}+\frac {\csc (c+d x)}{a b^5}+\frac {2 a+3 b}{4 b^4 (a+b)^2 (b-b \sin (c+d x))}-\frac {1}{a (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac {1}{4 b^3 (a+b) (b-b \sin (c+d x))^2}-\frac {1}{4 (a-b) b^3 (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^4 \left (-\frac {\log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )^2}+\frac {\log (b \sin (c+d x))}{a b^4}-\frac {(2 a+3 b) \log (b-b \sin (c+d x))}{4 b^4 (a+b)^2}-\frac {(2 a-3 b) \log (b \sin (c+d x)+b)}{4 b^4 (a-b)^2}+\frac {1}{4 b^3 (a+b) (b-b \sin (c+d x))}+\frac {1}{4 b^3 (a-b) (b \sin (c+d x)+b)}\right )}{d}\) |
(b^4*(Log[b*Sin[c + d*x]]/(a*b^4) - ((2*a + 3*b)*Log[b - b*Sin[c + d*x]])/ (4*b^4*(a + b)^2) - Log[a + b*Sin[c + d*x]]/(a*(a^2 - b^2)^2) - ((2*a - 3* b)*Log[b + b*Sin[c + d*x]])/(4*(a - b)^2*b^4) + 1/(4*b^3*(a + b)*(b - b*Si n[c + d*x])) + 1/(4*(a - b)*b^3*(b + b*Sin[c + d*x]))))/d
3.14.49.3.1 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 = 0.68 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-2 a +3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}}{d}\) | \(137\) |
default | \(\frac {\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-2 a +3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}}{d}\) | \(137\) |
parallelrisch | \(\frac {-b^{4} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (a -b \right )^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {3 b}{2}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-\left (a -\frac {3 b}{2}\right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (a -b \right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a \left (a \cos \left (2 d x +2 c \right )+2 b \sin \left (d x +c \right )-a \right )}{2}\right ) \left (a -b \right )\right ) \left (a +b \right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(218\) |
norman | \(\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right )}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (2 a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {b^{4} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(248\) |
risch | \(\frac {i a c}{\left (a^{2}-2 a b +b^{2}\right ) d}+\frac {i a c}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 i b c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {3 i b c}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i b^{4} x}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i x}{a}+\frac {i \left (-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left (a^{2}-b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 i b x}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 i b x}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b^{4} c}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i c}{d a}+\frac {i a x}{a^{2}+2 a b +b^{2}}+\frac {i a x}{a^{2}-2 a b +b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{\left (a^{2}-2 a b +b^{2}\right ) d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(502\) |
1/d*(1/a*ln(sin(d*x+c))+1/(4*a-4*b)/(1+sin(d*x+c))+1/4/(a-b)^2*(-2*a+3*b)* ln(1+sin(d*x+c))-1/(4*a+4*b)/(sin(d*x+c)-1)+1/4/(a+b)^2*(-2*a-3*b)*ln(sin( d*x+c)-1)-b^4/(a+b)^2/(a-b)^2/a*ln(a+b*sin(d*x+c)))
Time = 0.69 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.37 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {4 \, b^{4} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, a^{4} + 2 \, a^{2} b^{2} - 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]
-1/4*(4*b^4*cos(d*x + c)^2*log(b*sin(d*x + c) + a) - 2*a^4 + 2*a^2*b^2 - 4 *(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2*log(-1/2*sin(d*x + c)) + (2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2* (a^3*b - a*b^3)*sin(d*x + c))/((a^5 - 2*a^3*b^2 + a*b^4)*d*cos(d*x + c)^2)
\[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}} - \frac {4 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{4 \, d} \]
-1/4*(4*b^4*log(b*sin(d*x + c) + a)/(a^5 - 2*a^3*b^2 + a*b^4) + (2*a - 3*b )*log(sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) + (2*a + 3*b)*log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2) - 2*(b*sin(d*x + c) - a)/((a^2 - b^2)*sin(d*x + c)^2 - a^2 + b^2) - 4*log(sin(d*x + c))/a)/d
Time = 0.45 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.35 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b - 2 \, a^{3} b^{3} + a b^{5}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {2 \, {\left (a^{3} \sin \left (d x + c\right )^{2} - 2 \, a b^{2} \sin \left (d x + c\right )^{2} + a^{2} b \sin \left (d x + c\right ) - b^{3} \sin \left (d x + c\right ) - 2 \, a^{3} + 3 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]
-1/4*(4*b^5*log(abs(b*sin(d*x + c) + a))/(a^5*b - 2*a^3*b^3 + a*b^5) + (2* a - 3*b)*log(abs(sin(d*x + c) + 1))/(a^2 - 2*a*b + b^2) + (2*a + 3*b)*log( abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) - 4*log(abs(sin(d*x + c)))/a - 2*(a^3*sin(d*x + c)^2 - 2*a*b^2*sin(d*x + c)^2 + a^2*b*sin(d*x + c) - b^3* sin(d*x + c) - 2*a^3 + 3*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(sin(d*x + c)^2 - 1)))/d
Time = 13.00 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.09 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b}{4\,{\left (a-b\right )}^2}-\frac {1}{2\,\left (a-b\right )}\right )}{d}-\frac {\frac {a}{2\,\left (a^2-b^2\right )}-\frac {b\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )\right )}{a\,d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{2\,\left (a+b\right )}\right )}{d}-\frac {b^4\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a\,d\,{\left (a^2-b^2\right )}^2} \]