Integrand size = 27, antiderivative size = 132 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {b^4 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d} \] Output:
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 = 0.60 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \left (\frac {\csc (c+d x)}{a^2}-\frac {\csc ^2(c+d x)}{2 a b}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)}+\frac {\log (\sin (c+d x))}{a b}+\frac {b \log (\sin (c+d x))}{a^3}-\frac {\log (1+\sin (c+d x))}{2 (a-b) b}+\frac {b^3 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )}\right )}{d} \] Input:
Integrate[(Csc[c + d*x]^3*Sec[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(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 = 0.41 (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 |
| \(\displaystyle \frac {b^4 \left (\frac {\csc (c+d x)}{a^2 b^3}+\frac {\log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )}+\frac {\left (a^2+b^2\right ) \log (b \sin (c+d x))}{a^3 b^4}-\frac {\csc ^2(c+d x)}{2 a b^4}-\frac {\log (b-b \sin (c+d x))}{2 b^4 (a+b)}-\frac {\log (b \sin (c+d x)+b)}{2 b^4 (a-b)}\right )}{d}\) |
Input:
Int[(Csc[c + d*x]^3*Sec[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(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
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 = 0.67 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{3}}-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\) | \(120\) |
| default | \(\frac {\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{3}}-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\) | \(120\) |
| parallelrisch | \(\frac {8 b^{4} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-8 a^{3} \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-8 \left (a +b \right ) \left (\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}+\frac {\left (\left (-8 a^{2}-8 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -4 \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) b +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 b \right )\right ) a \right ) \left (a -b \right )}{8}\right )}{8 a^{5} d -8 a^{3} b^{2} d}\) | \(180\) |
| norman | \(\frac {-\frac {1}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 a d}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2} d}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a^{2} d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,a^{3} \left (a^{2}-b^{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \left (a +b \right )}\) | \(199\) |
| risch | \(\frac {i x}{a +b}-\frac {2 i x}{a}-\frac {2 i c}{a d}-\frac {2 i b^{2} c}{d \,a^{3}}-\frac {2 i b^{2} x}{a^{3}}-\frac {2 i b^{4} x}{a^{3} \left (a^{2}-b^{2}\right )}+\frac {i c}{d \left (a +b \right )}-\frac {2 i b^{4} c}{d \,a^{3} \left (a^{2}-b^{2}\right )}+\frac {i x}{a -b}+\frac {2 i \left (-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {i c}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{d \,a^{3}}+\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \,a^{3} \left (a^{2}-b^{2}\right )}\) | \(330\) |
Input:
int(csc(d*x+c)^3*sec(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
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(sin(d*x+c)-1)-1/(2*a-2 *b)*ln(1+sin(d*x+c)))
Time = 0.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.70 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^{4} - a^{2} b^{2} + 2 \, {\left (b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (a^{4} - b^{4} - {\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (a^{4} + a^{3} b - {\left (a^{4} + a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} - a^{3} b - {\left (a^{4} - a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{5} - a^{3} b^{2}\right )} d\right )}} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
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)
\[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)**3*sec(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)**3*sec(c + d*x)/(a + b*sin(c + d*x)), x)
Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5} - a^{3} b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} + \frac {2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
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 = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b d - a^{3} b^{3} d} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a d - b d\right )}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a d + b d\right )}} + \frac {{\left (a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3} d} + \frac {2 \, a b \sin \left (d x + c\right ) - a^{2}}{2 \, a^{3} d \sin \left (d x + c\right )^{2}} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
b^5*log(abs(b*sin(d*x + c) + a))/(a^5*b*d - a^3*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) + ( a^2 + b^2)*log(abs(sin(d*x + c)))/(a^3*d) + 1/2*(2*a*b*sin(d*x + c) - a^2) /(a^3*d*sin(d*x + c)^2)
Time = 19.78 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )\right )\,\left (a^2+b^2\right )}{a^3\,d}-\frac {\frac {1}{2\,a}-\frac {b\,\sin \left (c+d\,x\right )}{a^2}}{d\,{\sin \left (c+d\,x\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,\left (a+b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,\left (a-b\right )}+\frac {b^4\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^5-a^3\,b^2\right )} \] Input:
int(1/(cos(c + d*x)*sin(c + d*x)^3*(a + b*sin(c + d*x))),x)
Output:
(log(sin(c + d*x))*(a^2 + b^2))/(a^3*d) - (1/(2*a) - (b*sin(c + d*x))/a^2) /(d*sin(c + d*x)^2) - log(sin(c + d*x) - 1)/(2*d*(a + b)) - log(sin(c + d* x) + 1)/(2*d*(a - b)) + (b^4*log(a + b*sin(c + d*x)))/(d*(a^5 - a^3*b^2))
Time = 0.19 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.14 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{4}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3} b -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{4}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3} b +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{2} b^{4}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{4}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{4}+\sin \left (d x +c \right )^{2} a^{4}-\sin \left (d x +c \right )^{2} a^{2} b^{2}+4 \sin \left (d x +c \right ) a^{3} b -4 \sin \left (d x +c \right ) a \,b^{3}-2 a^{4}+2 a^{2} b^{2}}{4 \sin \left (d x +c \right )^{2} a^{3} d \left (a^{2}-b^{2}\right )} \] Input:
int(csc(d*x+c)^3*sec(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
( - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 + 4*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**2*a**3*b - 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x) **2*a**4 - 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b + 4*log(tan( (c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*b**4 + 4*log (tan((c + d*x)/2))*sin(c + d*x)**2*a**4 - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**4 + sin(c + d*x)**2*a**4 - sin(c + d*x)**2*a**2*b**2 + 4*sin(c + d*x)*a**3*b - 4*sin(c + d*x)*a*b**3 - 2*a**4 + 2*a**2*b**2)/(4*sin(c + d *x)**2*a**3*d*(a**2 - b**2))