Integrand size = 29, antiderivative size = 171 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {(3 a+4 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {(3 a-4 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {b^5 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sin (c+d x))}-\frac {1}{4 (a-b) d (1+\sin (c+d x))} \] Output:
-csc(d*x+c)/a/d-1/4*(3*a+4*b)*ln(1-sin(d*x+c))/(a+b)^2/d-b*ln(sin(d*x+c))/ a^2/d+1/4*(3*a-4*b)*ln(1+sin(d*x+c))/(a-b)^2/d+b^5*ln(a+b*sin(d*x+c))/a^2/ (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 = 0.88 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x) (a+b \sin (c+d x)) \left (\frac {4 \csc (c+d x)}{a}+\frac {(3 a+4 b) \log (1-\sin (c+d x))}{(a+b)^2}+\frac {4 b \log (\sin (c+d x))}{a^2}-\frac {(3 a-4 b) \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 b^5 \log (a+b \sin (c+d x))}{a^2 (a-b)^2 (a+b)^2}+\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}\right )}{4 d (b+a \csc (c+d x))} \] Input:
Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^3)/(a + b*Sin[c + d*x]),x]
Output:
-1/4*(Csc[c + d*x]*(a + b*Sin[c + d*x])*((4*Csc[c + d*x])/a + ((3*a + 4*b) *Log[1 - Sin[c + d*x]])/(a + b)^2 + (4*b*Log[Sin[c + d*x]])/a^2 - ((3*a - 4*b)*Log[1 + Sin[c + d*x]])/(a - b)^2 - (4*b^5*Log[a + b*Sin[c + d*x]])/(a ^2*(a - b)^2*(a + b)^2) + 1/((a + b)*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x]))))/(d*(b + a*Csc[c + d*x]))
Time = 0.48 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
\(\displaystyle \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x)^3 (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b^3 \int \frac {\csc ^2(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^5 \int \frac {\csc ^2(c+d x)}{b^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \frac {b^5 \int \left (\frac {\csc ^2(c+d x)}{a b^6}-\frac {\csc (c+d x)}{a^2 b^5}+\frac {3 a+4 b}{4 b^5 (a+b)^2 (b-b \sin (c+d x))}+\frac {1}{a^2 (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac {3 a-4 b}{4 (a-b)^2 b^5 (\sin (c+d x) b+b)}+\frac {1}{4 b^4 (a+b) (b-b \sin (c+d x))^2}+\frac {1}{4 (a-b) b^4 (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^5 \left (-\frac {\log (b \sin (c+d x))}{a^2 b^4}+\frac {\log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )^2}-\frac {\csc (c+d x)}{a b^5}-\frac {(3 a+4 b) \log (b-b \sin (c+d x))}{4 b^5 (a+b)^2}+\frac {(3 a-4 b) \log (b \sin (c+d x)+b)}{4 b^5 (a-b)^2}+\frac {1}{4 b^4 (a+b) (b-b \sin (c+d x))}-\frac {1}{4 b^4 (a-b) (b \sin (c+d x)+b)}\right )}{d}\) |
Input:
Int[(Csc[c + d*x]^2*Sec[c + d*x]^3)/(a + b*Sin[c + d*x]),x]
Output:
(b^5*(-(Csc[c + d*x]/(a*b^5)) - Log[b*Sin[c + d*x]]/(a^2*b^4) - ((3*a + 4* b)*Log[b - b*Sin[c + d*x]])/(4*b^5*(a + b)^2) + Log[a + b*Sin[c + d*x]]/(a ^2*(a^2 - b^2)^2) + ((3*a - 4*b)*Log[b + b*Sin[c + d*x]])/(4*(a - b)^2*b^5 ) + 1/(4*b^4*(a + b)*(b - b*Sin[c + d*x])) - 1/(4*(a - b)*b^4*(b + b*Sin[c + d*x]))))/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 = 1.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a -4 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a -4 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\) | \(152\) |
default | \(\frac {-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a -4 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a -4 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\) | \(152\) |
parallelrisch | \(\frac {4 \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) b^{5}-6 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right )^{2} \left (a +\frac {4 b}{3}\right ) a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (a +b \right ) \left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \left (a -\frac {4 b}{3}\right ) a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\left (a -b \right ) \left (\frac {4 b \left (a -b \right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (a^{2}-\frac {2 b^{2}}{3}\right ) \cos \left (2 d x +2 c \right )+\frac {a^{2}}{3}-\frac {2 b^{2}}{3}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 b a \left (-1+\cos \left (2 d x +2 c \right )\right )}{3}\right ) a \right )}{2}\right )}{4 \left (a -b \right )^{2} \left (a +b \right )^{2} a^{2} d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(266\) |
norman | \(\frac {-\frac {1}{2 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a d}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d \left (a^{2}-b^{2}\right )}+\frac {\left (3 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d a \left (a^{2}-b^{2}\right )}+\frac {\left (3 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 d a \left (a^{2}-b^{2}\right )}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {b^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}+\frac {\left (3 a -4 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (3 a +4 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}\) | \(315\) |
risch | \(\frac {2 i b x}{a^{2}+2 a b +b^{2}}+\frac {3 i a c}{2 d \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 i a x}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b c}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 i a x}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i b x}{a^{2}-2 a b +b^{2}}-\frac {3 i a c}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 i b^{5} x}{a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 i b c}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {i \left (3 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-2 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+2 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-2 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{i \left (d x +c \right )}-2 b^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {2 i b c}{a^{2} d}+\frac {2 i b x}{a^{2}}-\frac {2 i b^{5} c}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(597\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/(4*a+4*b)/(sin(d*x+c)-1)+1/4/(a+b)^2*(-3*a-4*b)*ln(sin(d*x+c)-1)-1 /a/sin(d*x+c)-1/a^2*b*ln(sin(d*x+c))+b^5/(a+b)^2/(a-b)^2/a^2*ln(a+b*sin(d* x+c))-1/(4*a-4*b)/(1+sin(d*x+c))+1/4*(3*a-4*b)/(a-b)^2*ln(1+sin(d*x+c)))
Time = 0.67 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.68 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {4 \, b^{5} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} - 4 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (3 \, a^{5} + 2 \, a^{4} b - 5 \, a^{3} b^{2} - 4 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - {\left (3 \, a^{5} - 2 \, a^{4} b - 5 \, a^{3} b^{2} + 4 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 2 \, {\left (3 \, a^{5} - 5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/4*(4*b^5*cos(d*x + c)^2*log(b*sin(d*x + c) + a)*sin(d*x + c) + 2*a^5 - 2 *a^3*b^2 - 4*(a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)^2*log(1/2*sin(d*x + c) )*sin(d*x + c) + (3*a^5 + 2*a^4*b - 5*a^3*b^2 - 4*a^2*b^3)*cos(d*x + c)^2* log(sin(d*x + c) + 1)*sin(d*x + c) - (3*a^5 - 2*a^4*b - 5*a^3*b^2 + 4*a^2* b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1)*sin(d*x + c) - 2*(3*a^5 - 5*a^3 *b^2 + 2*a*b^4)*cos(d*x + c)^2 - 2*(a^4*b - a^2*b^3)*sin(d*x + c))/((a^6 - 2*a^4*b^2 + a^2*b^4)*d*cos(d*x + c)^2*sin(d*x + c))
\[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)**3/(a+b*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)**2*sec(c + d*x)**3/(a + b*sin(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.17 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {4 \, b^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}} + \frac {{\left (3 \, a - 4 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (3 \, a + 4 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a b \sin \left (d x + c\right ) - {\left (3 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} + 2 \, a^{2} - 2 \, b^{2}\right )}}{{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )} - \frac {4 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{4 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/4*(4*b^5*log(b*sin(d*x + c) + a)/(a^6 - 2*a^4*b^2 + a^2*b^4) + (3*a - 4* b)*log(sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) - (3*a + 4*b)*log(sin(d*x + c ) - 1)/(a^2 + 2*a*b + b^2) + 2*(a*b*sin(d*x + c) - (3*a^2 - 2*b^2)*sin(d*x + c)^2 + 2*a^2 - 2*b^2)/((a^3 - a*b^2)*sin(d*x + c)^3 - (a^3 - a*b^2)*sin (d*x + c)) - 4*b*log(sin(d*x + c))/a^2)/d
Time = 0.21 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.50 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 2 \, a^{4} b^{3} d + a^{2} b^{5} d} - \frac {{\left (3 \, a + 4 \, b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} + \frac {{\left (3 \, a - 4 \, b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} - \frac {b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2} d} + \frac {2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} - {\left (3 \, a^{5} - 5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} a^{2} d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
b^6*log(abs(b*sin(d*x + c) + a))/(a^6*b*d - 2*a^4*b^3*d + a^2*b^5*d) - 1/4 *(3*a + 4*b)*log(abs(-sin(d*x + c) + 1))/(a^2*d + 2*a*b*d + b^2*d) + 1/4*( 3*a - 4*b)*log(abs(-sin(d*x + c) - 1))/(a^2*d - 2*a*b*d + b^2*d) - b*log(a bs(sin(d*x + c)))/(a^2*d) + 1/2*(2*a^5 - 4*a^3*b^2 + 2*a*b^4 - (3*a^5 - 5* a^3*b^2 + 2*a*b^4)*sin(d*x + c)^2 + (a^4*b - a^2*b^3)*sin(d*x + c))/((a + b)^2*(a - b)^2*a^2*d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)*sin(d*x + c))
Time = 19.96 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.14 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (3\,a-4\,b\right )}{4\,d\,{\left (a-b\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {3}{4\,\left (a+b\right )}\right )}{d}-\frac {\frac {1}{a}+\frac {b\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (3\,a^2-2\,b^2\right )}{2\,a\,\left (a^2-b^2\right )}}{d\,\left (\sin \left (c+d\,x\right )-{\sin \left (c+d\,x\right )}^3\right )}-\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{a^2\,d}+\frac {b^5\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a^2\,d\,{\left (a^2-b^2\right )}^2} \] Input:
int(1/(cos(c + d*x)^3*sin(c + d*x)^2*(a + b*sin(c + d*x))),x)
Output:
(log(sin(c + d*x) + 1)*(3*a - 4*b))/(4*d*(a - b)^2) - (log(sin(c + d*x) - 1)*(b/(4*(a + b)^2) + 3/(4*(a + b))))/d - (1/a + (b*sin(c + d*x))/(2*(a^2 - b^2)) - (sin(c + d*x)^2*(3*a^2 - 2*b^2))/(2*a*(a^2 - b^2)))/(d*(sin(c + d*x) - sin(c + d*x)^3)) - (b*log(sin(c + d*x)))/(a^2*d) + (b^5*log(a + b*s in(c + d*x)))/(a^2*d*(a^2 - b^2)^2)
Time = 0.20 (sec) , antiderivative size = 827, normalized size of antiderivative = 4.84 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x)
Output:
( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**5 + 2*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**3*a**4*b + 5*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **3*a**3*b**2 - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**2*b**3 + 3* log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**5 - 2*log(tan((c + d*x)/2) - 1)* sin(c + d*x)*a**4*b - 5*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**3*b**2 + 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**2*b**3 + 3*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**3*a**5 + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 *a**4*b - 5*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a**3*b**2 - 4*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)**3*a**2*b**3 - 3*log(tan((c + d*x)/2) + 1 )*sin(c + d*x)*a**5 - 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**4*b + 5* log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**3*b**2 + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**2*b**3 + 2*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x )/2)*b + a)*sin(c + d*x)**3*b**5 - 2*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*b**5 - 2*log(tan((c + d*x)/2))*sin(c + d*x)* *3*a**4*b + 4*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**2*b**3 - 2*log(tan( (c + d*x)/2))*sin(c + d*x)**3*b**5 + 2*log(tan((c + d*x)/2))*sin(c + d*x)* a**4*b - 4*log(tan((c + d*x)/2))*sin(c + d*x)*a**2*b**3 + 2*log(tan((c + d *x)/2))*sin(c + d*x)*b**5 - sin(c + d*x)**3*a**4*b + sin(c + d*x)**3*a**2* b**3 - 3*sin(c + d*x)**2*a**5 + 5*sin(c + d*x)**2*a**3*b**2 - 2*sin(c + d* x)**2*a*b**4 + 2*sin(c + d*x)*a**4*b - 2*sin(c + d*x)*a**2*b**3 + 2*a**...