Integrand size = 27, antiderivative size = 229 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 b^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac {2 b^3 \left (3 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {b^4 \cos (c+d x)}{a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \]
2*b^3*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)/d+2 *b^3*(3*a^2-b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2 -b^2)^(5/2)/d-arctanh(cos(d*x+c))/a^2/d+1/2*cos(d*x+c)/(a+b)^2/d/(1-sin(d* x+c))+1/2*cos(d*x+c)/(a-b)^2/d/(1+sin(d*x+c))+b^4*cos(d*x+c)/a/(a^2-b^2)^2 /d/(a+b*sin(d*x+c))
Time = 2.16 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.89 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\frac {2 \left (-4 a^2 b^3+b^5\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2}}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {1}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+\frac {-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a b^4 \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}}{a^2}}{d} \]
((-2*(-4*a^2*b^3 + b^5)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/ (a^2*(a^2 - b^2)^(5/2)) + Sin[(c + d*x)/2]*(1/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - 1/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) )) + (-Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]] + (a*b^4*Cos[c + d*x] )/((a - b)^2*(a + b)^2*(a + b*Sin[c + d*x])))/a^2)/d
Time = 0.58 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3376, 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.
| \(\displaystyle \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x) \cos (c+d x)^2 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3376 |
| \(\displaystyle \int \left (-\frac {b^3}{a \left (b^2-a^2\right ) (a+b \sin (c+d x))^2}+\frac {3 a^2 b^3-b^5}{a^2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\csc (c+d x)}{a^2}-\frac {1}{2 (a+b)^2 (\sin (c+d x)-1)}-\frac {1}{2 (a-b)^2 (\sin (c+d x)+1)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b^3 \left (3 a^2-b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}+\frac {2 b^3 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {b^4 \cos (c+d x)}{a d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)}\) |
(2*b^3*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2 )*d) + (2*b^3*(3*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2 ]])/(a^2*(a^2 - b^2)^(5/2)*d) - ArcTanh[Cos[c + d*x]]/(a^2*d) + Cos[c + d* x]/(2*(a + b)^2*d*(1 - Sin[c + d*x])) + Cos[c + d*x]/(2*(a - b)^2*d*(1 + S in[c + d*x])) + (b^4*Cos[c + d*x])/(a*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]) )
3.15.68.3.1 Defintions of rubi rules used
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
Time = 1.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {4 b^{3} \left (\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}+\frac {a b}{2}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (4 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a^{2}}}{d}\) | \(185\) |
| default | \(\frac {-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {4 b^{3} \left (\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}+\frac {a b}{2}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (4 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a^{2}}}{d}\) | \(185\) |
| risch | \(\frac {4 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-2 i a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-6 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-4 i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+4 a^{2} b^{2}+2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right ) \left (a^{2}-b^{2}\right )^{2} d}+\frac {4 i b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}-\frac {4 i b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\) | \(526\) |
1/d*(-1/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)+1/a^2*ln(tan(1/2*d*x+1/2*c))+1/(a-b )^2/(tan(1/2*d*x+1/2*c)+1)+4*b^3/(a-b)^2/(a+b)^2/a^2*((1/2*tan(1/2*d*x+1/2 *c)*b^2+1/2*a*b)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)+1/2*(4* a^2-b^2)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2) ^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (213) = 426\).
Time = 0.79 (sec) , antiderivative size = 957, normalized size of antiderivative = 4.18 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \]
[1/2*(2*a^7 - 4*a^5*b^2 + 2*a^3*b^4 + 2*(2*a^5*b^2 - a^3*b^4 - a*b^6)*cos( d*x + c)^2 + ((4*a^2*b^4 - b^6)*cos(d*x + c)*sin(d*x + c) + (4*a^3*b^3 - a *b^5)*cos(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d* x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - ((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + ((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^6*b - 2*a^4*b^3 + a^2*b^5)*sin(d*x + c))/((a^8*b - 3*a^6* b^3 + 3*a^4*b^5 - a^2*b^7)*d*cos(d*x + c)*sin(d*x + c) + (a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)), 1/2*(2*a^7 - 4*a^5*b^2 + 2*a^3*b^4 + 2*(2*a^5*b^2 - a^3*b^4 - a*b^6)*cos(d*x + c)^2 - 2*((4*a^2*b^4 - b^6)*c os(d*x + c)*sin(d*x + c) + (4*a^3*b^3 - a*b^5)*cos(d*x + c))*sqrt(a^2 - b^ 2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - ((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)*sin(d*x + c) + (a^7 - 3*a^5*b^ 2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + ((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)*sin(d*x + c) + (a^7 - 3*a^5*b ^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^ 6*b - 2*a^4*b^3 + a^2*b^5)*sin(d*x + c))/((a^8*b - 3*a^6*b^3 + 3*a^4*b^...
\[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\csc {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.35 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.37 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (4 \, a^{2} b^{3} - b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (2 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{5} - a^{3} b^{2} - a b^{4}\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}}}{d} \]
(2*(4*a^2*b^3 - b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a* tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^6 - 2*a^4*b^2 + a^2*b^4)*s qrt(a^2 - b^2)) + 2*(2*a^4*b*tan(1/2*d*x + 1/2*c)^3 + b^5*tan(1/2*d*x + 1/ 2*c)^3 - a^5*tan(1/2*d*x + 1/2*c)^2 + 3*a^3*b^2*tan(1/2*d*x + 1/2*c)^2 + a *b^4*tan(1/2*d*x + 1/2*c)^2 - 2*a^2*b^3*tan(1/2*d*x + 1/2*c) - b^5*tan(1/2 *d*x + 1/2*c) - a^5 - a^3*b^2 - a*b^4)/((a^6 - 2*a^4*b^2 + a^2*b^4)*(a*tan (1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 1/2*c ) - a)) + log(abs(tan(1/2*d*x + 1/2*c)))/a^2)/d
Time = 14.78 (sec) , antiderivative size = 2076, normalized size of antiderivative = 9.07 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \]
log(tan(c/2 + (d*x)/2))/(a^2*d) + ((2*(a^4 + b^4 + a^2*b^2))/(a*(a^4 + b^4 - 2*a^2*b^2)) + (2*tan(c/2 + (d*x)/2)*(b^5 + 2*a^2*b^3))/(a^2*(a^4 + b^4 - 2*a^2*b^2)) - (2*tan(c/2 + (d*x)/2)^2*(b^4 - a^4 + 3*a^2*b^2))/(a*(a^4 + b^4 - 2*a^2*b^2)) - (2*b*tan(c/2 + (d*x)/2)^3*(2*a^4 + b^4))/(a^2*(a^4 + b^4 - 2*a^2*b^2)))/(d*(a + 2*b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^4 - 2*b*tan(c/2 + (d*x)/2)^3)) - (b^3*atan(((b^3*(2*a + b)*(-(a + b)^5*(a - b)^5)^(1/2)*(2*a - b)*(tan(c/2 + (d*x)/2)*(2*a^16 - 8*a^2*b^14 + 58*a^4*b ^12 - 160*a^6*b^10 + 222*a^8*b^8 - 168*a^10*b^6 + 70*a^12*b^4 - 16*a^14*b^ 2) - 2*a^15*b - 4*a^3*b^13 + 28*a^5*b^11 - 74*a^7*b^9 + 96*a^9*b^7 - 64*a^ 11*b^5 + 20*a^13*b^3 + (b^3*(2*a + b)*(-(a + b)^5*(a - b)^5)^(1/2)*(2*a - b)*(2*a^17*b - tan(c/2 + (d*x)/2)*(6*a^18 - 8*a^4*b^14 + 54*a^6*b^12 - 156 *a^8*b^10 + 250*a^10*b^8 - 240*a^12*b^6 + 138*a^14*b^4 - 44*a^16*b^2) + 2* a^5*b^13 - 12*a^7*b^11 + 30*a^9*b^9 - 40*a^11*b^7 + 30*a^13*b^5 - 12*a^15* b^3))/(a^12 - a^2*b^10 + 5*a^4*b^8 - 10*a^6*b^6 + 10*a^8*b^4 - 5*a^10*b^2) )*1i)/(a^12 - a^2*b^10 + 5*a^4*b^8 - 10*a^6*b^6 + 10*a^8*b^4 - 5*a^10*b^2) - (b^3*(2*a + b)*(-(a + b)^5*(a - b)^5)^(1/2)*(2*a - b)*(2*a^15*b - tan(c /2 + (d*x)/2)*(2*a^16 - 8*a^2*b^14 + 58*a^4*b^12 - 160*a^6*b^10 + 222*a^8* b^8 - 168*a^10*b^6 + 70*a^12*b^4 - 16*a^14*b^2) + 4*a^3*b^13 - 28*a^5*b^11 + 74*a^7*b^9 - 96*a^9*b^7 + 64*a^11*b^5 - 20*a^13*b^3 + (b^3*(2*a + b)*(- (a + b)^5*(a - b)^5)^(1/2)*(2*a - b)*(2*a^17*b - tan(c/2 + (d*x)/2)*(6*...