Integrand size = 27, antiderivative size = 78 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-b^3 x-\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {3 a^2 b \tan (c+d x)}{d}+\frac {b^3 \tan (c+d x)}{d} \]
-b^3*x-a^3*arctanh(cos(d*x+c))/d+a^3*sec(d*x+c)/d+3*a*b^2*sec(d*x+c)/d+3*a ^2*b*tan(d*x+c)/d+b^3*tan(d*x+c)/d
Time = 0.92 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.06 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-b^3 c-b^3 d x-a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+a^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a \left (a^2+3 b^2\right ) \sec (c+d x)+b \left (3 a^2+b^2\right ) \tan (c+d x)}{d} \]
(-(b^3*c) - b^3*d*x - a^3*Log[Cos[(c + d*x)/2]] + a^3*Log[Sin[(c + d*x)/2] ] + a*(a^2 + 3*b^2)*Sec[c + d*x] + b*(3*a^2 + b^2)*Tan[c + d*x])/d
Time = 0.34 (sec) , antiderivative size = 78, 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, 3391, 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.
\(\displaystyle \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\sin (c+d x) \cos (c+d x)^2}dx\) |
\(\Big \downarrow \) 3391 |
\(\displaystyle \int \left (a^3 \csc (c+d x) \sec ^2(c+d x)+3 a^2 b \sec ^2(c+d x)+3 a b^2 \tan (c+d x) \sec (c+d x)+b^3 \tan ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {3 a^2 b \tan (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \tan (c+d x)}{d}-b^3 x\) |
-(b^3*x) - (a^3*ArcTanh[Cos[c + d*x]])/d + (a^3*Sec[c + d*x])/d + (3*a*b^2 *Sec[c + d*x])/d + (3*a^2*b*Tan[c + d*x])/d + (b^3*Tan[c + d*x])/d
3.15.60.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (G tQ[m, 0] || IntegerQ[n])
Time = 0.63 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01
method | result | size |
derivativedivides | \(\frac {a^{3} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{2} b \tan \left (d x +c \right )+\frac {3 a \,b^{2}}{\cos \left (d x +c \right )}+b^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(79\) |
default | \(\frac {a^{3} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{2} b \tan \left (d x +c \right )+\frac {3 a \,b^{2}}{\cos \left (d x +c \right )}+b^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(79\) |
parallelrisch | \(\frac {a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3} d +2 \left (-3 a^{2} b -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+b^{3} d x -2 a^{3}-6 a \,b^{2}}{d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(106\) |
risch | \(-b^{3} x +\frac {2 i \left (-i a^{3} {\mathrm e}^{i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+3 a^{2} b +b^{3}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\) | \(107\) |
norman | \(\frac {b^{3} x -\frac {2 a^{3}+6 a \,b^{2}}{d}+2 b^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 b^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (a^{3}+3 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (3 a^{3}+9 a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (3 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {6 b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(306\) |
1/d*(a^3*(1/cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+3*a^2*b*tan(d*x+c)+3*a*b ^2/cos(d*x+c)+b^3*(tan(d*x+c)-d*x-c))
Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.27 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {2 \, b^{3} d x \cos \left (d x + c\right ) + a^{3} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - a^{3} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, a^{3} - 6 \, a b^{2} - 2 \, {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
-1/2*(2*b^3*d*x*cos(d*x + c) + a^3*cos(d*x + c)*log(1/2*cos(d*x + c) + 1/2 ) - a^3*cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2) - 2*a^3 - 6*a*b^2 - 2*(3 *a^2*b + b^3)*sin(d*x + c))/(d*cos(d*x + c))
Timed out. \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} b^{3} - a^{3} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, a^{2} b \tan \left (d x + c\right ) - \frac {6 \, a b^{2}}{\cos \left (d x + c\right )}}{2 \, d} \]
-1/2*(2*(d*x + c - tan(d*x + c))*b^3 - a^3*(2/cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 6*a^2*b*tan(d*x + c) - 6*a*b^2/cos(d*x + c))/d
Time = 0.49 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {{\left (d x + c\right )} b^{3} - a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]
-((d*x + c)*b^3 - a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 2*(3*a^2*b*tan(1/2* d*x + 1/2*c) + b^3*tan(1/2*d*x + 1/2*c) + a^3 + 3*a*b^2)/(tan(1/2*d*x + 1/ 2*c)^2 - 1))/d
Time = 11.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.97 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2\,b+2\,b^3\right )+6\,a\,b^2+2\,a^3}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {2\,b^3\,\mathrm {atan}\left (\frac {4\,b^6}{4\,a^3\,b^3+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^6}-\frac {4\,a^3\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,b^3+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^6}\right )}{d} \]