Integrand size = 21, antiderivative size = 93 \[ \int (a+a \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {15 a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {15 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \tan ^3(c+d x)}{d} \]
15/8*a^3*arctanh(sin(d*x+c))/d+4*a^3*tan(d*x+c)/d+15/8*a^3*sec(d*x+c)*tan( d*x+c)/d+1/4*a^3*sec(d*x+c)^3*tan(d*x+c)/d+a^3*tan(d*x+c)^3/d
Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int (a+a \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {15 a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {15 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \tan ^3(c+d x)}{d} \]
(15*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a^3*Tan[c + d*x])/d + (15*a^3*Se c[c + d*x]*Tan[c + d*x])/(8*d) + (a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^3*Tan[c + d*x]^3)/d
Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3236, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3236 |
| \(\displaystyle \int \left (a^3 \sec ^5(c+d x)+3 a^3 \sec ^4(c+d x)+3 a^3 \sec ^3(c+d x)+a^3 \sec ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {15 a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 \tan ^3(c+d x)}{d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {15 a^3 \tan (c+d x) \sec (c+d x)}{8 d}\) |
(15*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a^3*Tan[c + d*x])/d + (15*a^3*Se c[c + d*x]*Tan[c + d*x])/(8*d) + (a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^3*Tan[c + d*x]^3)/d
3.1.31.3.1 Defintions of rubi rules used
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Time = 3.79 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.32
| method | result | size |
| derivativedivides | \(\frac {a^{3} \tan \left (d x +c \right )+3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(123\) |
| default | \(\frac {a^{3} \tan \left (d x +c \right )+3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(123\) |
| parts | \(\frac {a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a^{3} \tan \left (d x +c \right )}{d}-\frac {3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(131\) |
| risch | \(-\frac {i a^{3} \left (15 \,{\mathrm e}^{7 i \left (d x +c \right )}-8 \,{\mathrm e}^{6 i \left (d x +c \right )}+23 \,{\mathrm e}^{5 i \left (d x +c \right )}-72 \,{\mathrm e}^{4 i \left (d x +c \right )}-23 \,{\mathrm e}^{3 i \left (d x +c \right )}-88 \,{\mathrm e}^{2 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}-24\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\) | \(145\) |
| parallelrisch | \(\frac {a^{3} \left (15 \left (-\cos \left (4 d x +4 c \right )-4 \cos \left (2 d x +2 c \right )-3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 \sin \left (4 d x +4 c \right )+46 \sin \left (d x +c \right )+80 \sin \left (2 d x +2 c \right )+30 \sin \left (3 d x +3 c \right )\right )}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(149\) |
| norman | \(\frac {\frac {49 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {37 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {17 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {5 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {47 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {15 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {15 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {15 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(205\) |
1/d*(a^3*tan(d*x+c)+3*a^3*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan (d*x+c)))-3*a^3*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a^3*(-(-1/4*sec(d*x+c)^ 3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.19 \[ \int (a+a \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {15 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, a^{3} \cos \left (d x + c\right )^{3} + 15 \, a^{3} \cos \left (d x + c\right )^{2} + 8 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
1/16*(15*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*a^3*cos(d*x + c)^4* log(-sin(d*x + c) + 1) + 2*(24*a^3*cos(d*x + c)^3 + 15*a^3*cos(d*x + c)^2 + 8*a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int (a+a \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.68 \[ \int (a+a \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, a^{3} \tan \left (d x + c\right )}{16 \, d} \]
1/16*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*a^3*(2*sin(d*x + c)/(sin(d*x + c )^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 16*a^3*tan(d*x + c))/d
Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.31 \[ \int (a+a \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 55 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 73 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 49 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{8 \, d} \]
1/8*(15*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a^3*log(abs(tan(1/2*d* x + 1/2*c) - 1)) - 2*(15*a^3*tan(1/2*d*x + 1/2*c)^7 - 55*a^3*tan(1/2*d*x + 1/2*c)^5 + 73*a^3*tan(1/2*d*x + 1/2*c)^3 - 49*a^3*tan(1/2*d*x + 1/2*c))/( tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 16.97 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.52 \[ \int (a+a \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {15\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {55\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {73\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {49\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]