Integrand size = 21, antiderivative size = 111 \[ \int (a+a \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {7 a^4 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {7 a^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{d}+\frac {8 a^4 \tan ^3(c+d x)}{3 d}+\frac {a^4 \tan ^5(c+d x)}{5 d} \]
7/2*a^4*arctanh(sin(d*x+c))/d+8*a^4*tan(d*x+c)/d+7/2*a^4*sec(d*x+c)*tan(d* x+c)/d+a^4*sec(d*x+c)^3*tan(d*x+c)/d+8/3*a^4*tan(d*x+c)^3/d+1/5*a^4*tan(d* x+c)^5/d
Time = 1.57 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00 \[ \int (a+a \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {7 a^4 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {7 a^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{d}+\frac {8 a^4 \tan ^3(c+d x)}{3 d}+\frac {a^4 \tan ^5(c+d x)}{5 d} \]
(7*a^4*ArcTanh[Sin[c + d*x]])/(2*d) + (8*a^4*Tan[c + d*x])/d + (7*a^4*Sec[ c + d*x]*Tan[c + d*x])/(2*d) + (a^4*Sec[c + d*x]^3*Tan[c + d*x])/d + (8*a^ 4*Tan[c + d*x]^3)/(3*d) + (a^4*Tan[c + d*x]^5)/(5*d)
Time = 0.34 (sec) , antiderivative size = 111, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3236 |
| \(\displaystyle \int \left (a^4 \sec ^6(c+d x)+4 a^4 \sec ^5(c+d x)+6 a^4 \sec ^4(c+d x)+4 a^4 \sec ^3(c+d x)+a^4 \sec ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {7 a^4 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^4 \tan ^5(c+d x)}{5 d}+\frac {8 a^4 \tan ^3(c+d x)}{3 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {a^4 \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {7 a^4 \tan (c+d x) \sec (c+d x)}{2 d}\) |
(7*a^4*ArcTanh[Sin[c + d*x]])/(2*d) + (8*a^4*Tan[c + d*x])/d + (7*a^4*Sec[ c + d*x]*Tan[c + d*x])/(2*d) + (a^4*Sec[c + d*x]^3*Tan[c + d*x])/d + (8*a^ 4*Tan[c + d*x]^3)/(3*d) + (a^4*Tan[c + d*x]^5)/(5*d)
3.1.41.3.1 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]
Result contains complex when optimal does not.
Time = 4.38 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.41
| method | result | size |
| risch | \(-\frac {i a^{4} \left (105 \,{\mathrm e}^{9 i \left (d x +c \right )}-30 \,{\mathrm e}^{8 i \left (d x +c \right )}+330 \,{\mathrm e}^{7 i \left (d x +c \right )}-480 \,{\mathrm e}^{6 i \left (d x +c \right )}-1180 \,{\mathrm e}^{4 i \left (d x +c \right )}-330 \,{\mathrm e}^{3 i \left (d x +c \right )}-800 \,{\mathrm e}^{2 i \left (d x +c \right )}-105 \,{\mathrm e}^{i \left (d x +c \right )}-166\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {7 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {7 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(156\) |
| derivativedivides | \(\frac {a^{4} \tan \left (d x +c \right )+4 a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{4} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(157\) |
| default | \(\frac {a^{4} \tan \left (d x +c \right )+4 a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{4} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(157\) |
| parts | \(-\frac {a^{4} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{4} \tan \left (d x +c \right )}{d}+\frac {4 a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {6 a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {2 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {2 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(171\) |
| parallelrisch | \(\frac {77 \left (\left (-\frac {15 \cos \left (d x +c \right )}{11}-\frac {15 \cos \left (3 d x +3 c \right )}{22}-\frac {3 \cos \left (5 d x +5 c \right )}{22}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\frac {15 \cos \left (d x +c \right )}{11}+\frac {15 \cos \left (3 d x +3 c \right )}{22}+\frac {3 \cos \left (5 d x +5 c \right )}{22}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (3 d x +3 c \right )+\frac {3 \sin \left (4 d x +4 c \right )}{11}+\frac {83 \sin \left (5 d x +5 c \right )}{385}+\frac {10 \sin \left (d x +c \right )}{11}+\frac {6 \sin \left (2 d x +2 c \right )}{7}\right ) a^{4}}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(179\) |
-1/15*I*a^4*(105*exp(9*I*(d*x+c))-30*exp(8*I*(d*x+c))+330*exp(7*I*(d*x+c)) -480*exp(6*I*(d*x+c))-1180*exp(4*I*(d*x+c))-330*exp(3*I*(d*x+c))-800*exp(2 *I*(d*x+c))-105*exp(I*(d*x+c))-166)/d/(exp(2*I*(d*x+c))+1)^5+7/2*a^4/d*ln( exp(I*(d*x+c))+I)-7/2*a^4/d*ln(exp(I*(d*x+c))-I)
Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {105 \, a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (166 \, a^{4} \cos \left (d x + c\right )^{4} + 105 \, a^{4} \cos \left (d x + c\right )^{3} + 68 \, a^{4} \cos \left (d x + c\right )^{2} + 30 \, a^{4} \cos \left (d x + c\right ) + 6 \, a^{4}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \]
1/60*(105*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 105*a^4*cos(d*x + c)^ 5*log(-sin(d*x + c) + 1) + 2*(166*a^4*cos(d*x + c)^4 + 105*a^4*cos(d*x + c )^3 + 68*a^4*cos(d*x + c)^2 + 30*a^4*cos(d*x + c) + 6*a^4)*sin(d*x + c))/( d*cos(d*x + c)^5)
Timed out. \[ \int (a+a \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.71 \[ \int (a+a \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {4 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{4} + 120 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} - 15 \, a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 60 \, a^{4} \tan \left (d x + c\right )}{60 \, d} \]
1/60*(4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^4 + 120 *(tan(d*x + c)^3 + 3*tan(d*x + c))*a^4 - 15*a^4*(2*(3*sin(d*x + c)^3 - 5*s in(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)) - 60*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 60*a^4*tan(d*x + c) )/d
Time = 0.38 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.24 \[ \int (a+a \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {105 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 490 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 896 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 790 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{30 \, d} \]
1/30*(105*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*a^4*log(abs(tan(1/2 *d*x + 1/2*c) - 1)) - 2*(105*a^4*tan(1/2*d*x + 1/2*c)^9 - 490*a^4*tan(1/2* d*x + 1/2*c)^7 + 896*a^4*tan(1/2*d*x + 1/2*c)^5 - 790*a^4*tan(1/2*d*x + 1/ 2*c)^3 + 375*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 19.71 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.53 \[ \int (a+a \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {7\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {7\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {98\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {896\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {158\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(7*a^4*atanh(tan(c/2 + (d*x)/2)))/d - ((896*a^4*tan(c/2 + (d*x)/2)^5)/15 - (158*a^4*tan(c/2 + (d*x)/2)^3)/3 - (98*a^4*tan(c/2 + (d*x)/2)^7)/3 + 7*a^ 4*tan(c/2 + (d*x)/2)^9 + 25*a^4*tan(c/2 + (d*x)/2))/(d*(5*tan(c/2 + (d*x)/ 2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d* x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))