Integrand size = 21, antiderivative size = 114 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {13 a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {13 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a^3 \tan ^3(c+d x)}{3 d}+\frac {a^3 \tan ^5(c+d x)}{5 d} \]
13/8*a^3*arctanh(sin(d*x+c))/d+4*a^3*tan(d*x+c)/d+13/8*a^3*sec(d*x+c)*tan( d*x+c)/d+3/4*a^3*sec(d*x+c)^3*tan(d*x+c)/d+5/3*a^3*tan(d*x+c)^3/d+1/5*a^3* tan(d*x+c)^5/d
Time = 0.77 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.07 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {13 a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {19 a^3 \tan (c+d x)}{5 d}+\frac {13 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {19 a^3 \tan ^3(c+d x)}{15 d} \]
(13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (19*a^3*Tan[c + d*x])/(5*d) + (13*a ^3*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (3*a^3*Sec[c + d*x]^3*Tan[c + d*x])/ (4*d) + (a^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (19*a^3*Tan[c + d*x]^3)/ (15*d)
Time = 0.34 (sec) , antiderivative size = 114, 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, 4278, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \sec (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3dx\) |
| \(\Big \downarrow \) 4278 |
| \(\displaystyle \int \left (a^3 \sec ^6(c+d x)+3 a^3 \sec ^5(c+d x)+3 a^3 \sec ^4(c+d x)+a^3 \sec ^3(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {13 a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 \tan ^5(c+d x)}{5 d}+\frac {5 a^3 \tan ^3(c+d x)}{3 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {13 a^3 \tan (c+d x) \sec (c+d x)}{8 d}\) |
(13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a^3*Tan[c + d*x])/d + (13*a^3*Se c[c + d*x]*Tan[c + d*x])/(8*d) + (3*a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^3*Tan[c + d*x]^3)/(3*d) + (a^3*Tan[c + d*x]^5)/(5*d)
3.1.20.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f *x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I GtQ[m, 0] && RationalQ[n]
Result contains complex when optimal does not.
Time = 1.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.27
| method | result | size |
| risch | \(-\frac {i a^{3} \left (195 \,{\mathrm e}^{9 i \left (d x +c \right )}+750 \,{\mathrm e}^{7 i \left (d x +c \right )}-720 \,{\mathrm e}^{6 i \left (d x +c \right )}-2320 \,{\mathrm e}^{4 i \left (d x +c \right )}-750 \,{\mathrm e}^{3 i \left (d x +c \right )}-1520 \,{\mathrm e}^{2 i \left (d x +c \right )}-195 \,{\mathrm e}^{i \left (d x +c \right )}-304\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\) | \(145\) |
| derivativedivides | \(\frac {-a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+3 a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(146\) |
| default | \(\frac {-a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+3 a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(146\) |
| norman | \(\frac {-\frac {51 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {133 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {416 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {91 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {13 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {13 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {13 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(152\) |
| parts | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {3 a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {3 a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(154\) |
| parallelrisch | \(\frac {a^{3} \left (-1950 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+1950 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-195 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (5 d x +5 c \right )-975 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (3 d x +3 c \right )+195 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (5 d x +5 c \right )+975 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (3 d x +3 c \right )+304 \sin \left (5 d x +5 c \right )+390 \sin \left (4 d x +4 c \right )+1520 \sin \left (3 d x +3 c \right )+1500 \sin \left (2 d x +2 c \right )+1600 \sin \left (d x +c \right )\right )}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(225\) |
-1/60*I*a^3*(195*exp(9*I*(d*x+c))+750*exp(7*I*(d*x+c))-720*exp(6*I*(d*x+c) )-2320*exp(4*I*(d*x+c))-750*exp(3*I*(d*x+c))-1520*exp(2*I*(d*x+c))-195*exp (I*(d*x+c))-304)/d/(exp(2*I*(d*x+c))+1)^5-13/8*a^3/d*ln(exp(I*(d*x+c))-I)+ 13/8*a^3/d*ln(exp(I*(d*x+c))+I)
Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.09 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {195 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 195 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (304 \, a^{3} \cos \left (d x + c\right )^{4} + 195 \, a^{3} \cos \left (d x + c\right )^{3} + 152 \, a^{3} \cos \left (d x + c\right )^{2} + 90 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
1/240*(195*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 195*a^3*cos(d*x + c) ^5*log(-sin(d*x + c) + 1) + 2*(304*a^3*cos(d*x + c)^4 + 195*a^3*cos(d*x + c)^3 + 152*a^3*cos(d*x + c)^2 + 90*a^3*cos(d*x + c) + 24*a^3)*sin(d*x + c) )/(d*cos(d*x + c)^5)
\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(sec(c + d*x)**3, x) + Integral(3*sec(c + d*x)**4, x) + Inte gral(3*sec(c + d*x)**5, x) + Integral(sec(c + d*x)**6, x))
Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.57 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - 45 \, a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3 + 2 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - 45*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)) - 60*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)))/d
Time = 0.35 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.21 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {195 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 195 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (195 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 910 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1330 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
1/120*(195*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 195*a^3*log(abs(tan(1/ 2*d*x + 1/2*c) - 1)) - 2*(195*a^3*tan(1/2*d*x + 1/2*c)^9 - 910*a^3*tan(1/2 *d*x + 1/2*c)^7 + 1664*a^3*tan(1/2*d*x + 1/2*c)^5 - 1330*a^3*tan(1/2*d*x + 1/2*c)^3 + 765*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/ d
Time = 17.77 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.49 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {13\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}-\frac {91\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {416\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {133\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {51\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(13*a^3*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((416*a^3*tan(c/2 + (d*x)/2)^5) /15 - (133*a^3*tan(c/2 + (d*x)/2)^3)/6 - (91*a^3*tan(c/2 + (d*x)/2)^7)/6 + (13*a^3*tan(c/2 + (d*x)/2)^9)/4 + (51*a^3*tan(c/2 + (d*x)/2))/4)/(d*(5*ta n(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))