Integrand size = 19, antiderivative size = 96 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {35 a^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {27 a^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d} \]
35/8*a^4*arctanh(sin(d*x+c))/d+8*a^4*tan(d*x+c)/d+27/8*a^4*sec(d*x+c)*tan( d*x+c)/d+1/4*a^4*sec(d*x+c)^3*tan(d*x+c)/d+4/3*a^4*tan(d*x+c)^3/d
Time = 2.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {35 a^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {27 a^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d} \]
(35*a^4*ArcTanh[Sin[c + d*x]])/(8*d) + (8*a^4*Tan[c + d*x])/d + (27*a^4*Se c[c + d*x]*Tan[c + d*x])/(8*d) + (a^4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (4*a^4*Tan[c + d*x]^3)/(3*d)
Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, 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.
| \(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a)^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4dx\) |
| \(\Big \downarrow \) 4278 |
| \(\displaystyle \int \left (a^4 \sec ^5(c+d x)+4 a^4 \sec ^4(c+d x)+6 a^4 \sec ^3(c+d x)+4 a^4 \sec ^2(c+d x)+a^4 \sec (c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {35 a^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {a^4 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {27 a^4 \tan (c+d x) \sec (c+d x)}{8 d}\) |
(35*a^4*ArcTanh[Sin[c + d*x]])/(8*d) + (8*a^4*Tan[c + d*x])/d + (27*a^4*Se c[c + d*x]*Tan[c + d*x])/(8*d) + (a^4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (4*a^4*Tan[c + d*x]^3)/(3*d)
3.1.32.3.1 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]
Time = 0.82 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.39
| method | result | size |
| norman | \(\frac {\frac {93 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {511 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {385 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {35 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {35 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {35 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(133\) |
| parts | \(\frac {4 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {4 a^{4} \tan \left (d x +c \right )}{d}+\frac {3 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}-\frac {4 a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(135\) |
| derivativedivides | \(\frac {a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} \tan \left (d x +c \right )+a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(142\) |
| default | \(\frac {a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} \tan \left (d x +c \right )+a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(142\) |
| risch | \(-\frac {i a^{4} \left (81 \,{\mathrm e}^{7 i \left (d x +c \right )}-96 \,{\mathrm e}^{6 i \left (d x +c \right )}+105 \,{\mathrm e}^{5 i \left (d x +c \right )}-480 \,{\mathrm e}^{4 i \left (d x +c \right )}-105 \,{\mathrm e}^{3 i \left (d x +c \right )}-544 \,{\mathrm e}^{2 i \left (d x +c \right )}-81 \,{\mathrm e}^{i \left (d x +c \right )}-160\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\) | \(145\) |
| parallelrisch | \(\frac {a^{4} \left (105 \left (-\cos \left (4 d x +4 c \right )-4 \cos \left (2 d x +2 c \right )-3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+105 \left (3+\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+160 \sin \left (4 d x +4 c \right )+210 \sin \left (d x +c \right )+448 \sin \left (2 d x +2 c \right )+162 \sin \left (3 d x +3 c \right )\right )}{24 d \left (3+\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )\right )}\) | \(149\) |
(93/4*a^4/d*tan(1/2*d*x+1/2*c)-511/12*a^4/d*tan(1/2*d*x+1/2*c)^3+385/12*a^ 4/d*tan(1/2*d*x+1/2*c)^5-35/4*a^4/d*tan(1/2*d*x+1/2*c)^7)/(-1+tan(1/2*d*x+ 1/2*c)^2)^4-35/8*a^4/d*ln(tan(1/2*d*x+1/2*c)-1)+35/8*a^4/d*ln(tan(1/2*d*x+ 1/2*c)+1)
Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {105 \, a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (160 \, a^{4} \cos \left (d x + c\right )^{3} + 81 \, a^{4} \cos \left (d x + c\right )^{2} + 32 \, a^{4} \cos \left (d x + c\right ) + 6 \, a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(105*a^4*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 105*a^4*cos(d*x + c)^ 4*log(-sin(d*x + c) + 1) + 2*(160*a^4*cos(d*x + c)^3 + 81*a^4*cos(d*x + c) ^2 + 32*a^4*cos(d*x + c) + 6*a^4)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx=a^{4} \left (\int \sec {\left (c + d x \right )}\, dx + \int 4 \sec ^{2}{\left (c + d x \right )}\, dx + \int 6 \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 \sec ^{4}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]
a**4*(Integral(sec(c + d*x), x) + Integral(4*sec(c + d*x)**2, x) + Integra l(6*sec(c + d*x)**3, x) + Integral(4*sec(c + d*x)**4, x) + Integral(sec(c + d*x)**5, x))
Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.82 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} - 3 \, a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 192 \, a^{4} \tan \left (d x + c\right )}{48 \, d} \]
1/48*(64*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^4 - 3*a^4*(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)) - 72*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)) + 48*a^4*log(s ec(d*x + c) + tan(d*x + c)) + 192*a^4*tan(d*x + c))/d
Time = 0.34 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {105 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 385 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 279 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
1/24*(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)^7 - 385*a^4*tan(1/2* d*x + 1/2*c)^5 + 511*a^4*tan(1/2*d*x + 1/2*c)^3 - 279*a^4*tan(1/2*d*x + 1/ 2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 16.53 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.47 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {35\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {35\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {385\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {511\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}-\frac {93\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]