Integrand size = 8, antiderivative size = 76 \[ \int \sec ^7(a+b x) \, dx=\frac {5 \text {arctanh}(\sin (a+b x))}{16 b}+\frac {5 \sec (a+b x) \tan (a+b x)}{16 b}+\frac {5 \sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \]
5/16*arctanh(sin(b*x+a))/b+5/16*sec(b*x+a)*tan(b*x+a)/b+5/24*sec(b*x+a)^3* tan(b*x+a)/b+1/6*sec(b*x+a)^5*tan(b*x+a)/b
Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \sec ^7(a+b x) \, dx=\frac {5 \text {arctanh}(\sin (a+b x))}{16 b}+\frac {5 \sec (a+b x) \tan (a+b x)}{16 b}+\frac {5 \sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \]
(5*ArcTanh[Sin[a + b*x]])/(16*b) + (5*Sec[a + b*x]*Tan[a + b*x])/(16*b) + (5*Sec[a + b*x]^3*Tan[a + b*x])/(24*b) + (Sec[a + b*x]^5*Tan[a + b*x])/(6* b)
Time = 0.44 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4255, 3042, 4255, 3042, 4255, 3042, 4257}
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.
\(\displaystyle \int \sec ^7(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (a+b x+\frac {\pi }{2}\right )^7dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {5}{6} \int \sec ^5(a+b x)dx+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \int \csc \left (a+b x+\frac {\pi }{2}\right )^5dx+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \sec ^3(a+b x)dx+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \csc \left (a+b x+\frac {\pi }{2}\right )^3dx+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (a+b x)dx+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (a+b x+\frac {\pi }{2}\right )dx+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (a+b x))}{2 b}+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
(Sec[a + b*x]^5*Tan[a + b*x])/(6*b) + (5*((Sec[a + b*x]^3*Tan[a + b*x])/(4 *b) + (3*(ArcTanh[Sin[a + b*x]]/(2*b) + (Sec[a + b*x]*Tan[a + b*x])/(2*b)) )/4))/6
3.1.7.3.1 Defintions of rubi rules used
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.73 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {-\left (-\frac {\sec \left (b x +a \right )^{5}}{6}-\frac {5 \sec \left (b x +a \right )^{3}}{24}-\frac {5 \sec \left (b x +a \right )}{16}\right ) \tan \left (b x +a \right )+\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\) | \(59\) |
default | \(\frac {-\left (-\frac {\sec \left (b x +a \right )^{5}}{6}-\frac {5 \sec \left (b x +a \right )^{3}}{24}-\frac {5 \sec \left (b x +a \right )}{16}\right ) \tan \left (b x +a \right )+\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\) | \(59\) |
risch | \(-\frac {i \left (15 \,{\mathrm e}^{11 i \left (b x +a \right )}+85 \,{\mathrm e}^{9 i \left (b x +a \right )}+198 \,{\mathrm e}^{7 i \left (b x +a \right )}-198 \,{\mathrm e}^{5 i \left (b x +a \right )}-85 \,{\mathrm e}^{3 i \left (b x +a \right )}-15 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{24 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{6}}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{16 b}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{16 b}\) | \(124\) |
norman | \(\frac {\frac {11 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}+\frac {5 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}{24 b}+\frac {15 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{5}}{4 b}+\frac {15 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{7}}{4 b}+\frac {5 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{9}}{24 b}+\frac {11 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{11}}{8 b}}{\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{6}}-\frac {5 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{16 b}+\frac {5 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{16 b}\) | \(147\) |
parallelrisch | \(\frac {\left (-225 \cos \left (2 b x +2 a \right )-90 \cos \left (4 b x +4 a \right )-15 \cos \left (6 b x +6 a \right )-150\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (225 \cos \left (2 b x +2 a \right )+90 \cos \left (4 b x +4 a \right )+15 \cos \left (6 b x +6 a \right )+150\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+396 \sin \left (b x +a \right )+170 \sin \left (3 b x +3 a \right )+30 \sin \left (5 b x +5 a \right )}{48 b \left (10+\cos \left (6 b x +6 a \right )+6 \cos \left (4 b x +4 a \right )+15 \cos \left (2 b x +2 a \right )\right )}\) | \(168\) |
1/b*(-(-1/6*sec(b*x+a)^5-5/24*sec(b*x+a)^3-5/16*sec(b*x+a))*tan(b*x+a)+5/1 6*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11 \[ \int \sec ^7(a+b x) \, dx=\frac {15 \, \cos \left (b x + a\right )^{6} \log \left (\sin \left (b x + a\right ) + 1\right ) - 15 \, \cos \left (b x + a\right )^{6} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (15 \, \cos \left (b x + a\right )^{4} + 10 \, \cos \left (b x + a\right )^{2} + 8\right )} \sin \left (b x + a\right )}{96 \, b \cos \left (b x + a\right )^{6}} \]
1/96*(15*cos(b*x + a)^6*log(sin(b*x + a) + 1) - 15*cos(b*x + a)^6*log(-sin (b*x + a) + 1) + 2*(15*cos(b*x + a)^4 + 10*cos(b*x + a)^2 + 8)*sin(b*x + a ))/(b*cos(b*x + a)^6)
\[ \int \sec ^7(a+b x) \, dx=\int \sec ^{7}{\left (a + b x \right )}\, dx \]
Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.20 \[ \int \sec ^7(a+b x) \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (b x + a\right )^{5} - 40 \, \sin \left (b x + a\right )^{3} + 33 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{96 \, b} \]
-1/96*(2*(15*sin(b*x + a)^5 - 40*sin(b*x + a)^3 + 33*sin(b*x + a))/(sin(b* x + a)^6 - 3*sin(b*x + a)^4 + 3*sin(b*x + a)^2 - 1) - 15*log(sin(b*x + a) + 1) + 15*log(sin(b*x + a) - 1))/b
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \sec ^7(a+b x) \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (b x + a\right )^{5} - 40 \, \sin \left (b x + a\right )^{3} + 33 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{3}} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{96 \, b} \]
-1/96*(2*(15*sin(b*x + a)^5 - 40*sin(b*x + a)^3 + 33*sin(b*x + a))/(sin(b* x + a)^2 - 1)^3 - 15*log(abs(sin(b*x + a) + 1)) + 15*log(abs(sin(b*x + a) - 1)))/b
Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04 \[ \int \sec ^7(a+b x) \, dx=\frac {5\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{16\,b}-\frac {\frac {5\,{\sin \left (a+b\,x\right )}^5}{16}-\frac {5\,{\sin \left (a+b\,x\right )}^3}{6}+\frac {11\,\sin \left (a+b\,x\right )}{16}}{b\,\left ({\sin \left (a+b\,x\right )}^6-3\,{\sin \left (a+b\,x\right )}^4+3\,{\sin \left (a+b\,x\right )}^2-1\right )} \]