Integrand size = 19, antiderivative size = 47 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \]
Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \]
Time = 0.36 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 4274, 3042, 4254, 24, 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.
\(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )dx\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle a \int \sec ^3(c+d x)dx+a \int \sec ^2(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {a \int 1d(-\tan (c+d x))}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle a \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle a \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \tan (c+d x)}{d}\) |
3.1.3.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Time = 0.48 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a \tan \left (d x +c \right )}{d}\) | \(47\) |
default | \(\frac {a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a \tan \left (d x +c \right )}{d}\) | \(47\) |
parts | \(\frac {a \tan \left (d x +c \right )}{d}+\frac {a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(49\) |
norman | \(\frac {\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(87\) |
parallelrisch | \(\frac {\left (\left (-1-\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \sin \left (d x +c \right )+2 \sin \left (2 d x +2 c \right )\right ) a}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(90\) |
risch | \(-\frac {i a \left ({\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-2\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(93\) |
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.57 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*(a*cos(d*x + c)^2*log(sin(d*x + c) + 1) - a*cos(d*x + c)^2*log(-sin(d* x + c) + 1) + 2*(2*a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \, dx=a \left (\int \sec ^{2}{\left (c + d x \right )}\, dx + \int \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.23 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, a \tan \left (d x + c\right )}{4 \, d} \]
-1/4*(a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log (sin(d*x + c) - 1)) - 4*a*tan(d*x + c))/d
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.70 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(a*tan(1/2*d*x + 1/2*c)^3 - 3*a*tan(1/2*d*x + 1/2*c))/(tan(1/2* d*x + 1/2*c)^2 - 1)^2)/d
Time = 13.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.60 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \, dx=\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]