Integrand size = 26, antiderivative size = 47 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*C*arctanh(sin(d*x+c))/d+B*tan(d*x+c)/d+1/2*C*sec(d*x+c)*tan(d*x+c)/d
Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[Sec[c + d*x]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(C*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Tan[c + d*x])/d + (C*Sec[c + d*x]*Tan [c + d*x])/(2*d)
Time = 0.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 4535, 27, 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.
\(\displaystyle \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle B \int \sec ^2(c+d x)dx+\int C \sec ^3(c+d x)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle B \int \sec ^2(c+d x)dx+C \int \sec ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+C \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle C \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {B \int 1d(-\tan (c+d x))}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle C \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {B \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle C \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle C \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle C \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x)}{d}\) |
Input:
Int[Sec[c + d*x]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(B*Tan[c + d*x])/d + C*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {B \tan \left (d x +c \right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(47\) |
default | \(\frac {B \tan \left (d x +c \right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(47\) |
parts | \(\frac {B \tan \left (d x +c \right )}{d}+\frac {C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(49\) |
parallelrisch | \(\frac {-C \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 B \sin \left (2 d x +2 c \right )+2 C \sin \left (d x +c \right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(92\) |
norman | \(\frac {\frac {\left (2 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (2 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(96\) |
risch | \(-\frac {i \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\) | \(98\) |
Input:
int(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(B*tan(d*x+c)+C*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c ))))
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.57 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - C \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + C\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/4*(C*cos(d*x + c)^2*log(sin(d*x + c) + 1) - C*cos(d*x + c)^2*log(-sin(d* x + c) + 1) + 2*(2*B*cos(d*x + c) + C)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)
Output:
Integral((B + C*sec(c + d*x))*sec(c + d*x)**2, x)
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.23 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {C {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, B \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
-1/4*(C*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log (sin(d*x + c) - 1)) - 4*B*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (43) = 86\).
Time = 0.34 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.28 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/2*(C*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - C*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2*B*tan(1/2*d*x + 1/2*c)^3 - C*tan(1/2*d*x + 1/2*c)^3 - 2*B*ta n(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^ 2)/d
Time = 12.73 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.81 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B+C\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((B/cos(c + d*x) + C/cos(c + d*x)^2)/cos(c + d*x),x)
Output:
(C*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)^3*(2*B - C) - tan(c/ 2 + (d*x)/2)*(2*B + C))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1))
Time = 0.15 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.47 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -\sin \left (d x +c \right ) c}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)*b - log(tan((c + d*x)/2) - 1)*sin(c + d*x) **2*c + log(tan((c + d*x)/2) - 1)*c + log(tan((c + d*x)/2) + 1)*sin(c + d* x)**2*c - log(tan((c + d*x)/2) + 1)*c - sin(c + d*x)*c)/(2*d*(sin(c + d*x) **2 - 1))