Integrand size = 21, antiderivative size = 70 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {(3 A+4 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
1/8*(3*A+4*C)*arctanh(sin(d*x+c))/d+1/8*(3*A+4*C)*sec(d*x+c)*tan(d*x+c)/d+ 1/4*A*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.33 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 A \text {arctanh}(\sin (c+d x))}{8 d}+\frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 A \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {A \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
(3*A*ArcTanh[Sin[c + d*x]])/(8*d) + (C*ArcTanh[Sin[c + d*x]])/(2*d) + (3*A *Sec[c + d*x]*Tan[c + d*x])/(8*d) + (C*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
Time = 0.35 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3491, 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.
| \(\displaystyle \int \sec ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3491 |
| \(\displaystyle \frac {1}{4} (3 A+4 C) \int \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} (3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} (3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} (3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
(A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((3*A + 4*C)*(ArcTanh[Sin[c + d*x] ]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4
3.1.7.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x _)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)) Int[(b*Sin[e + f* x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
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 = 4.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(85\) |
| default | \(\frac {A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(85\) |
| parts | \(\frac {A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(87\) |
| parallelrisch | \(\frac {-6 \left (A +\frac {4 C}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (A +\frac {4 C}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (3 A +4 C \right ) \sin \left (3 d x +3 c \right )+11 \sin \left (d x +c \right ) \left (A +\frac {4 C}{11}\right )}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(143\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (3 A \,{\mathrm e}^{6 i \left (d x +c \right )}+4 C \,{\mathrm e}^{6 i \left (d x +c \right )}+11 A \,{\mathrm e}^{4 i \left (d x +c \right )}+4 C \,{\mathrm e}^{4 i \left (d x +c \right )}-11 A \,{\mathrm e}^{2 i \left (d x +c \right )}-4 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A -4 C \right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) | \(199\) |
| norman | \(\frac {\frac {\left (5 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (5 A +4 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (7 A -4 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (7 A -4 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (13 A +4 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (13 A +4 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (3 A +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 A +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(218\) |
1/d*(A*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+t an(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 = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, A\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
1/16*((3*A + 4*C)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*A + 4*C)*cos(d *x + c)^4*log(-sin(d*x + c) + 1) + 2*((3*A + 4*C)*cos(d*x + c)^2 + 2*A)*si n(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \]
Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.39 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {{\left (3 \, A + 4 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A + 4 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (3 \, A + 4 \, C\right )} \sin \left (d x + c\right )^{3} - {\left (5 \, A + 4 \, C\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
1/16*((3*A + 4*C)*log(sin(d*x + c) + 1) - (3*A + 4*C)*log(sin(d*x + c) - 1 ) - 2*((3*A + 4*C)*sin(d*x + c)^3 - (5*A + 4*C)*sin(d*x + c))/(sin(d*x + c )^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.40 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {{\left (3 \, A + 4 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, A + 4 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A \sin \left (d x + c\right )^{3} + 4 \, C \sin \left (d x + c\right )^{3} - 5 \, A \sin \left (d x + c\right ) - 4 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
1/16*((3*A + 4*C)*log(abs(sin(d*x + c) + 1)) - (3*A + 4*C)*log(abs(sin(d*x + c) - 1)) - 2*(3*A*sin(d*x + c)^3 + 4*C*sin(d*x + c)^3 - 5*A*sin(d*x + c ) - 4*C*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d
Time = 0.79 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {5\,A}{8}+\frac {C}{2}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {3\,A}{8}+\frac {C}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {3\,A}{8}+\frac {C}{2}\right )}{d} \]