Integrand size = 21, antiderivative size = 85 \[ \int \frac {\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {3 \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {2 \tan (c+d x)}{a d}+\frac {3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))} \]
3/2*arctanh(sin(d*x+c))/a/d-2*tan(d*x+c)/a/d+3/2*sec(d*x+c)*tan(d*x+c)/a/d -sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {3 \text {arctanh}(\sin (c+d x)) (1+\sec (c+d x))-(1+\cos (c+d x)+2 \cos (2 (c+d x))) \sec ^2(c+d x) \tan (c+d x)}{2 a d (1+\sec (c+d x))} \]
(3*ArcTanh[Sin[c + d*x]]*(1 + Sec[c + d*x]) - (1 + Cos[c + d*x] + 2*Cos[2* (c + d*x)])*Sec[c + d*x]^2*Tan[c + d*x])/(2*a*d*(1 + Sec[c + d*x]))
Time = 0.55 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4305, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 4305 |
| \(\displaystyle -\frac {\int \sec ^2(c+d x) (2 a-3 a \sec (c+d x))dx}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a-3 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle -\frac {2 a \int \sec ^2(c+d x)dx-3 a \int \sec ^3(c+d x)dx}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-3 a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {-\frac {2 a \int 1d(-\tan (c+d x))}{d}-3 a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {\frac {2 a \tan (c+d x)}{d}-3 a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle -\frac {\frac {2 a \tan (c+d x)}{d}-3 a \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {2 a \tan (c+d x)}{d}-3 a \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {2 a \tan (c+d x)}{d}-3 a \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
-((Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x]))) - ((2*a*Tan[c + d*x])/d - 3*a*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/( 2*d)))/a^2
3.1.43.3.1 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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[d^2*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 2)/(f*(a + b*Csc[e + f*x]))), x] - Simp[d^2/(a*b) Int[(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ [a^2 - b^2, 0] && GtQ[n, 1]
Time = 0.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24
| method | result | size |
| parallelrisch | \(\frac {\left (-3 \cos \left (2 d x +2 c \right )-3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (3 \cos \left (2 d x +2 c \right )+3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \left (1+2 \cos \left (2 d x +2 c \right )+\cos \left (d x +c \right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(105\) |
| derivativedivides | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}}{d a}\) | \(108\) |
| default | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}}{d a}\) | \(108\) |
| risch | \(-\frac {i \left (3 \,{\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{3 i \left (d x +c \right )}+5 \,{\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}+4\right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}\) | \(123\) |
| norman | \(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\) | \(133\) |
1/2*((-3*cos(2*d*x+2*c)-3)*ln(tan(1/2*d*x+1/2*c)-1)+(3*cos(2*d*x+2*c)+3)*l n(tan(1/2*d*x+1/2*c)+1)-2*(1+2*cos(2*d*x+2*c)+cos(d*x+c))*tan(1/2*d*x+1/2* c))/a/d/(1+cos(2*d*x+2*c))
Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {3 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]
1/4*(3*(cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 3*(cos(d* x + c)^3 + cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*cos(d*x + c)^2 + cos(d*x + c) - 1)*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
Time = 0.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.91 \[ \int \frac {\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{2 \, d} \]
-1/2*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(c os(d*x + c) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*log (sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d
Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.19 \[ \int \frac {\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \]
1/2*(3*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*log(abs(tan(1/2*d*x + 1/2* c) - 1))/a - 2*tan(1/2*d*x + 1/2*c)/a + 2*(3*tan(1/2*d*x + 1/2*c)^3 - tan( 1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a))/d
Time = 13.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int \frac {\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )} \]