Integrand size = 21, antiderivative size = 123 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7 \text {arctanh}(\sin (c+d x))}{2 a^2 d}-\frac {16 \tan (c+d x)}{3 a^2 d}+\frac {7 \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {8 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
7/2*arctanh(sin(d*x+c))/a^2/d-16/3*tan(d*x+c)/a^2/d+7/2*sec(d*x+c)*tan(d*x +c)/a^2/d-8/3*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*sec(d*x+c)^ 3*tan(d*x+c)/d/(a+a*sec(d*x+c))^2
Time = 0.86 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.60 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {42 \text {arctanh}(\sin (c+d x))-\frac {(37+60 \cos (c+d x)+43 \cos (2 (c+d x))+16 \cos (3 (c+d x))) \sec (c+d x) \tan (c+d x)}{(1+\cos (c+d x))^2}}{12 a^2 d} \]
(42*ArcTanh[Sin[c + d*x]] - ((37 + 60*Cos[c + d*x] + 43*Cos[2*(c + d*x)] + 16*Cos[3*(c + d*x)])*Sec[c + d*x]*Tan[c + d*x])/(1 + Cos[c + d*x])^2)/(12 *a^2*d)
Time = 0.84 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4303, 3042, 4507, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4303 |
| \(\displaystyle -\frac {\int \frac {\sec ^3(c+d x) (3 a-5 a \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a-5 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle -\frac {\frac {\int \sec ^2(c+d x) \left (16 a^2-21 a^2 \sec (c+d x)\right )dx}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (16 a^2-21 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle -\frac {\frac {16 a^2 \int \sec ^2(c+d x)dx-21 a^2 \int \sec ^3(c+d x)dx}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {16 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-21 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {\frac {-\frac {16 a^2 \int 1d(-\tan (c+d x))}{d}-21 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {\frac {\frac {16 a^2 \tan (c+d x)}{d}-21 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle -\frac {\frac {\frac {16 a^2 \tan (c+d x)}{d}-21 a^2 \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {16 a^2 \tan (c+d x)}{d}-21 a^2 \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {\frac {16 a^2 \tan (c+d x)}{d}-21 a^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {8 \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
-1/3*(Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) - ((8*Sec[c + d*x]^2*Tan[c + d*x])/(d*(1 + Sec[c + d*x])) + ((16*a^2*Tan[c + d*x])/d - 21*a^2*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d))) /a^2)/(3*a^2)
3.1.52.3.1 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_))^(m_), x_Symbol] :> Simp[(-d^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d *Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Simp[d^2/(a*b*(2*m + 1)) Int[ (a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Time = 0.40 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {5}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}\) | \(120\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {5}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}\) | \(120\) |
| parallelrisch | \(\frac {\left (-42 \cos \left (2 d x +2 c \right )-42\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (42 \cos \left (2 d x +2 c \right )+42\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-60 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\cos \left (d x +c \right )+\frac {43 \cos \left (2 d x +2 c \right )}{60}+\frac {4 \cos \left (3 d x +3 c \right )}{15}+\frac {37}{60}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{12 a^{2} d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(127\) |
| risch | \(-\frac {i \left (21 \,{\mathrm e}^{6 i \left (d x +c \right )}+63 \,{\mathrm e}^{5 i \left (d x +c \right )}+98 \,{\mathrm e}^{4 i \left (d x +c \right )}+126 \,{\mathrm e}^{3 i \left (d x +c \right )}+97 \,{\mathrm e}^{2 i \left (d x +c \right )}+75 \,{\mathrm e}^{i \left (d x +c \right )}+32\right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{2} d}\) | \(147\) |
| norman | \(\frac {-\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {149 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {100 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}+\frac {18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\) | \(174\) |
1/2/d/a^2*(-1/3*tan(1/2*d*x+1/2*c)^3-7*tan(1/2*d*x+1/2*c)-1/(tan(1/2*d*x+1 /2*c)+1)^2+5/(tan(1/2*d*x+1/2*c)+1)+7*ln(tan(1/2*d*x+1/2*c)+1)+1/(tan(1/2* d*x+1/2*c)-1)^2+5/(tan(1/2*d*x+1/2*c)-1)-7*ln(tan(1/2*d*x+1/2*c)-1))
Time = 0.31 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {21 \, {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \, {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (32 \, \cos \left (d x + c\right )^{3} + 43 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) - 3\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]
1/12*(21*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 21*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*log(-s in(d*x + c) + 1) - 2*(32*cos(d*x + c)^3 + 43*cos(d*x + c)^2 + 6*cos(d*x + c) - 3)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2 *d*cos(d*x + c)^2)
\[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.36 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.54 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{6 \, d} \]
-1/6*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c ) + 1)^3)/(a^2 - 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d* x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1 ) + 1)/a^2 + 21*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d
Time = 0.31 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {21 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {21 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
1/6*(21*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 21*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 6*(5*tan(1/2*d*x + 1/2*c)^3 - 3*tan(1/2*d*x + 1/2*c))/( (tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2) - (a^4*tan(1/2*d*x + 1/2*c)^3 + 21*a^4 *tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 13.36 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,a^2\,d}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d} \]