Integrand size = 21, antiderivative size = 159 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}-\frac {8 \tan (c+d x)}{63 a d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {\tan (c+d x)}{9 d \left (a^5+a^5 \sec (c+d x)\right )} \]
-1/9*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^5+5/63*sec(d*x+c)^3*tan(d* x+c)/a/d/(a+a*sec(d*x+c))^4+1/21*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^3-8/63* tan(d*x+c)/a/d/(a^2+a^2*sec(d*x+c))^2+1/9*tan(d*x+c)/d/(a^5+a^5*sec(d*x+c) )
Time = 0.41 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.40 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {(65+130 \cos (c+d x)+46 \cos (2 (c+d x))+10 \cos (3 (c+d x))+\cos (4 (c+d x))) \sin (c+d x)}{252 a^5 d (1+\cos (c+d x))^5} \]
((65 + 130*Cos[c + d*x] + 46*Cos[2*(c + d*x)] + 10*Cos[3*(c + d*x)] + Cos[ 4*(c + d*x)])*Sin[c + d*x])/(252*a^5*d*(1 + Cos[c + d*x])^5)
Time = 0.83 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4298, 3042, 4297, 3042, 4286, 25, 3042, 4488, 3042, 4281}
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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a \sec (c+d x)+a)^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^5}dx\) |
| \(\Big \downarrow \) 4298 |
| \(\displaystyle \frac {5 \int \frac {\sec ^4(c+d x)}{(\sec (c+d x) a+a)^4}dx}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}dx}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4297 |
| \(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sec ^3(c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4286 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\int -\frac {\sec (c+d x) (3 a-5 a \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {\int \frac {\sec (c+d x) (3 a-5 a \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}\right )}{7 a}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-5 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}\right )}{7 a}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4488 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {8 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {7}{3} \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{5 a^2}\right )}{7 a}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {8 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {7}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a^2}\right )}{7 a}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {8 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {7 \tan (c+d x)}{3 d (a \sec (c+d x)+a)}}{5 a^2}\right )}{7 a}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
-1/9*(Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^5) + (5*((Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + (3*(Tan[c + d*x]/(5 *d*(a + a*Sec[c + d*x])^3) - ((8*a*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x]) ^2) - (7*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])))/(5*a^2)))/(7*a)))/(9*a)
3.1.83.3.1 Defintions of rubi rules used
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1) *(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[b*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[d*((m + 1)/(b*(2*m + 1))) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ [{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && EqQ[m + n, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc [e + f*x])^n/(f*(2*m + 1))), x] + Simp[m/(a*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(a*b*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]
Time = 0.35 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36
| method | result | size |
| parallelrisch | \(-\frac {\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{7}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{144 d \,a^{5}}\) | \(57\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\) | \(58\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\) | \(58\) |
| risch | \(\frac {4 i \left (63 \,{\mathrm e}^{5 i \left (d x +c \right )}+63 \,{\mathrm e}^{4 i \left (d x +c \right )}+84 \,{\mathrm e}^{3 i \left (d x +c \right )}+36 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{63 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\) | \(80\) |
| norman | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{48 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{16 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{112 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{1008 a d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{336 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{336 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{144 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{4}}\) | \(171\) |
-1/144*(tan(1/2*d*x+1/2*c)^8+18/7*tan(1/2*d*x+1/2*c)^6-6*tan(1/2*d*x+1/2*c )^2-9)*tan(1/2*d*x+1/2*c)/d/a^5
Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.77 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {{\left (2 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 25 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{63 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]
1/63*(2*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 25*cos(d* x + c) + 5)*sin(d*x + c)/(a^5*d*cos(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c) + a^5*d)
\[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]
Integral(sec(c + d*x)**4/(sec(c + d*x)**5 + 5*sec(c + d*x)**4 + 10*sec(c + d*x)**3 + 10*sec(c + d*x)**2 + 5*sec(c + d*x) + 1), x)/a**5
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.55 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\frac {63 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {18 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{1008 \, a^{5} d} \]
1/1008*(63*sin(d*x + c)/(cos(d*x + c) + 1) + 42*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 18*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 7*sin(d*x + c)^9/(cos (d*x + c) + 1)^9)/(a^5*d)
Time = 0.35 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.37 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{1008 \, a^{5} d} \]
-1/1008*(7*tan(1/2*d*x + 1/2*c)^9 + 18*tan(1/2*d*x + 1/2*c)^7 - 42*tan(1/2 *d*x + 1/2*c)^3 - 63*tan(1/2*d*x + 1/2*c))/(a^5*d)
Time = 13.62 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.36 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+63\right )}{1008\,a^5\,d} \]