Integrand size = 21, antiderivative size = 143 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{63 a d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {2 \tan (c+d x)}{63 d \left (a^5+a^5 \sec (c+d x)\right )} \] Output:
-1/9*tan(d*x+c)/d/(a+a*sec(d*x+c))^5+5/63*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+2/63*tan(d*x+c)/a/d/(a^2+a^2*se c(d*x+c))^2+2/63*tan(d*x+c)/d/(a^5+a^5*sec(d*x+c))
Time = 4.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^9\left (\frac {1}{2} (c+d x)\right ) \left (315 \sin \left (\frac {d x}{2}\right )-315 \sin \left (c+\frac {d x}{2}\right )+273 \sin \left (c+\frac {3 d x}{2}\right )-147 \sin \left (2 c+\frac {3 d x}{2}\right )+117 \sin \left (2 c+\frac {5 d x}{2}\right )-63 \sin \left (3 c+\frac {5 d x}{2}\right )+45 \sin \left (3 c+\frac {7 d x}{2}\right )+5 \sin \left (4 c+\frac {9 d x}{2}\right )\right )}{16128 a^5 d} \] Input:
Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^5,x]
Output:
(Sec[c/2]*Sec[(c + d*x)/2]^9*(315*Sin[(d*x)/2] - 315*Sin[c + (d*x)/2] + 27 3*Sin[c + (3*d*x)/2] - 147*Sin[2*c + (3*d*x)/2] + 117*Sin[2*c + (5*d*x)/2] - 63*Sin[3*c + (5*d*x)/2] + 45*Sin[3*c + (7*d*x)/2] + 5*Sin[4*c + (9*d*x) /2]))/(16128*a^5*d)
Time = 0.76 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4284, 3042, 4283, 3042, 4283, 3042, 4283, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x)}{(a \sec (c+d x)+a)^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^5}dx\) |
| \(\Big \downarrow \) 4284 |
| \(\displaystyle \frac {5 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^4}dx}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}dx}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4283 |
| \(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4283 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4283 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {5 \left (\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {3 \left (\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {2 \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}\right )}{7 a}\right )}{9 a}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
Input:
Int[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^5,x]
Output:
-1/9*Tan[c + d*x]/(d*(a + a*Sec[c + d*x])^5) + (5*(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) + (2*(T an[c + d*x]/(3*d*(a + a*Sec[c + d*x])^2) + Tan[c + d*x]/(3*a*d*(a + a*Sec[ c + d*x]))))/(5*a)))/(7*a)))/(9*a)
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_)]*(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[(m + 1)/(a*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) ] && IntegerQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^m/(f*(2*m + 1))), x ] + Simp[m/(b*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x ], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.38 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.40
| method | result | size |
| parallelrisch | \(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{7}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9\right )}{144 d \,a^{5}}\) | \(57\) |
| 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 | \(\frac {2 i \left (63 \,{\mathrm e}^{7 i \left (d x +c \right )}+147 \,{\mathrm e}^{6 i \left (d x +c \right )}+315 \,{\mathrm e}^{5 i \left (d x +c \right )}+315 \,{\mathrm e}^{4 i \left (d x +c \right )}+273 \,{\mathrm e}^{3 i \left (d x +c \right )}+117 \,{\mathrm e}^{2 i \left (d x +c \right )}+45 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}{63 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\) | \(102\) |
| norman | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{48 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{24 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{56 a d}+\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{1008 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{144 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{4}}\) | \(133\) |
Input:
int(sec(d*x+c)^2/(a+a*sec(d*x+c))^5,x,method=_RETURNVERBOSE)
Output:
-1/144*tan(1/2*d*x+1/2*c)*(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)/d/a^5
Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {{\left (5 \, \cos \left (d x + c\right )^{4} + 25 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{63 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \] Input:
integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^5,x, algorithm="fricas")
Output:
1/63*(5*cos(d*x + c)^4 + 25*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 10*cos(d* x + c) + 2)*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)
\[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\int \frac {\sec ^{2}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \] Input:
integrate(sec(d*x+c)**2/(a+a*sec(d*x+c))**5,x)
Output:
Integral(sec(c + d*x)**2/(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 = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.61 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\frac {63 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {18 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{1008 \, a^{5} d} \] Input:
integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^5,x, algorithm="maxima")
Output:
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 = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.41 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{1008 \, a^{5} d} \] Input:
integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^5,x, algorithm="giac")
Output:
-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 = 9.39 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.41 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-63\right )}{1008\,a^5\,d} \] Input:
int(1/(cos(c + d*x)^2*(a + a/cos(c + d*x))^5),x)
Output:
-(tan(c/2 + (d*x)/2)*(42*tan(c/2 + (d*x)/2)^2 - 18*tan(c/2 + (d*x)/2)^6 + 7*tan(c/2 + (d*x)/2)^8 - 63))/(1008*a^5*d)
Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.41 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+63\right )}{1008 a^{5} d} \] Input:
int(sec(d*x+c)^2/(a+a*sec(d*x+c))^5,x)
Output:
(tan((c + d*x)/2)*( - 7*tan((c + d*x)/2)**8 + 18*tan((c + d*x)/2)**6 - 42* tan((c + d*x)/2)**2 + 63))/(1008*a**5*d)