Integrand size = 19, antiderivative size = 112 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {3 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{35 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {2 \tan (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )} \]
1/7*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+3/35*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3 +2/35*tan(d*x+c)/d/(a^2+a^2*sec(d*x+c))^2+2/35*tan(d*x+c)/d/(a^4+a^4*sec(d *x+c))
Time = 2.52 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (210 \sin \left (\frac {d x}{2}\right )-210 \sin \left (c+\frac {d x}{2}\right )+147 \sin \left (c+\frac {3 d x}{2}\right )-105 \sin \left (2 c+\frac {3 d x}{2}\right )+49 \sin \left (2 c+\frac {5 d x}{2}\right )-35 \sin \left (3 c+\frac {5 d x}{2}\right )+12 \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{2240 a^4 d} \]
(Sec[c/2]*Sec[(c + d*x)/2]^7*(210*Sin[(d*x)/2] - 210*Sin[c + (d*x)/2] + 14 7*Sin[c + (3*d*x)/2] - 105*Sin[2*c + (3*d*x)/2] + 49*Sin[2*c + (5*d*x)/2] - 35*Sin[3*c + (5*d*x)/2] + 12*Sin[3*c + (7*d*x)/2]))/(2240*a^4*d)
Time = 0.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a \sec (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
\(\Big \downarrow \) 4283 |
\(\displaystyle \frac {3 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4283 |
\(\displaystyle \frac {3 \left (\frac {2 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4283 |
\(\displaystyle \frac {3 \left (\frac {2 \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {2 \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4281 |
\(\displaystyle \frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {3 \left (\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {2 \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}\right )}{7 a}\) |
Tan[c + d*x]/(7*d*(a + a*Sec[c + d*x])^4) + (3*(Tan[c + d*x]/(5*d*(a + a*S ec[c + d*x])^3) + (2*(Tan[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)
3.1.76.3.1 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_)]*(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]
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.51
method | result | size |
parallelrisch | \(-\frac {\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a^{4} d}\) | \(57\) |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(58\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(58\) |
norman | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{40 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{56 a d}}{a^{3}}\) | \(80\) |
risch | \(\frac {2 i \left (35 \,{\mathrm e}^{6 i \left (d x +c \right )}+105 \,{\mathrm e}^{5 i \left (d x +c \right )}+210 \,{\mathrm e}^{4 i \left (d x +c \right )}+210 \,{\mathrm e}^{3 i \left (d x +c \right )}+147 \,{\mathrm e}^{2 i \left (d x +c \right )}+49 \,{\mathrm e}^{i \left (d x +c \right )}+12\right )}{35 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(91\) |
-1/56*(tan(1/2*d*x+1/2*c)^6-21/5*tan(1/2*d*x+1/2*c)^4+7*tan(1/2*d*x+1/2*c) ^2-7)*tan(1/2*d*x+1/2*c)/a^4/d
Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {{\left (12 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{35 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
1/35*(12*cos(d*x + c)^3 + 13*cos(d*x + c)^2 + 8*cos(d*x + c) + 2)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c) ^2 + 4*a^4*d*cos(d*x + c) + a^4*d)
\[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
Integral(sec(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x )**2 + 4*sec(c + d*x) + 1), x)/a**4
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{280 \, a^{4} d} \]
1/280*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c ) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos( d*x + c) + 1)^7)/(a^4*d)
Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{280 \, a^{4} d} \]
-1/280*(5*tan(1/2*d*x + 1/2*c)^7 - 21*tan(1/2*d*x + 1/2*c)^5 + 35*tan(1/2* d*x + 1/2*c)^3 - 35*tan(1/2*d*x + 1/2*c))/(a^4*d)
Time = 13.46 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.52 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-35\right )}{280\,a^4\,d} \]