Integrand size = 12, antiderivative size = 111 \[ \int \frac {1}{(a+a \sec (c+d x))^4} \, dx=\frac {x}{a^4}-\frac {11 \tan (c+d x)}{21 a^4 d (1+\sec (c+d x))^2}-\frac {32 \tan (c+d x)}{21 a^4 d (1+\sec (c+d x))}-\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 \tan (c+d x)}{7 a d (a+a \sec (c+d x))^3} \]
x/a^4-11/21*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-32/21*tan(d*x+c)/a^4/d/(1+se c(d*x+c))-1/7*tan(d*x+c)/d/(a+a*sec(d*x+c))^4-2/7*tan(d*x+c)/a/d/(a+a*sec( d*x+c))^3
Leaf count is larger than twice the leaf count of optimal. \(224\) vs. \(2(111)=222\).
Time = 2.53 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.02 \[ \int \frac {1}{(a+a \sec (c+d x))^4} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (735 d x \cos \left (\frac {d x}{2}\right )+735 d x \cos \left (c+\frac {d x}{2}\right )+441 d x \cos \left (c+\frac {3 d x}{2}\right )+441 d x \cos \left (2 c+\frac {3 d x}{2}\right )+147 d x \cos \left (2 c+\frac {5 d x}{2}\right )+147 d x \cos \left (3 c+\frac {5 d x}{2}\right )+21 d x \cos \left (3 c+\frac {7 d x}{2}\right )+21 d x \cos \left (4 c+\frac {7 d x}{2}\right )-1988 \sin \left (\frac {d x}{2}\right )+1652 \sin \left (c+\frac {d x}{2}\right )-1428 \sin \left (c+\frac {3 d x}{2}\right )+756 \sin \left (2 c+\frac {3 d x}{2}\right )-560 \sin \left (2 c+\frac {5 d x}{2}\right )+168 \sin \left (3 c+\frac {5 d x}{2}\right )-104 \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{2688 a^4 d} \]
(Sec[c/2]*Sec[(c + d*x)/2]^7*(735*d*x*Cos[(d*x)/2] + 735*d*x*Cos[c + (d*x) /2] + 441*d*x*Cos[c + (3*d*x)/2] + 441*d*x*Cos[2*c + (3*d*x)/2] + 147*d*x* Cos[2*c + (5*d*x)/2] + 147*d*x*Cos[3*c + (5*d*x)/2] + 21*d*x*Cos[3*c + (7* d*x)/2] + 21*d*x*Cos[4*c + (7*d*x)/2] - 1988*Sin[(d*x)/2] + 1652*Sin[c + ( d*x)/2] - 1428*Sin[c + (3*d*x)/2] + 756*Sin[2*c + (3*d*x)/2] - 560*Sin[2*c + (5*d*x)/2] + 168*Sin[3*c + (5*d*x)/2] - 104*Sin[3*c + (7*d*x)/2]))/(268 8*a^4*d)
Time = 0.74 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {3042, 4264, 25, 3042, 4410, 27, 3042, 4410, 25, 3042, 4407, 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.
| \(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 4264 |
| \(\displaystyle -\frac {\int -\frac {7 a-3 a \sec (c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {7 a-3 a \sec (c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {7 a-3 a \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4410 |
| \(\displaystyle \frac {-\frac {\int -\frac {5 \left (7 a^2-4 a^2 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {7 a^2-4 a^2 \sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {7 a^2-4 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4410 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {21 a^3-11 a^3 \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {21 a^3-11 a^3 \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {21 a^3-11 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\frac {\frac {21 a^2 x-32 a^3 \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {21 a^2 x-32 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {\frac {\frac {21 a^2 x-\frac {32 a^3 \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}-\frac {2 a \tan (c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
-1/7*Tan[c + d*x]/(d*(a + a*Sec[c + d*x])^4) + ((-2*a*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + ((-11*Tan[c + d*x])/(3*d*(1 + Sec[c + d*x])^2) + (2 1*a^2*x - (32*a^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/a^2)/(7 *a^2)
3.1.77.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c + d*x])*((a + b*Csc[c + d*x])^n/(d*(2*n + 1))), x] + Simp[1/(a^2*(2*n + 1)) Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && Int egerQ[2*n]
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_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-(b*c - a*d))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.58
| method | result | size |
| parallelrisch | \(\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+168 d x -315 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{168 a^{4} d}\) | \(64\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\) | \(72\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\) | \(72\) |
| norman | \(\frac {\frac {x}{a}-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{56 a d}}{a^{3}}\) | \(85\) |
| risch | \(\frac {x}{a^{4}}-\frac {4 i \left (42 \,{\mathrm e}^{6 i \left (d x +c \right )}+189 \,{\mathrm e}^{5 i \left (d x +c \right )}+413 \,{\mathrm e}^{4 i \left (d x +c \right )}+497 \,{\mathrm e}^{3 i \left (d x +c \right )}+357 \,{\mathrm e}^{2 i \left (d x +c \right )}+140 \,{\mathrm e}^{i \left (d x +c \right )}+26\right )}{21 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(97\) |
1/168*(3*tan(1/2*d*x+1/2*c)^7-21*tan(1/2*d*x+1/2*c)^5+77*tan(1/2*d*x+1/2*c )^3+168*d*x-315*tan(1/2*d*x+1/2*c))/a^4/d
Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.37 \[ \int \frac {1}{(a+a \sec (c+d x))^4} \, dx=\frac {21 \, d x \cos \left (d x + c\right )^{4} + 84 \, d x \cos \left (d x + c\right )^{3} + 126 \, d x \cos \left (d x + c\right )^{2} + 84 \, d x \cos \left (d x + c\right ) + 21 \, d x - {\left (52 \, \cos \left (d x + c\right )^{3} + 124 \, \cos \left (d x + c\right )^{2} + 107 \, \cos \left (d x + c\right ) + 32\right )} \sin \left (d x + c\right )}{21 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
1/21*(21*d*x*cos(d*x + c)^4 + 84*d*x*cos(d*x + c)^3 + 126*d*x*cos(d*x + c) ^2 + 84*d*x*cos(d*x + c) + 21*d*x - (52*cos(d*x + c)^3 + 124*cos(d*x + c)^ 2 + 107*cos(d*x + c) + 32)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*c os(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)
\[ \int \frac {1}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {1}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
Integral(1/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*se c(c + d*x) + 1), x)/a**4
Time = 0.37 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}}{168 \, d} \]
-1/168*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c)^7/(c os(d*x + c) + 1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) /d
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {168 \, {\left (d x + c\right )}}{a^{4}} + \frac {3 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 77 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 315 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{168 \, d} \]
1/168*(168*(d*x + c)/a^4 + (3*a^24*tan(1/2*d*x + 1/2*c)^7 - 21*a^24*tan(1/ 2*d*x + 1/2*c)^5 + 77*a^24*tan(1/2*d*x + 1/2*c)^3 - 315*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d
Time = 13.58 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+a \sec (c+d x))^4} \, dx=\frac {x}{a^4}+\frac {-\frac {52\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{21}+\frac {16\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{21}-\frac {5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{28}+\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]