Integrand size = 15, antiderivative size = 56 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=a^3 x-\frac {a^3 \tan (c+d x)}{d}+\frac {a^3 \tan ^3(c+d x)}{3 d}-\frac {a^3 \tan ^5(c+d x)}{5 d} \] Output:
a^3*x-a^3*tan(d*x+c)/d+1/3*a^3*tan(d*x+c)^3/d-1/5*a^3*tan(d*x+c)^5/d
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.04 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=-a^3 \left (-\frac {\arctan (\tan (c+d x))}{d}+\frac {\tan (c+d x)}{d}-\frac {\tan ^3(c+d x)}{3 d}+\frac {\tan ^5(c+d x)}{5 d}\right ) \] Input:
Integrate[(a - a*Sec[c + d*x]^2)^3,x]
Output:
-(a^3*(-(ArcTan[Tan[c + d*x]]/d) + Tan[c + d*x]/d - Tan[c + d*x]^3/(3*d) + Tan[c + d*x]^5/(5*d)))
Time = 0.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4608, 3042, 3954, 3042, 3954, 3042, 3954, 24}
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.
\(\displaystyle \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \sec (c+d x)^2\right )^3dx\) |
\(\Big \downarrow \) 4608 |
\(\displaystyle -a^3 \int \tan ^6(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a^3 \int \tan (c+d x)^6dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -a^3 \left (\frac {\tan ^5(c+d x)}{5 d}-\int \tan ^4(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a^3 \left (\frac {\tan ^5(c+d x)}{5 d}-\int \tan (c+d x)^4dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -a^3 \left (\int \tan ^2(c+d x)dx+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a^3 \left (\int \tan (c+d x)^2dx+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -a^3 \left (-\int 1dx+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}+\frac {\tan (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -a^3 \left (\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}+\frac {\tan (c+d x)}{d}-x\right )\) |
Input:
Int[(a - a*Sec[c + d*x]^2)^3,x]
Output:
-(a^3*(-x + Tan[c + d*x]/d - Tan[c + d*x]^3/(3*d) + Tan[c + d*x]^5/(5*d)))
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ b^p Int[ActivateTrig[u*tan[e + f*x]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]
Result contains complex when optimal does not.
Time = 2.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.34
method | result | size |
risch | \(a^{3} x -\frac {2 i a^{3} \left (45 \,{\mathrm e}^{8 i \left (d x +c \right )}+90 \,{\mathrm e}^{6 i \left (d x +c \right )}+140 \,{\mathrm e}^{4 i \left (d x +c \right )}+70 \,{\mathrm e}^{2 i \left (d x +c \right )}+23\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\) | \(75\) |
derivativedivides | \(\frac {a^{3} \left (d x +c \right )-3 a^{3} \tan \left (d x +c \right )-3 a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(81\) |
default | \(\frac {a^{3} \left (d x +c \right )-3 a^{3} \tan \left (d x +c \right )-3 a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(81\) |
parts | \(a^{3} x +\frac {a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {3 a^{3} \tan \left (d x +c \right )}{d}-\frac {3 a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(82\) |
parallelrisch | \(\frac {\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} x d -5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} x d +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} x d -\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3}-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} x d +\frac {356 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x d -\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-d x +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(176\) |
norman | \(\frac {a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-a^{3} x +\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {32 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {356 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {32 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+10 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}\) | \(201\) |
Input:
int((a-sec(d*x+c)^2*a)^3,x,method=_RETURNVERBOSE)
Output:
a^3*x-2/15*I*a^3*(45*exp(8*I*(d*x+c))+90*exp(6*I*(d*x+c))+140*exp(4*I*(d*x +c))+70*exp(2*I*(d*x+c))+23)/d/(exp(2*I*(d*x+c))+1)^5
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.23 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=\frac {15 \, a^{3} d x \cos \left (d x + c\right )^{5} - {\left (23 \, a^{3} \cos \left (d x + c\right )^{4} - 11 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3}\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a-a*sec(d*x+c)^2)^3,x, algorithm="fricas")
Output:
1/15*(15*a^3*d*x*cos(d*x + c)^5 - (23*a^3*cos(d*x + c)^4 - 11*a^3*cos(d*x + c)^2 + 3*a^3)*sin(d*x + c))/(d*cos(d*x + c)^5)
\[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=- a^{3} \left (\int \left (-1\right )\, dx + \int 3 \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 3 \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a-a*sec(d*x+c)**2)**3,x)
Output:
-a**3*(Integral(-1, x) + Integral(3*sec(c + d*x)**2, x) + Integral(-3*sec( c + d*x)**4, x) + Integral(sec(c + d*x)**6, x))
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.45 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=a^{3} x - \frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3}}{15 \, d} + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3}}{d} - \frac {3 \, a^{3} \tan \left (d x + c\right )}{d} \] Input:
integrate((a-a*sec(d*x+c)^2)^3,x, algorithm="maxima")
Output:
a^3*x - 1/15*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3/ d + (tan(d*x + c)^3 + 3*tan(d*x + c))*a^3/d - 3*a^3*tan(d*x + c)/d
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=-\frac {3 \, a^{3} \tan \left (d x + c\right )^{5} - 5 \, a^{3} \tan \left (d x + c\right )^{3} - 15 \, {\left (d x + c\right )} a^{3} + 15 \, a^{3} \tan \left (d x + c\right )}{15 \, d} \] Input:
integrate((a-a*sec(d*x+c)^2)^3,x, algorithm="giac")
Output:
-1/15*(3*a^3*tan(d*x + c)^5 - 5*a^3*tan(d*x + c)^3 - 15*(d*x + c)*a^3 + 15 *a^3*tan(d*x + c))/d
Time = 15.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=-\frac {\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}-\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+a^3\,\mathrm {tan}\left (c+d\,x\right )-d\,x\,a^3}{d} \] Input:
int((a - a/cos(c + d*x)^2)^3,x)
Output:
-(a^3*tan(c + d*x) - (a^3*tan(c + d*x)^3)/3 + (a^3*tan(c + d*x)^5)/5 - a^3 *d*x)/d
Time = 0.17 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.02 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=\frac {a^{3} \left (15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} d x -30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x +15 \cos \left (d x +c \right ) d x -23 \sin \left (d x +c \right )^{5}+35 \sin \left (d x +c \right )^{3}-15 \sin \left (d x +c \right )\right )}{15 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((a-a*sec(d*x+c)^2)^3,x)
Output:
(a**3*(15*cos(c + d*x)*sin(c + d*x)**4*d*x - 30*cos(c + d*x)*sin(c + d*x)* *2*d*x + 15*cos(c + d*x)*d*x - 23*sin(c + d*x)**5 + 35*sin(c + d*x)**3 - 1 5*sin(c + d*x)))/(15*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))