\(\int (a-a \sec ^2(c+d x))^3 \, dx\) [145]

Optimal result

Integrand size = 15, antiderivative size = 56 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=a^3 x-\frac {a^3 \tan (c+d x)}{d}+\frac {a^3 \tan ^3(c+d x)}{3 d}-\frac {a^3 \tan ^5(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4205, 3554, 8} \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=-\frac {a^3 \tan ^5(c+d x)}{5 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}-\frac {a^3 \tan (c+d x)}{d}+a^3 x \]

[In]

Int[(a - a*Sec[c + d*x]^2)^3,x]

[Out]

a^3*x - (a^3*Tan[c + d*x])/d + (a^3*Tan[c + d*x]^3)/(3*d) - (a^3*Tan[c + d*x]^5)/(5*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 4205

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\left (a^3 \int \tan ^6(c+d x) \, dx\right ) \\ & = -\frac {a^3 \tan ^5(c+d x)}{5 d}+a^3 \int \tan ^4(c+d x) \, dx \\ & = \frac {a^3 \tan ^3(c+d x)}{3 d}-\frac {a^3 \tan ^5(c+d x)}{5 d}-a^3 \int \tan ^2(c+d x) \, dx \\ & = -\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}+a^3 \int 1 \, 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.04 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=-a^3 \left (-\frac {\arctan (\tan (c+d x))}{d}+\frac {\tan (c+d x)}{d}-\frac {\tan ^3(c+d x)}{3 d}+\frac {\tan ^5(c+d x)}{5 d}\right ) \]

[In]

Integrate[(a - a*Sec[c + d*x]^2)^3,x]

[Out]

-(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)))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.55 (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	\(a^{3} x -\frac {3 a^{3} \tan \left (d x +c \right )}{d}-\frac {3 a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(82\)
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	\(\frac {a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-a^{3} x +\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {32 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {356 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {32 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+10 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}\)	\(201\)

[In]

int((a-a*sec(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.23 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=\frac {15 \, a^{3} d x \cos \left (d x + c\right )^{5} - {\left (23 \, a^{3} \cos \left (d x + c\right )^{4} - 11 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3}\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate((a-a*sec(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

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)

Sympy [F]

\[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=- a^{3} \left (\int \left (-1\right )\, dx + \int 3 \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 3 \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a-a*sec(d*x+c)**2)**3,x)

[Out]

-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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.45 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=a^{3} x - \frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3}}{15 \, d} + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3}}{d} - \frac {3 \, a^{3} \tan \left (d x + c\right )}{d} \]

[In]

integrate((a-a*sec(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=-\frac {3 \, a^{3} \tan \left (d x + c\right )^{5} - 5 \, a^{3} \tan \left (d x + c\right )^{3} - 15 \, {\left (d x + c\right )} a^{3} + 15 \, a^{3} \tan \left (d x + c\right )}{15 \, d} \]

[In]

integrate((a-a*sec(d*x+c)^2)^3,x, algorithm="giac")

[Out]

-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

Mupad [B] (verification not implemented)

Time = 19.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \left (a-a \sec ^2(c+d x)\right )^3 \, dx=-\frac {\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}-\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+a^3\,\mathrm {tan}\left (c+d\,x\right )-d\,x\,a^3}{d} \]

[In]

int((a - a/cos(c + d*x)^2)^3,x)

[Out]

-(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