3.150 \(\int \frac{1}{(a-a \sec ^2(c+d x))^3} \, dx\)

Optimal. Leaf size=55 \[ \frac{\cot ^5(c+d x)}{5 a^3 d}-\frac{\cot ^3(c+d x)}{3 a^3 d}+\frac{\cot (c+d x)}{a^3 d}+\frac{x}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.0387751, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4120, 3473, 8} \[ \frac{\cot ^5(c+d x)}{5 a^3 d}-\frac{\cot ^3(c+d x)}{3 a^3 d}+\frac{\cot (c+d x)}{a^3 d}+\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sec[c + d*x]^2)^(-3),x]

[Out]

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

Rule 4120

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]

Rule 3473

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 8

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

Rubi steps

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

Mathematica [C]  time = 0.0473571, size = 36, normalized size = 0.65 \[ \frac{\cot ^5(c+d x) \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},-\tan ^2(c+d x)\right )}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sec[c + d*x]^2)^(-3),x]

[Out]

(Cot[c + d*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2])/(5*a^3*d)

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Maple [A]  time = 0.049, size = 63, normalized size = 1.2 \begin{align*}{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d{a}^{3}}}-{\frac{1}{3\,d{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}+{\frac{1}{5\,d{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{5}}}+{\frac{1}{d{a}^{3}\tan \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-a*sec(d*x+c)^2)^3,x)

[Out]

1/d/a^3*arctan(tan(d*x+c))-1/3/d/a^3/tan(d*x+c)^3+1/5/d/a^3/tan(d*x+c)^5+1/d/a^3/tan(d*x+c)

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Maxima [A]  time = 1.51894, size = 68, normalized size = 1.24 \begin{align*} \frac{\frac{15 \,{\left (d x + c\right )}}{a^{3}} + \frac{15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/15*(15*(d*x + c)/a^3 + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/(a^3*tan(d*x + c)^5))/d

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Fricas [B]  time = 0.484161, size = 274, normalized size = 4.98 \begin{align*} \frac{23 \, \cos \left (d x + c\right )^{5} - 35 \, \cos \left (d x + c\right )^{3} + 15 \,{\left (d x \cos \left (d x + c\right )^{4} - 2 \, d x \cos \left (d x + c\right )^{2} + d x\right )} \sin \left (d x + c\right ) + 15 \, \cos \left (d x + c\right )}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/15*(23*cos(d*x + c)^5 - 35*cos(d*x + c)^3 + 15*(d*x*cos(d*x + c)^4 - 2*d*x*cos(d*x + c)^2 + d*x)*sin(d*x + c
) + 15*cos(d*x + c))/((a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{1}{\sec ^{6}{\left (c + d x \right )} - 3 \sec ^{4}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} - 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)**2)**3,x)

[Out]

-Integral(1/(sec(c + d*x)**6 - 3*sec(c + d*x)**4 + 3*sec(c + d*x)**2 - 1), x)/a**3

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Giac [B]  time = 1.30958, size = 150, normalized size = 2.73 \begin{align*} \frac{\frac{480 \,{\left (d x + c\right )}}{a^{3}} + \frac{330 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 35 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}} - \frac{3 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 35 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 330 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/480*(480*(d*x + c)/a^3 + (330*tan(1/2*d*x + 1/2*c)^4 - 35*tan(1/2*d*x + 1/2*c)^2 + 3)/(a^3*tan(1/2*d*x + 1/2
*c)^5) - (3*a^12*tan(1/2*d*x + 1/2*c)^5 - 35*a^12*tan(1/2*d*x + 1/2*c)^3 + 330*a^12*tan(1/2*d*x + 1/2*c))/a^15
)/d