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

Optimal. Leaf size=19 \[ \frac{\cot (c+d x)}{a d}+\frac{x}{a} \]

[Out]

x/a + Cot[c + d*x]/(a*d)

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Rubi [A]  time = 0.0214506, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4120, 3473, 8} \[ \frac{\cot (c+d x)}{a d}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sec[c + d*x]^2)^(-1),x]

[Out]

x/a + Cot[c + d*x]/(a*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}{a-a \sec ^2(c+d x)} \, dx &=-\frac{\int \cot ^2(c+d x) \, dx}{a}\\ &=\frac{\cot (c+d x)}{a d}+\frac{\int 1 \, dx}{a}\\ &=\frac{x}{a}+\frac{\cot (c+d x)}{a d}\\ \end{align*}

Mathematica [C]  time = 0.0279704, size = 31, normalized size = 1.63 \[ \frac{\cot (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-\tan ^2(c+d x)\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sec[c + d*x]^2)^(-1),x]

[Out]

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

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Maple [A]  time = 0.047, size = 31, normalized size = 1.6 \begin{align*}{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{ad}}+{\frac{1}{ad\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),x)

[Out]

1/a/d*arctan(tan(d*x+c))+1/a/d/tan(d*x+c)

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Maxima [A]  time = 1.52573, size = 35, normalized size = 1.84 \begin{align*} \frac{\frac{d x + c}{a} + \frac{1}{a \tan \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)/a + 1/(a*tan(d*x + c)))/d

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Fricas [A]  time = 0.463929, size = 73, normalized size = 3.84 \begin{align*} \frac{d x \sin \left (d x + c\right ) + \cos \left (d x + c\right )}{a d \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),x, algorithm="fricas")

[Out]

(d*x*sin(d*x + c) + cos(d*x + c))/(a*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 ^{2}{\left (c + d x \right )} - 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.29523, size = 61, normalized size = 3.21 \begin{align*} \frac{\frac{2 \,{\left (d x + c\right )}}{a} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} + \frac{1}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*x + c)/a - tan(1/2*d*x + 1/2*c)/a + 1/(a*tan(1/2*d*x + 1/2*c)))/d