3.47 \(\int \frac{1}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=29 \[ \frac{x}{a}-\frac{\tan (c+d x)}{d (a \sec (c+d x)+a)} \]

[Out]

x/a - Tan[c + d*x]/(d*(a + a*Sec[c + d*x]))

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Rubi [A]  time = 0.0135317, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3777, 8} \[ \frac{x}{a}-\frac{\tan (c+d x)}{d (a \sec (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(-1),x]

[Out]

x/a - Tan[c + d*x]/(d*(a + a*Sec[c + d*x]))

Rule 3777

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] + Dist[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] && IntegerQ[2*n]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{1}{a+a \sec (c+d x)} \, dx &=-\frac{\tan (c+d x)}{d (a+a \sec (c+d x))}+\frac{\int a \, dx}{a^2}\\ &=\frac{x}{a}-\frac{\tan (c+d x)}{d (a+a \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.122824, size = 58, normalized size = 2. \[ \frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (d x \cos \left (c+\frac{d x}{2}\right )-2 \sin \left (\frac{d x}{2}\right )+d x \cos \left (\frac{d x}{2}\right )\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]*(d*x*Cos[(d*x)/2] + d*x*Cos[c + (d*x)/2] - 2*Sin[(d*x)/2]))/(2*a*d)

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Maple [A]  time = 0.032, size = 37, normalized size = 1.3 \begin{align*} -{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.6481, size = 66, normalized size = 2.28 \begin{align*} \frac{\frac{2 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.59883, size = 89, normalized size = 3.07 \begin{align*} \frac{d x \cos \left (d x + c\right ) + d x - \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x*cos(d*x + c) + d*x - sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sec{\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)),x)

[Out]

Integral(1/(sec(c + d*x) + 1), x)/a

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Giac [A]  time = 1.29995, size = 38, normalized size = 1.31 \begin{align*} \frac{\frac{d x + c}{a} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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