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

Optimal. Leaf size=22 \[ \frac {\tan (c+d x)}{d (a \sec (c+d x)+a)} \]

[Out]

tan(d*x+c)/d/(a+a*sec(d*x+c))

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {3794} \[ \frac {\tan (c+d x)}{d (a \sec (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a + a*Sec[c + d*x]),x]

[Out]

Tan[c + d*x]/(d*(a + a*Sec[c + d*x]))

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.77 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a + a*Sec[c + d*x]),x]

[Out]

Tan[(c + d*x)/2]/(a*d)

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fricas [A]  time = 0.69, size = 22, normalized size = 1.00 \[ \frac {\sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(d*x + c)/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 3.89, size = 16, normalized size = 0.73 \[ \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(1/2*d*x + 1/2*c)/(a*d)

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maple [A]  time = 0.36, size = 17, normalized size = 0.77 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.39, size = 23, normalized size = 1.05 \[ \frac {\sin \left (d x + c\right )}{a d {\left (\cos \left (d x + c\right ) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(d*x + c)/(a*d*(cos(d*x + c) + 1))

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mupad [B]  time = 0.59, size = 16, normalized size = 0.73 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)/(a*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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