3.117 \(\int \sec ^2(a+b x) \, dx\)

Optimal. Leaf size=10 \[ \frac{\tan (a+b x)}{b} \]

[Out]

Tan[a + b*x]/b

________________________________________________________________________________________

Rubi [A]  time = 0.0084205, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3767, 8} \[ \frac{\tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^2,x]

[Out]

Tan[a + b*x]/b

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0029589, size = 10, normalized size = 1. \[ \frac{\tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^2,x]

[Out]

Tan[a + b*x]/b

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 11, normalized size = 1.1 \begin{align*}{\frac{\tan \left ( bx+a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(b*x+a)^2,x)

[Out]

tan(b*x+a)/b

________________________________________________________________________________________

Maxima [A]  time = 0.934283, size = 14, normalized size = 1.4 \begin{align*} \frac{\tan \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)^2,x, algorithm="maxima")

[Out]

tan(b*x + a)/b

________________________________________________________________________________________

Fricas [A]  time = 1.7253, size = 42, normalized size = 4.2 \begin{align*} \frac{\sin \left (b x + a\right )}{b \cos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)^2,x, algorithm="fricas")

[Out]

sin(b*x + a)/(b*cos(b*x + a))

________________________________________________________________________________________

Sympy [A]  time = 1.27562, size = 58, normalized size = 5.8 \begin{align*} \begin{cases} \tilde{\infty } x & \text{for}\: \left (a = - \frac{\pi }{2} \vee a = - b x - \frac{\pi }{2}\right ) \wedge \left (a = - b x - \frac{\pi }{2} \vee b = 0\right ) \\\frac{x}{\cos ^{2}{\left (a \right )}} & \text{for}\: b = 0 \\- \frac{2 \tan{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} - b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)**2,x)

[Out]

Piecewise((zoo*x, (Eq(b, 0) | Eq(a, -b*x - pi/2)) & (Eq(a, -pi/2) | Eq(a, -b*x - pi/2))), (x/cos(a)**2, Eq(b,
0)), (-2*tan(a/2 + b*x/2)/(b*tan(a/2 + b*x/2)**2 - b), True))

________________________________________________________________________________________

Giac [A]  time = 1.08762, size = 14, normalized size = 1.4 \begin{align*} \frac{\tan \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)^2,x, algorithm="giac")

[Out]

tan(b*x + a)/b