3.46 \(\int \sec (a+b x) \tan (a+b x) \, dx\)

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

[Out]

sec(b*x+a)/b

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2606, 8} \[ \frac {\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]*Tan[a + b*x],x]

[Out]

Sec[a + b*x]/b

Rule 8

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \[ \frac {\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]*Tan[a + b*x],x]

[Out]

Sec[a + b*x]/b

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fricas [A]  time = 0.40, size = 12, normalized size = 1.20 \[ \frac {1}{b \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*sin(b*x+a),x, algorithm="fricas")

[Out]

1/(b*cos(b*x + a))

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giac [A]  time = 1.18, size = 12, normalized size = 1.20 \[ \frac {1}{b \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*sin(b*x+a),x, algorithm="giac")

[Out]

1/(b*cos(b*x + a))

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maple [A]  time = 0.02, size = 11, normalized size = 1.10 \[ \frac {\sec \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^2*sin(b*x+a),x)

[Out]

sec(b*x+a)/b

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maxima [A]  time = 0.32, size = 12, normalized size = 1.20 \[ \frac {1}{b \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*sin(b*x+a),x, algorithm="maxima")

[Out]

1/(b*cos(b*x + a))

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mupad [B]  time = 0.52, size = 20, normalized size = 2.00 \[ -\frac {2}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/cos(a + b*x)^2,x)

[Out]

-2/(b*(tan(a/2 + (b*x)/2)^2 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin {\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**2*sin(b*x+a),x)

[Out]

Integral(sin(a + b*x)*sec(a + b*x)**2, x)

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