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

Optimal. Leaf size=15 \[ \frac {\sec ^2(a+b x)}{2 b} \]

[Out]

1/2*sec(b*x+a)^2/b

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2606, 30} \[ \frac {\sec ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^2*Tan[a + b*x],x]

[Out]

Sec[a + b*x]^2/(2*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \[ \frac {\sec ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^2*Tan[a + b*x],x]

[Out]

Sec[a + b*x]^2/(2*b)

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fricas [A]  time = 0.45, size = 13, normalized size = 0.87 \[ \frac {1}{2 \, b \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a),x, algorithm="fricas")

[Out]

1/2/(b*cos(b*x + a)^2)

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giac [A]  time = 0.22, size = 13, normalized size = 0.87 \[ \frac {1}{2 \, b \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a),x, algorithm="giac")

[Out]

1/2/(b*cos(b*x + a)^2)

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maple [A]  time = 0.02, size = 14, normalized size = 0.93 \[ \frac {\sec ^{2}\left (b x +a \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3*sin(b*x+a),x)

[Out]

1/2*sec(b*x+a)^2/b

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maxima [A]  time = 0.32, size = 17, normalized size = 1.13 \[ -\frac {1}{2 \, {\left (\sin \left (b x + a\right )^{2} - 1\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/2/((sin(b*x + a)^2 - 1)*b)

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mupad [B]  time = 0.38, size = 13, normalized size = 0.87 \[ \frac {{\mathrm {tan}\left (a+b\,x\right )}^2}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/cos(a + b*x)^3,x)

[Out]

tan(a + b*x)^2/(2*b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin {\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**3*sin(b*x+a),x)

[Out]

Integral(sin(a + b*x)*sec(a + b*x)**3, x)

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