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

Optimal. Leaf size=34 \[ \frac{\tan (a+b x) \sec (a+b x)}{2 b}-\frac{\tanh ^{-1}(\sin (a+b x))}{2 b} \]

[Out]

-ArcTanh[Sin[a + b*x]]/(2*b) + (Sec[a + b*x]*Tan[a + b*x])/(2*b)

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Rubi [A]  time = 0.0230989, antiderivative size = 34, normalized size of antiderivative = 1., 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 = {2611, 3770} \[ \frac{\tan (a+b x) \sec (a+b x)}{2 b}-\frac{\tanh ^{-1}(\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]*Tan[a + b*x]^2,x]

[Out]

-ArcTanh[Sin[a + b*x]]/(2*b) + (Sec[a + b*x]*Tan[a + b*x])/(2*b)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0141463, size = 34, normalized size = 1. \[ \frac{\tan (a+b x) \sec (a+b x)}{2 b}-\frac{\tanh ^{-1}(\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Sin[a + b*x]]/(2*b) + (Sec[a + b*x]*Tan[a + b*x])/(2*b)

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Maple [A]  time = 0.019, size = 53, normalized size = 1.6 \begin{align*}{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{2\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\frac{\sin \left ( bx+a \right ) }{2\,b}}-{\frac{\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3*sin(b*x+a)^2,x)

[Out]

1/2/b*sin(b*x+a)^3/cos(b*x+a)^2+1/2*sin(b*x+a)/b-1/2/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [A]  time = 0.979486, size = 62, normalized size = 1.82 \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + \log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (\sin \left (b x + a\right ) - 1\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/4*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1))/b

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Fricas [B]  time = 1.63399, size = 163, normalized size = 4.79 \begin{align*} -\frac{\cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, \sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/4*(cos(b*x + a)^2*log(sin(b*x + a) + 1) - cos(b*x + a)^2*log(-sin(b*x + a) + 1) - 2*sin(b*x + a))/(b*cos(b*
x + a)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin ^{2}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**3*sin(b*x+a)**2,x)

[Out]

Integral(sin(a + b*x)**2*sec(a + b*x)**3, x)

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Giac [A]  time = 1.29343, size = 65, normalized size = 1.91 \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) - \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/4*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + log(abs(sin(b*x + a) + 1)) - log(abs(sin(b*x + a) - 1)))/b