### 3.97 $$\int \sec (x) \tan ^2(x) \, dx$$

Optimal. Leaf size=16 $\frac{1}{2} \tan (x) \sec (x)-\frac{1}{2} \tanh ^{-1}(\sin (x))$

[Out]

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

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Rubi [A]  time = 0.0152064, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 7, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {2611, 3770} $\frac{1}{2} \tan (x) \sec (x)-\frac{1}{2} \tanh ^{-1}(\sin (x))$

Antiderivative was successfully veriﬁed.

[In]

Int[Sec[x]*Tan[x]^2,x]

[Out]

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

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

Mathematica [A]  time = 0.005882, size = 16, normalized size = 1. $\frac{1}{2} \tan (x) \sec (x)-\frac{1}{2} \tanh ^{-1}(\sin (x))$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sec[x]*Tan[x]^2,x]

[Out]

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

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Maple [A]  time = 0.008, size = 24, normalized size = 1.5 \begin{align*}{\frac{ \left ( \sin \left ( x \right ) \right ) ^{3}}{2\, \left ( \cos \left ( x \right ) \right ) ^{2}}}+{\frac{\sin \left ( x \right ) }{2}}-{\frac{\ln \left ( \sec \left ( x \right ) +\tan \left ( x \right ) \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)*tan(x)^2,x)

[Out]

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

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Maxima [B]  time = 0.928819, size = 36, normalized size = 2.25 \begin{align*} -\frac{\sin \left (x\right )}{2 \,{\left (\sin \left (x\right )^{2} - 1\right )}} - \frac{1}{4} \, \log \left (\sin \left (x\right ) + 1\right ) + \frac{1}{4} \, \log \left (\sin \left (x\right ) - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.17193, size = 109, normalized size = 6.81 \begin{align*} -\frac{\cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) - \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, \sin \left (x\right )}{4 \, \cos \left (x\right )^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.109246, size = 27, normalized size = 1.69 \begin{align*} \frac{\log{\left (\sin{\left (x \right )} - 1 \right )}}{4} - \frac{\log{\left (\sin{\left (x \right )} + 1 \right )}}{4} - \frac{\sin{\left (x \right )}}{2 \sin ^{2}{\left (x \right )} - 2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)**2,x)

[Out]

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

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Giac [B]  time = 1.05921, size = 39, normalized size = 2.44 \begin{align*} -\frac{\sin \left (x\right )}{2 \,{\left (\sin \left (x\right )^{2} - 1\right )}} - \frac{1}{4} \, \log \left (\sin \left (x\right ) + 1\right ) + \frac{1}{4} \, \log \left (-\sin \left (x\right ) + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)^2,x, algorithm="giac")

[Out]

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