3.705 \(\int \frac{a^2-b^2 \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=17 \[ a x-\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

a*x - (b*ArcTanh[Sin[c + d*x]])/d

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Rubi [A]  time = 0.0442866, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {4042, 3770} \[ a x-\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x]),x]

[Out]

a*x - (b*ArcTanh[Sin[c + d*x]])/d

Rule 4042

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Dist[
C/b^2, Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[-a + b*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x
] && EqQ[A*b^2 + a^2*C, 0]

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

Mathematica [A]  time = 0.0066265, size = 17, normalized size = 1. \[ a x-\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x]),x]

[Out]

a*x - (b*ArcTanh[Sin[c + d*x]])/d

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Maple [A]  time = 0.043, size = 31, normalized size = 1.8 \begin{align*} ax-{\frac{b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{ac}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)

[Out]

a*x-1/d*b*ln(sec(d*x+c)+tan(d*x+c))+1/d*a*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.497741, size = 95, normalized size = 5.59 \begin{align*} \frac{2 \, a d x - b \log \left (\sin \left (d x + c\right ) + 1\right ) + b \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*a*d*x - b*log(sin(d*x + c) + 1) + b*log(-sin(d*x + c) + 1))/d

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Sympy [A]  time = 3.25023, size = 41, normalized size = 2.41 \begin{align*} a x - b \left (\begin{cases} \frac{x \left (\tan{\left (c \right )} \sec{\left (c \right )} + \sec ^{2}{\left (c \right )}\right )}{\tan{\left (c \right )} + \sec{\left (c \right )}} & \text{for}\: d = 0 \\\frac{\log{\left (\tan{\left (c + d x \right )} + \sec{\left (c + d x \right )} \right )}}{d} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2-b**2*sec(d*x+c)**2)/(a+b*sec(d*x+c)),x)

[Out]

a*x - b*Piecewise((x*(tan(c)*sec(c) + sec(c)**2)/(tan(c) + sec(c)), Eq(d, 0)), (log(tan(c + d*x) + sec(c + d*x
))/d, True))

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Giac [B]  time = 1.19349, size = 58, normalized size = 3.41 \begin{align*} \frac{{\left (d x + c\right )} a - b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

((d*x + c)*a - b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + b*log(abs(tan(1/2*d*x + 1/2*c) - 1)))/d