3.4 \(\int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx\)

Optimal. Leaf size=38 \[ \frac{a c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac{a c \tan (e+f x) \sec (e+f x)}{2 f} \]

[Out]

(a*c*ArcTanh[Sin[e + f*x]])/(2*f) - (a*c*Sec[e + f*x]*Tan[e + f*x])/(2*f)

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Rubi [A]  time = 0.0497687, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3958, 2611, 3770} \[ \frac{a c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac{a c \tan (e+f x) \sec (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x]),x]

[Out]

(a*c*ArcTanh[Sin[e + f*x]])/(2*f) - (a*c*Sec[e + f*x]*Tan[e + f*x])/(2*f)

Rule 3958

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
)^(n_.), x_Symbol] :> Dist[(-(a*c))^m, Int[ExpandTrig[csc[e + f*x]*cot[e + f*x]^(2*m), (c + d*csc[e + f*x])^(n
- m), x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegersQ[m,
n] && GeQ[n - m, 0] && GtQ[m*n, 0]

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

Mathematica [A]  time = 0.0288633, size = 38, normalized size = 1. \[ -a c \left (\frac{\tan (e+f x) \sec (e+f x)}{2 f}-\frac{\tanh ^{-1}(\sin (e+f x))}{2 f}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x]),x]

[Out]

-(a*c*(-ArcTanh[Sin[e + f*x]]/(2*f) + (Sec[e + f*x]*Tan[e + f*x])/(2*f)))

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Maple [A]  time = 0.015, size = 42, normalized size = 1.1 \begin{align*}{\frac{ac\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}-{\frac{ac\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x)

[Out]

1/2/f*a*c*ln(sec(f*x+e)+tan(f*x+e))-1/2*a*c*sec(f*x+e)*tan(f*x+e)/f

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Maxima [A]  time = 0.951144, size = 92, normalized size = 2.42 \begin{align*} \frac{a c{\left (\frac{2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 4 \, a c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x, algorithm="maxima")

[Out]

1/4*(a*c*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) + 4*a*c*log(sec
(f*x + e) + tan(f*x + e)))/f

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Fricas [A]  time = 0.476123, size = 178, normalized size = 4.68 \begin{align*} \frac{a c \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - a c \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, a c \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x, algorithm="fricas")

[Out]

1/4*(a*c*cos(f*x + e)^2*log(sin(f*x + e) + 1) - a*c*cos(f*x + e)^2*log(-sin(f*x + e) + 1) - 2*a*c*sin(f*x + e)
)/(f*cos(f*x + e)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x)

[Out]

-a*c*(Integral(-sec(e + f*x), x) + Integral(sec(e + f*x)**3, x))

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Giac [A]  time = 1.82412, size = 80, normalized size = 2.11 \begin{align*} \frac{a c \log \left (\sin \left (f x + e\right ) + 1\right ) - a c \log \left (-\sin \left (f x + e\right ) + 1\right ) + \frac{2 \, a c \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x, algorithm="giac")

[Out]

1/4*(a*c*log(sin(f*x + e) + 1) - a*c*log(-sin(f*x + e) + 1) + 2*a*c*sin(f*x + e)/(sin(f*x + e)^2 - 1))/f