3.37 \(\int \frac{\sec (e+f x) (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx\)

Optimal. Leaf size=41 \[ \frac{2 c \tan (e+f x)}{f (a \sec (e+f x)+a)}-\frac{c \tanh ^{-1}(\sin (e+f x))}{a f} \]

[Out]

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

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Rubi [A]  time = 0.0533329, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3957, 3770} \[ \frac{2 c \tan (e+f x)}{f (a \sec (e+f x)+a)}-\frac{c \tanh ^{-1}(\sin (e+f x))}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3957

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1))/(b*f*(2*m +
1)), x] - Dist[(d*(2*n - 1))/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x]
)^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && L
tQ[m, -2^(-1)] && IntegerQ[2*m]

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

Mathematica [A]  time = 0.0686182, size = 77, normalized size = 1.88 \[ -\frac{c \left (-\frac{2 \tan \left (\frac{1}{2} (e+f x)\right )}{f}-\frac{\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}+\frac{\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 61, normalized size = 1.5 \begin{align*} 2\,{\frac{c\tan \left ( 1/2\,fx+e/2 \right ) }{fa}}+{\frac{c}{fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }-{\frac{c}{fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.954331, size = 136, normalized size = 3.32 \begin{align*} -\frac{c{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - \frac{c \sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.469585, size = 190, normalized size = 4.63 \begin{align*} -\frac{{\left (c \cos \left (f x + e\right ) + c\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (c \cos \left (f x + e\right ) + c\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 4 \, c \sin \left (f x + e\right )}{2 \,{\left (a f \cos \left (f x + e\right ) + a f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{c \left (\int - \frac{\sec{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{2}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*(Integral(-sec(e + f*x)/(sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**2/(sec(e + f*x) + 1), x))/a

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Giac [A]  time = 1.30494, size = 82, normalized size = 2. \begin{align*} -\frac{\frac{c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{2 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a - c*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a - 2*c*tan(1/2*f*x + 1/2*e)/a
)/f