3.171 \(\int \frac{\sec ^2(e+f x) (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx\)

Optimal. Leaf size=56 \[ -\frac{c \tan (e+f x)}{a f}+\frac{2 c \tanh ^{-1}(\sin (e+f x))}{a f}-\frac{2 c \tan (e+f x)}{f (a \sec (e+f x)+a)} \]

[Out]

(2*c*ArcTanh[Sin[e + f*x]])/(a*f) - (c*Tan[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 = 0.108156, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {4008, 3787, 3770, 3767, 8} \[ -\frac{c \tan (e+f x)}{a f}+\frac{2 c \tanh ^{-1}(\sin (e+f x))}{a f}-\frac{2 c \tan (e+f x)}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4008

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [B]  time = 0.607151, size = 154, normalized size = 2.75 \[ -\frac{c \left (\frac{2 \tan \left (\frac{1}{2} (e+f x)\right )}{f}+\frac{\sin \left (\frac{1}{2} (e+f x)\right )}{f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{\sin \left (\frac{1}{2} (e+f x)\right )}{f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}-\frac{2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.058, size = 104, normalized size = 1.9 \begin{align*} -2\,{\frac{c\tan \left ( 1/2\,fx+e/2 \right ) }{fa}}+2\,{\frac{c\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{fa}}+{\frac{c}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}+{\frac{c}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{c\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{fa}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [B]  time = 1.02407, size = 262, normalized size = 4.68 \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{2 \, \sin \left (f x + e\right )}{{\left (a - \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 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 )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(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 - 2*sin(f*x + e)/(
(a - a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)) - sin(f*x + e)/(a*(cos(f*x + e) + 1))) + c*(lo
g(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))))/f

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Fricas [A]  time = 0.476078, size = 270, normalized size = 4.82 \begin{align*} \frac{{\left (c \cos \left (f x + e\right )^{2} + c \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (c \cos \left (f x + e\right )^{2} + c \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) -{\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right )^{2} + a f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c*cos(f*x + e)^2 + c*cos(f*x + e))*log(sin(f*x + e) + 1) - (c*cos(f*x + e)^2 + c*cos(f*x + e))*log(-sin(f*x
+ e) + 1) - (3*c*cos(f*x + e) + c)*sin(f*x + e))/(a*f*cos(f*x + e)^2 + a*f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{c \left (\int - \frac{\sec ^{2}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{3}{\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)**2*(c-c*sec(f*x+e))/(a+a*sec(f*x+e)),x)

[Out]

-c*(Integral(-sec(e + f*x)**2/(sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**3/(sec(e + f*x) + 1), x))/a

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Giac [A]  time = 1.20791, size = 124, normalized size = 2.21 \begin{align*} \frac{2 \,{\left (\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{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a} + \frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} a}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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