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

Optimal. Leaf size=59 \[ \frac{\cot ^3(e+f x)}{3 a^2 c f}-\frac{\csc ^3(e+f x)}{3 a^2 c f}+\frac{\csc (e+f x)}{a^2 c f} \]

[Out]

Cot[e + f*x]^3/(3*a^2*c*f) + Csc[e + f*x]/(a^2*c*f) - Csc[e + f*x]^3/(3*a^2*c*f)

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Rubi [A]  time = 0.136156, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3958, 2606, 2607, 30} \[ \frac{\cot ^3(e+f x)}{3 a^2 c f}-\frac{\csc ^3(e+f x)}{3 a^2 c f}+\frac{\csc (e+f x)}{a^2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cot[e + f*x]^3/(3*a^2*c*f) + Csc[e + f*x]/(a^2*c*f) - Csc[e + f*x]^3/(3*a^2*c*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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))} \, dx &=\frac{\int \left (c \cot ^3(e+f x) \csc (e+f x)-c \cot ^2(e+f x) \csc ^2(e+f x)\right ) \, dx}{a^2 c^2}\\ &=\frac{\int \cot ^3(e+f x) \csc (e+f x) \, dx}{a^2 c}-\frac{\int \cot ^2(e+f x) \csc ^2(e+f x) \, dx}{a^2 c}\\ &=-\frac{\operatorname{Subst}\left (\int x^2 \, dx,x,-\cot (e+f x)\right )}{a^2 c f}-\frac{\operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\csc (e+f x)\right )}{a^2 c f}\\ &=\frac{\cot ^3(e+f x)}{3 a^2 c f}+\frac{\csc (e+f x)}{a^2 c f}-\frac{\csc ^3(e+f x)}{3 a^2 c f}\\ \end{align*}

Mathematica [A]  time = 0.570834, size = 83, normalized size = 1.41 \[ -\frac{\csc (e) \sin ^2\left (\frac{1}{2} (e+f x)\right ) (10 \sin (e+f x)+5 \sin (2 (e+f x))-6 \sin (2 e+f x)-2 \sin (e+2 f x)-6 \sin (e)+2 \sin (f x)) \csc ^3(e+f x)}{6 a^2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[e]*Csc[e + f*x]^3*Sin[(e + f*x)/2]^2*(-6*Sin[e] + 2*Sin[f*x] + 10*Sin[e + f*x] + 5*Sin[2*(e + f*x)] - 6*
Sin[2*e + f*x] - 2*Sin[e + 2*f*x]))/(6*a^2*c*f)

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Maple [A]  time = 0.048, size = 48, normalized size = 0.8 \begin{align*}{\frac{1}{4\,f{a}^{2}c} \left ( -{\frac{1}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+2\,\tan \left ( 1/2\,fx+e/2 \right ) + \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e)),x)

[Out]

1/4/f/a^2/c*(-1/3*tan(1/2*f*x+1/2*e)^3+2*tan(1/2*f*x+1/2*e)+1/tan(1/2*f*x+1/2*e))

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Maxima [A]  time = 0.988689, size = 103, normalized size = 1.75 \begin{align*} \frac{\frac{\frac{6 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2} c} + \frac{3 \,{\left (\cos \left (f x + e\right ) + 1\right )}}{a^{2} c \sin \left (f x + e\right )}}{12 \, 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))^2/(c-c*sec(f*x+e)),x, algorithm="maxima")

[Out]

1/12*((6*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*c) + 3*(cos(f*x + e) + 1)
/(a^2*c*sin(f*x + e)))/f

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Fricas [A]  time = 0.438719, size = 124, normalized size = 2.1 \begin{align*} -\frac{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) - 2}{3 \,{\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\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))^2/(c-c*sec(f*x+e)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22861, size = 97, normalized size = 1.64 \begin{align*} \frac{\frac{3}{a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} - \frac{a^{4} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 \, a^{4} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{6} c^{3}}}{12 \, 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))^2/(c-c*sec(f*x+e)),x, algorithm="giac")

[Out]

1/12*(3/(a^2*c*tan(1/2*f*x + 1/2*e)) - (a^4*c^2*tan(1/2*f*x + 1/2*e)^3 - 6*a^4*c^2*tan(1/2*f*x + 1/2*e))/(a^6*
c^3))/f