3.27 \(\int \frac{1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=46 \[ -\frac{\cot ^3(e+f x)}{3 a^2 c^2 f}+\frac{\cot (e+f x)}{a^2 c^2 f}+\frac{x}{a^2 c^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0713255, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3904, 3473, 8} \[ -\frac{\cot ^3(e+f x)}{3 a^2 c^2 f}+\frac{\cot (e+f x)}{a^2 c^2 f}+\frac{x}{a^2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^2),x]

[Out]

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

Rule 3904

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^2} \, dx &=\frac{\int \cot ^4(e+f x) \, dx}{a^2 c^2}\\ &=-\frac{\cot ^3(e+f x)}{3 a^2 c^2 f}-\frac{\int \cot ^2(e+f x) \, dx}{a^2 c^2}\\ &=\frac{\cot (e+f x)}{a^2 c^2 f}-\frac{\cot ^3(e+f x)}{3 a^2 c^2 f}+\frac{\int 1 \, dx}{a^2 c^2}\\ &=\frac{x}{a^2 c^2}+\frac{\cot (e+f x)}{a^2 c^2 f}-\frac{\cot ^3(e+f x)}{3 a^2 c^2 f}\\ \end{align*}

Mathematica [C]  time = 0.0504444, size = 39, normalized size = 0.85 \[ -\frac{\cot ^3(e+f x) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-\tan ^2(e+f x)\right )}{3 a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^2),x]

[Out]

-(Cot[e + f*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[e + f*x]^2])/(3*a^2*c^2*f)

________________________________________________________________________________________

Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+a\sec \left ( fx+e \right ) \right ) ^{2} \left ( c-c\sec \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.52462, size = 62, normalized size = 1.35 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )}}{a^{2} c^{2}} + \frac{3 \, \tan \left (f x + e\right )^{2} - 1}{a^{2} c^{2} \tan \left (f x + e\right )^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.05791, size = 188, normalized size = 4.09 \begin{align*} \frac{4 \, \cos \left (f x + e\right )^{3} + 3 \,{\left (f x \cos \left (f x + e\right )^{2} - f x\right )} \sin \left (f x + e\right ) - 3 \, \cos \left (f x + e\right )}{3 \,{\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sec ^{4}{\left (e + f x \right )} - 2 \sec ^{2}{\left (e + f x \right )} + 1}\, dx}{a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sec(e + f*x)**4 - 2*sec(e + f*x)**2 + 1), x)/(a**2*c**2)

________________________________________________________________________________________

Giac [B]  time = 1.41077, size = 135, normalized size = 2.93 \begin{align*} \frac{\frac{24 \,{\left (f x + e\right )}}{a^{2} c^{2}} + \frac{15 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1}{a^{2} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}} + \frac{a^{4} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 15 \, a^{4} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{6} c^{6}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^2,x, algorithm="giac")

[Out]

1/24*(24*(f*x + e)/(a^2*c^2) + (15*tan(1/2*f*x + 1/2*e)^2 - 1)/(a^2*c^2*tan(1/2*f*x + 1/2*e)^3) + (a^4*c^4*tan
(1/2*f*x + 1/2*e)^3 - 15*a^4*c^4*tan(1/2*f*x + 1/2*e))/(a^6*c^6))/f