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

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Rubi [A]  time = 0.07, antiderivative size = 46, normalized size of antiderivative = 1.00, 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 8

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

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 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])

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*}

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Mathematica [C]  time = 0.05, size = 39, normalized size = 0.85 \[ -\frac {\cot ^3(e+f x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\tan ^2(e+f x)\right )}{3 a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 81, normalized size = 1.76 \[ \frac {4 \, \cos \left (f x + e\right )^{3} + 3 \, {\left (f x \cos \left (f x + e\right )^{2} - f x\right )} \sin \left (f x + e\right ) - 3 \, \cos \left (f x + e\right )}{3 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )} \]

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))

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giac [B]  time = 0.29, size = 100, normalized size = 2.17 \[ \frac {\frac {24 \, {\left (f x + e\right )}}{a^{2} c^{2}} + \frac {15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1}{a^{2} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}} + \frac {a^{4} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 15 \, a^{4} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{6} c^{6}}}{24 \, f} \]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +a \sec \left (f x +e \right )\right )^{2} \left (c -c \sec \left (f x +e \right )\right )^{2}}\, dx \]

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)

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maxima [A]  time = 0.42, size = 46, normalized size = 1.00 \[ \frac {\frac {3 \, {\left (f x + e\right )}}{a^{2} c^{2}} + \frac {3 \, \tan \left (f x + e\right )^{2} - 1}{a^{2} c^{2} \tan \left (f x + e\right )^{3}}}{3 \, f} \]

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

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mupad [B]  time = 1.47, size = 58, normalized size = 1.26 \[ -\frac {\cos \left (3\,e+3\,f\,x\right )+\frac {3\,\sin \left (3\,e+3\,f\,x\right )\,\left (e+f\,x\right )}{4}-\frac {9\,\sin \left (e+f\,x\right )\,\left (e+f\,x\right )}{4}}{3\,a^2\,c^2\,f\,{\sin \left (e+f\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(3*e + 3*f*x) + (3*sin(3*e + 3*f*x)*(e + f*x))/4 - (9*sin(e + f*x)*(e + f*x))/4)/(3*a^2*c^2*f*sin(e + f*x
)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\sec ^{4}{\left (e + f x \right )} - 2 \sec ^{2}{\left (e + f x \right )} + 1}\, dx}{a^{2} c^{2}} \]

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)

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