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

Optimal. Leaf size=38 \[ \frac {\csc (e+f x)}{a^2 c^2 f}-\frac {\csc ^3(e+f x)}{3 a^2 c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {3958, 2606} \[ \frac {\csc (e+f x)}{a^2 c^2 f}-\frac {\csc ^3(e+f x)}{3 a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 33, normalized size = 0.87 \[ \frac {\frac {\csc (e+f x)}{f}-\frac {\csc ^3(e+f x)}{3 f}}{a^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 50, normalized size = 1.32 \[ \frac {3 \, \cos \left (f x + e\right )^{2} - 2}{3 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 33, normalized size = 0.87 \[ \frac {3 \, \sin \left (f x + e\right )^{2} - 1}{3 \, a^{2} c^{2} f \sin \left (f x + e\right )^{3}} \]

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))^2,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 31, normalized size = 0.82 \[ \frac {3 \, \sin \left (f x + e\right )^{2} - 1}{3 \, a^{2} c^{2} f \sin \left (f x + e\right )^{3}} \]

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))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.57, size = 28, normalized size = 0.74 \[ \frac {{\sin \left (e+f\,x\right )}^2-\frac {1}{3}}{a^2\,c^2\,f\,{\sin \left (e+f\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(e + f*x)^2 - 1/3)/(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 \[ \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 2 \sec ^{2}{\left (e + f x \right )} + 1}\, dx}{a^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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