3.60 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=59 \[ \frac {\csc ^5(e+f x)}{5 a^3 c^3 f}-\frac {2 \csc ^3(e+f x)}{3 a^3 c^3 f}+\frac {\csc (e+f x)}{a^3 c^3 f} \]

[Out]

csc(f*x+e)/a^3/c^3/f-2/3*csc(f*x+e)^3/a^3/c^3/f+1/5*csc(f*x+e)^5/a^3/c^3/f

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Rubi [A]  time = 0.11, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3958, 2606, 194} \[ \frac {\csc ^5(e+f x)}{5 a^3 c^3 f}-\frac {2 \csc ^3(e+f x)}{3 a^3 c^3 f}+\frac {\csc (e+f x)}{a^3 c^3 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^3),x]

[Out]

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

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 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 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))^3 (c-c \sec (e+f x))^3} \, dx &=-\frac {\int \cot ^5(e+f x) \csc (e+f x) \, dx}{a^3 c^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (-1+x^2\right )^2 \, dx,x,\csc (e+f x)\right )}{a^3 c^3 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\csc (e+f x)\right )}{a^3 c^3 f}\\ &=\frac {\csc (e+f x)}{a^3 c^3 f}-\frac {2 \csc ^3(e+f x)}{3 a^3 c^3 f}+\frac {\csc ^5(e+f x)}{5 a^3 c^3 f}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 50, normalized size = 0.85 \[ -\frac {-\frac {\csc ^5(e+f x)}{5 f}+\frac {2 \csc ^3(e+f x)}{3 f}-\frac {\csc (e+f x)}{f}}{a^3 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^3),x]

[Out]

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

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fricas [A]  time = 0.43, size = 76, normalized size = 1.29 \[ \frac {15 \, \cos \left (f x + e\right )^{4} - 20 \, \cos \left (f x + e\right )^{2} + 8}{15 \, {\left (a^{3} c^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{3} f \cos \left (f x + e\right )^{2} + a^{3} c^{3} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(15*cos(f*x + e)^4 - 20*cos(f*x + e)^2 + 8)/((a^3*c^3*f*cos(f*x + e)^4 - 2*a^3*c^3*f*cos(f*x + e)^2 + a^3
*c^3*f)*sin(f*x + e))

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giac [A]  time = 0.70, size = 44, normalized size = 0.75 \[ \frac {15 \, \sin \left (f x + e\right )^{4} - 10 \, \sin \left (f x + e\right )^{2} + 3}{15 \, a^{3} c^{3} f \sin \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(15*sin(f*x + e)^4 - 10*sin(f*x + e)^2 + 3)/(a^3*c^3*f*sin(f*x + e)^5)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x +e \right )}{\left (a +a \sec \left (f x +e \right )\right )^{3} \left (c -c \sec \left (f x +e \right )\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x)

[Out]

int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x)

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maxima [A]  time = 0.38, size = 41, normalized size = 0.69 \[ \frac {15 \, \sin \left (f x + e\right )^{4} - 10 \, \sin \left (f x + e\right )^{2} + 3}{15 \, a^{3} c^{3} f \sin \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/15*(15*sin(f*x + e)^4 - 10*sin(f*x + e)^2 + 3)/(a^3*c^3*f*sin(f*x + e)^5)

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mupad [B]  time = 1.63, size = 38, normalized size = 0.64 \[ \frac {{\sin \left (e+f\,x\right )}^4-\frac {2\,{\sin \left (e+f\,x\right )}^2}{3}+\frac {1}{5}}{a^3\,c^3\,f\,{\sin \left (e+f\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^3),x)

[Out]

(sin(e + f*x)^4 - (2*sin(e + f*x)^2)/3 + 1/5)/(a^3*c^3*f*sin(e + f*x)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{6}{\left (e + f x \right )} - 3 \sec ^{4}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} - 1}\, dx}{a^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**3,x)

[Out]

-Integral(sec(e + f*x)/(sec(e + f*x)**6 - 3*sec(e + f*x)**4 + 3*sec(e + f*x)**2 - 1), x)/(a**3*c**3)

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