3.38 \(\int \frac{1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=67 \[ \frac{\cot ^5(e+f x)}{5 a^3 c^3 f}-\frac{\cot ^3(e+f x)}{3 a^3 c^3 f}+\frac{\cot (e+f x)}{a^3 c^3 f}+\frac{x}{a^3 c^3} \]

[Out]

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

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Rubi [A]  time = 0.0805624, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3904, 3473, 8} \[ \frac{\cot ^5(e+f x)}{5 a^3 c^3 f}-\frac{\cot ^3(e+f x)}{3 a^3 c^3 f}+\frac{\cot (e+f x)}{a^3 c^3 f}+\frac{x}{a^3 c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^3),x]

[Out]

x/(a^3*c^3) + Cot[e + f*x]/(a^3*c^3*f) - Cot[e + f*x]^3/(3*a^3*c^3*f) + Cot[e + f*x]^5/(5*a^3*c^3*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))^3 (c-c \sec (e+f x))^3} \, dx &=-\frac{\int \cot ^6(e+f x) \, dx}{a^3 c^3}\\ &=\frac{\cot ^5(e+f x)}{5 a^3 c^3 f}+\frac{\int \cot ^4(e+f x) \, dx}{a^3 c^3}\\ &=-\frac{\cot ^3(e+f x)}{3 a^3 c^3 f}+\frac{\cot ^5(e+f x)}{5 a^3 c^3 f}-\frac{\int \cot ^2(e+f x) \, dx}{a^3 c^3}\\ &=\frac{\cot (e+f x)}{a^3 c^3 f}-\frac{\cot ^3(e+f x)}{3 a^3 c^3 f}+\frac{\cot ^5(e+f x)}{5 a^3 c^3 f}+\frac{\int 1 \, dx}{a^3 c^3}\\ &=\frac{x}{a^3 c^3}+\frac{\cot (e+f x)}{a^3 c^3 f}-\frac{\cot ^3(e+f x)}{3 a^3 c^3 f}+\frac{\cot ^5(e+f x)}{5 a^3 c^3 f}\\ \end{align*}

Mathematica [C]  time = 0.0737104, size = 39, normalized size = 0.58 \[ \frac{\cot ^5(e+f x) \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},-\tan ^2(e+f x)\right )}{5 a^3 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^3),x]

[Out]

(Cot[e + f*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[e + f*x]^2])/(5*a^3*c^3*f)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+a\sec \left ( fx+e \right ) \right ) ^{3} \left ( c-c\sec \left ( fx+e \right ) \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.57528, size = 76, normalized size = 1.13 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )}}{a^{3} c^{3}} + \frac{15 \, \tan \left (f x + e\right )^{4} - 5 \, \tan \left (f x + e\right )^{2} + 3}{a^{3} c^{3} \tan \left (f x + e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.11877, size = 290, normalized size = 4.33 \begin{align*} \frac{23 \, \cos \left (f x + e\right )^{5} - 35 \, \cos \left (f x + e\right )^{3} + 15 \,{\left (f x \cos \left (f x + e\right )^{4} - 2 \, f x \cos \left (f x + e\right )^{2} + f x\right )} \sin \left (f x + e\right ) + 15 \, \cos \left (f x + e\right )}{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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(23*cos(f*x + e)^5 - 35*cos(f*x + e)^3 + 15*(f*x*cos(f*x + e)^4 - 2*f*x*cos(f*x + e)^2 + f*x)*sin(f*x + e
) + 15*cos(f*x + e))/((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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{1}{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(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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Giac [B]  time = 1.45574, size = 184, normalized size = 2.75 \begin{align*} \frac{\frac{480 \,{\left (f x + e\right )}}{a^{3} c^{3}} + \frac{330 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 35 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3}{a^{3} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5}} - \frac{3 \, a^{12} c^{12} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 35 \, a^{12} c^{12} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 330 \, a^{12} c^{12} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{15} c^{15}}}{480 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/480*(480*(f*x + e)/(a^3*c^3) + (330*tan(1/2*f*x + 1/2*e)^4 - 35*tan(1/2*f*x + 1/2*e)^2 + 3)/(a^3*c^3*tan(1/2
*f*x + 1/2*e)^5) - (3*a^12*c^12*tan(1/2*f*x + 1/2*e)^5 - 35*a^12*c^12*tan(1/2*f*x + 1/2*e)^3 + 330*a^12*c^12*t
an(1/2*f*x + 1/2*e))/(a^15*c^15))/f