3.39 \(\int \frac{1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=129 \[ -\frac{\cot ^7(e+f x) (\sec (e+f x)+1)}{7 a^3 c^4 f}+\frac{\cot ^5(e+f x) (6 \sec (e+f x)+7)}{35 a^3 c^4 f}-\frac{\cot ^3(e+f x) (24 \sec (e+f x)+35)}{105 a^3 c^4 f}+\frac{\cot (e+f x) (16 \sec (e+f x)+35)}{35 a^3 c^4 f}+\frac{x}{a^3 c^4} \]

[Out]

x/(a^3*c^4) - (Cot[e + f*x]^7*(1 + Sec[e + f*x]))/(7*a^3*c^4*f) + (Cot[e + f*x]^5*(7 + 6*Sec[e + f*x]))/(35*a^
3*c^4*f) + (Cot[e + f*x]*(35 + 16*Sec[e + f*x]))/(35*a^3*c^4*f) - (Cot[e + f*x]^3*(35 + 24*Sec[e + f*x]))/(105
*a^3*c^4*f)

________________________________________________________________________________________

Rubi [A]  time = 0.172945, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3904, 3882, 8} \[ -\frac{\cot ^7(e+f x) (\sec (e+f x)+1)}{7 a^3 c^4 f}+\frac{\cot ^5(e+f x) (6 \sec (e+f x)+7)}{35 a^3 c^4 f}-\frac{\cot ^3(e+f x) (24 \sec (e+f x)+35)}{105 a^3 c^4 f}+\frac{\cot (e+f x) (16 \sec (e+f x)+35)}{35 a^3 c^4 f}+\frac{x}{a^3 c^4} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4),x]

[Out]

x/(a^3*c^4) - (Cot[e + f*x]^7*(1 + Sec[e + f*x]))/(7*a^3*c^4*f) + (Cot[e + f*x]^5*(7 + 6*Sec[e + f*x]))/(35*a^
3*c^4*f) + (Cot[e + f*x]*(35 + 16*Sec[e + f*x]))/(35*a^3*c^4*f) - (Cot[e + f*x]^3*(35 + 24*Sec[e + f*x]))/(105
*a^3*c^4*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 3882

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[((e*Cot[c
+ d*x])^(m + 1)*(a + b*Csc[c + d*x]))/(d*e*(m + 1)), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)*(
a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -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))^4} \, dx &=\frac{\int \cot ^8(e+f x) (a+a \sec (e+f x)) \, dx}{a^4 c^4}\\ &=-\frac{\cot ^7(e+f x) (1+\sec (e+f x))}{7 a^3 c^4 f}+\frac{\int \cot ^6(e+f x) (-7 a-6 a \sec (e+f x)) \, dx}{7 a^4 c^4}\\ &=-\frac{\cot ^7(e+f x) (1+\sec (e+f x))}{7 a^3 c^4 f}+\frac{\cot ^5(e+f x) (7+6 \sec (e+f x))}{35 a^3 c^4 f}+\frac{\int \cot ^4(e+f x) (35 a+24 a \sec (e+f x)) \, dx}{35 a^4 c^4}\\ &=-\frac{\cot ^7(e+f x) (1+\sec (e+f x))}{7 a^3 c^4 f}+\frac{\cot ^5(e+f x) (7+6 \sec (e+f x))}{35 a^3 c^4 f}-\frac{\cot ^3(e+f x) (35+24 \sec (e+f x))}{105 a^3 c^4 f}+\frac{\int \cot ^2(e+f x) (-105 a-48 a \sec (e+f x)) \, dx}{105 a^4 c^4}\\ &=-\frac{\cot ^7(e+f x) (1+\sec (e+f x))}{7 a^3 c^4 f}+\frac{\cot ^5(e+f x) (7+6 \sec (e+f x))}{35 a^3 c^4 f}+\frac{\cot (e+f x) (35+16 \sec (e+f x))}{35 a^3 c^4 f}-\frac{\cot ^3(e+f x) (35+24 \sec (e+f x))}{105 a^3 c^4 f}+\frac{\int 105 a \, dx}{105 a^4 c^4}\\ &=\frac{x}{a^3 c^4}-\frac{\cot ^7(e+f x) (1+\sec (e+f x))}{7 a^3 c^4 f}+\frac{\cot ^5(e+f x) (7+6 \sec (e+f x))}{35 a^3 c^4 f}+\frac{\cot (e+f x) (35+16 \sec (e+f x))}{35 a^3 c^4 f}-\frac{\cot ^3(e+f x) (35+24 \sec (e+f x))}{105 a^3 c^4 f}\\ \end{align*}

Mathematica [B]  time = 1.37081, size = 362, normalized size = 2.81 \[ \frac{\csc \left (\frac{e}{2}\right ) \sec \left (\frac{e}{2}\right ) \csc ^7\left (\frac{1}{2} (e+f x)\right ) \sec ^5\left (\frac{1}{2} (e+f x)\right ) (-22860 \sin (e+f x)+5715 \sin (2 (e+f x))+11430 \sin (3 (e+f x))-4572 \sin (4 (e+f x))-2286 \sin (5 (e+f x))+1143 \sin (6 (e+f x))-26208 \sin (2 e+f x)+14080 \sin (e+2 f x)+16400 \sin (2 e+3 f x)+11760 \sin (4 e+3 f x)-7904 \sin (3 e+4 f x)-3360 \sin (5 e+4 f x)-3952 \sin (4 e+5 f x)-1680 \sin (6 e+5 f x)+2816 \sin (5 e+6 f x)-16800 f x \cos (2 e+f x)-4200 f x \cos (e+2 f x)+4200 f x \cos (3 e+2 f x)-8400 f x \cos (2 e+3 f x)+8400 f x \cos (4 e+3 f x)+3360 f x \cos (3 e+4 f x)-3360 f x \cos (5 e+4 f x)+1680 f x \cos (4 e+5 f x)-1680 f x \cos (6 e+5 f x)-840 f x \cos (5 e+6 f x)+840 f x \cos (7 e+6 f x)+3136 \sin (e)-30112 \sin (f x)+16800 f x \cos (f x))}{6881280 a^3 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4),x]

[Out]

(Csc[e/2]*Csc[(e + f*x)/2]^7*Sec[e/2]*Sec[(e + f*x)/2]^5*(16800*f*x*Cos[f*x] - 16800*f*x*Cos[2*e + f*x] - 4200
*f*x*Cos[e + 2*f*x] + 4200*f*x*Cos[3*e + 2*f*x] - 8400*f*x*Cos[2*e + 3*f*x] + 8400*f*x*Cos[4*e + 3*f*x] + 3360
*f*x*Cos[3*e + 4*f*x] - 3360*f*x*Cos[5*e + 4*f*x] + 1680*f*x*Cos[4*e + 5*f*x] - 1680*f*x*Cos[6*e + 5*f*x] - 84
0*f*x*Cos[5*e + 6*f*x] + 840*f*x*Cos[7*e + 6*f*x] + 3136*Sin[e] - 30112*Sin[f*x] - 22860*Sin[e + f*x] + 5715*S
in[2*(e + f*x)] + 11430*Sin[3*(e + f*x)] - 4572*Sin[4*(e + f*x)] - 2286*Sin[5*(e + f*x)] + 1143*Sin[6*(e + f*x
)] - 26208*Sin[2*e + f*x] + 14080*Sin[e + 2*f*x] + 16400*Sin[2*e + 3*f*x] + 11760*Sin[4*e + 3*f*x] - 7904*Sin[
3*e + 4*f*x] - 3360*Sin[5*e + 4*f*x] - 3952*Sin[4*e + 5*f*x] - 1680*Sin[6*e + 5*f*x] + 2816*Sin[5*e + 6*f*x]))
/(6881280*a^3*c^4*f)

________________________________________________________________________________________

Maple [A]  time = 0.069, size = 174, normalized size = 1.4 \begin{align*} -{\frac{1}{320\,f{a}^{3}{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{1}{24\,f{a}^{3}{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{29}{64\,f{a}^{3}{c}^{4}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}{c}^{4}}}-{\frac{1}{448\,f{a}^{3}{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-7}}+{\frac{1}{40\,f{a}^{3}{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}-{\frac{29}{192\,f{a}^{3}{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+{\frac{1}{f{a}^{3}{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-1}} \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))^4,x)

[Out]

-1/320/f/a^3/c^4*tan(1/2*f*x+1/2*e)^5+1/24/f/a^3/c^4*tan(1/2*f*x+1/2*e)^3-29/64/f/a^3/c^4*tan(1/2*f*x+1/2*e)+2
/f/a^3/c^4*arctan(tan(1/2*f*x+1/2*e))-1/448/f/a^3/c^4/tan(1/2*f*x+1/2*e)^7+1/40/f/a^3/c^4/tan(1/2*f*x+1/2*e)^5
-29/192/f/a^3/c^4/tan(1/2*f*x+1/2*e)^3+1/f/a^3/c^4/tan(1/2*f*x+1/2*e)

________________________________________________________________________________________

Maxima [A]  time = 1.56434, size = 252, normalized size = 1.95 \begin{align*} -\frac{\frac{7 \,{\left (\frac{435 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{40 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3} c^{4}} - \frac{13440 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3} c^{4}} - \frac{{\left (\frac{168 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{1015 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{6720 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 15\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{a^{3} c^{4} \sin \left (f x + e\right )^{7}}}{6720 \, 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))^4,x, algorithm="maxima")

[Out]

-1/6720*(7*(435*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(c
os(f*x + e) + 1)^5)/(a^3*c^4) - 13440*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/(a^3*c^4) - (168*sin(f*x + e)^2/
(cos(f*x + e) + 1)^2 - 1015*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 6720*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 1
5)*(cos(f*x + e) + 1)^7/(a^3*c^4*sin(f*x + e)^7))/f

________________________________________________________________________________________

Fricas [A]  time = 1.06171, size = 581, normalized size = 4.5 \begin{align*} \frac{176 \, \cos \left (f x + e\right )^{6} - 71 \, \cos \left (f x + e\right )^{5} - 335 \, \cos \left (f x + e\right )^{4} + 125 \, \cos \left (f x + e\right )^{3} + 225 \, \cos \left (f x + e\right )^{2} + 105 \,{\left (f x \cos \left (f x + e\right )^{5} - f x \cos \left (f x + e\right )^{4} - 2 \, f x \cos \left (f x + e\right )^{3} + 2 \, f x \cos \left (f x + e\right )^{2} + f x \cos \left (f x + e\right ) - f x\right )} \sin \left (f x + e\right ) - 57 \, \cos \left (f x + e\right ) - 48}{105 \,{\left (a^{3} c^{4} f \cos \left (f x + e\right )^{5} - a^{3} c^{4} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{4} f \cos \left (f x + e\right )^{3} + 2 \, a^{3} c^{4} f \cos \left (f x + e\right )^{2} + a^{3} c^{4} f \cos \left (f x + e\right ) - a^{3} c^{4} 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))^4,x, algorithm="fricas")

[Out]

1/105*(176*cos(f*x + e)^6 - 71*cos(f*x + e)^5 - 335*cos(f*x + e)^4 + 125*cos(f*x + e)^3 + 225*cos(f*x + e)^2 +
105*(f*x*cos(f*x + e)^5 - f*x*cos(f*x + e)^4 - 2*f*x*cos(f*x + e)^3 + 2*f*x*cos(f*x + e)^2 + f*x*cos(f*x + e)
- f*x)*sin(f*x + e) - 57*cos(f*x + e) - 48)/((a^3*c^4*f*cos(f*x + e)^5 - a^3*c^4*f*cos(f*x + e)^4 - 2*a^3*c^4
*f*cos(f*x + e)^3 + 2*a^3*c^4*f*cos(f*x + e)^2 + a^3*c^4*f*cos(f*x + e) - a^3*c^4*f)*sin(f*x + e))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sec ^{7}{\left (e + f x \right )} - \sec ^{6}{\left (e + f x \right )} - 3 \sec ^{5}{\left (e + f x \right )} + 3 \sec ^{4}{\left (e + f x \right )} + 3 \sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} - \sec{\left (e + f x \right )} + 1}\, dx}{a^{3} c^{4}} \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))**4,x)

[Out]

Integral(1/(sec(e + f*x)**7 - sec(e + f*x)**6 - 3*sec(e + f*x)**5 + 3*sec(e + f*x)**4 + 3*sec(e + f*x)**3 - 3*
sec(e + f*x)**2 - sec(e + f*x) + 1), x)/(a**3*c**4)

________________________________________________________________________________________

Giac [A]  time = 1.51067, size = 203, normalized size = 1.57 \begin{align*} \frac{\frac{6720 \,{\left (f x + e\right )}}{a^{3} c^{4}} + \frac{6720 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 1015 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 168 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 15}{a^{3} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7}} - \frac{7 \,{\left (3 \, a^{12} c^{16} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 40 \, a^{12} c^{16} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 435 \, a^{12} c^{16} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{15} c^{20}}}{6720 \, 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))^4,x, algorithm="giac")

[Out]

1/6720*(6720*(f*x + e)/(a^3*c^4) + (6720*tan(1/2*f*x + 1/2*e)^6 - 1015*tan(1/2*f*x + 1/2*e)^4 + 168*tan(1/2*f*
x + 1/2*e)^2 - 15)/(a^3*c^4*tan(1/2*f*x + 1/2*e)^7) - 7*(3*a^12*c^16*tan(1/2*f*x + 1/2*e)^5 - 40*a^12*c^16*tan
(1/2*f*x + 1/2*e)^3 + 435*a^12*c^16*tan(1/2*f*x + 1/2*e))/(a^15*c^20))/f