\(\int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx\) [39]

Optimal result

Integrand size = 26, antiderivative size = 129 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {x}{a^3 c^4}-\frac {\cot ^7(e+f x) (1+\sec (e+f x))}{7 a^3 c^4 f}+\frac {\cot ^5(e+f x) (7+6 \sec (e+f x))}{35 a^3 c^4 f}+\frac {\cot (e+f x) (35+16 \sec (e+f x))}{35 a^3 c^4 f}-\frac {\cot ^3(e+f x) (35+24 \sec (e+f x))}{105 a^3 c^4 f} \]

[Out]

x/a^3/c^4-1/7*cot(f*x+e)^7*(1+sec(f*x+e))/a^3/c^4/f+1/35*cot(f*x+e)^5*(7+6*sec(f*x+e))/a^3/c^4/f+1/35*cot(f*x+
e)*(35+16*sec(f*x+e))/a^3/c^4/f-1/105*cot(f*x+e)^3*(35+24*sec(f*x+e))/a^3/c^4/f

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3989, 3967, 8} \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=-\frac {\cot ^7(e+f x) (\sec (e+f x)+1)}{7 a^3 c^4 f}+\frac {\cot ^5(e+f x) (6 \sec (e+f x)+7)}{35 a^3 c^4 f}-\frac {\cot ^3(e+f x) (24 \sec (e+f x)+35)}{105 a^3 c^4 f}+\frac {\cot (e+f x) (16 \sec (e+f x)+35)}{35 a^3 c^4 f}+\frac {x}{a^3 c^4} \]

[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 8

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

Rule 3967

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 3989

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*} \text {integral}& = \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 [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 5.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=-\frac {\csc ^7(e+f x) \left (-106+301 \cos (2 (e+f x))-70 \cos (4 (e+f x))+35 \cos (6 (e+f x))+160 \cos ^7(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},1,-\frac {5}{2},-\tan ^2(e+f x)\right )\right )}{1120 a^3 c^4 f} \]

[In]

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

[Out]

-1/1120*(Csc[e + f*x]^7*(-106 + 301*Cos[2*(e + f*x)] - 70*Cos[4*(e + f*x)] + 35*Cos[6*(e + f*x)] + 160*Cos[e +
f*x]^7*Hypergeometric2F1[-7/2, 1, -5/2, -Tan[e + f*x]^2]))/(a^3*c^4*f)

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(\frac {-15 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}-21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+168 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+280 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-1015 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+6720 f x -3045 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+6720 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{6720 f \,a^{3} c^{4}}\)	\(104\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-29 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {8}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {29}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {64}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+128 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{64 f \,c^{4} a^{3}}\)	\(114\)
default	\(\frac {-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-29 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {8}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {29}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {64}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+128 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{64 f \,c^{4} a^{3}}\)	\(114\)
risch	\(\frac {x}{a^{3} c^{4}}+\frac {2 i \left (105 \,{\mathrm e}^{11 i \left (f x +e \right )}+210 \,{\mathrm e}^{10 i \left (f x +e \right )}-735 \,{\mathrm e}^{9 i \left (f x +e \right )}+1638 \,{\mathrm e}^{7 i \left (f x +e \right )}-196 \,{\mathrm e}^{6 i \left (f x +e \right )}-1882 \,{\mathrm e}^{5 i \left (f x +e \right )}+880 \,{\mathrm e}^{4 i \left (f x +e \right )}+1025 \,{\mathrm e}^{3 i \left (f x +e \right )}-494 \,{\mathrm e}^{2 i \left (f x +e \right )}-247 \,{\mathrm e}^{i \left (f x +e \right )}+176\right )}{105 f \,c^{4} a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{7} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\)	\(160\)
norman	\(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a c f}+\frac {x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c a}-\frac {1}{448 a c f}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{40 a c f}-\frac {29 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{192 a c f}-\frac {29 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{64 a c f}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{24 a c f}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}}{320 a c f}}{a^{2} c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\)	\(181\)

[In]

int(1/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

1/6720*(-15*cot(1/2*f*x+1/2*e)^7-21*tan(1/2*f*x+1/2*e)^5+168*cot(1/2*f*x+1/2*e)^5+280*tan(1/2*f*x+1/2*e)^3-101
5*cot(1/2*f*x+1/2*e)^3+6720*f*x-3045*tan(1/2*f*x+1/2*e)+6720*cot(1/2*f*x+1/2*e))/f/a^3/c^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.80 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {176 \, \cos \left (f x + e\right )^{6} - 71 \, \cos \left (f x + e\right )^{5} - 335 \, \cos \left (f x + e\right )^{4} + 125 \, \cos \left (f x + e\right )^{3} + 225 \, \cos \left (f x + e\right )^{2} + 105 \, {\left (f x \cos \left (f x + e\right )^{5} - f x \cos \left (f x + e\right )^{4} - 2 \, f x \cos \left (f x + e\right )^{3} + 2 \, f x \cos \left (f x + e\right )^{2} + f x \cos \left (f x + e\right ) - f x\right )} \sin \left (f x + e\right ) - 57 \, \cos \left (f x + e\right ) - 48}{105 \, {\left (a^{3} c^{4} f \cos \left (f x + e\right )^{5} - a^{3} c^{4} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{4} f \cos \left (f x + e\right )^{3} + 2 \, a^{3} c^{4} f \cos \left (f x + e\right )^{2} + a^{3} c^{4} f \cos \left (f x + e\right ) - a^{3} c^{4} f\right )} \sin \left (f x + e\right )} \]

[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]

\[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {\int \frac {1}{\sec ^{7}{\left (e + f x \right )} - \sec ^{6}{\left (e + f x \right )} - 3 \sec ^{5}{\left (e + f x \right )} + 3 \sec ^{4}{\left (e + f x \right )} + 3 \sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} - \sec {\left (e + f x \right )} + 1}\, dx}{a^{3} c^{4}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=-\frac {\frac {7 \, {\left (\frac {435 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {40 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3} c^{4}} - \frac {13440 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3} c^{4}} - \frac {{\left (\frac {168 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {1015 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {6720 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 15\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{a^{3} c^{4} \sin \left (f x + e\right )^{7}}}{6720 \, f} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {\frac {6720 \, {\left (f x + e\right )}}{a^{3} c^{4}} + \frac {6720 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1015 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 168 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15}{a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}} - \frac {7 \, {\left (3 \, a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 40 \, a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 435 \, a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15} c^{20}}}{6720 \, f} \]

[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

Mupad [B] (verification not implemented)

Time = 15.21 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.62 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=-\frac {15\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}+21\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}-280\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+3045\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-6720\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+1015\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-168\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-6720\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (e+f\,x\right )}{6720\,a^3\,c^4\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7} \]

[In]

int(1/((a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^4),x)

[Out]

-(15*cos(e/2 + (f*x)/2)^12 + 21*sin(e/2 + (f*x)/2)^12 - 280*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^10 + 3045*
cos(e/2 + (f*x)/2)^4*sin(e/2 + (f*x)/2)^8 - 6720*cos(e/2 + (f*x)/2)^6*sin(e/2 + (f*x)/2)^6 + 1015*cos(e/2 + (f
*x)/2)^8*sin(e/2 + (f*x)/2)^4 - 168*cos(e/2 + (f*x)/2)^10*sin(e/2 + (f*x)/2)^2 - 6720*cos(e/2 + (f*x)/2)^5*sin
(e/2 + (f*x)/2)^7*(e + f*x))/(6720*a^3*c^4*f*cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2)^7)