\(\int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx\) [28]

Optimal result

Integrand size = 26, antiderivative size = 98 \[ \int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx=\frac {x}{a^2 c^3}+\frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (5+4 \sec (e+f x))}{15 a^2 c^3 f}+\frac {\cot (e+f x) (15+8 \sec (e+f x))}{15 a^2 c^3 f} \]

[Out]

x/a^2/c^3+1/5*cot(f*x+e)^5*(1+sec(f*x+e))/a^2/c^3/f-1/15*cot(f*x+e)^3*(5+4*sec(f*x+e))/a^2/c^3/f+1/15*cot(f*x+
e)*(15+8*sec(f*x+e))/a^2/c^3/f

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3989, 3967, 8} \[ \int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx=\frac {\cot ^5(e+f x) (\sec (e+f x)+1)}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (4 \sec (e+f x)+5)}{15 a^2 c^3 f}+\frac {\cot (e+f x) (8 \sec (e+f x)+15)}{15 a^2 c^3 f}+\frac {x}{a^2 c^3} \]

[In]

Int[1/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3),x]

[Out]

x/(a^2*c^3) + (Cot[e + f*x]^5*(1 + Sec[e + f*x]))/(5*a^2*c^3*f) - (Cot[e + f*x]^3*(5 + 4*Sec[e + f*x]))/(15*a^
2*c^3*f) + (Cot[e + f*x]*(15 + 8*Sec[e + f*x]))/(15*a^2*c^3*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 ^6(e+f x) (a+a \sec (e+f x)) \, dx}{a^3 c^3} \\ & = \frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\int \cot ^4(e+f x) (-5 a-4 a \sec (e+f x)) \, dx}{5 a^3 c^3} \\ & = \frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (5+4 \sec (e+f x))}{15 a^2 c^3 f}-\frac {\int \cot ^2(e+f x) (15 a+8 a \sec (e+f x)) \, dx}{15 a^3 c^3} \\ & = \frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (5+4 \sec (e+f x))}{15 a^2 c^3 f}+\frac {\cot (e+f x) (15+8 \sec (e+f x))}{15 a^2 c^3 f}-\frac {\int -15 a \, dx}{15 a^3 c^3} \\ & = \frac {x}{a^2 c^3}+\frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (5+4 \sec (e+f x))}{15 a^2 c^3 f}+\frac {\cot (e+f x) (15+8 \sec (e+f x))}{15 a^2 c^3 f} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 1.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx=\frac {\cot ^5(e+f x) \left (3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(e+f x)\right )+15 \sec (e+f x)-20 \sec ^3(e+f x)+8 \sec ^5(e+f x)\right )}{15 a^2 c^3 f} \]

[In]

Integrate[1/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3),x]

[Out]

(Cot[e + f*x]^5*(3*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[e + f*x]^2] + 15*Sec[e + f*x] - 20*Sec[e + f*x]^3 + 8
*Sec[e + f*x]^5))/(15*a^2*c^3*f)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80

method	result	size
parallelrisch	\(\frac {3 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-30 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+240 f x +240 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )-90 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{240 f \,a^{2} c^{3}}\)	\(78\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {16}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+32 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{16 f \,a^{2} c^{3}}\)	\(88\)
default	\(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {16}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+32 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{16 f \,a^{2} c^{3}}\)	\(88\)
risch	\(\frac {x}{a^{2} c^{3}}+\frac {2 i \left (15 \,{\mathrm e}^{7 i \left (f x +e \right )}+15 \,{\mathrm e}^{6 i \left (f x +e \right )}-65 \,{\mathrm e}^{5 i \left (f x +e \right )}+25 \,{\mathrm e}^{4 i \left (f x +e \right )}+73 \,{\mathrm e}^{3 i \left (f x +e \right )}-31 \,{\mathrm e}^{2 i \left (f x +e \right )}-31 \,{\mathrm e}^{i \left (f x +e \right )}+23\right )}{15 f \,a^{2} c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}\)	\(127\)
norman	\(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a c f}+\frac {x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c a}+\frac {1}{80 a c f}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{8 a c f}-\frac {3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{8 a c f}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{48 a c f}}{a \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\)	\(137\)

[In]

int(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/240*(3*cot(1/2*f*x+1/2*e)^5-30*cot(1/2*f*x+1/2*e)^3+5*tan(1/2*f*x+1/2*e)^3+240*f*x+240*cot(1/2*f*x+1/2*e)-90
*tan(1/2*f*x+1/2*e))/f/a^2/c^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.57 \[ \int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx=\frac {23 \, \cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{3} - 27 \, \cos \left (f x + e\right )^{2} + 15 \, {\left (f x \cos \left (f x + e\right )^{3} - f x \cos \left (f x + e\right )^{2} - f x \cos \left (f x + e\right ) + f x\right )} \sin \left (f x + e\right ) + 7 \, \cos \left (f x + e\right ) + 8}{15 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} - a^{2} c^{3} f \cos \left (f x + e\right )^{2} - a^{2} c^{3} f \cos \left (f x + e\right ) + a^{2} c^{3} f\right )} \sin \left (f x + e\right )} \]

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(23*cos(f*x + e)^4 - 8*cos(f*x + e)^3 - 27*cos(f*x + e)^2 + 15*(f*x*cos(f*x + e)^3 - f*x*cos(f*x + e)^2 -
f*x*cos(f*x + e) + f*x)*sin(f*x + e) + 7*cos(f*x + e) + 8)/((a^2*c^3*f*cos(f*x + e)^3 - a^2*c^3*f*cos(f*x + e
)^2 - a^2*c^3*f*cos(f*x + e) + a^2*c^3*f)*sin(f*x + e))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.50 \[ \int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx=-\frac {\frac {5 \, {\left (\frac {18 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2} c^{3}} - \frac {480 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2} c^{3}} + \frac {3 \, {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {80 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{a^{2} c^{3} \sin \left (f x + e\right )^{5}}}{240 \, f} \]

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/240*(5*(18*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*c^3) - 480*arctan(si
n(f*x + e)/(cos(f*x + e) + 1))/(a^2*c^3) + 3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 80*sin(f*x + e)^4/(cos(
f*x + e) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(a^2*c^3*sin(f*x + e)^5))/f

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx=\frac {\frac {240 \, {\left (f x + e\right )}}{a^{2} c^{3}} + \frac {3 \, {\left (80 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 10 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}}{a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}} + \frac {5 \, {\left (a^{4} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 18 \, a^{4} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6} c^{9}}}{240 \, f} \]

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/240*(240*(f*x + e)/(a^2*c^3) + 3*(80*tan(1/2*f*x + 1/2*e)^4 - 10*tan(1/2*f*x + 1/2*e)^2 + 1)/(a^2*c^3*tan(1/
2*f*x + 1/2*e)^5) + 5*(a^4*c^6*tan(1/2*f*x + 1/2*e)^3 - 18*a^4*c^6*tan(1/2*f*x + 1/2*e))/(a^6*c^9))/f

Mupad [B] (verification not implemented)

Time = 14.17 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.64 \[ \int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx=\frac {3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-90\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+240\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-30\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+240\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (e+f\,x\right )}{240\,a^2\,c^3\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]

[In]

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

[Out]

(3*cos(e/2 + (f*x)/2)^8 + 5*sin(e/2 + (f*x)/2)^8 - 90*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^6 + 240*cos(e/2
+ (f*x)/2)^4*sin(e/2 + (f*x)/2)^4 - 30*cos(e/2 + (f*x)/2)^6*sin(e/2 + (f*x)/2)^2 + 240*cos(e/2 + (f*x)/2)^3*si
n(e/2 + (f*x)/2)^5*(e + f*x))/(240*a^2*c^3*f*cos(e/2 + (f*x)/2)^3*sin(e/2 + (f*x)/2)^5)