3.1.61 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx\) [61]

3.1.61.1 Optimal result
3.1.61.2 Mathematica [A] (verified)
3.1.61.3 Rubi [A] (verified)
3.1.61.4 Maple [A] (verified)
3.1.61.5 Fricas [A] (verification not implemented)
3.1.61.6 Sympy [F]
3.1.61.7 Maxima [A] (verification not implemented)
3.1.61.8 Giac [A] (verification not implemented)
3.1.61.9 Mupad [B] (verification not implemented)

3.1.61.1 Optimal result

Integrand size = 32, antiderivative size = 99 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=-\frac {\cot ^7(e+f x)}{7 a^3 c^4 f}+\frac {\csc (e+f x)}{a^3 c^4 f}-\frac {\csc ^3(e+f x)}{a^3 c^4 f}+\frac {3 \csc ^5(e+f x)}{5 a^3 c^4 f}-\frac {\csc ^7(e+f x)}{7 a^3 c^4 f} \]

output
-1/7*cot(f*x+e)^7/a^3/c^4/f+csc(f*x+e)/a^3/c^4/f-csc(f*x+e)^3/a^3/c^4/f+3/ 
5*csc(f*x+e)^5/a^3/c^4/f-1/7*csc(f*x+e)^7/a^3/c^4/f
 
3.1.61.2 Mathematica [A] (verified)

Time = 3.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {\left (-5-30 \sec (e+f x)+30 \sec ^2(e+f x)+40 \sec ^3(e+f x)-40 \sec ^4(e+f x)-16 \sec ^5(e+f x)+16 \sec ^6(e+f x)\right ) \tan (e+f x)}{35 a^3 c^4 f (-1+\sec (e+f x))^4 (1+\sec (e+f x))^3} \]

input
Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4),x]
 
output
((-5 - 30*Sec[e + f*x] + 30*Sec[e + f*x]^2 + 40*Sec[e + f*x]^3 - 40*Sec[e 
+ f*x]^4 - 16*Sec[e + f*x]^5 + 16*Sec[e + f*x]^6)*Tan[e + f*x])/(35*a^3*c^ 
4*f*(-1 + Sec[e + f*x])^4*(1 + Sec[e + f*x])^3)
 
3.1.61.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.82, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3042, 4446, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 4446

\(\displaystyle \frac {\int \left (a \csc (e+f x) \cot ^7(e+f x)+a \csc ^2(e+f x) \cot ^6(e+f x)\right )dx}{a^4 c^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {a \cot ^7(e+f x)}{7 f}-\frac {a \csc ^7(e+f x)}{7 f}+\frac {3 a \csc ^5(e+f x)}{5 f}-\frac {a \csc ^3(e+f x)}{f}+\frac {a \csc (e+f x)}{f}}{a^4 c^4}\)

input
Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4),x]
 
output
(-1/7*(a*Cot[e + f*x]^7)/f + (a*Csc[e + f*x])/f - (a*Csc[e + f*x]^3)/f + ( 
3*a*Csc[e + f*x]^5)/(5*f) - (a*Csc[e + f*x]^7)/(7*f))/(a^4*c^4)
 

3.1.61.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4446
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c 
sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-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] && Eq 
Q[a^2 - b^2, 0] && IntegersQ[m, n] && GeQ[n - m, 0] && GtQ[m*n, 0]
 
3.1.61.4 Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00

method result size
parallelrisch \(-\frac {\left (-90 \cos \left (4 f x +4 e \right )+5 \cos \left (6 f x +6 e \right )+152 \cos \left (f x +e \right )+60 \cos \left (5 f x +5 e \right )-182-20 \cos \left (3 f x +3 e \right )+235 \cos \left (2 f x +2 e \right )\right ) \sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \csc \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{71680 f \,a^{3} c^{4}}\) \(99\)
derivativedivides \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {20}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {6}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {5}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}}{64 f \,c^{4} a^{3}}\) \(102\)
default \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {20}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {6}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {5}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}}{64 f \,c^{4} a^{3}}\) \(102\)
risch \(\frac {2 i \left (35 \,{\mathrm e}^{11 i \left (f x +e \right )}-35 \,{\mathrm e}^{10 i \left (f x +e \right )}-35 \,{\mathrm e}^{9 i \left (f x +e \right )}+105 \,{\mathrm e}^{8 i \left (f x +e \right )}+126 \,{\mathrm e}^{7 i \left (f x +e \right )}-182 \,{\mathrm e}^{6 i \left (f x +e \right )}+26 \,{\mathrm e}^{5 i \left (f x +e \right )}+130 \,{\mathrm e}^{4 i \left (f x +e \right )}+15 \,{\mathrm e}^{3 i \left (f x +e \right )}-55 \,{\mathrm e}^{2 i \left (f x +e \right )}+25 \,{\mathrm e}^{i \left (f x +e \right )}+5\right )}{35 f \,c^{4} a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{7}}\) \(162\)
norman \(\frac {-\frac {1}{448 a c f}+\frac {3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{160 a c f}-\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{64 a c f}+\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{16 a c f}+\frac {15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{64 a c f}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{32 a c f}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}}{320 a c f}}{a^{2} c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\) \(163\)

input
int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x,method=_RETURNVERBO 
SE)
 
output
-1/71680*(-90*cos(4*f*x+4*e)+5*cos(6*f*x+6*e)+152*cos(f*x+e)+60*cos(5*f*x+ 
5*e)-182-20*cos(3*f*x+3*e)+235*cos(2*f*x+2*e))*sec(1/2*f*x+1/2*e)^5*csc(1/ 
2*f*x+1/2*e)^7/f/a^3/c^4
 
3.1.61.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.65 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {5 \, \cos \left (f x + e\right )^{6} + 30 \, \cos \left (f x + e\right )^{5} - 30 \, \cos \left (f x + e\right )^{4} - 40 \, \cos \left (f x + e\right )^{3} + 40 \, \cos \left (f x + e\right )^{2} + 16 \, \cos \left (f x + e\right ) - 16}{35 \, {\left (a^{3} c^{4} f \cos \left (f x + e\right )^{5} - a^{3} c^{4} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{4} f \cos \left (f x + e\right )^{3} + 2 \, a^{3} c^{4} f \cos \left (f x + e\right )^{2} + a^{3} c^{4} f \cos \left (f x + e\right ) - a^{3} c^{4} f\right )} \sin \left (f x + e\right )} \]

input
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x, algorithm="f 
ricas")
 
output
1/35*(5*cos(f*x + e)^6 + 30*cos(f*x + e)^5 - 30*cos(f*x + e)^4 - 40*cos(f* 
x + e)^3 + 40*cos(f*x + e)^2 + 16*cos(f*x + e) - 16)/((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))
 
3.1.61.6 Sympy [F]

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

input
integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**4,x)
 
output
Integral(sec(e + f*x)/(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)
 
3.1.61.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.61 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {\frac {7 \, {\left (\frac {75 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3} c^{4}} + \frac {{\left (\frac {42 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {175 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {700 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{a^{3} c^{4} \sin \left (f x + e\right )^{7}}}{2240 \, f} \]

input
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x, algorithm="m 
axima")
 
output
1/2240*(7*(75*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x 
 + e) + 1)^3 + sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^3*c^4) + (42*sin(f* 
x + e)^2/(cos(f*x + e) + 1)^2 - 175*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 
700*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 5)*(cos(f*x + e) + 1)^7/(a^3*c^4 
*sin(f*x + e)^7))/f
 
3.1.61.8 Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {\frac {700 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 175 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 42 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5}{a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}} + \frac {7 \, {\left (a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 75 \, a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15} c^{20}}}{2240 \, f} \]

input
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x, algorithm="g 
iac")
 
output
1/2240*((700*tan(1/2*f*x + 1/2*e)^6 - 175*tan(1/2*f*x + 1/2*e)^4 + 42*tan( 
1/2*f*x + 1/2*e)^2 - 5)/(a^3*c^4*tan(1/2*f*x + 1/2*e)^7) + 7*(a^12*c^16*ta 
n(1/2*f*x + 1/2*e)^5 - 10*a^12*c^16*tan(1/2*f*x + 1/2*e)^3 + 75*a^12*c^16* 
tan(1/2*f*x + 1/2*e))/(a^15*c^20))/f
 
3.1.61.9 Mupad [B] (verification not implemented)

Time = 14.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.30 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx=\frac {\left (2\,{\sin \left (\frac {e}{4}+\frac {f\,x}{4}\right )}^2-1\right )\,\left (\frac {235\,{\sin \left (e+f\,x\right )}^2}{16}-\frac {45\,{\sin \left (2\,e+2\,f\,x\right )}^2}{8}+\frac {19\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{2}+\frac {5\,{\sin \left (3\,e+3\,f\,x\right )}^2}{16}-\frac {5\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2}{4}+\frac {15\,{\sin \left (\frac {5\,e}{2}+\frac {5\,f\,x}{2}\right )}^2}{4}-5\right )}{2240\,a^3\,c^4\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,{\left ({\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}^3} \]

input
int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^4),x)
 
output
((2*sin(e/4 + (f*x)/4)^2 - 1)*((19*sin(e/2 + (f*x)/2)^2)/2 - (45*sin(2*e + 
 2*f*x)^2)/8 + (5*sin(3*e + 3*f*x)^2)/16 - (5*sin((3*e)/2 + (3*f*x)/2)^2)/ 
4 + (15*sin((5*e)/2 + (5*f*x)/2)^2)/4 + (235*sin(e + f*x)^2)/16 - 5))/(224 
0*a^3*c^4*f*sin(e/2 + (f*x)/2)^7*(sin(e/2 + (f*x)/2)^2 - 1)^3)