3.1.63 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^6} \, dx\) [63]

3.1.63.1 Optimal result
3.1.63.2 Mathematica [A] (verified)
3.1.63.3 Rubi [A] (verified)
3.1.63.4 Maple [A] (verified)
3.1.63.5 Fricas [A] (verification not implemented)
3.1.63.6 Sympy [F]
3.1.63.7 Maxima [A] (verification not implemented)
3.1.63.8 Giac [A] (verification not implemented)
3.1.63.9 Mupad [B] (verification not implemented)

3.1.63.1 Optimal result

Integrand size = 32, antiderivative size = 162 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^6} \, dx=-\frac {\cot ^9(e+f x)}{9 a^3 c^6 f}-\frac {4 \cot ^{11}(e+f x)}{11 a^3 c^6 f}+\frac {\csc (e+f x)}{a^3 c^6 f}-\frac {8 \csc ^3(e+f x)}{3 a^3 c^6 f}+\frac {22 \csc ^5(e+f x)}{5 a^3 c^6 f}-\frac {4 \csc ^7(e+f x)}{a^3 c^6 f}+\frac {17 \csc ^9(e+f x)}{9 a^3 c^6 f}-\frac {4 \csc ^{11}(e+f x)}{11 a^3 c^6 f} \]

output
-1/9*cot(f*x+e)^9/a^3/c^6/f-4/11*cot(f*x+e)^11/a^3/c^6/f+csc(f*x+e)/a^3/c^ 
6/f-8/3*csc(f*x+e)^3/a^3/c^6/f+22/5*csc(f*x+e)^5/a^3/c^6/f-4*csc(f*x+e)^7/ 
a^3/c^6/f+17/9*csc(f*x+e)^9/a^3/c^6/f-4/11*csc(f*x+e)^11/a^3/c^6/f
 
3.1.63.2 Mathematica [A] (verified)

Time = 5.14 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.73 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^6} \, dx=\frac {\left (-125-120 \sec (e+f x)+680 \sec ^2(e+f x)-400 \sec ^3(e+f x)-720 \sec ^4(e+f x)+832 \sec ^5(e+f x)+64 \sec ^6(e+f x)-384 \sec ^7(e+f x)+128 \sec ^8(e+f x)\right ) \tan (e+f x)}{495 a^3 c^6 f (-1+\sec (e+f x))^6 (1+\sec (e+f x))^3} \]

input
Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^6),x]
 
output
((-125 - 120*Sec[e + f*x] + 680*Sec[e + f*x]^2 - 400*Sec[e + f*x]^3 - 720* 
Sec[e + f*x]^4 + 832*Sec[e + f*x]^5 + 64*Sec[e + f*x]^6 - 384*Sec[e + f*x] 
^7 + 128*Sec[e + f*x]^8)*Tan[e + f*x])/(495*a^3*c^6*f*(-1 + Sec[e + f*x])^ 
6*(1 + Sec[e + f*x])^3)
 
3.1.63.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.90, 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.

\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^6} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^6}dx\)

\(\Big \downarrow \) 4446

\(\displaystyle \frac {\int \left (a^3 \csc (e+f x) \cot ^{11}(e+f x)+3 a^3 \csc ^2(e+f x) \cot ^{10}(e+f x)+3 a^3 \csc ^3(e+f x) \cot ^9(e+f x)+a^3 \csc ^4(e+f x) \cot ^8(e+f x)\right )dx}{a^6 c^6}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {4 a^3 \cot ^{11}(e+f x)}{11 f}-\frac {a^3 \cot ^9(e+f x)}{9 f}-\frac {4 a^3 \csc ^{11}(e+f x)}{11 f}+\frac {17 a^3 \csc ^9(e+f x)}{9 f}-\frac {4 a^3 \csc ^7(e+f x)}{f}+\frac {22 a^3 \csc ^5(e+f x)}{5 f}-\frac {8 a^3 \csc ^3(e+f x)}{3 f}+\frac {a^3 \csc (e+f x)}{f}}{a^6 c^6}\)

input
Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^6),x]
 
output
(-1/9*(a^3*Cot[e + f*x]^9)/f - (4*a^3*Cot[e + f*x]^11)/(11*f) + (a^3*Csc[e 
 + f*x])/f - (8*a^3*Csc[e + f*x]^3)/(3*f) + (22*a^3*Csc[e + f*x]^5)/(5*f) 
- (4*a^3*Csc[e + f*x]^7)/f + (17*a^3*Csc[e + f*x]^9)/(9*f) - (4*a^3*Csc[e 
+ f*x]^11)/(11*f))/(a^6*c^6)
 

3.1.63.3.1 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]
 
3.1.63.4 Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.75

method result size
parallelrisch \(-\frac {\left (240 \cos \left (7 f x +7 e \right )-1300 \cos \left (4 f x +4 e \right )-1720 \cos \left (6 f x +6 e \right )+9680 \cos \left (f x +e \right )+4880 \cos \left (5 f x +5 e \right )-5584 \cos \left (3 f x +3 e \right )+8184 \cos \left (2 f x +2 e \right )-8745+125 \cos \left (8 f x +8 e \right )\right ) \sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \csc \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{16220160 f \,a^{3} c^{6}}\) \(121\)
derivativedivides \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+28 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{11 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}+\frac {8}{9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}+\frac {56}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {70}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {56}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}}{256 f \,a^{3} c^{6}}\) \(128\)
default \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+28 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{11 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}+\frac {8}{9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}+\frac {56}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {70}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {56}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}}{256 f \,a^{3} c^{6}}\) \(128\)
risch \(\frac {2 i \left (495 \,{\mathrm e}^{15 i \left (f x +e \right )}-1485 \,{\mathrm e}^{14 i \left (f x +e \right )}+1815 \,{\mathrm e}^{13 i \left (f x +e \right )}+2475 \,{\mathrm e}^{12 i \left (f x +e \right )}-4917 \,{\mathrm e}^{11 i \left (f x +e \right )}-33 \,{\mathrm e}^{10 i \left (f x +e \right )}+11715 \,{\mathrm e}^{9 i \left (f x +e \right )}-8745 \,{\mathrm e}^{8 i \left (f x +e \right )}-2035 \,{\mathrm e}^{7 i \left (f x +e \right )}+8217 \,{\mathrm e}^{6 i \left (f x +e \right )}-667 \,{\mathrm e}^{5 i \left (f x +e \right )}-3775 \,{\mathrm e}^{4 i \left (f x +e \right )}+3065 \,{\mathrm e}^{3 i \left (f x +e \right )}-235 \,{\mathrm e}^{2 i \left (f x +e \right )}-255 \,{\mathrm e}^{i \left (f x +e \right )}+125\right )}{495 f \,a^{3} c^{6} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{11} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\) \(206\)

input
int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^6,x,method=_RETURNVERBO 
SE)
 
output
-1/16220160*(240*cos(7*f*x+7*e)-1300*cos(4*f*x+4*e)-1720*cos(6*f*x+6*e)+96 
80*cos(f*x+e)+4880*cos(5*f*x+5*e)-5584*cos(3*f*x+3*e)+8184*cos(2*f*x+2*e)- 
8745+125*cos(8*f*x+8*e))*sec(1/2*f*x+1/2*e)^5*csc(1/2*f*x+1/2*e)^11/f/a^3/ 
c^6
 
3.1.63.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.34 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^6} \, dx=\frac {125 \, \cos \left (f x + e\right )^{8} + 120 \, \cos \left (f x + e\right )^{7} - 680 \, \cos \left (f x + e\right )^{6} + 400 \, \cos \left (f x + e\right )^{5} + 720 \, \cos \left (f x + e\right )^{4} - 832 \, \cos \left (f x + e\right )^{3} - 64 \, \cos \left (f x + e\right )^{2} + 384 \, \cos \left (f x + e\right ) - 128}{495 \, {\left (a^{3} c^{6} f \cos \left (f x + e\right )^{7} - 3 \, a^{3} c^{6} f \cos \left (f x + e\right )^{6} + a^{3} c^{6} f \cos \left (f x + e\right )^{5} + 5 \, a^{3} c^{6} f \cos \left (f x + e\right )^{4} - 5 \, a^{3} c^{6} f \cos \left (f x + e\right )^{3} - a^{3} c^{6} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} c^{6} f \cos \left (f x + e\right ) - a^{3} c^{6} f\right )} \sin \left (f x + e\right )} \]

input
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^6,x, algorithm="f 
ricas")
 
output
1/495*(125*cos(f*x + e)^8 + 120*cos(f*x + e)^7 - 680*cos(f*x + e)^6 + 400* 
cos(f*x + e)^5 + 720*cos(f*x + e)^4 - 832*cos(f*x + e)^3 - 64*cos(f*x + e) 
^2 + 384*cos(f*x + e) - 128)/((a^3*c^6*f*cos(f*x + e)^7 - 3*a^3*c^6*f*cos( 
f*x + e)^6 + a^3*c^6*f*cos(f*x + e)^5 + 5*a^3*c^6*f*cos(f*x + e)^4 - 5*a^3 
*c^6*f*cos(f*x + e)^3 - a^3*c^6*f*cos(f*x + e)^2 + 3*a^3*c^6*f*cos(f*x + e 
) - a^3*c^6*f)*sin(f*x + e))
 
3.1.63.6 Sympy [F]

\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^6} \, dx=\frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{9}{\left (e + f x \right )} - 3 \sec ^{8}{\left (e + f x \right )} + 8 \sec ^{6}{\left (e + f x \right )} - 6 \sec ^{5}{\left (e + f x \right )} - 6 \sec ^{4}{\left (e + f x \right )} + 8 \sec ^{3}{\left (e + f x \right )} - 3 \sec {\left (e + f x \right )} + 1}\, dx}{a^{3} c^{6}} \]

input
integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**6,x)
 
output
Integral(sec(e + f*x)/(sec(e + f*x)**9 - 3*sec(e + f*x)**8 + 8*sec(e + f*x 
)**6 - 6*sec(e + f*x)**5 - 6*sec(e + f*x)**4 + 8*sec(e + f*x)**3 - 3*sec(e 
 + f*x) + 1), x)/(a**3*c**6)
 
3.1.63.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.23 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^6} \, dx=\frac {\frac {33 \, {\left (\frac {420 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {40 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3} c^{6}} + \frac {{\left (\frac {440 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {1980 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {5544 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {11550 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac {27720 \, \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}} - 45\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{11}}{a^{3} c^{6} \sin \left (f x + e\right )^{11}}}{126720 \, f} \]

input
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^6,x, algorithm="m 
axima")
 
output
1/126720*(33*(420*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/(cos(f*x + e) + 1)^5)/(a^3*c^6) + (440 
*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1980*sin(f*x + e)^4/(cos(f*x + e) + 
 1)^4 + 5544*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 11550*sin(f*x + e)^8/(c 
os(f*x + e) + 1)^8 + 27720*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 45)*(co 
s(f*x + e) + 1)^11/(a^3*c^6*sin(f*x + e)^11))/f
 
3.1.63.8 Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.96 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^6} \, dx=\frac {\frac {27720 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 11550 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} + 5544 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1980 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 440 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 45}{a^{3} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{11}} + \frac {33 \, {\left (3 \, a^{12} c^{24} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 40 \, a^{12} c^{24} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 420 \, a^{12} c^{24} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15} c^{30}}}{126720 \, f} \]

input
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^6,x, algorithm="g 
iac")
 
output
1/126720*((27720*tan(1/2*f*x + 1/2*e)^10 - 11550*tan(1/2*f*x + 1/2*e)^8 + 
5544*tan(1/2*f*x + 1/2*e)^6 - 1980*tan(1/2*f*x + 1/2*e)^4 + 440*tan(1/2*f* 
x + 1/2*e)^2 - 45)/(a^3*c^6*tan(1/2*f*x + 1/2*e)^11) + 33*(3*a^12*c^24*tan 
(1/2*f*x + 1/2*e)^5 - 40*a^12*c^24*tan(1/2*f*x + 1/2*e)^3 + 420*a^12*c^24* 
tan(1/2*f*x + 1/2*e))/(a^15*c^30))/f
 
3.1.63.9 Mupad [B] (verification not implemented)

Time = 14.68 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.74 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^6} \, dx=-\frac {\frac {605\,\cos \left (e+f\,x\right )}{8}+\frac {1023\,\cos \left (2\,e+2\,f\,x\right )}{16}-\frac {349\,\cos \left (3\,e+3\,f\,x\right )}{8}-\frac {325\,\cos \left (4\,e+4\,f\,x\right )}{32}+\frac {305\,\cos \left (5\,e+5\,f\,x\right )}{8}-\frac {215\,\cos \left (6\,e+6\,f\,x\right )}{16}+\frac {15\,\cos \left (7\,e+7\,f\,x\right )}{8}+\frac {125\,\cos \left (8\,e+8\,f\,x\right )}{128}-\frac {8745}{128}}{126720\,a^3\,c^6\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}} \]

input
int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^6),x)
 
output
-((605*cos(e + f*x))/8 + (1023*cos(2*e + 2*f*x))/16 - (349*cos(3*e + 3*f*x 
))/8 - (325*cos(4*e + 4*f*x))/32 + (305*cos(5*e + 5*f*x))/8 - (215*cos(6*e 
 + 6*f*x))/16 + (15*cos(7*e + 7*f*x))/8 + (125*cos(8*e + 8*f*x))/128 - 874 
5/128)/(126720*a^3*c^6*f*cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2)^11)