\(\int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx\) [286]

Optimal result

Integrand size = 28, antiderivative size = 89 \[ \int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx=\frac {\cot ^5\left (\frac {1}{2} (e+f x)\right )}{40 c^8 f}-\frac {3 \cot ^7\left (\frac {1}{2} (e+f x)\right )}{56 c^8 f}+\frac {\cot ^9\left (\frac {1}{2} (e+f x)\right )}{24 c^8 f}-\frac {\cot ^{11}\left (\frac {1}{2} (e+f x)\right )}{88 c^8 f} \]

[Out]

1/40*cot(1/2*f*x+1/2*e)^5/c^8/f-3/56*cot(1/2*f*x+1/2*e)^7/c^8/f+1/24*cot(1/2*f*x+1/2*e)^9/c^8/f-1/88*cot(1/2*f
*x+1/2*e)^11/c^8/f

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 276} \[ \int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx=-\frac {\cot ^{11}\left (\frac {1}{2} (e+f x)\right )}{88 c^8 f}+\frac {\cot ^9\left (\frac {1}{2} (e+f x)\right )}{24 c^8 f}-\frac {3 \cot ^7\left (\frac {1}{2} (e+f x)\right )}{56 c^8 f}+\frac {\cot ^5\left (\frac {1}{2} (e+f x)\right )}{40 c^8 f} \]

[In]

Int[(Sec[e + f*x]*Tan[e + f*x]^4)/(c - c*Sec[e + f*x])^8,x]

[Out]

Cot[(e + f*x)/2]^5/(40*c^8*f) - (3*Cot[(e + f*x)/2]^7)/(56*c^8*f) + Cot[(e + f*x)/2]^9/(24*c^8*f) - Cot[(e + f
*x)/2]^11/(88*c^8*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{16 c^8 x^{12}} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^{12}} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{8 c^8 f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{x^{12}}-\frac {3}{x^{10}}+\frac {3}{x^8}-\frac {1}{x^6}\right ) \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{8 c^8 f} \\ & = \frac {\cot ^5\left (\frac {1}{2} (e+f x)\right )}{40 c^8 f}-\frac {3 \cot ^7\left (\frac {1}{2} (e+f x)\right )}{56 c^8 f}+\frac {\cot ^9\left (\frac {1}{2} (e+f x)\right )}{24 c^8 f}-\frac {\cot ^{11}\left (\frac {1}{2} (e+f x)\right )}{88 c^8 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.02 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.97 \[ \int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx=-\frac {\csc \left (\frac {e}{2}\right ) \csc ^{11}\left (\frac {1}{2} (e+f x)\right ) \left (425964 \sin \left (\frac {f x}{2}\right )+486024 \sin \left (e+\frac {f x}{2}\right )-351450 \sin \left (e+\frac {3 f x}{2}\right )-299970 \sin \left (2 e+\frac {3 f x}{2}\right )+145695 \sin \left (2 e+\frac {5 f x}{2}\right )+180015 \sin \left (3 e+\frac {5 f x}{2}\right )-63580 \sin \left (3 e+\frac {7 f x}{2}\right )-44990 \sin \left (4 e+\frac {7 f x}{2}\right )+6710 \sin \left (4 e+\frac {9 f x}{2}\right )+15004 \sin \left (5 e+\frac {9 f x}{2}\right )-1975 \sin \left (5 e+\frac {11 f x}{2}\right )+\sin \left (6 e+\frac {11 f x}{2}\right )\right )}{15375360 c^8 f} \]

[In]

Integrate[(Sec[e + f*x]*Tan[e + f*x]^4)/(c - c*Sec[e + f*x])^8,x]

[Out]

-1/15375360*(Csc[e/2]*Csc[(e + f*x)/2]^11*(425964*Sin[(f*x)/2] + 486024*Sin[e + (f*x)/2] - 351450*Sin[e + (3*f
*x)/2] - 299970*Sin[2*e + (3*f*x)/2] + 145695*Sin[2*e + (5*f*x)/2] + 180015*Sin[3*e + (5*f*x)/2] - 63580*Sin[3
*e + (7*f*x)/2] - 44990*Sin[4*e + (7*f*x)/2] + 6710*Sin[4*e + (9*f*x)/2] + 15004*Sin[5*e + (9*f*x)/2] - 1975*S
in[5*e + (11*f*x)/2] + Sin[6*e + (11*f*x)/2]))/(c^8*f)

Maple [A] (verified)

Time = 17.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {-\frac {3}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}-\frac {1}{11 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}}{8 f \,c^{8}}\)	\(62\)
default	\(\frac {-\frac {3}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}-\frac {1}{11 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}}{8 f \,c^{8}}\)	\(62\)
risch	\(\frac {2 i \left (1155 \,{\mathrm e}^{10 i \left (f x +e \right )}-3465 \,{\mathrm e}^{9 i \left (f x +e \right )}+13860 \,{\mathrm e}^{8 i \left (f x +e \right )}-23100 \,{\mathrm e}^{7 i \left (f x +e \right )}+37422 \,{\mathrm e}^{6 i \left (f x +e \right )}-32802 \,{\mathrm e}^{5 i \left (f x +e \right )}+27060 \,{\mathrm e}^{4 i \left (f x +e \right )}-11220 \,{\mathrm e}^{3 i \left (f x +e \right )}+4895 \,{\mathrm e}^{2 i \left (f x +e \right )}-517 \,{\mathrm e}^{i \left (f x +e \right )}+152\right )}{1155 f \,c^{8} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{11}}\)	\(135\)

[In]

int(sec(f*x+e)*tan(f*x+e)^4/(c-c*sec(f*x+e))^8,x,method=_RETURNVERBOSE)

[Out]

1/8/f/c^8*(-3/7/tan(1/2*f*x+1/2*e)^7-1/11/tan(1/2*f*x+1/2*e)^11+1/5/tan(1/2*f*x+1/2*e)^5+1/3/tan(1/2*f*x+1/2*e
)^9)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.64 \[ \int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx=\frac {152 \, \cos \left (f x + e\right )^{6} + 395 \, \cos \left (f x + e\right )^{5} + 289 \, \cos \left (f x + e\right )^{4} + 15 \, \cos \left (f x + e\right )^{3} - 19 \, \cos \left (f x + e\right )^{2} + 10 \, \cos \left (f x + e\right ) - 2}{1155 \, {\left (c^{8} f \cos \left (f x + e\right )^{5} - 5 \, c^{8} f \cos \left (f x + e\right )^{4} + 10 \, c^{8} f \cos \left (f x + e\right )^{3} - 10 \, c^{8} f \cos \left (f x + e\right )^{2} + 5 \, c^{8} f \cos \left (f x + e\right ) - c^{8} f\right )} \sin \left (f x + e\right )} \]

[In]

integrate(sec(f*x+e)*tan(f*x+e)^4/(c-c*sec(f*x+e))^8,x, algorithm="fricas")

[Out]

1/1155*(152*cos(f*x + e)^6 + 395*cos(f*x + e)^5 + 289*cos(f*x + e)^4 + 15*cos(f*x + e)^3 - 19*cos(f*x + e)^2 +
10*cos(f*x + e) - 2)/((c^8*f*cos(f*x + e)^5 - 5*c^8*f*cos(f*x + e)^4 + 10*c^8*f*cos(f*x + e)^3 - 10*c^8*f*cos
(f*x + e)^2 + 5*c^8*f*cos(f*x + e) - c^8*f)*sin(f*x + e))

Sympy [F]

\[ \int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx=\frac {\int \frac {\tan ^{4}{\left (e + f x \right )} \sec {\left (e + f x \right )}}{\sec ^{8}{\left (e + f x \right )} - 8 \sec ^{7}{\left (e + f x \right )} + 28 \sec ^{6}{\left (e + f x \right )} - 56 \sec ^{5}{\left (e + f x \right )} + 70 \sec ^{4}{\left (e + f x \right )} - 56 \sec ^{3}{\left (e + f x \right )} + 28 \sec ^{2}{\left (e + f x \right )} - 8 \sec {\left (e + f x \right )} + 1}\, dx}{c^{8}} \]

[In]

integrate(sec(f*x+e)*tan(f*x+e)**4/(c-c*sec(f*x+e))**8,x)

[Out]

Integral(tan(e + f*x)**4*sec(e + f*x)/(sec(e + f*x)**8 - 8*sec(e + f*x)**7 + 28*sec(e + f*x)**6 - 56*sec(e + f
*x)**5 + 70*sec(e + f*x)**4 - 56*sec(e + f*x)**3 + 28*sec(e + f*x)**2 - 8*sec(e + f*x) + 1), x)/c**8

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx=\frac {{\left (\frac {385 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {495 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {231 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 105\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{11}}{9240 \, c^{8} f \sin \left (f x + e\right )^{11}} \]

[In]

integrate(sec(f*x+e)*tan(f*x+e)^4/(c-c*sec(f*x+e))^8,x, algorithm="maxima")

[Out]

1/9240*(385*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 495*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 231*sin(f*x + e)^6
/(cos(f*x + e) + 1)^6 - 105)*(cos(f*x + e) + 1)^11/(c^8*f*sin(f*x + e)^11)

Giac [A] (verification not implemented)

none

Time = 1.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx=\frac {231 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 495 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 385 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 105}{9240 \, c^{8} f \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{11}} \]

[In]

integrate(sec(f*x+e)*tan(f*x+e)^4/(c-c*sec(f*x+e))^8,x, algorithm="giac")

[Out]

1/9240*(231*tan(1/2*f*x + 1/2*e)^6 - 495*tan(1/2*f*x + 1/2*e)^4 + 385*tan(1/2*f*x + 1/2*e)^2 - 105)/(c^8*f*tan
(1/2*f*x + 1/2*e)^11)

Mupad [B] (verification not implemented)

Time = 14.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int \frac {\sec (e+f x) \tan ^4(e+f x)}{(c-c \sec (e+f x))^8} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6}{5}-\frac {3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{7}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{3}-\frac {1}{11}}{8\,c^8\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}} \]

[In]

int(tan(e + f*x)^4/(cos(e + f*x)*(c - c/cos(e + f*x))^8),x)

[Out]

(tan(e/2 + (f*x)/2)^2/3 - (3*tan(e/2 + (f*x)/2)^4)/7 + tan(e/2 + (f*x)/2)^6/5 - 1/11)/(8*c^8*f*tan(e/2 + (f*x)
/2)^11)