3.10 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^5 \, dx\)

Optimal. Leaf size=171 \[ -\frac {a^2 c^5 \tan ^7(e+f x)}{7 f}-\frac {4 a^2 c^5 \tan ^5(e+f x)}{5 f}+\frac {9 a^2 c^5 \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac {a^2 c^5 \tan ^3(e+f x) \sec ^3(e+f x)}{2 f}-\frac {3 a^2 c^5 \tan (e+f x) \sec ^3(e+f x)}{8 f}+\frac {a^2 c^5 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac {3 a^2 c^5 \tan (e+f x) \sec (e+f x)}{16 f} \]

[Out]

9/16*a^2*c^5*arctanh(sin(f*x+e))/f-3/16*a^2*c^5*sec(f*x+e)*tan(f*x+e)/f-3/8*a^2*c^5*sec(f*x+e)^3*tan(f*x+e)/f+
1/4*a^2*c^5*sec(f*x+e)*tan(f*x+e)^3/f+1/2*a^2*c^5*sec(f*x+e)^3*tan(f*x+e)^3/f-4/5*a^2*c^5*tan(f*x+e)^5/f-1/7*a
^2*c^5*tan(f*x+e)^7/f

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3958, 2611, 3770, 2607, 30, 3768, 14} \[ -\frac {a^2 c^5 \tan ^7(e+f x)}{7 f}-\frac {4 a^2 c^5 \tan ^5(e+f x)}{5 f}+\frac {9 a^2 c^5 \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac {a^2 c^5 \tan ^3(e+f x) \sec ^3(e+f x)}{2 f}-\frac {3 a^2 c^5 \tan (e+f x) \sec ^3(e+f x)}{8 f}+\frac {a^2 c^5 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac {3 a^2 c^5 \tan (e+f x) \sec (e+f x)}{16 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^5,x]

[Out]

(9*a^2*c^5*ArcTanh[Sin[e + f*x]])/(16*f) - (3*a^2*c^5*Sec[e + f*x]*Tan[e + f*x])/(16*f) - (3*a^2*c^5*Sec[e + f
*x]^3*Tan[e + f*x])/(8*f) + (a^2*c^5*Sec[e + f*x]*Tan[e + f*x]^3)/(4*f) + (a^2*c^5*Sec[e + f*x]^3*Tan[e + f*x]
^3)/(2*f) - (4*a^2*c^5*Tan[e + f*x]^5)/(5*f) - (a^2*c^5*Tan[e + f*x]^7)/(7*f)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3958

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
)^(n_.), x_Symbol] :> Dist[(-(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] && EqQ[a^2 - b^2, 0] && IntegersQ[m,
n] && GeQ[n - m, 0] && GtQ[m*n, 0]

Rubi steps

\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^5 \, dx &=\left (a^2 c^2\right ) \int \left (c^3 \sec (e+f x) \tan ^4(e+f x)-3 c^3 \sec ^2(e+f x) \tan ^4(e+f x)+3 c^3 \sec ^3(e+f x) \tan ^4(e+f x)-c^3 \sec ^4(e+f x) \tan ^4(e+f x)\right ) \, dx\\ &=\left (a^2 c^5\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx-\left (a^2 c^5\right ) \int \sec ^4(e+f x) \tan ^4(e+f x) \, dx-\left (3 a^2 c^5\right ) \int \sec ^2(e+f x) \tan ^4(e+f x) \, dx+\left (3 a^2 c^5\right ) \int \sec ^3(e+f x) \tan ^4(e+f x) \, dx\\ &=\frac {a^2 c^5 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^5 \sec ^3(e+f x) \tan ^3(e+f x)}{2 f}-\frac {1}{4} \left (3 a^2 c^5\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx-\frac {1}{2} \left (3 a^2 c^5\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx-\frac {\left (a^2 c^5\right ) \operatorname {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}-\frac {\left (3 a^2 c^5\right ) \operatorname {Subst}\left (\int x^4 \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {3 a^2 c^5 \sec (e+f x) \tan (e+f x)}{8 f}-\frac {3 a^2 c^5 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^5 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^5 \sec ^3(e+f x) \tan ^3(e+f x)}{2 f}-\frac {3 a^2 c^5 \tan ^5(e+f x)}{5 f}+\frac {1}{8} \left (3 a^2 c^5\right ) \int \sec (e+f x) \, dx+\frac {1}{8} \left (3 a^2 c^5\right ) \int \sec ^3(e+f x) \, dx-\frac {\left (a^2 c^5\right ) \operatorname {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {3 a^2 c^5 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {3 a^2 c^5 \sec (e+f x) \tan (e+f x)}{16 f}-\frac {3 a^2 c^5 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^5 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^5 \sec ^3(e+f x) \tan ^3(e+f x)}{2 f}-\frac {4 a^2 c^5 \tan ^5(e+f x)}{5 f}-\frac {a^2 c^5 \tan ^7(e+f x)}{7 f}+\frac {1}{16} \left (3 a^2 c^5\right ) \int \sec (e+f x) \, dx\\ &=\frac {9 a^2 c^5 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {3 a^2 c^5 \sec (e+f x) \tan (e+f x)}{16 f}-\frac {3 a^2 c^5 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^5 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^5 \sec ^3(e+f x) \tan ^3(e+f x)}{2 f}-\frac {4 a^2 c^5 \tan ^5(e+f x)}{5 f}-\frac {a^2 c^5 \tan ^7(e+f x)}{7 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.85, size = 102, normalized size = 0.60 \[ \frac {a^2 c^5 \left (10080 \tanh ^{-1}(\sin (e+f x))-(2520 \sin (e+f x)-455 \sin (2 (e+f x))-616 \sin (3 (e+f x))+2380 \sin (4 (e+f x))-392 \sin (5 (e+f x))+245 \sin (6 (e+f x))+184 \sin (7 (e+f x))) \sec ^7(e+f x)\right )}{17920 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^5,x]

[Out]

(a^2*c^5*(10080*ArcTanh[Sin[e + f*x]] - Sec[e + f*x]^7*(2520*Sin[e + f*x] - 455*Sin[2*(e + f*x)] - 616*Sin[3*(
e + f*x)] + 2380*Sin[4*(e + f*x)] - 392*Sin[5*(e + f*x)] + 245*Sin[6*(e + f*x)] + 184*Sin[7*(e + f*x)])))/(179
20*f)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 177, normalized size = 1.04 \[ \frac {315 \, a^{2} c^{5} \cos \left (f x + e\right )^{7} \log \left (\sin \left (f x + e\right ) + 1\right ) - 315 \, a^{2} c^{5} \cos \left (f x + e\right )^{7} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (368 \, a^{2} c^{5} \cos \left (f x + e\right )^{6} + 245 \, a^{2} c^{5} \cos \left (f x + e\right )^{5} - 656 \, a^{2} c^{5} \cos \left (f x + e\right )^{4} + 350 \, a^{2} c^{5} \cos \left (f x + e\right )^{3} + 208 \, a^{2} c^{5} \cos \left (f x + e\right )^{2} - 280 \, a^{2} c^{5} \cos \left (f x + e\right ) + 80 \, a^{2} c^{5}\right )} \sin \left (f x + e\right )}{1120 \, f \cos \left (f x + e\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^5,x, algorithm="fricas")

[Out]

1/1120*(315*a^2*c^5*cos(f*x + e)^7*log(sin(f*x + e) + 1) - 315*a^2*c^5*cos(f*x + e)^7*log(-sin(f*x + e) + 1) -
2*(368*a^2*c^5*cos(f*x + e)^6 + 245*a^2*c^5*cos(f*x + e)^5 - 656*a^2*c^5*cos(f*x + e)^4 + 350*a^2*c^5*cos(f*x
+ e)^3 + 208*a^2*c^5*cos(f*x + e)^2 - 280*a^2*c^5*cos(f*x + e) + 80*a^2*c^5)*sin(f*x + e))/(f*cos(f*x + e)^7)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^5,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)-2/f*(9*a^2*c^5/32*ln(abs(tan((f*x+exp(1))/2)-1))-9*a^2*c^5/32
*ln(abs(tan((f*x+exp(1))/2)+1))+(315*tan((f*x+exp(1))/2)^13*a^2*c^5-2100*tan((f*x+exp(1))/2)^11*a^2*c^5-8393*t
an((f*x+exp(1))/2)^9*a^2*c^5+9216*tan((f*x+exp(1))/2)^7*a^2*c^5-5943*tan((f*x+exp(1))/2)^5*a^2*c^5+2100*tan((f
*x+exp(1))/2)^3*a^2*c^5-315*tan((f*x+exp(1))/2)*a^2*c^5)*1/560/(tan((f*x+exp(1))/2)^2-1)^7)

________________________________________________________________________________________

maple [A]  time = 1.94, size = 192, normalized size = 1.12 \[ -\frac {23 a^{2} c^{5} \tan \left (f x +e \right )}{35 f}-\frac {13 a^{2} c^{5} \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{35 f}+\frac {41 a^{2} c^{5} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{35 f}-\frac {5 a^{2} c^{5} \left (\sec ^{3}\left (f x +e \right )\right ) \tan \left (f x +e \right )}{8 f}-\frac {7 a^{2} c^{5} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{16 f}+\frac {9 a^{2} c^{5} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16 f}+\frac {a^{2} c^{5} \tan \left (f x +e \right ) \left (\sec ^{5}\left (f x +e \right )\right )}{2 f}-\frac {a^{2} c^{5} \tan \left (f x +e \right ) \left (\sec ^{6}\left (f x +e \right )\right )}{7 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^5,x)

[Out]

-23/35*a^2*c^5*tan(f*x+e)/f-13/35/f*a^2*c^5*tan(f*x+e)*sec(f*x+e)^4+41/35/f*a^2*c^5*tan(f*x+e)*sec(f*x+e)^2-5/
8*a^2*c^5*sec(f*x+e)^3*tan(f*x+e)/f-7/16*a^2*c^5*sec(f*x+e)*tan(f*x+e)/f+9/16/f*a^2*c^5*ln(sec(f*x+e)+tan(f*x+
e))+1/2/f*a^2*c^5*tan(f*x+e)*sec(f*x+e)^5-1/7/f*a^2*c^5*tan(f*x+e)*sec(f*x+e)^6

________________________________________________________________________________________

maxima [B]  time = 0.34, size = 368, normalized size = 2.15 \[ -\frac {96 \, {\left (5 \, \tan \left (f x + e\right )^{7} + 21 \, \tan \left (f x + e\right )^{5} + 35 \, \tan \left (f x + e\right )^{3} + 35 \, \tan \left (f x + e\right )\right )} a^{2} c^{5} + 224 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{2} c^{5} - 5600 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{5} + 105 \, a^{2} c^{5} {\left (\frac {2 \, {\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 1050 \, a^{2} c^{5} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 840 \, a^{2} c^{5} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 3360 \, a^{2} c^{5} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 10080 \, a^{2} c^{5} \tan \left (f x + e\right )}{3360 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^5,x, algorithm="maxima")

[Out]

-1/3360*(96*(5*tan(f*x + e)^7 + 21*tan(f*x + e)^5 + 35*tan(f*x + e)^3 + 35*tan(f*x + e))*a^2*c^5 + 224*(3*tan(
f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*c^5 - 5600*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c^5 + 1
05*a^2*c^5*(2*(15*sin(f*x + e)^5 - 40*sin(f*x + e)^3 + 33*sin(f*x + e))/(sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3
*sin(f*x + e)^2 - 1) - 15*log(sin(f*x + e) + 1) + 15*log(sin(f*x + e) - 1)) - 1050*a^2*c^5*(2*(3*sin(f*x + e)^
3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1
)) + 840*a^2*c^5*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 3360*
a^2*c^5*log(sec(f*x + e) + tan(f*x + e)) + 10080*a^2*c^5*tan(f*x + e))/f

________________________________________________________________________________________

mupad [B]  time = 5.76, size = 251, normalized size = 1.47 \[ \frac {-\frac {9\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{13}}{8}+\frac {15\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}}{2}+\frac {1199\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{40}-\frac {1152\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{35}+\frac {849\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{40}-\frac {15\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{2}+\frac {9\,a^2\,c^5\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}+\frac {9\,a^2\,c^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{8\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^5)/cos(e + f*x),x)

[Out]

((849*a^2*c^5*tan(e/2 + (f*x)/2)^5)/40 - (15*a^2*c^5*tan(e/2 + (f*x)/2)^3)/2 - (1152*a^2*c^5*tan(e/2 + (f*x)/2
)^7)/35 + (1199*a^2*c^5*tan(e/2 + (f*x)/2)^9)/40 + (15*a^2*c^5*tan(e/2 + (f*x)/2)^11)/2 - (9*a^2*c^5*tan(e/2 +
(f*x)/2)^13)/8 + (9*a^2*c^5*tan(e/2 + (f*x)/2))/8)/(f*(7*tan(e/2 + (f*x)/2)^2 - 21*tan(e/2 + (f*x)/2)^4 + 35*
tan(e/2 + (f*x)/2)^6 - 35*tan(e/2 + (f*x)/2)^8 + 21*tan(e/2 + (f*x)/2)^10 - 7*tan(e/2 + (f*x)/2)^12 + tan(e/2
+ (f*x)/2)^14 - 1)) + (9*a^2*c^5*atanh(tan(e/2 + (f*x)/2)))/(8*f)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} c^{5} \left (\int \left (- \sec {\left (e + f x \right )}\right )\, dx + \int 3 \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (- \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- 5 \sec ^{4}{\left (e + f x \right )}\right )\, dx + \int 5 \sec ^{5}{\left (e + f x \right )}\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx + \int \left (- 3 \sec ^{7}{\left (e + f x \right )}\right )\, dx + \int \sec ^{8}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2*(c-c*sec(f*x+e))**5,x)

[Out]

-a**2*c**5*(Integral(-sec(e + f*x), x) + Integral(3*sec(e + f*x)**2, x) + Integral(-sec(e + f*x)**3, x) + Inte
gral(-5*sec(e + f*x)**4, x) + Integral(5*sec(e + f*x)**5, x) + Integral(sec(e + f*x)**6, x) + Integral(-3*sec(
e + f*x)**7, x) + Integral(sec(e + f*x)**8, x))

________________________________________________________________________________________