Integrand size = 32, antiderivative size = 150 \[ \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx=\frac {7 a^2 c^4 \text {arctanh}(\sin (e+f x))}{16 f}-\frac {5 a^2 c^4 \sec (e+f x) \tan (e+f x)}{16 f}-\frac {a^2 c^4 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac {2 a^2 c^4 \tan ^5(e+f x)}{5 f} \] Output:
7/16*a^2*c^4*arctanh(sin(f*x+e))/f-5/16*a^2*c^4*sec(f*x+e)*tan(f*x+e)/f-1/ 8*a^2*c^4*sec(f*x+e)^3*tan(f*x+e)/f+1/4*a^2*c^4*sec(f*x+e)*tan(f*x+e)^3/f+ 1/6*a^2*c^4*sec(f*x+e)^3*tan(f*x+e)^3/f-2/5*a^2*c^4*tan(f*x+e)^5/f
Time = 0.79 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.61 \[ \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx=\frac {a^2 c^4 \left (1680 \text {arctanh}(\sin (e+f x))+\sec ^6(e+f x) (330 \sin (e+f x)-240 \sin (2 (e+f x))-445 \sin (3 (e+f x))+192 \sin (4 (e+f x))-135 \sin (5 (e+f x))-48 \sin (6 (e+f x)))\right )}{3840 f} \] Input:
Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^4,x]
Output:
(a^2*c^4*(1680*ArcTanh[Sin[e + f*x]] + Sec[e + f*x]^6*(330*Sin[e + f*x] - 240*Sin[2*(e + f*x)] - 445*Sin[3*(e + f*x)] + 192*Sin[4*(e + f*x)] - 135*S in[5*(e + f*x)] - 48*Sin[6*(e + f*x)])))/(3840*f)
Time = 0.45 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.93, 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.
| \(\displaystyle \int \sec (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4dx\) |
| \(\Big \downarrow \) 4446 |
| \(\displaystyle a^2 c^2 \int \left (c^2 \sec ^3(e+f x) \tan ^4(e+f x)-2 c^2 \sec ^2(e+f x) \tan ^4(e+f x)+c^2 \sec (e+f x) \tan ^4(e+f x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a^2 c^2 \left (\frac {7 c^2 \text {arctanh}(\sin (e+f x))}{16 f}-\frac {2 c^2 \tan ^5(e+f x)}{5 f}+\frac {c^2 \tan ^3(e+f x) \sec ^3(e+f x)}{6 f}-\frac {c^2 \tan (e+f x) \sec ^3(e+f x)}{8 f}+\frac {c^2 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac {5 c^2 \tan (e+f x) \sec (e+f x)}{16 f}\right )\) |
Input:
Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^4,x]
Output:
a^2*c^2*((7*c^2*ArcTanh[Sin[e + f*x]])/(16*f) - (5*c^2*Sec[e + f*x]*Tan[e + f*x])/(16*f) - (c^2*Sec[e + f*x]^3*Tan[e + f*x])/(8*f) + (c^2*Sec[e + f* x]*Tan[e + f*x]^3)/(4*f) + (c^2*Sec[e + f*x]^3*Tan[e + f*x]^3)/(6*f) - (2* c^2*Tan[e + f*x]^5)/(5*f))
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]
Time = 0.51 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.30
| method | result | size |
| norman | \(\frac {-\frac {7 a^{2} c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{8 f}+\frac {119 a^{2} c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{24 f}-\frac {231 a^{2} c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{20 f}+\frac {281 a^{2} c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{20 f}+\frac {119 a^{2} c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{24 f}-\frac {7 a^{2} c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{8 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{6}}-\frac {7 a^{2} c^{4} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{16 f}+\frac {7 a^{2} c^{4} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{16 f}\) | \(195\) |
| risch | \(\frac {i c^{4} a^{2} \left (135 \,{\mathrm e}^{11 i \left (f x +e \right )}-480 \,{\mathrm e}^{10 i \left (f x +e \right )}+445 \,{\mathrm e}^{9 i \left (f x +e \right )}-480 \,{\mathrm e}^{8 i \left (f x +e \right )}-330 \,{\mathrm e}^{7 i \left (f x +e \right )}-960 \,{\mathrm e}^{6 i \left (f x +e \right )}+330 \,{\mathrm e}^{5 i \left (f x +e \right )}-960 \,{\mathrm e}^{4 i \left (f x +e \right )}-445 \,{\mathrm e}^{3 i \left (f x +e \right )}-96 \,{\mathrm e}^{2 i \left (f x +e \right )}-135 \,{\mathrm e}^{i \left (f x +e \right )}-96\right )}{120 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{6}}-\frac {7 a^{2} c^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{16 f}+\frac {7 a^{2} c^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{16 f}\) | \(198\) |
| parallelrisch | \(-\frac {2 a^{2} \left (\frac {7 \left (5+\frac {\cos \left (6 f x +6 e \right )}{2}+3 \cos \left (4 f x +4 e \right )+\frac {15 \cos \left (2 f x +2 e \right )}{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{16}+\frac {7 \left (-5-\frac {15 \cos \left (2 f x +2 e \right )}{2}-3 \cos \left (4 f x +4 e \right )-\frac {\cos \left (6 f x +6 e \right )}{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{16}+\sin \left (2 f x +2 e \right )+\frac {89 \sin \left (3 f x +3 e \right )}{48}-\frac {4 \sin \left (4 f x +4 e \right )}{5}+\frac {9 \sin \left (5 f x +5 e \right )}{16}+\frac {\sin \left (6 f x +6 e \right )}{5}-\frac {11 \sin \left (f x +e \right )}{8}\right ) c^{4}}{f \left (10+\cos \left (6 f x +6 e \right )+6 \cos \left (4 f x +4 e \right )+15 \cos \left (2 f x +2 e \right )\right )}\) | \(207\) |
| derivativedivides | \(\frac {a^{2} c^{4} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-2 a^{2} c^{4} \tan \left (f x +e \right )-a^{2} c^{4} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-4 a^{2} c^{4} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )-a^{2} c^{4} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a^{2} c^{4} \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )+a^{2} c^{4} \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}\) | \(255\) |
| default | \(\frac {a^{2} c^{4} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-2 a^{2} c^{4} \tan \left (f x +e \right )-a^{2} c^{4} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-4 a^{2} c^{4} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )-a^{2} c^{4} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a^{2} c^{4} \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )+a^{2} c^{4} \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}\) | \(255\) |
| parts | \(\frac {a^{2} c^{4} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {a^{2} c^{4} \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}-\frac {2 a^{2} c^{4} \tan \left (f x +e \right )}{f}-\frac {a^{2} c^{4} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}-\frac {4 a^{2} c^{4} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )}{f}-\frac {a^{2} c^{4} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}+\frac {2 a^{2} c^{4} \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) | \(272\) |
Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x,method=_RETURNVERBO SE)
Output:
(-7/8*a^2*c^4/f*tan(1/2*f*x+1/2*e)+119/24*a^2*c^4/f*tan(1/2*f*x+1/2*e)^3-2 31/20*a^2*c^4/f*tan(1/2*f*x+1/2*e)^5+281/20*a^2*c^4/f*tan(1/2*f*x+1/2*e)^7 +119/24*a^2*c^4/f*tan(1/2*f*x+1/2*e)^9-7/8*a^2*c^4/f*tan(1/2*f*x+1/2*e)^11 )/(tan(1/2*f*x+1/2*e)^2-1)^6-7/16*a^2*c^4/f*ln(tan(1/2*f*x+1/2*e)-1)+7/16* a^2*c^4/f*ln(tan(1/2*f*x+1/2*e)+1)
Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.07 \[ \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx=\frac {105 \, a^{2} c^{4} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \, a^{2} c^{4} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (96 \, a^{2} c^{4} \cos \left (f x + e\right )^{5} + 135 \, a^{2} c^{4} \cos \left (f x + e\right )^{4} - 192 \, a^{2} c^{4} \cos \left (f x + e\right )^{3} + 10 \, a^{2} c^{4} \cos \left (f x + e\right )^{2} + 96 \, a^{2} c^{4} \cos \left (f x + e\right ) - 40 \, a^{2} c^{4}\right )} \sin \left (f x + e\right )}{480 \, f \cos \left (f x + e\right )^{6}} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x, algorithm="f ricas")
Output:
1/480*(105*a^2*c^4*cos(f*x + e)^6*log(sin(f*x + e) + 1) - 105*a^2*c^4*cos( f*x + e)^6*log(-sin(f*x + e) + 1) - 2*(96*a^2*c^4*cos(f*x + e)^5 + 135*a^2 *c^4*cos(f*x + e)^4 - 192*a^2*c^4*cos(f*x + e)^3 + 10*a^2*c^4*cos(f*x + e) ^2 + 96*a^2*c^4*cos(f*x + e) - 40*a^2*c^4)*sin(f*x + e))/(f*cos(f*x + e)^6 )
\[ \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx=a^{2} c^{4} \left (\int \sec {\left (e + f x \right )}\, dx + \int \left (- 2 \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int 4 \sec ^{4}{\left (e + f x \right )}\, dx + \int \left (- \sec ^{5}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 \sec ^{6}{\left (e + f x \right )}\right )\, dx + \int \sec ^{7}{\left (e + f x \right )}\, dx\right ) \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2*(c-c*sec(f*x+e))**4,x)
Output:
a**2*c**4*(Integral(sec(e + f*x), x) + Integral(-2*sec(e + f*x)**2, x) + I ntegral(-sec(e + f*x)**3, x) + Integral(4*sec(e + f*x)**4, x) + Integral(- sec(e + f*x)**5, x) + Integral(-2*sec(e + f*x)**6, x) + Integral(sec(e + f *x)**7, x))
Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (138) = 276\).
Time = 0.04 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.14 \[ \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx=-\frac {64 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{2} c^{4} - 640 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{4} + 5 \, a^{2} c^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 30 \, a^{2} c^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 120 \, a^{2} c^{4} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 480 \, a^{2} c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 960 \, a^{2} c^{4} \tan \left (f x + e\right )}{480 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x, algorithm="m axima")
Output:
-1/480*(64*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*c^ 4 - 640*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c^4 + 5*a^2*c^4*(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)) - 30*a^2*c^4*(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)) - 120*a^2*c^4*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f *x + e) + 1) + log(sin(f*x + e) - 1)) - 480*a^2*c^4*log(sec(f*x + e) + tan (f*x + e)) + 960*a^2*c^4*tan(f*x + e))/f
Time = 0.20 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.19 \[ \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx=\frac {105 \, a^{2} c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 105 \, a^{2} c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{11} - 595 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 1686 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 1386 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 595 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 105 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{6}}}{240 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x, algorithm="g iac")
Output:
1/240*(105*a^2*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 105*a^2*c^4*log(ab s(tan(1/2*f*x + 1/2*e) - 1)) - 2*(105*a^2*c^4*tan(1/2*f*x + 1/2*e)^11 - 59 5*a^2*c^4*tan(1/2*f*x + 1/2*e)^9 - 1686*a^2*c^4*tan(1/2*f*x + 1/2*e)^7 + 1 386*a^2*c^4*tan(1/2*f*x + 1/2*e)^5 - 595*a^2*c^4*tan(1/2*f*x + 1/2*e)^3 + 105*a^2*c^4*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^6)/f
Time = 13.99 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.46 \[ \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx=\frac {-\frac {7\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}}{8}+\frac {119\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{24}+\frac {281\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{20}-\frac {231\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{20}+\frac {119\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{24}-\frac {7\,a^2\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {7\,a^2\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{8\,f} \] Input:
int(((a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^4)/cos(e + f*x),x)
Output:
((119*a^2*c^4*tan(e/2 + (f*x)/2)^3)/24 - (231*a^2*c^4*tan(e/2 + (f*x)/2)^5 )/20 + (281*a^2*c^4*tan(e/2 + (f*x)/2)^7)/20 + (119*a^2*c^4*tan(e/2 + (f*x )/2)^9)/24 - (7*a^2*c^4*tan(e/2 + (f*x)/2)^11)/8 - (7*a^2*c^4*tan(e/2 + (f *x)/2))/8)/(f*(15*tan(e/2 + (f*x)/2)^4 - 6*tan(e/2 + (f*x)/2)^2 - 20*tan(e /2 + (f*x)/2)^6 + 15*tan(e/2 + (f*x)/2)^8 - 6*tan(e/2 + (f*x)/2)^10 + tan( e/2 + (f*x)/2)^12 + 1)) + (7*a^2*c^4*atanh(tan(e/2 + (f*x)/2)))/(8*f)
Time = 0.16 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.65 \[ \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx=\frac {a^{2} c^{4} \left (96 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5}-105 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{6}+315 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{4}-315 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{2}+105 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+105 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{6}-315 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{4}+315 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{2}-105 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+135 \sin \left (f x +e \right )^{5}-280 \sin \left (f x +e \right )^{3}+105 \sin \left (f x +e \right )\right )}{240 f \left (\sin \left (f x +e \right )^{6}-3 \sin \left (f x +e \right )^{4}+3 \sin \left (f x +e \right )^{2}-1\right )} \] Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x)
Output:
(a**2*c**4*(96*cos(e + f*x)*sin(e + f*x)**5 - 105*log(tan((e + f*x)/2) - 1 )*sin(e + f*x)**6 + 315*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**4 - 315*lo g(tan((e + f*x)/2) - 1)*sin(e + f*x)**2 + 105*log(tan((e + f*x)/2) - 1) + 105*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**6 - 315*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**4 + 315*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**2 - 105* log(tan((e + f*x)/2) + 1) + 135*sin(e + f*x)**5 - 280*sin(e + f*x)**3 + 10 5*sin(e + f*x)))/(240*f*(sin(e + f*x)**6 - 3*sin(e + f*x)**4 + 3*sin(e + f *x)**2 - 1))