3.1.12 \(\int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx\) [12]

3.1.12.1 Optimal result
3.1.12.2 Mathematica [A] (verified)
3.1.12.3 Rubi [A] (verified)
3.1.12.4 Maple [C] (verified)
3.1.12.5 Fricas [A] (verification not implemented)
3.1.12.6 Sympy [F]
3.1.12.7 Maxima [B] (verification not implemented)
3.1.12.8 Giac [A] (verification not implemented)
3.1.12.9 Mupad [B] (verification not implemented)

3.1.12.1 Optimal result

Integrand size = 26, antiderivative size = 132 \[ \int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx=a^3 c^4 x-\frac {5 a^3 c^4 \text {arctanh}(\sin (e+f x))}{16 f}-\frac {a^3 \left (16 c^4-5 c^4 \sec (e+f x)\right ) \tan (e+f x)}{16 f}+\frac {a^3 \left (8 c^4-5 c^4 \sec (e+f x)\right ) \tan ^3(e+f x)}{24 f}-\frac {a^3 \left (6 c^4-5 c^4 \sec (e+f x)\right ) \tan ^5(e+f x)}{30 f} \]

output
a^3*c^4*x-5/16*a^3*c^4*arctanh(sin(f*x+e))/f-1/16*a^3*(16*c^4-5*c^4*sec(f* 
x+e))*tan(f*x+e)/f+1/24*a^3*(8*c^4-5*c^4*sec(f*x+e))*tan(f*x+e)^3/f-1/30*a 
^3*(6*c^4-5*c^4*sec(f*x+e))*tan(f*x+e)^5/f
 
3.1.12.2 Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.25 \[ \int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx=\frac {a^3 c^4 \sec ^6(e+f x) \left (1200 e+1200 f x-1200 \text {arctanh}(\sin (e+f x)) \cos ^6(e+f x)+1800 (e+f x) \cos (2 (e+f x))+720 e \cos (4 (e+f x))+720 f x \cos (4 (e+f x))+120 e \cos (6 (e+f x))+120 f x \cos (6 (e+f x))+450 \sin (e+f x)-600 \sin (2 (e+f x))-25 \sin (3 (e+f x))-384 \sin (4 (e+f x))+165 \sin (5 (e+f x))-184 \sin (6 (e+f x))\right )}{3840 f} \]

input
Integrate[(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4,x]
 
output
(a^3*c^4*Sec[e + f*x]^6*(1200*e + 1200*f*x - 1200*ArcTanh[Sin[e + f*x]]*Co 
s[e + f*x]^6 + 1800*(e + f*x)*Cos[2*(e + f*x)] + 720*e*Cos[4*(e + f*x)] + 
720*f*x*Cos[4*(e + f*x)] + 120*e*Cos[6*(e + f*x)] + 120*f*x*Cos[6*(e + f*x 
)] + 450*Sin[e + f*x] - 600*Sin[2*(e + f*x)] - 25*Sin[3*(e + f*x)] - 384*S 
in[4*(e + f*x)] + 165*Sin[5*(e + f*x)] - 184*Sin[6*(e + f*x)]))/(3840*f)
 
3.1.12.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4392, 3042, 4369, 3042, 4369, 27, 3042, 4369, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4392

\(\displaystyle -a^3 c^3 \int (c-c \sec (e+f x)) \tan ^6(e+f x)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4369

\(\displaystyle -a^3 c^3 \left (\frac {\tan ^5(e+f x) (6 c-5 c \sec (e+f x))}{30 f}-\frac {1}{6} \int (6 c-5 c \sec (e+f x)) \tan ^4(e+f x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -a^3 c^3 \left (\frac {\tan ^5(e+f x) (6 c-5 c \sec (e+f x))}{30 f}-\frac {1}{6} \int \cot \left (e+f x+\frac {\pi }{2}\right )^4 \left (6 c-5 c \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx\right )\)

\(\Big \downarrow \) 4369

\(\displaystyle -a^3 c^3 \left (\frac {1}{6} \left (\frac {1}{4} \int 3 (8 c-5 c \sec (e+f x)) \tan ^2(e+f x)dx-\frac {\tan ^3(e+f x) (8 c-5 c \sec (e+f x))}{4 f}\right )+\frac {\tan ^5(e+f x) (6 c-5 c \sec (e+f x))}{30 f}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -a^3 c^3 \left (\frac {1}{6} \left (\frac {3}{4} \int (8 c-5 c \sec (e+f x)) \tan ^2(e+f x)dx-\frac {\tan ^3(e+f x) (8 c-5 c \sec (e+f x))}{4 f}\right )+\frac {\tan ^5(e+f x) (6 c-5 c \sec (e+f x))}{30 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -a^3 c^3 \left (\frac {1}{6} \left (\frac {3}{4} \int \cot \left (e+f x+\frac {\pi }{2}\right )^2 \left (8 c-5 c \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx-\frac {\tan ^3(e+f x) (8 c-5 c \sec (e+f x))}{4 f}\right )+\frac {\tan ^5(e+f x) (6 c-5 c \sec (e+f x))}{30 f}\right )\)

\(\Big \downarrow \) 4369

\(\displaystyle -a^3 c^3 \left (\frac {1}{6} \left (\frac {3}{4} \left (\frac {\tan (e+f x) (16 c-5 c \sec (e+f x))}{2 f}-\frac {1}{2} \int (16 c-5 c \sec (e+f x))dx\right )-\frac {\tan ^3(e+f x) (8 c-5 c \sec (e+f x))}{4 f}\right )+\frac {\tan ^5(e+f x) (6 c-5 c \sec (e+f x))}{30 f}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle -a^3 c^3 \left (\frac {1}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {5 c \text {arctanh}(\sin (e+f x))}{f}-16 c x\right )+\frac {\tan (e+f x) (16 c-5 c \sec (e+f x))}{2 f}\right )-\frac {\tan ^3(e+f x) (8 c-5 c \sec (e+f x))}{4 f}\right )+\frac {\tan ^5(e+f x) (6 c-5 c \sec (e+f x))}{30 f}\right )\)

input
Int[(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4,x]
 
output
-(a^3*c^3*(((6*c - 5*c*Sec[e + f*x])*Tan[e + f*x]^5)/(30*f) + (-1/4*((8*c 
- 5*c*Sec[e + f*x])*Tan[e + f*x]^3)/f + (3*((-16*c*x + (5*c*ArcTanh[Sin[e 
+ f*x]])/f)/2 + ((16*c - 5*c*Sec[e + f*x])*Tan[e + f*x])/(2*f)))/4)/6))
 

3.1.12.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 4369
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[(-e)*(e*Cot[c + d*x])^(m - 1)*((a*m + b*(m - 1)*Csc 
[c + d*x])/(d*m*(m - 1))), x] - Simp[e^2/m   Int[(e*Cot[c + d*x])^(m - 2)*( 
a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m 
, 1]
 

rule 4392
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( 
d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m   Int[Cot[e + f*x]^(2*m)*( 
c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E 
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !( 
IntegerQ[n] && GtQ[m - n, 0])
 
3.1.12.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.83 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.56

method result size
risch \(a^{3} c^{4} x -\frac {i c^{4} a^{3} \left (165 \,{\mathrm e}^{11 i \left (f x +e \right )}+720 \,{\mathrm e}^{10 i \left (f x +e \right )}-25 \,{\mathrm e}^{9 i \left (f x +e \right )}+2160 \,{\mathrm e}^{8 i \left (f x +e \right )}+450 \,{\mathrm e}^{7 i \left (f x +e \right )}+3680 \,{\mathrm e}^{6 i \left (f x +e \right )}-450 \,{\mathrm e}^{5 i \left (f x +e \right )}+3360 \,{\mathrm e}^{4 i \left (f x +e \right )}+25 \,{\mathrm e}^{3 i \left (f x +e \right )}+1488 \,{\mathrm e}^{2 i \left (f x +e \right )}-165 \,{\mathrm e}^{i \left (f x +e \right )}+368\right )}{120 f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{6}}+\frac {5 c^{4} a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{16 f}-\frac {5 c^{4} a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{16 f}\) \(206\)
parallelrisch \(-\frac {5 a^{3} \left (\frac {\left (-5-\frac {15 \cos \left (2 f x +2 e \right )}{2}-3 \cos \left (4 f x +4 e \right )-\frac {\cos \left (6 f x +6 e \right )}{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{8}+\frac {\left (5+\frac {\cos \left (6 f x +6 e \right )}{2}+3 \cos \left (4 f x +4 e \right )+\frac {15 \cos \left (2 f x +2 e \right )}{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{8}-3 f x \cos \left (2 f x +2 e \right )-\frac {6 f x \cos \left (4 f x +4 e \right )}{5}-\frac {f x \cos \left (6 f x +6 e \right )}{5}-2 f x +\sin \left (2 f x +2 e \right )+\frac {\sin \left (3 f x +3 e \right )}{24}+\frac {16 \sin \left (4 f x +4 e \right )}{25}-\frac {11 \sin \left (5 f x +5 e \right )}{40}+\frac {23 \sin \left (6 f x +6 e \right )}{75}-\frac {3 \sin \left (f x +e \right )}{4}\right ) c^{4}}{f \left (6 \cos \left (4 f x +4 e \right )+10+15 \cos \left (2 f x +2 e \right )+\cos \left (6 f x +6 e \right )\right )}\) \(250\)
derivativedivides \(\frac {c^{4} a^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )+c^{4} a^{3} \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )-3 c^{4} a^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )-3 c^{4} a^{3} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )+3 c^{4} a^{3} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-3 c^{4} a^{3} \tan \left (f x +e \right )-c^{4} a^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+c^{4} a^{3} \left (f x +e \right )}{f}\) \(267\)
default \(\frac {c^{4} a^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )+c^{4} a^{3} \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )-3 c^{4} a^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )-3 c^{4} a^{3} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )+3 c^{4} a^{3} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-3 c^{4} a^{3} \tan \left (f x +e \right )-c^{4} a^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+c^{4} a^{3} \left (f x +e \right )}{f}\) \(267\)
parts \(a^{3} c^{4} x +\frac {c^{4} a^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}-\frac {c^{4} a^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}-\frac {3 c^{4} a^{3} \tan \left (f x +e \right )}{f}+\frac {3 c^{4} a^{3} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}-\frac {3 c^{4} a^{3} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )}{f}-\frac {3 c^{4} a^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}+\frac {c^{4} a^{3} \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) \(280\)
norman \(\frac {a^{3} c^{4} x +a^{3} c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}-6 a^{3} c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+15 a^{3} c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-20 a^{3} c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+15 a^{3} c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-6 a^{3} c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}-\frac {11 c^{4} a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{8 f}+\frac {73 c^{4} a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{8 f}-\frac {523 c^{4} a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{20 f}+\frac {853 c^{4} a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{20 f}-\frac {389 c^{4} a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{24 f}+\frac {21 c^{4} a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{8 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{6}}+\frac {5 c^{4} a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{16 f}-\frac {5 c^{4} a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{16 f}\) \(322\)

input
int((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^4,x,method=_RETURNVERBOSE)
 
output
a^3*c^4*x-1/120*I*c^4*a^3*(165*exp(11*I*(f*x+e))+720*exp(10*I*(f*x+e))-25* 
exp(9*I*(f*x+e))+2160*exp(8*I*(f*x+e))+450*exp(7*I*(f*x+e))+3680*exp(6*I*( 
f*x+e))-450*exp(5*I*(f*x+e))+3360*exp(4*I*(f*x+e))+25*exp(3*I*(f*x+e))+148 
8*exp(2*I*(f*x+e))-165*exp(I*(f*x+e))+368)/f/(1+exp(2*I*(f*x+e)))^6+5/16*c 
^4*a^3/f*ln(exp(I*(f*x+e))-I)-5/16*c^4*a^3/f*ln(exp(I*(f*x+e))+I)
 
3.1.12.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.36 \[ \int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx=\frac {480 \, a^{3} c^{4} f x \cos \left (f x + e\right )^{6} - 75 \, a^{3} c^{4} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) + 75 \, a^{3} c^{4} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (368 \, a^{3} c^{4} \cos \left (f x + e\right )^{5} - 165 \, a^{3} c^{4} \cos \left (f x + e\right )^{4} - 176 \, a^{3} c^{4} \cos \left (f x + e\right )^{3} + 130 \, a^{3} c^{4} \cos \left (f x + e\right )^{2} + 48 \, a^{3} c^{4} \cos \left (f x + e\right ) - 40 \, a^{3} c^{4}\right )} \sin \left (f x + e\right )}{480 \, f \cos \left (f x + e\right )^{6}} \]

input
integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^4,x, algorithm="fricas")
 
output
1/480*(480*a^3*c^4*f*x*cos(f*x + e)^6 - 75*a^3*c^4*cos(f*x + e)^6*log(sin( 
f*x + e) + 1) + 75*a^3*c^4*cos(f*x + e)^6*log(-sin(f*x + e) + 1) - 2*(368* 
a^3*c^4*cos(f*x + e)^5 - 165*a^3*c^4*cos(f*x + e)^4 - 176*a^3*c^4*cos(f*x 
+ e)^3 + 130*a^3*c^4*cos(f*x + e)^2 + 48*a^3*c^4*cos(f*x + e) - 40*a^3*c^4 
)*sin(f*x + e))/(f*cos(f*x + e)^6)
 
3.1.12.6 Sympy [F]

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

input
integrate((a+a*sec(f*x+e))**3*(c-c*sec(f*x+e))**4,x)
 
output
a**3*c**4*(Integral(1, x) + Integral(-sec(e + f*x), x) + Integral(-3*sec(e 
 + f*x)**2, x) + Integral(3*sec(e + f*x)**3, x) + Integral(3*sec(e + f*x)* 
*4, x) + Integral(-3*sec(e + f*x)**5, x) + Integral(-sec(e + f*x)**6, x) + 
 Integral(sec(e + f*x)**7, x))
 
3.1.12.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (124) = 248\).

Time = 0.20 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.53 \[ \int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx=-\frac {32 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{3} c^{4} - 480 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c^{4} - 480 \, {\left (f x + e\right )} a^{3} c^{4} + 5 \, a^{3} c^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 90 \, a^{3} c^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 360 \, a^{3} c^{4} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 480 \, a^{3} c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 1440 \, a^{3} c^{4} \tan \left (f x + e\right )}{480 \, f} \]

input
integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^4,x, algorithm="maxima")
 
output
-1/480*(32*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^3*c^ 
4 - 480*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^3*c^4 - 480*(f*x + e)*a^3*c^4 
+ 5*a^3*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)) - 90*a^3*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)) + 360*a^3*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^3*c^ 
4*log(sec(f*x + e) + tan(f*x + e)) + 1440*a^3*c^4*tan(f*x + e))/f
 
3.1.12.8 Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.45 \[ \int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx=\frac {240 \, {\left (f x + e\right )} a^{3} c^{4} - 75 \, a^{3} c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) + 75 \, a^{3} c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) + \frac {2 \, {\left (315 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{11} - 1945 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 5118 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 3138 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 1095 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 165 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{6}}}{240 \, f} \]

input
integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^4,x, algorithm="giac")
 
output
1/240*(240*(f*x + e)*a^3*c^4 - 75*a^3*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1 
)) + 75*a^3*c^4*log(abs(tan(1/2*f*x + 1/2*e) - 1)) + 2*(315*a^3*c^4*tan(1/ 
2*f*x + 1/2*e)^11 - 1945*a^3*c^4*tan(1/2*f*x + 1/2*e)^9 + 5118*a^3*c^4*tan 
(1/2*f*x + 1/2*e)^7 - 3138*a^3*c^4*tan(1/2*f*x + 1/2*e)^5 + 1095*a^3*c^4*t 
an(1/2*f*x + 1/2*e)^3 - 165*a^3*c^4*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1 
/2*e)^2 - 1)^6)/f
 
3.1.12.9 Mupad [B] (verification not implemented)

Time = 15.34 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.72 \[ \int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx=a^3\,c^4\,x+\frac {\frac {21\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}}{8}-\frac {389\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{24}+\frac {853\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{20}-\frac {523\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{20}+\frac {73\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{8}-\frac {11\,a^3\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {5\,a^3\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{8\,f} \]

input
int((a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^4,x)
 
output
a^3*c^4*x + ((73*a^3*c^4*tan(e/2 + (f*x)/2)^3)/8 - (523*a^3*c^4*tan(e/2 + 
(f*x)/2)^5)/20 + (853*a^3*c^4*tan(e/2 + (f*x)/2)^7)/20 - (389*a^3*c^4*tan( 
e/2 + (f*x)/2)^9)/24 + (21*a^3*c^4*tan(e/2 + (f*x)/2)^11)/8 - (11*a^3*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)) - (5*a^3*c^4*atanh(tan(e/2 + (f*x)/2)) 
)/(8*f)