3.1.3 \(\int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx\) [3]

3.1.3.1 Optimal result
3.1.3.2 Mathematica [A] (verified)
3.1.3.3 Rubi [A] (verified)
3.1.3.4 Maple [A] (verified)
3.1.3.5 Fricas [A] (verification not implemented)
3.1.3.6 Sympy [F]
3.1.3.7 Maxima [B] (verification not implemented)
3.1.3.8 Giac [A] (verification not implemented)
3.1.3.9 Mupad [B] (verification not implemented)

3.1.3.1 Optimal result

Integrand size = 26, antiderivative size = 97 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx=a^2 c^3 x-\frac {3 a^2 c^3 \text {arctanh}(\sin (e+f x))}{8 f}-\frac {a^2 \left (8 c^3-3 c^3 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {a^2 \left (4 c^3-3 c^3 \sec (e+f x)\right ) \tan ^3(e+f x)}{12 f} \]

output
a^2*c^3*x-3/8*a^2*c^3*arctanh(sin(f*x+e))/f-1/8*a^2*(8*c^3-3*c^3*sec(f*x+e 
))*tan(f*x+e)/f+1/12*a^2*(4*c^3-3*c^3*sec(f*x+e))*tan(f*x+e)^3/f
 
3.1.3.2 Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx=\frac {a^2 c^3 \sec ^4(e+f x) \left (72 e+72 f x-72 \text {arctanh}(\sin (e+f x)) \cos ^4(e+f x)+96 (e+f x) \cos (2 (e+f x))+24 e \cos (4 (e+f x))+24 f x \cos (4 (e+f x))-18 \sin (e+f x)-32 \sin (2 (e+f x))+30 \sin (3 (e+f x))-32 \sin (4 (e+f x))\right )}{192 f} \]

input
Integrate[(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3,x]
 
output
(a^2*c^3*Sec[e + f*x]^4*(72*e + 72*f*x - 72*ArcTanh[Sin[e + f*x]]*Cos[e + 
f*x]^4 + 96*(e + f*x)*Cos[2*(e + f*x)] + 24*e*Cos[4*(e + f*x)] + 24*f*x*Co 
s[4*(e + f*x)] - 18*Sin[e + f*x] - 32*Sin[2*(e + f*x)] + 30*Sin[3*(e + f*x 
)] - 32*Sin[4*(e + f*x)]))/(192*f)
 
3.1.3.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 4392, 3042, 4369, 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.

\(\displaystyle \int (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^3 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4392

\(\displaystyle a^2 c^2 \int (c-c \sec (e+f x)) \tan ^4(e+f x)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4369

\(\displaystyle a^2 c^2 \left (\frac {\tan ^3(e+f x) (4 c-3 c \sec (e+f x))}{12 f}-\frac {1}{4} \int (4 c-3 c \sec (e+f x)) \tan ^2(e+f x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {\tan ^3(e+f x) (4 c-3 c \sec (e+f x))}{12 f}-\frac {1}{4} \int \cot \left (e+f x+\frac {\pi }{2}\right )^2 \left (4 c-3 c \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx\right )\)

\(\Big \downarrow \) 4369

\(\displaystyle a^2 c^2 \left (\frac {1}{4} \left (\frac {1}{2} \int (8 c-3 c \sec (e+f x))dx-\frac {\tan (e+f x) (8 c-3 c \sec (e+f x))}{2 f}\right )+\frac {\tan ^3(e+f x) (4 c-3 c \sec (e+f x))}{12 f}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle a^2 c^2 \left (\frac {1}{4} \left (\frac {1}{2} \left (8 c x-\frac {3 c \text {arctanh}(\sin (e+f x))}{f}\right )-\frac {\tan (e+f x) (8 c-3 c \sec (e+f x))}{2 f}\right )+\frac {\tan ^3(e+f x) (4 c-3 c \sec (e+f x))}{12 f}\right )\)

input
Int[(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3,x]
 
output
a^2*c^2*(((4*c - 3*c*Sec[e + f*x])*Tan[e + f*x]^3)/(12*f) + ((8*c*x - (3*c 
*ArcTanh[Sin[e + f*x]])/f)/2 - ((8*c - 3*c*Sec[e + f*x])*Tan[e + f*x])/(2* 
f))/4)
 

3.1.3.3.1 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 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])
 
3.1.3.4 Maple [A] (verified)

Time = 3.32 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.37

method result size
parts \(a^{2} c^{3} x -\frac {a^{2} c^{3} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )}{f}-\frac {2 a^{2} c^{3} \tan \left (f x +e \right )}{f}+\frac {a^{2} c^{3} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{f}-\frac {a^{2} c^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}\) \(133\)
risch \(a^{2} c^{3} x -\frac {i a^{2} c^{3} \left (15 \,{\mathrm e}^{7 i \left (f x +e \right )}+48 \,{\mathrm e}^{6 i \left (f x +e \right )}-9 \,{\mathrm e}^{5 i \left (f x +e \right )}+96 \,{\mathrm e}^{4 i \left (f x +e \right )}+9 \,{\mathrm e}^{3 i \left (f x +e \right )}+80 \,{\mathrm e}^{2 i \left (f x +e \right )}-15 \,{\mathrm e}^{i \left (f x +e \right )}+32\right )}{12 f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{4}}+\frac {3 a^{2} c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{8 f}-\frac {3 a^{2} c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{8 f}\) \(162\)
derivativedivides \(\frac {-a^{2} c^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )-a^{2} c^{3} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )+2 a^{2} c^{3} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-2 a^{2} c^{3} \tan \left (f x +e \right )-a^{2} c^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a^{2} c^{3} \left (f x +e \right )}{f}\) \(171\)
default \(\frac {-a^{2} c^{3} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )-a^{2} c^{3} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )+2 a^{2} c^{3} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-2 a^{2} c^{3} \tan \left (f x +e \right )-a^{2} c^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a^{2} c^{3} \left (f x +e \right )}{f}\) \(171\)
parallelrisch \(\frac {a^{2} c^{3} \left (9 \left (3+\cos \left (4 f x +4 e \right )+4 \cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+9 \left (-\cos \left (4 f x +4 e \right )-4 \cos \left (2 f x +2 e \right )-3\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+96 f x \cos \left (2 f x +2 e \right )+24 f x \cos \left (4 f x +4 e \right )+72 f x +30 \sin \left (3 f x +3 e \right )-32 \sin \left (4 f x +4 e \right )-18 \sin \left (f x +e \right )-32 \sin \left (2 f x +2 e \right )\right )}{24 f \left (3+\cos \left (4 f x +4 e \right )+4 \cos \left (2 f x +2 e \right )\right )}\) \(182\)
norman \(\frac {a^{2} c^{3} x +a^{2} c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-\frac {5 a^{2} c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {71 a^{2} c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{12 f}-\frac {137 a^{2} c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{12 f}+\frac {11 a^{2} c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{4 f}-4 a^{2} c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+6 a^{2} c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-4 a^{2} c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{4}}+\frac {3 a^{2} c^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{8 f}-\frac {3 a^{2} c^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{8 f}\) \(238\)

input
int((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^3,x,method=_RETURNVERBOSE)
 
output
a^2*c^3*x-a^2*c^3/f*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)-2*a^2*c^3/f*tan(f*x 
+e)+a^2*c^3/f*sec(f*x+e)*tan(f*x+e)-a^2*c^3/f*(-(-1/4*sec(f*x+e)^3-3/8*sec 
(f*x+e))*tan(f*x+e)+3/8*ln(sec(f*x+e)+tan(f*x+e)))
 
3.1.3.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.52 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx=\frac {48 \, a^{2} c^{3} f x \cos \left (f x + e\right )^{4} - 9 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) + 9 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (32 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 15 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} - 8 \, a^{2} c^{3} \cos \left (f x + e\right ) + 6 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \]

input
integrate((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^3,x, algorithm="fricas")
 
output
1/48*(48*a^2*c^3*f*x*cos(f*x + e)^4 - 9*a^2*c^3*cos(f*x + e)^4*log(sin(f*x 
 + e) + 1) + 9*a^2*c^3*cos(f*x + e)^4*log(-sin(f*x + e) + 1) - 2*(32*a^2*c 
^3*cos(f*x + e)^3 - 15*a^2*c^3*cos(f*x + e)^2 - 8*a^2*c^3*cos(f*x + e) + 6 
*a^2*c^3)*sin(f*x + e))/(f*cos(f*x + e)^4)
 
3.1.3.6 Sympy [F]

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

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

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (91) = 182\).

Time = 0.21 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.09 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx=\frac {16 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{3} + 48 \, {\left (f x + e\right )} a^{2} c^{3} + 3 \, a^{2} c^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 24 \, a^{2} c^{3} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 48 \, a^{2} c^{3} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 96 \, a^{2} c^{3} \tan \left (f x + e\right )}{48 \, f} \]

input
integrate((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^3,x, algorithm="maxima")
 
output
1/48*(16*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c^3 + 48*(f*x + e)*a^2*c^3 
+ 3*a^2*c^3*(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)) - 24 
*a^2*c^3*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + lo 
g(sin(f*x + e) - 1)) - 48*a^2*c^3*log(sec(f*x + e) + tan(f*x + e)) - 96*a^ 
2*c^3*tan(f*x + e))/f
 
3.1.3.8 Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.58 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx=\frac {24 \, {\left (f x + e\right )} a^{2} c^{3} - 9 \, a^{2} c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) + 9 \, a^{2} c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) + \frac {2 \, {\left (33 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 137 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 71 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 15 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{4}}}{24 \, f} \]

input
integrate((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^3,x, algorithm="giac")
 
output
1/24*(24*(f*x + e)*a^2*c^3 - 9*a^2*c^3*log(abs(tan(1/2*f*x + 1/2*e) + 1)) 
+ 9*a^2*c^3*log(abs(tan(1/2*f*x + 1/2*e) - 1)) + 2*(33*a^2*c^3*tan(1/2*f*x 
 + 1/2*e)^7 - 137*a^2*c^3*tan(1/2*f*x + 1/2*e)^5 + 71*a^2*c^3*tan(1/2*f*x 
+ 1/2*e)^3 - 15*a^2*c^3*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1) 
^4)/f
 
3.1.3.9 Mupad [B] (verification not implemented)

Time = 15.22 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.68 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx=\frac {\frac {11\,a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}-\frac {137\,a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{12}+\frac {71\,a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12}-\frac {5\,a^2\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+a^2\,c^3\,x-\frac {3\,a^2\,c^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{4\,f} \]

input
int((a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^3,x)
 
output
((71*a^2*c^3*tan(e/2 + (f*x)/2)^3)/12 - (137*a^2*c^3*tan(e/2 + (f*x)/2)^5) 
/12 + (11*a^2*c^3*tan(e/2 + (f*x)/2)^7)/4 - (5*a^2*c^3*tan(e/2 + (f*x)/2)) 
/4)/(f*(6*tan(e/2 + (f*x)/2)^4 - 4*tan(e/2 + (f*x)/2)^2 - 4*tan(e/2 + (f*x 
)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1)) + a^2*c^3*x - (3*a^2*c^3*atanh(tan(e/2 
 + (f*x)/2)))/(4*f)