\(\int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx\) [66]

Optimal result

Integrand size = 32, antiderivative size = 81 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx=-\frac {8 c^2 (a+a \sec (e+f x)) \tan (e+f x)}{15 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c (a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{5 f} \] Output:

-8/15*c^2*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)-2/5*c*(a+a* 
sec(f*x+e))*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx=\frac {2 a c^2 (1+\sec (e+f x)) (-7+3 \sec (e+f x)) \tan (e+f x)}{15 f \sqrt {c-c \sec (e+f x)}} \] Input:

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2),x]
 

Output:

(2*a*c^2*(1 + Sec[e + f*x])*(-7 + 3*Sec[e + f*x])*Tan[e + f*x])/(15*f*Sqrt 
[c - c*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3042, 4443, 3042, 4441}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4443

\(\displaystyle \frac {4}{5} c \int \sec (e+f x) (\sec (e+f x) a+a) \sqrt {c-c \sec (e+f x)}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right ) \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\)

\(\Big \downarrow \) 4441

\(\displaystyle -\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)}{15 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\)

Input:

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2),x]
 

Output:

(-8*c^2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(15*f*Sqrt[c - c*Sec[e + f*x]]) 
 - (2*c*(a + a*Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(5*f)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4441

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*Sq 
rt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Simp[2*a*c*Cot[e + f 
*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1)*Sqrt[c + d*Csc[e + f*x]])), x] / 
; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] 
 && NeQ[m, -2^(-1)]
 

rule 4443

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c 
sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[(-d)*Cot[e + f 
*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + 
 Simp[c*((2*n - 1)/(m + n))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(c + 
d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b 
*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0] &&  !LtQ[m, -2^(-1)] 
&&  !(IGtQ[m - 1/2, 0] && LtQ[m, n])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(182\) vs. \(2(73)=146\).

Time = 2.08 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.26

method	result	size
default	\(a \left (-\frac {4 \left (12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-15 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, \sqrt {2}\, c \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2}}+\frac {4 \left (-3+5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, \sqrt {2}\, c \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\right )\)	\(183\)
parts	\(\frac {4 a \left (-3+5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, \sqrt {2}\, c \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}-\frac {4 a \left (12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-15 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, \sqrt {2}\, c \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2}}\)	\(183\)

Input:

int(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(3/2),x,method=_RETURNVER 
BOSE)
 

Output:

a*(-4/5/f*(12*cos(1/2*f*x+1/2*e)^4-15*cos(1/2*f*x+1/2*e)^2+5)*(-c/(2*cos(1 
/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*2^(1/2)*c/(2*cos(1/2*f*x+1/ 
2*e)^2-1)^2*cot(1/2*f*x+1/2*e)+4/3/f*(-3+5*cos(1/2*f*x+1/2*e)^2)*(-c/(2*co 
s(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*2^(1/2)*c/(2*cos(1/2*f*x 
+1/2*e)^2-1)*cot(1/2*f*x+1/2*e))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx=\frac {2 \, {\left (7 \, a c \cos \left (f x + e\right )^{3} + 11 \, a c \cos \left (f x + e\right )^{2} + a c \cos \left (f x + e\right ) - 3 \, a c\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )} \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(3/2),x, algorithm= 
"fricas")
 

Output:

2/15*(7*a*c*cos(f*x + e)^3 + 11*a*c*cos(f*x + e)^2 + a*c*cos(f*x + e) - 3* 
a*c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^2*sin(f*x + e 
))
 

Sympy [F]

\[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx=a \left (\int c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )}\, dx + \int \left (- c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{3}{\left (e + f x \right )}\right )\, dx\right ) \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))**(3/2),x)
 

Output:

a*(Integral(c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x), x) + Integral(-c*sqr 
t(-c*sec(e + f*x) + c)*sec(e + f*x)**3, x))
 

Maxima [F]

\[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )} {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \sec \left (f x + e\right ) \,d x } \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(3/2),x, algorithm= 
"maxima")
 

Output:

-2/15*(15*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 
1)^(1/4)*((a*c*f*cos(2*f*x + 2*e)^2 + a*c*f*sin(2*f*x + 2*e)^2 + 2*a*c*f*c 
os(2*f*x + 2*e) + a*c*f)*integrate((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^ 
2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(((cos(6*f*x + 6*e)*cos(2*f*x + 2*e) + 2 
*cos(4*f*x + 4*e)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + sin(6*f*x + 6*e) 
*sin(2*f*x + 2*e) + 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + sin(2*f*x + 2*e) 
^2)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (cos(2*f*x + 2* 
e)*sin(6*f*x + 6*e) + 2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) - cos(6*f*x + 6* 
e)*sin(2*f*x + 2*e) - 2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e))*sin(5/2*arctan2 
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*cos(3/2*arctan2(sin(2*f*x + 2*e), c 
os(2*f*x + 2*e) + 1)) - ((cos(2*f*x + 2*e)*sin(6*f*x + 6*e) + 2*cos(2*f*x 
+ 2*e)*sin(4*f*x + 4*e) - cos(6*f*x + 6*e)*sin(2*f*x + 2*e) - 2*cos(4*f*x 
+ 4*e)*sin(2*f*x + 2*e))*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e 
))) - (cos(6*f*x + 6*e)*cos(2*f*x + 2*e) + 2*cos(4*f*x + 4*e)*cos(2*f*x + 
2*e) + cos(2*f*x + 2*e)^2 + sin(6*f*x + 6*e)*sin(2*f*x + 2*e) + 2*sin(4*f* 
x + 4*e)*sin(2*f*x + 2*e) + sin(2*f*x + 2*e)^2)*sin(5/2*arctan2(sin(2*f*x 
+ 2*e), cos(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 
2*e) + 1)))/((cos(2*f*x + 2*e)^4 + sin(2*f*x + 2*e)^4 + (cos(2*f*x + 2*e)^ 
2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e)^2 + 4*(c 
os(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(4*...
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.69 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx=\frac {8 \, \sqrt {2} {\left (5 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{3} + 3 \, c^{4}\right )} a}{15 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {5}{2}} f} \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(3/2),x, algorithm= 
"giac")
 

Output:

8/15*sqrt(2)*(5*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^3 + 3*c^4)*a/((c*tan(1/2* 
f*x + 1/2*e)^2 - c)^(5/2)*f)
 

Mupad [B] (verification not implemented)

Time = 14.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.48 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx=-\frac {2\,a\,c\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^3\,\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (7+7\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-6\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\right )}{15\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2} \] Input:

int(((a + a/cos(e + f*x))*(c - c/cos(e + f*x))^(3/2))/cos(e + f*x),x)
 

Output:

-(2*a*c*(exp(e*1i + f*x*1i)*1i + 1i)^3*(c - c/(exp(- e*1i - f*x*1i)/2 + ex 
p(e*1i + f*x*1i)/2))^(1/2)*(7*exp(e*2i + f*x*2i) - 6*exp(e*1i + f*x*1i) + 
7))/(15*f*(exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^2)
 

Reduce [F]

\[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx=\sqrt {c}\, a c \left (-\left (\int \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{3}d x \right )+\int \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )d x \right ) \] Input:

int(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(3/2),x)
 

Output:

sqrt(c)*a*c*( - int(sqrt( - sec(e + f*x) + 1)*sec(e + f*x)**3,x) + int(sqr 
t( - sec(e + f*x) + 1)*sec(e + f*x),x))