\(\int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx\) [122]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 30, antiderivative size = 101 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\cot (e+f x)}{2 a c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\log (\sin (e+f x)) \tan (e+f x)}{a c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

1/2*cot(f*x+e)/a/c/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)+ln(sin(f*x+e))*tan(f*x+e)/a/c/f/(a+a*sec(f*
x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3990, 3554, 3556} \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\cot (e+f x)}{2 a c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x) \log (\sin (e+f x))}{a c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

[In]

Int[1/((a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

Cot[e + f*x]/(2*a*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[Sin[e + f*x]]*Tan[e + f*x])/(a
*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3556

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

Rule 3990

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan (e+f x) \int \cot ^3(e+f x) \, dx}{a c \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ & = \frac {\cot (e+f x)}{2 a c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x) \int \cot (e+f x) \, dx}{a c \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ & = \frac {\cot (e+f x)}{2 a c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\log (\sin (e+f x)) \tan (e+f x)}{a c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\cot (e+f x)+2 (\log (\cos (e+f x))+\log (\tan (e+f x))) \tan (e+f x)}{2 a c f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

Integrate[1/((a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

(Cot[e + f*x] + 2*(Log[Cos[e + f*x]] + Log[Tan[e + f*x]])*Tan[e + f*x])/(2*a*c*f*Sqrt[a*(1 + Sec[e + f*x])]*Sq
rt[c - c*Sec[e + f*x]])

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.55

method result size
default \(-\frac {\left (4 \cos \left (f x +e \right )^{2} \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-4 \cos \left (f x +e \right )^{2} \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-\cos \left (f x +e \right )^{2}-4 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-1\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \tan \left (f x +e \right )}{4 f \,a^{2} \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, c \left (\sec \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )^{2}}\) \(157\)
risch \(-\frac {i {\mathrm e}^{4 i \left (f x +e \right )} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )+{\mathrm e}^{4 i \left (f x +e \right )} f x +2 \,{\mathrm e}^{4 i \left (f x +e \right )} e -2 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )-2 \,{\mathrm e}^{2 i \left (f x +e \right )} f x -4 \,{\mathrm e}^{2 i \left (f x +e \right )} e -2 i {\mathrm e}^{2 i \left (f x +e \right )}+i \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )+f x +2 e}{a c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\) \(242\)

[In]

int(1/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/f/a^2*(4*cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e))-4*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-cos(f*x+e)^2-4*ln(-c
ot(f*x+e)+csc(f*x+e))+4*ln(2/(cos(f*x+e)+1))-1)*(a*(sec(f*x+e)+1))^(1/2)/(-c*(sec(f*x+e)-1))^(1/2)/c/(sec(f*x+
e)-1)/(cos(f*x+e)+1)^2*tan(f*x+e)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (91) = 182\).

Time = 0.49 (sec) , antiderivative size = 492, normalized size of antiderivative = 4.87 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx=\left [-\frac {9 \, \sqrt {-a c} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {8 \, {\left ({\left (256 \, \cos \left (f x + e\right )^{5} - 512 \, \cos \left (f x + e\right )^{3} + 175 \, \cos \left (f x + e\right )\right )} \sqrt {-a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} - {\left (256 \, a c \cos \left (f x + e\right )^{4} - 512 \, a c \cos \left (f x + e\right )^{2} + 337 \, a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + {\left (16 \, \cos \left (f x + e\right )^{3} - 25 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{18 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}, -\frac {18 \, \sqrt {a c} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac {{\left (16 \, \cos \left (f x + e\right )^{3} - 7 \, \cos \left (f x + e\right )\right )} \sqrt {a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{{\left (16 \, a c \cos \left (f x + e\right )^{2} - 25 \, a c\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + {\left (16 \, \cos \left (f x + e\right )^{3} - 25 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{18 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]

[In]

integrate(1/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/18*(9*sqrt(-a*c)*(cos(f*x + e)^2 - 1)*log(-8*((256*cos(f*x + e)^5 - 512*cos(f*x + e)^3 + 175*cos(f*x + e))
*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) - (256*a*c*cos(f*x
 + e)^4 - 512*a*c*cos(f*x + e)^2 + 337*a*c)*sin(f*x + e))/((cos(f*x + e)^2 - 1)*sin(f*x + e)))*sin(f*x + e) +
(16*cos(f*x + e)^3 - 25*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*
x + e)))/((a^2*c^2*f*cos(f*x + e)^2 - a^2*c^2*f)*sin(f*x + e)), -1/18*(18*sqrt(a*c)*(cos(f*x + e)^2 - 1)*arcta
n((16*cos(f*x + e)^3 - 7*cos(f*x + e))*sqrt(a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e)
- c)/cos(f*x + e))/((16*a*c*cos(f*x + e)^2 - 25*a*c)*sin(f*x + e)))*sin(f*x + e) + (16*cos(f*x + e)^3 - 25*cos
(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^2*f*cos(f*
x + e)^2 - a^2*c^2*f)*sin(f*x + e))]

Sympy [F]

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

[In]

integrate(1/(a+a*sec(f*x+e))**(3/2)/(c-c*sec(f*x+e))**(3/2),x)

[Out]

Integral(1/((a*(sec(e + f*x) + 1))**(3/2)*(-c*(sec(e + f*x) - 1))**(3/2)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 486 vs. \(2 (91) = 182\).

Time = 0.38 (sec) , antiderivative size = 486, normalized size of antiderivative = 4.81 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx=-\frac {{\left ({\left (f x + e\right )} \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + {\left (f x + e\right )} \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + f x + {\left (2 \, {\left (2 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \cos \left (4 \, f x + 4 \, e\right ) - \cos \left (4 \, f x + 4 \, e\right )^{2} - 4 \, \cos \left (2 \, f x + 2 \, e\right )^{2} - \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) - 4 \, \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) - 1\right ) + 2 \, {\left (f x - 2 \, {\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right ) + e + \sin \left (2 \, f x + 2 \, e\right )\right )} \cos \left (4 \, f x + 4 \, e\right ) - 4 \, {\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right ) - 2 \, {\left (2 \, {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) + \cos \left (2 \, f x + 2 \, e\right )\right )} \sin \left (4 \, f x + 4 \, e\right ) + e + 2 \, \sin \left (2 \, f x + 2 \, e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (a^{2} c^{2} \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + a^{2} c^{2} \sin \left (4 \, f x + 4 \, e\right )^{2} - 4 \, a^{2} c^{2} \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, a^{2} c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} - 4 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right ) + a^{2} c^{2} - 2 \, {\left (2 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right ) - a^{2} c^{2}\right )} \cos \left (4 \, f x + 4 \, e\right )\right )} f} \]

[In]

integrate(1/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-((f*x + e)*cos(4*f*x + 4*e)^2 + 4*(f*x + e)*cos(2*f*x + 2*e)^2 + (f*x + e)*sin(4*f*x + 4*e)^2 + 4*(f*x + e)*s
in(2*f*x + 2*e)^2 + f*x + (2*(2*cos(2*f*x + 2*e) - 1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 - 4*cos(2*f*x + 2*
e)^2 - sin(4*f*x + 4*e)^2 + 4*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) - 4*sin(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*e) -
1)*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) - 1) + 2*(f*x - 2*(f*x + e)*cos(2*f*x + 2*e) + e + sin(2*f*x + 2
*e))*cos(4*f*x + 4*e) - 4*(f*x + e)*cos(2*f*x + 2*e) - 2*(2*(f*x + e)*sin(2*f*x + 2*e) + cos(2*f*x + 2*e))*sin
(4*f*x + 4*e) + e + 2*sin(2*f*x + 2*e))*sqrt(a)*sqrt(c)/((a^2*c^2*cos(4*f*x + 4*e)^2 + 4*a^2*c^2*cos(2*f*x + 2
*e)^2 + a^2*c^2*sin(4*f*x + 4*e)^2 - 4*a^2*c^2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*a^2*c^2*sin(2*f*x + 2*e)^
2 - 4*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2 - 2*(2*a^2*c^2*cos(2*f*x + 2*e) - a^2*c^2)*cos(4*f*x + 4*e))*f)

Giac [A] (verification not implemented)

none

Time = 2.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx=-\frac {\frac {4 \, \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{\sqrt {-a c} a {\left | c \right |}} - \frac {8 \, \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{\sqrt {-a c} a {\left | c \right |}} + \frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}{\sqrt {-a c} a c {\left | c \right |}} - \frac {4 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}{\sqrt {-a c} a c {\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}}{8 \, f} \]

[In]

integrate(1/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-1/8*(4*log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/(sqrt(-a*c)*a*abs(c)) - 8*log(abs(c*tan(1/2*f*x + 1/2*e)^2 + c))/(s
qrt(-a*c)*a*abs(c)) + (c*tan(1/2*f*x + 1/2*e)^2 - c)/(sqrt(-a*c)*a*c*abs(c)) - (4*c*tan(1/2*f*x + 1/2*e)^2 - c
)/(sqrt(-a*c)*a*c*abs(c)*tan(1/2*f*x + 1/2*e)^2))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int(1/((a + a/cos(e + f*x))^(3/2)*(c - c/cos(e + f*x))^(3/2)),x)

[Out]

int(1/((a + a/cos(e + f*x))^(3/2)*(c - c/cos(e + f*x))^(3/2)), x)