\(\int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx\) [120]

Optimal result

Integrand size = 30, antiderivative size = 94 \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx=-\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c \log (1+\cos (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

-c*tan(f*x+e)/f/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(1/2)+c*ln(1+cos(f*x+e))*tan(f*x+e)/a/f/(a+a*sec(f*x+e
))^(1/2)/(c-c*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 94, 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 = {3992, 3996, 31} \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx=\frac {c \tan (e+f x) \log (\cos (e+f x)+1)}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3992

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

Rule 3996

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

Rubi steps \begin{align*} \text {integral}& = -\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a} \\ & = -\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {(c \tan (e+f x)) \text {Subst}\left (\int \frac {1}{a+a x} \, dx,x,\cos (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ & = -\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c \log (1+\cos (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx=-\frac {c \left (-\log (\cos (e+f x))-\log (1+\sec (e+f x))+\frac {1}{1+\sec (e+f x)}\right ) \tan (e+f x)}{a f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

-((c*(-Log[Cos[e + f*x]] - Log[1 + Sec[e + f*x]] + (1 + Sec[e + f*x])^(-1))*Tan[e + f*x])/(a*f*Sqrt[a*(1 + Sec
[e + f*x])]*Sqrt[c - c*Sec[e + f*x]]))

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\left (2 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+\cos \left (f x +e \right )-1\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \cot \left (f x +e \right )}{2 f \,a^{2} \left (\cos \left (f x +e \right )+1\right )}\)	\(94\)
risch	\(-\frac {\sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (2 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+{\mathrm e}^{2 i \left (f x +e \right )} f x +4 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+2 \,{\mathrm e}^{2 i \left (f x +e \right )} e +2 \,{\mathrm e}^{i \left (f x +e \right )} f x +2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+2 i {\mathrm e}^{i \left (f x +e \right )}+4 \,{\mathrm e}^{i \left (f x +e \right )} e +f x +2 e \right )}{a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(226\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2), x
)

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(-c*(sec(e + f*x) - 1))/(a*(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. 395 vs. \(2 (86) = 172\).

Time = 0.37 (sec) , antiderivative size = 395, normalized size of antiderivative = 4.20 \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx=-\frac {{\left ({\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \cos \left (f x + e\right )^{2} + {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \sin \left (f x + e\right )^{2} + f x - 2 \, {\left (2 \, {\left (2 \, \cos \left (f x + e\right ) + 1\right )} \cos \left (2 \, f x + 2 \, e\right ) + \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (f x + e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, \sin \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) + 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) + 2 \, {\left (f x + 2 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + e - \sin \left (f x + e\right )\right )} \cos \left (2 \, f x + 2 \, e\right ) + 4 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + 2 \, {\left (2 \, {\left (f x + e\right )} \sin \left (f x + e\right ) + \cos \left (f x + e\right )\right )} \sin \left (2 \, f x + 2 \, e\right ) + e - 2 \, \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (a^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, a^{2} \sin \left (f x + e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right ) + a^{2} + 2 \, {\left (2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right )} f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} \sqrt {-a c} c \log \left ({\left | -2 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, c \right |}\right )}{a^{2} {\left | c \right |}} - \frac {\sqrt {2} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c}}{a^{2} {\left | c \right |}}\right )}}{4 \, f} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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