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

Optimal. Leaf size=168 \[ \frac{\tan (e+f x)}{2 c f (1-\cos (e+f x)) \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{3 \tan (e+f x) \log (1-\cos (e+f x))}{4 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\cos (e+f x)+1)}{4 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

Tan[e + f*x]/(2*c*f*(1 - Cos[e + f*x])*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (3*Log[1 - Cos[e +
f*x]]*Tan[e + f*x])/(4*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[1 + Cos[e + f*x]]*Tan[e
+ f*x])/(4*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.136506, antiderivative size = 217, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3912, 72} \[ -\frac{\tan (e+f x)}{2 c f (1-\sec (e+f x)) \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{3 \tan (e+f x) \log (1-\sec (e+f x))}{4 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\sec (e+f x)+1)}{4 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\cos (e+f x))}{c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[Cos[e + f*x]]*Tan[e + f*x])/(c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (3*Log[1 - Sec[e +
f*x]]*Tan[e + f*x])/(4*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[1 + Sec[e + f*x]]*Tan[e +
f*x])/(4*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(2*c*f*(1 - Sec[e + f*x])*Sqrt
[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3912

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{x (a+a x) (c-c x)^2} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \left (\frac{1}{2 a c^2 (-1+x)^2}-\frac{3}{4 a c^2 (-1+x)}+\frac{1}{a c^2 x}-\frac{1}{4 a c^2 (1+x)}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{\log (\cos (e+f x)) \tan (e+f x)}{c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{3 \log (1-\sec (e+f x)) \tan (e+f x)}{4 c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{\log (1+\sec (e+f x)) \tan (e+f x)}{4 c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{2 c f (1-\sec (e+f x)) \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 8.14501, size = 143, normalized size = 0.85 \[ \frac{\tan (e+f x) \left (-3 \log \left (1-e^{i (e+f x)}\right )-\log \left (1+e^{i (e+f x)}\right )+\left (3 \log \left (1-e^{i (e+f x)}\right )+\log \left (1+e^{i (e+f x)}\right )-2 i f x\right ) \cos (e+f x)+2 i f x-1\right )}{2 c f (\cos (e+f x)-1) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.294, size = 167, normalized size = 1. \begin{align*} -{\frac{-1+\cos \left ( fx+e \right ) }{4\,af\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) } \left ( 6\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -4\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -\cos \left ( fx+e \right ) -6\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +4\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -1 \right ) \sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/f/a*(-1+cos(f*x+e))*(6*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))-4*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-cos(f
*x+e)-6*ln(-(-1+cos(f*x+e))/sin(f*x+e))+4*ln(2/(1+cos(f*x+e)))-1)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/(c*(-1
+cos(f*x+e))/cos(f*x+e))^(3/2)/cos(f*x+e)/sin(f*x+e)

________________________________________________________________________________________

Maxima [B]  time = 1.9407, size = 1104, normalized size = 6.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(f*x + e)*cos(2*f*x + 2*e)^2 + 8*(f*x + e)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*
(f*x + e)*sin(2*f*x + 2*e)^2 + 8*(f*x + e)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*f*x - (c
os(2*f*x + 2*e)^2 - 4*(cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*cos(1/2*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(2*f*x + 2*e)^2 - 4*sin(2*f*x + 2*e)*sin(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e))) + 4*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*cos(2*f*x + 2*e
) + 1)*arctan2(sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))) + 1) - 3*(cos(2*f*x + 2*e)^2 - 4*(cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 4*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(2*f*x + 2*e)^2 - 4*sin(2*f*x + 2*e
)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
)^2 + 2*cos(2*f*x + 2*e) + 1)*arctan2(sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) - 1) + 4*(f*x + e)*cos(2*f*x + 2*e) - 2*(4*f*x + 4*(f*x + e)*cos(2*f*x + 2*
e) + 4*e + sin(2*f*x + 2*e))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*(4*(f*x + e)*sin(2*f*x +
2*e) - cos(2*f*x + 2*e) - 1)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*e)/((c*cos(2*f*x + 2*e)
^2 + 4*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + c*sin(2*f*x + 2*e)^2 - 4*c*sin(2*f*x + 2*e)*
sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
)^2 + 2*c*cos(2*f*x + 2*e) - 4*(c*cos(2*f*x + 2*e) + c)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
c)*sqrt(a)*sqrt(c)*f)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \sec \left (f x + e\right ) + a} \sqrt{-c \sec \left (f x + e\right ) + c}}{a c^{2} \sec \left (f x + e\right )^{3} - a c^{2} \sec \left (f x + e\right )^{2} - a c^{2} \sec \left (f x + e\right ) + a c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \left (\sec{\left (e + f x \right )} + 1\right )} \left (- c \left (\sec{\left (e + f x \right )} - 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out