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

Optimal. Leaf size=215 \[ -\frac{\tan (e+f x)}{2 a f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (1-\sec (e+f x))}{4 a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{3 \tan (e+f x) \log (\sec (e+f x)+1)}{4 a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\cos (e+f x))}{a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.146146, antiderivative size = 215, normalized size of antiderivative = 1., 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} \[ -\frac{\tan (e+f x)}{2 a f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (1-\sec (e+f x))}{4 a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{3 \tan (e+f x) \log (\sec (e+f x)+1)}{4 a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\cos (e+f x))}{a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[Cos[e + f*x]]*Tan[e + f*x])/(a*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*a*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*a*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(2*a*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}{(a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}} \, dx &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{x (a+a x)^2 (c-c x)} \, 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}{4 a^2 c (-1+x)}+\frac{1}{a^2 c x}-\frac{1}{2 a^2 c (1+x)^2}-\frac{3}{4 a^2 c (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)}{a 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 a 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 a f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{2 a 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 = 1.38374, size = 141, normalized size = 0.66 \[ \frac{\tan (e+f x) \left (\log \left (1-e^{i (e+f x)}\right )+3 \log \left (1+e^{i (e+f x)}\right )+\left (\log \left (1-e^{i (e+f x)}\right )+3 \log \left (1+e^{i (e+f x)}\right )-2 i f x\right ) \cos (e+f x)-2 i f x+1\right )}{2 a f (\cos (e+f x)+1) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.3, size = 161, normalized size = 0.8 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}{4\,f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( 2\,\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 ) +2\,\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 ){\frac{1}{\sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.91558, size = 1104, normalized size = 5.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^(3/2)/(c-c*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 - 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(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) + 1) - (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)/((a*cos(2*f*x + 2*e)
^2 + 4*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + a*sin(2*f*x + 2*e)^2 + 4*a*sin(2*f*x + 2*e)*
sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
)^2 + 2*a*cos(2*f*x + 2*e) + 4*(a*cos(2*f*x + 2*e) + a)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
a)*sqrt(a)*sqrt(c)*f)

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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^{2} c \sec \left (f x + e\right )^{3} + a^{2} c \sec \left (f x + e\right )^{2} - a^{2} c \sec \left (f x + e\right ) - a^{2} c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out