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

Optimal. Leaf size=95 \[ -\frac{\tan (e+f x)}{2 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac{\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{2 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.280026, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3960, 3959, 3770} \[ -\frac{\tan (e+f x)}{2 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac{\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{2 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3960

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

Rule 3959

Int[csc[(e_.) + (f_.)*(x_)]*(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[Csc[e + f*x]*Cot[e + f*x]^(2*m), 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]

Rule 3770

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx &=-\frac{\tan (e+f x)}{2 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac{\int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}} \, dx}{2 c}\\ &=-\frac{\tan (e+f x)}{2 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x) \int \csc (e+f x) \, dx}{2 c \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{\tan (e+f x)}{2 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}-\frac{\tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{2 c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.880798, size = 79, normalized size = 0.83 \[ -\frac{\tan (e+f x) \left (1+2 (\cos (e+f x)-1) \tanh ^{-1}\left (e^{i (e+f x)}\right )\right )}{2 c f (\cos (e+f x)-1) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.299, size = 131, normalized size = 1.4 \begin{align*} -{\frac{-1+\cos \left ( fx+e \right ) }{4\,af\cos \left ( fx+e \right ) \sin \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 ) -\cos \left ( fx+e \right ) -2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \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(sec(f*x+e)/(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))*(2*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))-cos(f*x+e)-2*ln(-(-1+cos(f*x+e))/sin(f*
x+e))-1)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/cos(f*x+e)/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x+e)

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Maxima [B]  time = 1.92862, size = 548, normalized size = 5.77 \begin{align*} \frac{{\left ({\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 ) -{\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 \, \cos \left (f x + e\right ) \sin \left (2 \, f x + 2 \, e\right ) - 2 \, \cos \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) - 2 \, \sin \left (f x + e\right )\right )} \sqrt{a} \sqrt{c}}{2 \,{\left (a c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a c^{2} \cos \left (f x + e\right )^{2} + a c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} - 4 \, a c^{2} \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, a c^{2} \sin \left (f x + e\right )^{2} - 4 \, a c^{2} \cos \left (f x + e\right ) + a c^{2} - 2 \,{\left (2 \, a c^{2} \cos \left (f x + e\right ) - a c^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.667688, size = 952, normalized size = 10.02 \begin{align*} \left [-\frac{\sqrt{-a c}{\left (\cos \left (f x + e\right ) - 1\right )} \log \left (-\frac{4 \,{\left (2 \, \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 )}} \cos \left (f x + e\right )^{2} +{\left (a c \cos \left (f x + e\right )^{2} + 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 ) - 2 \, \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 )}} \cos \left (f x + e\right )}{4 \,{\left (a c^{2} f \cos \left (f x + e\right ) - a c^{2} f\right )} \sin \left (f x + e\right )}, \frac{\sqrt{a c}{\left (\cos \left (f x + e\right ) - 1\right )} \arctan \left (\frac{\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 )}}}{a c \sin \left (f x + e\right )}\right ) \sin \left (f x + e\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 )}} \cos \left (f x + e\right )}{2 \,{\left (a c^{2} f \cos \left (f x + e\right ) - a c^{2} f\right )} \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(sqrt(-a*c)*(cos(f*x + e) - 1)*log(-4*(2*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))*cos(f*x + e)^2 + (a*c*cos(f*x + e)^2 + a*c)*sin(f*x + e))/((cos(f*x + e)^2 - 1)*si
n(f*x + e)))*sin(f*x + e) - 2*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*
cos(f*x + e))/((a*c^2*f*cos(f*x + e) - a*c^2*f)*sin(f*x + e)), 1/2*(sqrt(a*c)*(cos(f*x + e) - 1)*arctan(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))/(a*c*sin(f*x + e)))*sin(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))*cos(f*x + e))/((a*c^2
*f*cos(f*x + e) - a*c^2*f)*sin(f*x + e))]

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

[Out]

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

[Out]

Timed out