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

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

[Out]

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

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Rubi [A]  time = 0.263783, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {3955, 3952} \[ \frac{2 a^2 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{a \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{f \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3955

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

Rule 3952

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

Rubi steps

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

Mathematica [C]  time = 1.44814, size = 174, normalized size = 1.83 \[ \frac{\sqrt{2} a \sin \left (\frac{1}{2} (e+f x)\right ) \sec ^{\frac{3}{2}}(e+f x) \sqrt{a (\sec (e+f x)+1)} \left (1+\left (4 \log \left (1-e^{i (e+f x)}\right )-2 \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )+i \sin \left (\frac{1}{2} (e+f x)\right )\right )}{f \left (1+e^{i (e+f x)}\right ) \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*a*(1 + Cos[e + f*x]*(4*Log[1 - E^(I*(e + f*x))] - 2*Log[1 + E^((2*I)*(e + f*x))]))*Sec[e + f*x]^(3/2)
*Sqrt[a*(1 + Sec[e + f*x])]*(Cos[(e + f*x)/2] + I*Sin[(e + f*x)/2])*Sin[(e + f*x)/2])/((1 + E^(I*(e + f*x)))*S
qrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*f*Sqrt[c - c*Sec[e + f*x]])

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Maple [A]  time = 0.338, size = 162, normalized size = 1.7 \begin{align*}{\frac{a}{f\sin \left ( fx+e \right ) c} \left ( 2\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -4\,\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 ) -1 \right ) \sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\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(sec(f*x+e)*(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.90335, size = 371, normalized size = 3.91 \begin{align*} -\frac{2 \,{\left (a \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) \sin \left (2 \, f x + 2 \, e\right ) +{\left (a \cos \left (2 \, f x + 2 \, e\right )^{2} + a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 2 \,{\left (a \cos \left (2 \, f x + 2 \, e\right )^{2} + a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} \arctan \left (\sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 1\right ) -{\left (a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt{a} \sqrt{c}}{{\left (c \cos \left (2 \, f x + 2 \, e\right )^{2} + c \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, c \cos \left (2 \, f x + 2 \, e\right ) + c\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Timed out