3.127 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{\sqrt{c-c \sec (e+f x)}} \, dx\)

Optimal. Leaf size=141 \[ \frac{2 a^2 \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{f \sqrt{c-c \sec (e+f x)}}+\frac{4 a^3 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f \sqrt{c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.404526, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {3955, 3952} \[ \frac{2 a^2 \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{f \sqrt{c-c \sec (e+f x)}}+\frac{4 a^3 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/Sqrt[c - c*Sec[e + f*x]],x]

[Out]

(4*a^3*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]]) + (2*a^2*Sqrt
[a + a*Sec[e + f*x]]*Tan[e + f*x])/(f*Sqrt[c - c*Sec[e + f*x]]) + (a*(a + a*Sec[e + f*x])^(3/2)*Tan[e + f*x])/
(2*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))^{5/2}}{\sqrt{c-c \sec (e+f x)}} \, dx &=\frac{a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f \sqrt{c-c \sec (e+f x)}}+(2 a) \int \frac{\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{\sqrt{c-c \sec (e+f x)}} \, dx\\ &=\frac{2 a^2 \sqrt{a+a \sec (e+f x)} \tan (e+f x)}{f \sqrt{c-c \sec (e+f x)}}+\frac{a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f \sqrt{c-c \sec (e+f x)}}+\left (4 a^2\right ) \int \frac{\sec (e+f x) \sqrt{a+a \sec (e+f x)}}{\sqrt{c-c \sec (e+f x)}} \, dx\\ &=\frac{4 a^3 \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{2 a^2 \sqrt{a+a \sec (e+f x)} \tan (e+f x)}{f \sqrt{c-c \sec (e+f x)}}+\frac{a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 6.62073, size = 328, normalized size = 2.33 \[ \frac{\sin \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec (e+f x) (a (\sec (e+f x)+1))^{5/2} \sqrt{(\cos (e+f x)+1) \sec (e+f x)} \left (\frac{5 \sec \left (\frac{e}{2}+\frac{f x}{2}\right )}{2 f}+\frac{\cos \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec (e+f x)}{f}\right )}{(\sec (e+f x)+1)^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{4 \sqrt{2} e^{\frac{1}{2} i (e+f x)} \sqrt{\frac{\left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}} \left (2 \log \left (1-e^{i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )\right ) \sin \left (\frac{e}{2}+\frac{f x}{2}\right ) \sqrt{\sec (e+f x)} (a (\sec (e+f x)+1))^{5/2}}{f \left (1+e^{i (e+f x)}\right ) \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} (\sec (e+f x)+1)^{5/2} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/Sqrt[c - c*Sec[e + f*x]],x]

[Out]

(4*Sqrt[2]*E^((I/2)*(e + f*x))*Sqrt[(1 + E^(I*(e + f*x)))^2/(1 + E^((2*I)*(e + f*x)))]*(2*Log[1 - E^(I*(e + f*
x))] - Log[1 + E^((2*I)*(e + f*x))])*Sqrt[Sec[e + f*x]]*(a*(1 + Sec[e + f*x]))^(5/2)*Sin[e/2 + (f*x)/2])/((1 +
E^(I*(e + f*x)))*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*f*(1 + Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e
+ f*x]]) + (Sec[e + f*x]*Sqrt[(1 + Cos[e + f*x])*Sec[e + f*x]]*(a*(1 + Sec[e + f*x]))^(5/2)*((5*Sec[e/2 + (f*x
)/2])/(2*f) + (Cos[e/2 + (f*x)/2]*Sec[e + f*x])/f)*Sin[e/2 + (f*x)/2])/((1 + Sec[e + f*x])^(5/2)*Sqrt[c - c*Se
c[e + f*x]])

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Maple [A]  time = 0.319, size = 189, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2}}{2\,f\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) c} \left ( 16\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -8\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+6\,\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))^(5/2)/(c-c*sec(f*x+e))^(1/2),x)

[Out]

-1/2/f*a^2*(16*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-8*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(
f*x+e))-8*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+5*cos(f*x+e)^2+6*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)/cos(f*x+e)/c

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Maxima [B]  time = 1.99269, size = 995, normalized size = 7.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-2*(a^2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) - a^2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e) - a^2*sin(2*f*x + 2*e) + 2*(
a^2*cos(4*f*x + 4*e)^2 + 4*a^2*cos(2*f*x + 2*e)^2 + a^2*sin(4*f*x + 4*e)^2 + 4*a^2*sin(4*f*x + 4*e)*sin(2*f*x
+ 2*e) + 4*a^2*sin(2*f*x + 2*e)^2 + 4*a^2*cos(2*f*x + 2*e) + a^2 + 2*(2*a^2*cos(2*f*x + 2*e) + a^2)*cos(4*f*x
+ 4*e))*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*(a^2*cos(4*f*x + 4*e)^2 + 4*a^2*cos(2*f*x + 2*e)^2
+ a^2*sin(4*f*x + 4*e)^2 + 4*a^2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*a^2*sin(2*f*x + 2*e)^2 + 4*a^2*cos(2*f
*x + 2*e) + a^2 + 2*(2*a^2*cos(2*f*x + 2*e) + a^2)*cos(4*f*x + 4*e))*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*(a^2*sin(4*f*x + 4*e) + 2*a
^2*sin(2*f*x + 2*e))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 3*(a^2*sin(4*f*x + 4*e) + 2*a^2*si
n(2*f*x + 2*e))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 3*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f
*x + 2*e) + a^2)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 3*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*
f*x + 2*e) + a^2)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((c*cos(4*f*x + 4*e)^2
+ 4*c*cos(2*f*x + 2*e)^2 + c*sin(4*f*x + 4*e)^2 + 4*c*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*c*sin(2*f*x + 2*e
)^2 + 2*(2*c*cos(2*f*x + 2*e) + c)*cos(4*f*x + 4*e) + 4*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^{2} \sec \left (f x + e\right )^{3} + 2 \, a^{2} \sec \left (f x + e\right )^{2} + a^{2} \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))^(5/2)/(c-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-(a^2*sec(f*x + e)^3 + 2*a^2*sec(f*x + e)^2 + a^2*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))**(5/2)/(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))^(5/2)/(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out