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

Optimal. Leaf size=139 \[ -\frac{2 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{f \sqrt{a \sec (e+f x)+a}}-\frac{4 c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f \sqrt{a \sec (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.409218, antiderivative size = 139, 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} \[ -\frac{2 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{f \sqrt{a \sec (e+f x)+a}}-\frac{4 c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 1.49376, size = 141, normalized size = 1.01 \[ \frac{c^2 \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt{c-c \sec (e+f x)} \left (8 \log \left (1+e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )-6 \cos (e+f x)+\left (8 \log \left (1+e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))+1\right )}{2 f \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Cot[(e + f*x)/2]*(1 - 6*Cos[e + f*x] + 8*Log[1 + E^(I*(e + f*x))] + Cos[2*(e + f*x)]*(8*Log[1 + E^(I*(e +
f*x))] - 4*Log[1 + E^((2*I)*(e + f*x))]) - 4*Log[1 + E^((2*I)*(e + f*x))])*Sec[e + f*x]^2*Sqrt[c - c*Sec[e +
f*x]])/(2*f*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [A]  time = 0.315, size = 165, normalized size = 1.2 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{2\,af\sin \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}} \left ( 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}+7\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+6\,\cos \left ( fx+e \right ) -1 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}\sqrt{{\frac{a \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)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x)

[Out]

-1/2/f/a*(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+7*cos(f*x+e)^2+6*cos(f*x+e)-1)*cos(f*x+e)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*(1/cos(f*x+e)*a
*(1+cos(f*x+e)))^(1/2)/sin(f*x+e)/(-1+cos(f*x+e))^2

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Maxima [B]  time = 1.92371, size = 995, normalized size = 7.16 \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)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{2} \sec \left (f x + e\right )^{3} - 2 \, c^{2} \sec \left (f x + e\right )^{2} + c^{2} \sec \left (f x + e\right )\right )} \sqrt{-c \sec \left (f x + e\right ) + c}}{\sqrt{a \sec \left (f x + e\right ) + a}}, x\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))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral((c^2*sec(f*x + e)^3 - 2*c^2*sec(f*x + e)^2 + c^2*sec(f*x + e))*sqrt(-c*sec(f*x + e) + c)/sqrt(a*sec(f
*x + e) + a), 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)*(c-c*sec(f*x+e))**(5/2)/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 3.66539, size = 188, normalized size = 1.35 \begin{align*} \frac{2 \,{\left (2 \, c^{3} \log \left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right ) - \frac{3 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} c^{3} + 4 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{4} + c^{5}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2}}\right )} c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{\sqrt{-a c} f{\left | c \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))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

2*(2*c^3*log(c*tan(1/2*f*x + 1/2*e)^2 - c) - (3*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^3 + 4*(c*tan(1/2*f*x + 1/2*
e)^2 - c)*c^4 + c^5)/(c*tan(1/2*f*x + 1/2*e)^2 - c)^2)*c*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e))/(s
qrt(-a*c)*f*abs(c))