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

Optimal. Leaf size=140 \[ \frac{c \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x)}{a f (a \sec (e+f x)+a)^{3/2} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}} \]

[Out]

-(c*Tan[e + f*x])/(2*f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]]) - (c*Tan[e + f*x])/(a*f*(a + a*Sec
[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x]]) + (c*Log[1 + Cos[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e +
f*x]]*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.282233, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3907, 3911, 31} \[ \frac{c \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x)}{a f (a \sec (e+f x)+a)^{3/2} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*Tan[e + f*x])/(2*f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]]) - (c*Tan[e + f*x])/(a*f*(a + a*Sec
[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x]]) + (c*Log[1 + Cos[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e +
f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3907

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

Rule 3911

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> -Dis
t[(a*c*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), Subst[Int[((b + a*x)^(m - 1/2)*(d
+ c*x)^(n - 1/2))/x^(m + n), x], x, Sin[e + f*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] && EqQ[m + n, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac{c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{\int \frac{\sqrt{c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx}{a}\\ &=-\frac{c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{\int \frac{\sqrt{c-c \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx}{a^2}\\ &=-\frac{c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{(c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{a+a x} \, dx,x,\cos (e+f x)\right )}{a f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{c \log (1+\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.641076, size = 151, normalized size = 1.08 \[ \frac{i \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)} \left (6 i \log \left (1+e^{i (e+f x)}\right )+\left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right ) \cos (2 (e+f x))+4 \left (2 i \log \left (1+e^{i (e+f x)}\right )+f x+i\right ) \cos (e+f x)+3 f x+3 i\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*Cot[(e + f*x)/2]*(3*I + 3*f*x + Cos[2*(e + f*x)]*(f*x + (2*I)*Log[1 + E^(I*(e + f*x))]) + 4*Cos[e + f*x
]*(I + f*x + (2*I)*Log[1 + E^(I*(e + f*x))]) + (6*I)*Log[1 + E^(I*(e + f*x))])*Sqrt[c - c*Sec[e + f*x]])/(a^2*
f*(1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [A]  time = 0.291, size = 152, normalized size = 1.1 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) }{8\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( 8\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+7\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+16\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -2\,\cos \left ( fx+e \right ) +8\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -5 \right ) \sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\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((c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(5/2),x)

[Out]

1/8/f/a^3*(-1+cos(f*x+e))^2*(8*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2+7*cos(f*x+e)^2+16*cos(f*x+e)*ln(2/(1+cos(f*x+
e)))-2*cos(f*x+e)+8*ln(2/(1+cos(f*x+e)))-5)*cos(f*x+e)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(1/2)*(1/cos(f*x+e)*a*(1
+cos(f*x+e)))^(1/2)/sin(f*x+e)^5

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Maxima [B]  time = 2.43413, size = 1573, normalized size = 11.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((f*x + e)*cos(4*f*x + 4*e)^2 + 16*(f*x + e)*cos(3*f*x + 3*e)^2 + 36*(f*x + e)*cos(2*f*x + 2*e)^2 + 16*(f*x +
e)*cos(f*x + e)^2 + (f*x + e)*sin(4*f*x + 4*e)^2 + 16*(f*x + e)*sin(3*f*x + 3*e)^2 + 36*(f*x + e)*sin(2*f*x +
2*e)^2 + 16*(f*x + e)*sin(f*x + e)^2 + f*x - 2*(2*(4*cos(3*f*x + 3*e) + 6*cos(2*f*x + 2*e) + 4*cos(f*x + e) +
1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^2 + 8*(6*cos(2*f*x + 2*e) + 4*cos(f*x + e) + 1)*cos(3*f*x + 3*e) + 16*
cos(3*f*x + 3*e)^2 + 12*(4*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + 36*cos(2*f*x + 2*e)^2 + 16*cos(f*x + e)^2 + 4*
(2*sin(3*f*x + 3*e) + 3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*sin(4*f*x + 4*e) + sin(4*f*x + 4*e)^2 + 16*(3*sin(2
*f*x + 2*e) + 2*sin(f*x + e))*sin(3*f*x + 3*e) + 16*sin(3*f*x + 3*e)^2 + 36*sin(2*f*x + 2*e)^2 + 48*sin(2*f*x
+ 2*e)*sin(f*x + e) + 16*sin(f*x + e)^2 + 8*cos(f*x + e) + 1)*arctan2(sin(f*x + e), cos(f*x + e) + 1) + 2*(f*x
+ 4*(f*x + e)*cos(3*f*x + 3*e) + 6*(f*x + e)*cos(2*f*x + 2*e) + 4*(f*x + e)*cos(f*x + e) + e - 2*sin(3*f*x +
3*e) - 3*sin(2*f*x + 2*e) - 2*sin(f*x + e))*cos(4*f*x + 4*e) + 8*(f*x + 6*(f*x + e)*cos(2*f*x + 2*e) + 4*(f*x
+ e)*cos(f*x + e) + e)*cos(3*f*x + 3*e) + 12*(f*x + 4*(f*x + e)*cos(f*x + e) + e)*cos(2*f*x + 2*e) + 8*(f*x +
e)*cos(f*x + e) + 2*(4*(f*x + e)*sin(3*f*x + 3*e) + 6*(f*x + e)*sin(2*f*x + 2*e) + 4*(f*x + e)*sin(f*x + e) +
2*cos(3*f*x + 3*e) + 3*cos(2*f*x + 2*e) + 2*cos(f*x + e))*sin(4*f*x + 4*e) + 4*(12*(f*x + e)*sin(2*f*x + 2*e)
+ 8*(f*x + e)*sin(f*x + e) - 1)*sin(3*f*x + 3*e) + 6*(8*(f*x + e)*sin(f*x + e) - 1)*sin(2*f*x + 2*e) + e - 4*s
in(f*x + e))*sqrt(a)*sqrt(c)/((a^3*cos(4*f*x + 4*e)^2 + 16*a^3*cos(3*f*x + 3*e)^2 + 36*a^3*cos(2*f*x + 2*e)^2
+ 16*a^3*cos(f*x + e)^2 + a^3*sin(4*f*x + 4*e)^2 + 16*a^3*sin(3*f*x + 3*e)^2 + 36*a^3*sin(2*f*x + 2*e)^2 + 48*
a^3*sin(2*f*x + 2*e)*sin(f*x + e) + 16*a^3*sin(f*x + e)^2 + 8*a^3*cos(f*x + e) + a^3 + 2*(4*a^3*cos(3*f*x + 3*
e) + 6*a^3*cos(2*f*x + 2*e) + 4*a^3*cos(f*x + e) + a^3)*cos(4*f*x + 4*e) + 8*(6*a^3*cos(2*f*x + 2*e) + 4*a^3*c
os(f*x + e) + a^3)*cos(3*f*x + 3*e) + 12*(4*a^3*cos(f*x + e) + a^3)*cos(2*f*x + 2*e) + 4*(2*a^3*sin(3*f*x + 3*
e) + 3*a^3*sin(2*f*x + 2*e) + 2*a^3*sin(f*x + e))*sin(4*f*x + 4*e) + 16*(3*a^3*sin(2*f*x + 2*e) + 2*a^3*sin(f*
x + e))*sin(3*f*x + 3*e))*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^{3} \sec \left (f x + e\right )^{3} + 3 \, a^{3} \sec \left (f x + e\right )^{2} + 3 \, a^{3} \sec \left (f x + e\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 2.40347, size = 215, normalized size = 1.54 \begin{align*} \frac{\sqrt{2}{\left (\frac{8 \, \sqrt{2} \sqrt{-a c} c \log \left ({\left | c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c \right |}\right )}{a^{3}{\left | c \right |}} + \frac{\sqrt{2}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt{-a c} a^{3} c{\left | c \right |} - 4 \, \sqrt{2}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} \sqrt{-a c} a^{3} c^{2}{\left | c \right |}}{a^{6} c^{4}}\right )} \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 ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{16 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(2)*(8*sqrt(2)*sqrt(-a*c)*c*log(abs(c*tan(1/2*f*x + 1/2*e)^2 + c))/(a^3*abs(c)) + (sqrt(2)*(c*tan(1/2
*f*x + 1/2*e)^2 - c)^2*sqrt(-a*c)*a^3*c*abs(c) - 4*sqrt(2)*(c*tan(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*c)*a^3*c^2*a
bs(c))/(a^6*c^4))*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e))*sgn(cos(f*x + e))/f