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

Optimal. Leaf size=98 \[ \frac{c^3 \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{2 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}} \]

[Out]

(-2*c^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]]) + (c^3*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 = 0.187188, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3910, 3911, 31} \[ \frac{c^3 \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{2 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]]) + (c^3*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 3910

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(5/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[(-8*a^3*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[a^2/c^2, Int
[Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^(n + 2), 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{(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac{2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{c^2 \int \frac{\sqrt{c-c \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx}{a^2}\\ &=-\frac{2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{\left (c^3 \tan (e+f x)\right ) \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{2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{c^3 \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 = 1.51971, size = 154, normalized size = 1.57 \[ \frac{i c^2 \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)} \left (6 i \log \left (1+e^{i (e+f x)}\right )+\left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right ) \cos (2 (e+f x))+4 \left (2 i \log \left (1+e^{i (e+f x)}\right )+f x+2 i\right ) \cos (e+f x)+3 f x+4 i\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*c^2*Cot[(e + f*x)/2]*(4*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]*(2*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 = 0.252, size = 144, normalized size = 1.5 \begin{align*}{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{2\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( 2\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -2\,\cos \left ( fx+e \right ) +2\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \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((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(5/2),x)

[Out]

1/2/f/a^3*(2*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2+3*cos(f*x+e)^2+4*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-2*cos(f*x+e)+2
*ln(2/(1+cos(f*x+e)))-1)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*cos(f*x+e)^3*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/
2)/sin(f*x+e)^5

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Maxima [A]  time = 1.51093, size = 138, normalized size = 1.41 \begin{align*} \frac{\frac{2 \, c^{\frac{5}{2}} \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt{-a} a^{2}} + \frac{\frac{2 \, \sqrt{-a} c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{\sqrt{-a} c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}{a^{3}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 7.16082, size = 170, normalized size = 1.73 \begin{align*} -\frac{c{\left (\frac{2 \, \sqrt{-a c} c^{2} \log \left (2 \,{\left | c \right |}\right )}{a^{3}{\left | c \right |}} - \frac{2 \, \sqrt{-a c} c^{2} \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{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt{-a c}{\left | c \right |}}{a^{3} c^{2}}\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 )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c*(2*sqrt(-a*c)*c^2*log(2*abs(c))/(a^3*abs(c)) - 2*sqrt(-a*c)*c^2*log(abs(c*tan(1/2*f*x + 1/2*e)^2 + c))/
(a^3*abs(c)) - (c*tan(1/2*f*x + 1/2*e)^2 - c)^2*sqrt(-a*c)*abs(c)/(a^3*c^2))*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(
1/2*f*x + 1/2*e))/f