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

Optimal. Leaf size=145 \[ -\frac{c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{a f (a \sec (e+f x)+a)^{3/2}}+\frac{c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.435758, antiderivative size = 145, 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 = {3954, 3952} \[ -\frac{c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{a f (a \sec (e+f x)+a)^{3/2}}+\frac{c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3954

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

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

Mathematica [C]  time = 1.42275, size = 178, normalized size = 1.23 \[ \frac{c^2 \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)} \left (6 \log \left (1+e^{i (e+f x)}\right )-3 \log \left (1+e^{2 i (e+f x)}\right )+\left (8 \log \left (1+e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)+\left (2 \log \left (1+e^{i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))-4\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Cot[(e + f*x)/2]*(-4 + 6*Log[1 + E^(I*(e + f*x))] + Cos[e + f*x]*(8*Log[1 + E^(I*(e + f*x))] - 4*Log[1 +
E^((2*I)*(e + f*x))]) + Cos[2*(e + f*x)]*(2*Log[1 + E^(I*(e + f*x))] - Log[1 + E^((2*I)*(e + f*x))]) - 3*Log[1
+ E^((2*I)*(e + f*x))])*Sqrt[c - c*Sec[e + f*x]])/(2*a^2*f*(1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [B]  time = 0.261, size = 281, normalized size = 1.9 \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 ( -{\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}+2\, \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 ) +4\,\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 ) }{\sin \left ( fx+e \right ) }} \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \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 ) -2\,\cos \left ( fx+e \right ) +3 \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))^(5/2),x)

[Out]

-1/2/f/a^3*(2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+2*cos(f*x+e)^2*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))/sin(f*x+e))+4*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(
f*x+e))/sin(f*x+e))-cos(f*x+e)^2+2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2*ln(-(-1+cos(f*x+e)-sin(f*x+e))
/sin(f*x+e))-2*cos(f*x+e)+3)*(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 = 1.50696, size = 180, normalized size = 1.24 \begin{align*} -\frac{\frac{2 \, c^{\frac{5}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt{-a} a^{2}} + \frac{2 \, c^{\frac{5}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 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(sec(f*x+e)*(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)/(cos(f*x + e) + 1) + 1)/(sqrt(-a)*a^2) + 2*c^(5/2)*log(sin(f*x + e)/(cos(f*x
+ e) + 1) - 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 )^{3} - 2 \, c^{2} \sec \left (f x + e\right )^{2} + c^{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}}{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(sec(f*x+e)*(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)^3 - 2*c^2*sec(f*x + e)^2 + c^2*sec(f*x + e))*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(sec(f*x+e)*(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 = 7.3788, size = 153, normalized size = 1.06 \begin{align*} \frac{{\left (2 \, c^{3} \log \left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right ) +{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} c + 4 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{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 )}{2 \, \sqrt{-a c} a^{2} 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))^(5/2),x, algorithm="giac")

[Out]

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