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

Optimal. Leaf size=123 \[ -\frac{16 c^3 \tan (e+f x)}{3 a^2 f \sqrt{c-c \sec (e+f x)}}-\frac{8 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

(-16*c^3*Tan[e + f*x])/(3*a^2*f*Sqrt[c - c*Sec[e + f*x]]) - (8*c^2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(3*f
*(a^2 + a^2*Sec[e + f*x])) + (2*c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2)

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Rubi [A]  time = 0.274039, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3954, 3792} \[ -\frac{16 c^3 \tan (e+f x)}{3 a^2 f \sqrt{c-c \sec (e+f x)}}-\frac{8 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*c^3*Tan[e + f*x])/(3*a^2*f*Sqrt[c - c*Sec[e + f*x]]) - (8*c^2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(3*f
*(a^2 + a^2*Sec[e + f*x])) + (2*c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^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 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && 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))^2} \, dx &=\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{(4 c) \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{a+a \sec (e+f x)} \, dx}{3 a}\\ &=-\frac{8 c^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{\left (8 c^2\right ) \int \sec (e+f x) \sqrt{c-c \sec (e+f x)} \, dx}{3 a^2}\\ &=-\frac{16 c^3 \tan (e+f x)}{3 a^2 f \sqrt{c-c \sec (e+f x)}}-\frac{8 c^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 0.397296, size = 68, normalized size = 0.55 \[ \frac{c^2 (36 \cos (e+f x)+11 \cos (2 (e+f x))+17) \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)}}{3 a^2 f (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(17 + 36*Cos[e + f*x] + 11*Cos[2*(e + f*x)])*Cot[(e + f*x)/2]*Sqrt[c - c*Sec[e + f*x]])/(3*a^2*f*(1 + Cos
[e + f*x])^2)

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Maple [A]  time = 0.208, size = 75, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 22\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+36\,\cos \left ( fx+e \right ) +6 \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{3\,f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( -1+\cos \left ( fx+e \right ) \right ) } \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}} \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))^2,x)

[Out]

-2/3/a^2/f*(11*cos(f*x+e)^2+18*cos(f*x+e)+3)*cos(f*x+e)^2*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/sin(f*x+e)^3/(-
1+cos(f*x+e))

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Maxima [A]  time = 1.58908, size = 220, normalized size = 1.79 \begin{align*} \frac{2 \,{\left (8 \, \sqrt{2} c^{\frac{5}{2}} - \frac{20 \, \sqrt{2} c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{15 \, \sqrt{2} c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{2 \, \sqrt{2} c^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac{\sqrt{2} c^{\frac{5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )}}{3 \, a^{2} f{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac{5}{2}}{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac{5}{2}}} \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))^2,x, algorithm="maxima")

[Out]

2/3*(8*sqrt(2)*c^(5/2) - 20*sqrt(2)*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*sqrt(2)*c^(5/2)*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 - 2*sqrt(2)*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - sqrt(2)*c^(5/2)*sin(f*x +
e)^8/(cos(f*x + e) + 1)^8)/(a^2*f*(sin(f*x + e)/(cos(f*x + e) + 1) + 1)^(5/2)*(sin(f*x + e)/(cos(f*x + e) + 1
) - 1)^(5/2))

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Fricas [A]  time = 0.470874, size = 193, normalized size = 1.57 \begin{align*} \frac{2 \,{\left (11 \, c^{2} \cos \left (f x + e\right )^{2} + 18 \, c^{2} \cos \left (f x + e\right ) + 3 \, c^{2}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{3 \,{\left (a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )} \sin \left (f x + e\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))^2,x, algorithm="fricas")

[Out]

2/3*(11*c^2*cos(f*x + e)^2 + 18*c^2*cos(f*x + e) + 3*c^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/((a^2*f*cos(
f*x + e) + a^2*f)*sin(f*x + e))

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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))**2,x)

[Out]

Timed out

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Giac [A]  time = 2.9263, size = 109, normalized size = 0.89 \begin{align*} \frac{2 \, \sqrt{2}{\left ({\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} + 6 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c - \frac{3 \, c^{2}}{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}\right )} c}{3 \, a^{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))^2,x, algorithm="giac")

[Out]

2/3*sqrt(2)*((c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2) + 6*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c - 3*c^2/sqrt(c*tan(
1/2*f*x + 1/2*e)^2 - c))*c/(a^2*f)