3.102 \(\int \frac{\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=88 \[ \frac{2 c \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{5 f (a \sec (e+f x)+a)^3}-\frac{4 c^2 \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}} \]

[Out]

(-4*c^2*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]]) + (2*c*Sqrt[c - c*Sec[e + f*x]]
*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3)

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Rubi [A]  time = 0.222202, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3954, 3953} \[ \frac{2 c \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{5 f (a \sec (e+f x)+a)^3}-\frac{4 c^2 \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(3/2))/(a + a*Sec[e + f*x])^3,x]

[Out]

(-4*c^2*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]]) + (2*c*Sqrt[c - c*Sec[e + f*x]]
*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3)

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 3953

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^3} \, dx &=\frac{2 c \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{(2 c) \int \frac{\sec (e+f x) \sqrt{c-c \sec (e+f x)}}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=-\frac{4 c^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}}+\frac{2 c \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 0.317465, size = 60, normalized size = 0.68 \[ -\frac{2 c (\cos (e+f x)-5) \cot \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \sqrt{c-c \sec (e+f x)}}{15 a^3 f (\sec (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(3/2))/(a + a*Sec[e + f*x])^3,x]

[Out]

(-2*c*(-5 + Cos[e + f*x])*Cot[(e + f*x)/2]*Sec[e + f*x]*Sqrt[c - c*Sec[e + f*x]])/(15*a^3*f*(1 + Sec[e + f*x])
^3)

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Maple [A]  time = 0.237, size = 63, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+10-12\,\cos \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{15\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{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))^(3/2)/(a+a*sec(f*x+e))^3,x)

[Out]

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

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Maxima [B]  time = 1.51664, size = 220, normalized size = 2.5 \begin{align*} -\frac{2 \, \sqrt{2} c^{\frac{3}{2}} - \frac{3 \, \sqrt{2} c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{3 \, \sqrt{2} c^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{7 \, \sqrt{2} c^{\frac{3}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac{3 \, \sqrt{2} c^{\frac{3}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}}{30 \, a^{3} f{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac{3}{2}}{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac{3}{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))^(3/2)/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/30*(2*sqrt(2)*c^(3/2) - 3*sqrt(2)*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 3*sqrt(2)*c^(3/2)*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 + 7*sqrt(2)*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 3*sqrt(2)*c^(3/2)*sin(f*x
+ e)^8/(cos(f*x + e) + 1)^8)/(a^3*f*(sin(f*x + e)/(cos(f*x + e) + 1) + 1)^(3/2)*(sin(f*x + e)/(cos(f*x + e) +
1) - 1)^(3/2))

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Fricas [A]  time = 0.46934, size = 211, normalized size = 2.4 \begin{align*} -\frac{2 \,{\left (c \cos \left (f x + e\right )^{3} - 5 \, c \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{2} + 2 \, a^{3} f \cos \left (f x + e\right ) + a^{3} 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))^(3/2)/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/15*(c*cos(f*x + e)^3 - 5*c*cos(f*x + e)^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/((a^3*f*cos(f*x + e)^2 +
2*a^3*f*cos(f*x + e) + a^3*f)*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c \sqrt{- c \sec{\left (e + f x \right )} + c} \sec{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{c \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))**(3/2)/(a+a*sec(f*x+e))**3,x)

[Out]

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

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Giac [A]  time = 2.85786, size = 81, normalized size = 0.92 \begin{align*} -\frac{\sqrt{2}{\left (3 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{5}{2}} + 5 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c\right )}}{30 \, a^{3} c 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))^(3/2)/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

-1/30*sqrt(2)*(3*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2) + 5*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c)/(a^3*c*f)