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

Optimal. Leaf size=72 \[ \frac{4 c^2 \tan (e+f x)}{a f \sqrt{c-c \sec (e+f x)}}+\frac{2 c \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.152922, antiderivative size = 72, 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, 3792} \[ \frac{4 c^2 \tan (e+f x)}{a f \sqrt{c-c \sec (e+f x)}}+\frac{2 c \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.241361, size = 54, normalized size = 0.75 \[ -\frac{2 c (3 \cos (e+f x)+1) \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)}}{a f (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.174, size = 63, normalized size = 0.9 \begin{align*} -2\,{\frac{ \left ( 3\,\cos \left ( fx+e \right ) +1 \right ) \cos \left ( fx+e \right ) }{fa\sin \left ( fx+e \right ) \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 ) ^{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)),x)

[Out]

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

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Maxima [A]  time = 1.52169, size = 149, normalized size = 2.07 \begin{align*} \frac{2 \,{\left (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{\sqrt{2} c^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{a 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)),x, algorithm="maxima")

[Out]

2*(2*sqrt(2)*c^(3/2) - 3*sqrt(2)*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + sqrt(2)*c^(3/2)*sin(f*x + e)^4/
(cos(f*x + e) + 1)^4)/(a*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 = 0.462894, size = 116, normalized size = 1.61 \begin{align*} -\frac{2 \,{\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{a f \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)),x, algorithm="fricas")

[Out]

-2*(3*c*cos(f*x + e) + c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(a*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{\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{\left (e + f x \right )} + 1}\, dx}{a} \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)),x)

[Out]

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

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Giac [A]  time = 1.49003, size = 86, normalized size = 1.19 \begin{align*} -\frac{2 \, \sqrt{2}{\left (\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{2} - \frac{c^{3}}{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}\right )}}{a 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)),x, algorithm="giac")

[Out]

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