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

Optimal. Leaf size=108 \[ \frac{32 c^3 \tan (e+f x)}{3 a f \sqrt{c-c \sec (e+f x)}}+\frac{8 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{3 a f}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.190516, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3954, 3793, 3792} \[ \frac{32 c^3 \tan (e+f x)}{3 a f \sqrt{c-c \sec (e+f x)}}+\frac{8 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{3 a f}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*c^3*Tan[e + f*x])/(3*a*f*Sqrt[c - c*Sec[e + f*x]]) + (8*c^2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(3*a*f)
+ (2*c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(a + a*Sec[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 3793

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

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)} \, dx &=\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{(4 c) \int \sec (e+f x) (c-c \sec (e+f x))^{3/2} \, dx}{a}\\ &=\frac{8 c^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{3 a f}+\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left (16 c^2\right ) \int \sec (e+f x) \sqrt{c-c \sec (e+f x)} \, dx}{3 a}\\ &=\frac{32 c^3 \tan (e+f x)}{3 a f \sqrt{c-c \sec (e+f x)}}+\frac{8 c^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{3 a f}+\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (a+a \sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.429616, size = 74, normalized size = 0.69 \[ -\frac{c^2 (20 \cos (e+f x)+23 \cos (2 (e+f x))+21) \cot \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \sqrt{c-c \sec (e+f x)}}{3 a f (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2*(21 + 20*Cos[e + f*x] + 23*Cos[2*(e + f*x)])*Cot[(e + f*x)/2]*Sec[e + f*x]*Sqrt[c - c*Sec[e + f*x]])/(3*
a*f*(1 + Cos[e + f*x]))

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Maple [A]  time = 0.177, size = 73, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 46\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+20\,\cos \left ( fx+e \right ) -2 \right ) \cos \left ( fx+e \right ) }{3\,fa\sin \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}} \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)),x)

[Out]

-2/3/a/f*(23*cos(f*x+e)^2+10*cos(f*x+e)-1)*cos(f*x+e)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/sin(f*x+e)/(-1+cos(
f*x+e))^2

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Maxima [A]  time = 1.50242, size = 185, normalized size = 1.71 \begin{align*} -\frac{4 \,{\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{3 \, \sqrt{2} c^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{3 \, a 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)),x, algorithm="maxima")

[Out]

-4/3*(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 - 3*sqrt(2)*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/(a*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 = 0.469432, size = 176, normalized size = 1.63 \begin{align*} -\frac{2 \,{\left (23 \, c^{2} \cos \left (f x + e\right )^{2} + 10 \, c^{2} \cos \left (f x + e\right ) - c^{2}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{3 \, a f \cos \left (f x + e\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)),x, algorithm="fricas")

[Out]

-2/3*(23*c^2*cos(f*x + e)^2 + 10*c^2*cos(f*x + e) - c^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(a*f*cos(f*x
+ e)*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)),x)

[Out]

Timed out

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Giac [A]  time = 1.63695, size = 115, normalized size = 1.06 \begin{align*} -\frac{4 \, \sqrt{2}{\left (3 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c - \frac{6 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{2} + c^{3}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}}}\right )} c}{3 \, a 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)),x, algorithm="giac")

[Out]

-4/3*sqrt(2)*(3*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c - (6*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^2 + c^3)/(c*tan(1/2
*f*x + 1/2*e)^2 - c)^(3/2))*c/(a*f)