3.101 \(\int \frac{\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=135 \[ \frac{16 c^3 \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right ) \sqrt{c-c \sec (e+f x)}}-\frac{8 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{15 a f (a \sec (e+f x)+a)^2}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

(16*c^3*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]]) - (8*c^2*Sqrt[c - c*Sec[e + f*x
]]*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2) + (2*c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(5*f*(a + a*S
ec[e + f*x])^3)

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Rubi [A]  time = 0.348345, antiderivative size = 135, 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, 3953} \[ \frac{16 c^3 \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right ) \sqrt{c-c \sec (e+f x)}}-\frac{8 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{15 a f (a \sec (e+f x)+a)^2}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(5/2))/(a + a*Sec[e + f*x])^3,x]

[Out]

(16*c^3*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]]) - (8*c^2*Sqrt[c - c*Sec[e + f*x
]]*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2) + (2*c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(5*f*(a + a*S
ec[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))^{5/2}}{(a+a \sec (e+f x))^3} \, dx &=\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{(4 c) \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=-\frac{8 c^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\left (8 c^2\right ) \int \frac{\sec (e+f x) \sqrt{c-c \sec (e+f x)}}{a+a \sec (e+f x)} \, dx}{15 a^2}\\ &=\frac{16 c^3 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right ) \sqrt{c-c \sec (e+f x)}}-\frac{8 c^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{2 c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 0.343672, size = 74, normalized size = 0.55 \[ -\frac{c^2 \cos (e+f x) (20 \cos (e+f x)+7 \cos (2 (e+f x))+37) \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)}}{15 a^3 f (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(5/2))/(a + a*Sec[e + f*x])^3,x]

[Out]

-(c^2*Cos[e + f*x]*(37 + 20*Cos[e + f*x] + 7*Cos[2*(e + f*x)])*Cot[(e + f*x)/2]*Sqrt[c - c*Sec[e + f*x]])/(15*
a^3*f*(1 + Cos[e + f*x])^3)

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Maple [A]  time = 0.247, size = 65, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 14\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+20\,\cos \left ( fx+e \right ) +30 \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{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))^3,x)

[Out]

-2/15/a^3/f*(7*cos(f*x+e)^2+10*cos(f*x+e)+15)*cos(f*x+e)^3*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/sin(f*x+e)^5

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Maxima [A]  time = 1.50809, size = 255, normalized size = 1.89 \begin{align*} -\frac{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{5 \, \sqrt{2} c^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{5 \, \sqrt{2} c^{\frac{5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - \frac{3 \, \sqrt{2} c^{\frac{5}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}}{15 \, a^{3} 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))^3,x, algorithm="maxima")

[Out]

-1/15*(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 - 5*sqrt(2)*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 5*sqrt(2)*c^(5/2)*sin(f
*x + e)^8/(cos(f*x + e) + 1)^8 - 3*sqrt(2)*c^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)/(a^3*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 = 0.471848, size = 250, normalized size = 1.85 \begin{align*} -\frac{2 \,{\left (7 \, c^{2} \cos \left (f x + e\right )^{3} + 10 \, c^{2} \cos \left (f x + e\right )^{2} + 15 \, c^{2} \cos \left (f x + e\right )\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))^(5/2)/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/15*(7*c^2*cos(f*x + e)^3 + 10*c^2*cos(f*x + e)^2 + 15*c^2*cos(f*x + e))*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(-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))**3,x)

[Out]

Timed out

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Giac [A]  time = 4.376, size = 111, normalized size = 0.82 \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}} + 10 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c + 15 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{2}\right )}}{15 \, a^{3} 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))^3,x, algorithm="giac")

[Out]

-1/15*sqrt(2)*(3*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2) + 10*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c + 15*sqrt(c*
tan(1/2*f*x + 1/2*e)^2 - c)*c^2)/(a^3*f)