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

Optimal. Leaf size=102 \[ -\frac{26 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}+\frac{4 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac{8 a^3 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac{a^3 x}{c^3} \]

[Out]

(a^3*x)/c^3 - (8*a^3*Tan[e + f*x])/(5*c^3*f*(1 - Sec[e + f*x])^3) + (4*a^3*Tan[e + f*x])/(15*c^3*f*(1 - Sec[e
+ f*x])^2) - (26*a^3*Tan[e + f*x])/(15*c^3*f*(1 - Sec[e + f*x]))

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Rubi [A]  time = 0.451617, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {3903, 3777, 3922, 3919, 3794, 3796, 3797, 3799, 4000} \[ -\frac{26 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}+\frac{4 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac{8 a^3 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac{a^3 x}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x)/c^3 - (8*a^3*Tan[e + f*x])/(5*c^3*f*(1 - Sec[e + f*x])^3) + (4*a^3*Tan[e + f*x])/(15*c^3*f*(1 - Sec[e
+ f*x])^2) - (26*a^3*Tan[e + f*x])/(15*c^3*f*(1 - Sec[e + f*x]))

Rule 3903

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dis
t[c^n, Int[ExpandTrig[(1 + (d*csc[e + f*x])/c)^n, (a + b*csc[e + f*x])^m, x], x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0] && ILtQ[n, 0] && LtQ[m + n, 2]

Rule 3777

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(Cot[c + d*x]*(a + b*Csc[c + d*x])^n)/(d*(
2*n + 1)), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3922

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

Rule 3919

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

Rule 3794

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

Rule 3796

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)/(a*f*(2*m + 1)), x] + Dist[(m + 1)/(a*(2*m + 1)), 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] && LtQ[m, -2^(-1)] && IntegerQ[2*m]

Rule 3797

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(f*(2*m + 1)), x] + Dist[m/(b*(2*m + 1)), 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] && LtQ[m, -2^(-1)]

Rule 3799

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(
a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 4000

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

Rubi steps

\begin{align*} \int \frac{(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx &=\frac{\int \left (\frac{a^3}{(1-\sec (e+f x))^3}+\frac{3 a^3 \sec (e+f x)}{(1-\sec (e+f x))^3}+\frac{3 a^3 \sec ^2(e+f x)}{(1-\sec (e+f x))^3}+\frac{a^3 \sec ^3(e+f x)}{(1-\sec (e+f x))^3}\right ) \, dx}{c^3}\\ &=\frac{a^3 \int \frac{1}{(1-\sec (e+f x))^3} \, dx}{c^3}+\frac{a^3 \int \frac{\sec ^3(e+f x)}{(1-\sec (e+f x))^3} \, dx}{c^3}+\frac{\left (3 a^3\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{c^3}+\frac{\left (3 a^3\right ) \int \frac{\sec ^2(e+f x)}{(1-\sec (e+f x))^3} \, dx}{c^3}\\ &=-\frac{8 a^3 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac{a^3 \int \frac{-5-2 \sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}+\frac{a^3 \int \frac{(-3-5 \sec (e+f x)) \sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}+\frac{\left (6 a^3\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}-\frac{\left (9 a^3\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}\\ &=-\frac{8 a^3 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac{4 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}+\frac{a^3 \int \frac{15+7 \sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3}+\frac{\left (2 a^3\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{5 c^3}+\frac{\left (7 a^3\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3}-\frac{\left (3 a^3\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{5 c^3}\\ &=\frac{a^3 x}{c^3}-\frac{8 a^3 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac{4 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac{4 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}+\frac{\left (22 a^3\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3}\\ &=\frac{a^3 x}{c^3}-\frac{8 a^3 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac{4 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac{26 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}\\ \end{align*}

Mathematica [C]  time = 0.0847297, size = 53, normalized size = 0.52 \[ \frac{2 a^3 \cot ^5\left (\frac{e}{2}+\frac{f x}{2}\right ) \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},-\tan ^2\left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{5 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*Cot[e/2 + (f*x)/2]^5*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[e/2 + (f*x)/2]^2])/(5*c^3*f)

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Maple [A]  time = 0.104, size = 89, normalized size = 0.9 \begin{align*} 2\,{\frac{{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{3}}}-{\frac{2\,{a}^{3}}{3\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+{\frac{2\,{a}^{3}}{5\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}+2\,{\frac{{a}^{3}}{f{c}^{3}\tan \left ( 1/2\,fx+e/2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x)

[Out]

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

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Maxima [B]  time = 1.58716, size = 381, normalized size = 3.74 \begin{align*} \frac{a^{3}{\left (\frac{120 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{3}} - \frac{{\left (\frac{20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}\right )} + \frac{a^{3}{\left (\frac{10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac{3 \, a^{3}{\left (\frac{10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac{9 \, a^{3}{\left (\frac{5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/60*(a^3*(120*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^3 - (20*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 105*sin
(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5)) + a^3*(10*sin(f*x + e)^2/(cos
(f*x + e) + 1)^2 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) - 3*a
^3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/
(c^3*sin(f*x + e)^5) - 9*a^3*(5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e
)^5))/f

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Fricas [A]  time = 1.07334, size = 312, normalized size = 3.06 \begin{align*} \frac{46 \, a^{3} \cos \left (f x + e\right )^{3} - 2 \, a^{3} \cos \left (f x + e\right )^{2} - 22 \, a^{3} \cos \left (f x + e\right ) + 26 \, a^{3} + 15 \,{\left (a^{3} f x \cos \left (f x + e\right )^{2} - 2 \, a^{3} f x \cos \left (f x + e\right ) + a^{3} f x\right )} \sin \left (f x + e\right )}{15 \,{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(46*a^3*cos(f*x + e)^3 - 2*a^3*cos(f*x + e)^2 - 22*a^3*cos(f*x + e) + 26*a^3 + 15*(a^3*f*x*cos(f*x + e)^2
- 2*a^3*f*x*cos(f*x + e) + a^3*f*x)*sin(f*x + e))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(
f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.44092, size = 104, normalized size = 1.02 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )} a^{3}}{c^{3}} + \frac{2 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 5 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, a^{3}\right )}}{c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(15*(f*x + e)*a^3/c^3 + 2*(15*a^3*tan(1/2*f*x + 1/2*e)^4 - 5*a^3*tan(1/2*f*x + 1/2*e)^2 + 3*a^3)/(c^3*tan
(1/2*f*x + 1/2*e)^5))/f