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^3-8/5*a^3*tan(f*x+e)/c^3/f/(1-sec(f*x+e))^3+4/15*a^3*tan(f*x+e)/c^3/f/(1-sec(f*x+e))^2-26/15*a^3*tan(f
*x+e)/c^3/f/(1-sec(f*x+e))

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Rubi [A]  time = 0.45, antiderivative size = 102, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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*}

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Mathematica [C]  time = 0.10, size = 53, normalized size = 0.52 \[ \frac {2 a^3 \cot ^5\left (\frac {e}{2}+\frac {f x}{2}\right ) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\tan ^2\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5 c^3 f} \]

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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fricas [A]  time = 0.43, size = 128, normalized size = 1.25 \[ \frac {46 \, a^{3} \cos \left (f x + e\right )^{3} - 2 \, a^{3} \cos \left (f x + e\right )^{2} - 22 \, a^{3} \cos \left (f x + e\right ) + 26 \, a^{3} + 15 \, {\left (a^{3} f x \cos \left (f x + e\right )^{2} - 2 \, a^{3} f x \cos \left (f x + e\right ) + a^{3} f x\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \]

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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giac [A]  time = 0.40, size = 77, normalized size = 0.75 \[ \frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c^{3}} + \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 5 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{3}\right )}}{c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{15 \, f} \]

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

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maple [A]  time = 0.82, size = 89, normalized size = 0.87 \[ -\frac {2 a^{3}}{3 f \,c^{3} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {2 a^{3}}{5 f \,c^{3} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}+\frac {2 a^{3}}{f \,c^{3} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}+\frac {2 a^{3} \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,c^{3}} \]

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]

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

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maxima [B]  time = 0.44, size = 282, normalized size = 2.76 \[ \frac {a^{3} {\left (\frac {120 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{3}} - \frac {{\left (\frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}\right )} + \frac {a^{3} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac {3 \, a^{3} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac {9 \, a^{3} {\left (\frac {5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, f} \]

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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mupad [B]  time = 1.38, size = 96, normalized size = 0.94 \[ \frac {a^3\,x}{c^3}+\frac {\frac {2\,a^3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{5}-\frac {2\,a^3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{3}+2\,a^3\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{c^3\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*x)/c^3 + ((2*a^3*cos(e/2 + (f*x)/2)^5)/5 + 2*a^3*cos(e/2 + (f*x)/2)*sin(e/2 + (f*x)/2)^4 - (2*a^3*cos(e/2
+ (f*x)/2)^3*sin(e/2 + (f*x)/2)^2)/3)/(c^3*f*sin(e/2 + (f*x)/2)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {a^{3} \left (\int \frac {3 \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {1}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx\right )}{c^{3}} \]

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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