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

Optimal. Leaf size=164 \[ -\frac {496 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))}-\frac {181 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}-\frac {38 a^3 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}+\frac {4 a^3 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac {8 a^3 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac {a^3 x}{c^5} \]

[Out]

a^3*x/c^5-8/9*a^3*tan(f*x+e)/c^5/f/(1-sec(f*x+e))^5+4/63*a^3*tan(f*x+e)/c^5/f/(1-sec(f*x+e))^4-38/105*a^3*tan(
f*x+e)/c^5/f/(1-sec(f*x+e))^3-181/315*a^3*tan(f*x+e)/c^5/f/(1-sec(f*x+e))^2-496/315*a^3*tan(f*x+e)/c^5/f/(1-se
c(f*x+e))

________________________________________________________________________________________

Rubi [A]  time = 0.73, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 23, 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} \[ -\frac {496 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))}-\frac {181 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}-\frac {38 a^3 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}+\frac {4 a^3 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac {8 a^3 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac {a^3 x}{c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x)/c^5 - (8*a^3*Tan[e + f*x])/(9*c^5*f*(1 - Sec[e + f*x])^5) + (4*a^3*Tan[e + f*x])/(63*c^5*f*(1 - Sec[e
+ f*x])^4) - (38*a^3*Tan[e + f*x])/(105*c^5*f*(1 - Sec[e + f*x])^3) - (181*a^3*Tan[e + f*x])/(315*c^5*f*(1 - S
ec[e + f*x])^2) - (496*a^3*Tan[e + f*x])/(315*c^5*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))^5} \, dx &=\frac {\int \left (\frac {a^3}{(1-\sec (e+f x))^5}+\frac {3 a^3 \sec (e+f x)}{(1-\sec (e+f x))^5}+\frac {3 a^3 \sec ^2(e+f x)}{(1-\sec (e+f x))^5}+\frac {a^3 \sec ^3(e+f x)}{(1-\sec (e+f x))^5}\right ) \, dx}{c^5}\\ &=\frac {a^3 \int \frac {1}{(1-\sec (e+f x))^5} \, dx}{c^5}+\frac {a^3 \int \frac {\sec ^3(e+f x)}{(1-\sec (e+f x))^5} \, dx}{c^5}+\frac {\left (3 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^5} \, dx}{c^5}+\frac {\left (3 a^3\right ) \int \frac {\sec ^2(e+f x)}{(1-\sec (e+f x))^5} \, dx}{c^5}\\ &=-\frac {8 a^3 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}-\frac {a^3 \int \frac {-9-4 \sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{9 c^5}+\frac {a^3 \int \frac {(-5-9 \sec (e+f x)) \sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{9 c^5}+\frac {\left (4 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{3 c^5}-\frac {\left (5 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{3 c^5}\\ &=-\frac {8 a^3 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac {4 a^3 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}+\frac {a^3 \int \frac {63+39 \sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{63 c^5}+\frac {a^3 \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{3 c^5}+\frac {\left (4 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{7 c^5}-\frac {\left (5 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{7 c^5}\\ &=-\frac {8 a^3 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac {4 a^3 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac {38 a^3 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac {a^3 \int \frac {-315-204 \sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{315 c^5}+\frac {\left (2 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{15 c^5}+\frac {\left (8 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{35 c^5}-\frac {\left (2 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{7 c^5}\\ &=-\frac {8 a^3 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac {4 a^3 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac {38 a^3 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac {181 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}+\frac {a^3 \int \frac {945+519 \sec (e+f x)}{1-\sec (e+f x)} \, dx}{945 c^5}+\frac {\left (2 a^3\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{45 c^5}+\frac {\left (8 a^3\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{105 c^5}-\frac {\left (2 a^3\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{21 c^5}\\ &=\frac {a^3 x}{c^5}-\frac {8 a^3 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac {4 a^3 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac {38 a^3 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac {181 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}-\frac {8 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))}+\frac {\left (488 a^3\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{315 c^5}\\ &=\frac {a^3 x}{c^5}-\frac {8 a^3 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac {4 a^3 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac {38 a^3 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac {181 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}-\frac {496 a^3 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.90, size = 283, normalized size = 1.73 \[ \frac {a^3 \csc \left (\frac {e}{2}\right ) \csc ^9\left (\frac {1}{2} (e+f x)\right ) \left (-122850 \sin \left (e+\frac {f x}{2}\right )+103278 \sin \left (e+\frac {3 f x}{2}\right )+73290 \sin \left (2 e+\frac {3 f x}{2}\right )-51102 \sin \left (2 e+\frac {5 f x}{2}\right )-24570 \sin \left (3 e+\frac {5 f x}{2}\right )+13878 \sin \left (3 e+\frac {7 f x}{2}\right )+5040 \sin \left (4 e+\frac {7 f x}{2}\right )-2102 \sin \left (4 e+\frac {9 f x}{2}\right )-39690 f x \cos \left (e+\frac {f x}{2}\right )-26460 f x \cos \left (e+\frac {3 f x}{2}\right )+26460 f x \cos \left (2 e+\frac {3 f x}{2}\right )+11340 f x \cos \left (2 e+\frac {5 f x}{2}\right )-11340 f x \cos \left (3 e+\frac {5 f x}{2}\right )-2835 f x \cos \left (3 e+\frac {7 f x}{2}\right )+2835 f x \cos \left (4 e+\frac {7 f x}{2}\right )+315 f x \cos \left (4 e+\frac {9 f x}{2}\right )-315 f x \cos \left (5 e+\frac {9 f x}{2}\right )-142002 \sin \left (\frac {f x}{2}\right )+39690 f x \cos \left (\frac {f x}{2}\right )\right )}{161280 c^5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Csc[e/2]*Csc[(e + f*x)/2]^9*(39690*f*x*Cos[(f*x)/2] - 39690*f*x*Cos[e + (f*x)/2] - 26460*f*x*Cos[e + (3*f
*x)/2] + 26460*f*x*Cos[2*e + (3*f*x)/2] + 11340*f*x*Cos[2*e + (5*f*x)/2] - 11340*f*x*Cos[3*e + (5*f*x)/2] - 28
35*f*x*Cos[3*e + (7*f*x)/2] + 2835*f*x*Cos[4*e + (7*f*x)/2] + 315*f*x*Cos[4*e + (9*f*x)/2] - 315*f*x*Cos[5*e +
(9*f*x)/2] - 142002*Sin[(f*x)/2] - 122850*Sin[e + (f*x)/2] + 103278*Sin[e + (3*f*x)/2] + 73290*Sin[2*e + (3*f
*x)/2] - 51102*Sin[2*e + (5*f*x)/2] - 24570*Sin[3*e + (5*f*x)/2] + 13878*Sin[3*e + (7*f*x)/2] + 5040*Sin[4*e +
(7*f*x)/2] - 2102*Sin[4*e + (9*f*x)/2]))/(161280*c^5*f)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 212, normalized size = 1.29 \[ \frac {1051 \, a^{3} \cos \left (f x + e\right )^{5} - 1684 \, a^{3} \cos \left (f x + e\right )^{4} + 898 \, a^{3} \cos \left (f x + e\right )^{3} + 1468 \, a^{3} \cos \left (f x + e\right )^{2} - 1669 \, a^{3} \cos \left (f x + e\right ) + 496 \, a^{3} + 315 \, {\left (a^{3} f x \cos \left (f x + e\right )^{4} - 4 \, a^{3} f x \cos \left (f x + e\right )^{3} + 6 \, a^{3} f x \cos \left (f x + e\right )^{2} - 4 \, a^{3} f x \cos \left (f x + e\right ) + a^{3} f x\right )} \sin \left (f x + e\right )}{315 \, {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} + 6 \, c^{5} f \cos \left (f x + e\right )^{2} - 4 \, c^{5} f \cos \left (f x + e\right ) + c^{5} f\right )} \sin \left (f x + e\right )} \]

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))^5,x, algorithm="fricas")

[Out]

1/315*(1051*a^3*cos(f*x + e)^5 - 1684*a^3*cos(f*x + e)^4 + 898*a^3*cos(f*x + e)^3 + 1468*a^3*cos(f*x + e)^2 -
1669*a^3*cos(f*x + e) + 496*a^3 + 315*(a^3*f*x*cos(f*x + e)^4 - 4*a^3*f*x*cos(f*x + e)^3 + 6*a^3*f*x*cos(f*x +
e)^2 - 4*a^3*f*x*cos(f*x + e) + a^3*f*x)*sin(f*x + e))/((c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 + 6*c^
5*f*cos(f*x + e)^2 - 4*c^5*f*cos(f*x + e) + c^5*f)*sin(f*x + e))

________________________________________________________________________________________

giac [A]  time = 0.56, size = 110, normalized size = 0.67 \[ \frac {\frac {630 \, {\left (f x + e\right )} a^{3}}{c^{5}} + \frac {1260 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 420 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 252 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 135 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 35 \, a^{3}}{c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}}}{630 \, f} \]

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))^5,x, algorithm="giac")

[Out]

1/630*(630*(f*x + e)*a^3/c^5 + (1260*a^3*tan(1/2*f*x + 1/2*e)^8 - 420*a^3*tan(1/2*f*x + 1/2*e)^6 + 252*a^3*tan
(1/2*f*x + 1/2*e)^4 - 135*a^3*tan(1/2*f*x + 1/2*e)^2 + 35*a^3)/(c^5*tan(1/2*f*x + 1/2*e)^9))/f

________________________________________________________________________________________

maple [A]  time = 0.89, size = 133, normalized size = 0.81 \[ \frac {a^{3}}{18 f \,c^{5} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{9}}-\frac {3 a^{3}}{14 f \,c^{5} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{7}}+\frac {2 a^{3}}{5 f \,c^{5} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}-\frac {2 a^{3}}{3 f \,c^{5} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {2 a^{3}}{f \,c^{5} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}+\frac {2 a^{3} \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,c^{5}} \]

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))^5,x)

[Out]

1/18/f*a^3/c^5/tan(1/2*e+1/2*f*x)^9-3/14/f*a^3/c^5/tan(1/2*e+1/2*f*x)^7+2/5/f*a^3/c^5/tan(1/2*e+1/2*f*x)^5-2/3
/f*a^3/c^5/tan(1/2*e+1/2*f*x)^3+2/f*a^3/c^5/tan(1/2*e+1/2*f*x)+2/f*a^3/c^5*arctan(tan(1/2*e+1/2*f*x))

________________________________________________________________________________________

maxima [B]  time = 0.46, size = 403, normalized size = 2.46 \[ \frac {a^{3} {\left (\frac {10080 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{5}} - \frac {{\left (\frac {270 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {1008 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {2730 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {9765 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 35\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}}\right )} - \frac {3 \, a^{3} {\left (\frac {180 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {378 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {420 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {315 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 35\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} - \frac {15 \, a^{3} {\left (\frac {18 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {42 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {63 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 7\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} - \frac {7 \, a^{3} {\left (\frac {18 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {45 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}}}{5040 \, 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))^5,x, algorithm="maxima")

[Out]

1/5040*(a^3*(10080*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^5 - (270*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10
08*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 2730*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 9765*sin(f*x + e)^8/(cos(f
*x + e) + 1)^8 - 35)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x + e)^9)) - 3*a^3*(180*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 - 378*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 420*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 315*sin(f*x + e)^8/(c
os(f*x + e) + 1)^8 - 35)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x + e)^9) - 15*a^3*(18*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 - 42*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 63*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 7)*(cos(f*x + e) + 1
)^9/(c^5*sin(f*x + e)^9) - 7*a^3*(18*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 45*sin(f*x + e)^8/(cos(f*x + e) + 1
)^8 - 5)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x + e)^9))/f

________________________________________________________________________________________

mupad [B]  time = 1.46, size = 146, normalized size = 0.89 \[ \frac {a^3\,\left (\frac {{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{18}-\frac {3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{14}+\frac {2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{5}-\frac {2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6}{3}+2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+\left (e+f\,x\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9\right )}{c^5\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9} \]

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))^5,x)

[Out]

(a^3*(cos(e/2 + (f*x)/2)^9/18 + 2*cos(e/2 + (f*x)/2)*sin(e/2 + (f*x)/2)^8 + sin(e/2 + (f*x)/2)^9*(e + f*x) - (
2*cos(e/2 + (f*x)/2)^3*sin(e/2 + (f*x)/2)^6)/3 + (2*cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2)^4)/5 - (3*cos(e/2
+ (f*x)/2)^7*sin(e/2 + (f*x)/2)^2)/14))/(c^5*f*sin(e/2 + (f*x)/2)^9)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {a^{3} \left (\int \frac {3 \sec {\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {1}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx\right )}{c^{5}} \]

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))**5,x)

[Out]

-a**3*(Integral(3*sec(e + f*x)/(sec(e + f*x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2
+ 5*sec(e + f*x) - 1), x) + Integral(3*sec(e + f*x)**2/(sec(e + f*x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)*
*3 - 10*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x) + Integral(sec(e + f*x)**3/(sec(e + f*x)**5 - 5*sec(e + f*x)
**4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x) + Integral(1/(sec(e + f*x)**5 - 5*sec(
e + f*x)**4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x))/c**5

________________________________________________________________________________________