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

Optimal. Leaf size=164 \[ -\frac{494 a^2 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))}-\frac{179 a^2 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}-\frac{37 a^2 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac{16 a^2 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac{4 a^2 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac{a^2 x}{c^5} \]

[Out]

(a^2*x)/c^5 - (4*a^2*Tan[e + f*x])/(9*c^5*f*(1 - Sec[e + f*x])^5) - (16*a^2*Tan[e + f*x])/(63*c^5*f*(1 - Sec[e
+ f*x])^4) - (37*a^2*Tan[e + f*x])/(105*c^5*f*(1 - Sec[e + f*x])^3) - (179*a^2*Tan[e + f*x])/(315*c^5*f*(1 -
Sec[e + f*x])^2) - (494*a^2*Tan[e + f*x])/(315*c^5*f*(1 - Sec[e + f*x]))

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Rubi [A]  time = 0.542792, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {3903, 3777, 3922, 3919, 3794, 3796, 3797} \[ -\frac{494 a^2 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))}-\frac{179 a^2 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}-\frac{37 a^2 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac{16 a^2 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac{4 a^2 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}+\frac{a^2 x}{c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/c^5 - (4*a^2*Tan[e + f*x])/(9*c^5*f*(1 - Sec[e + f*x])^5) - (16*a^2*Tan[e + f*x])/(63*c^5*f*(1 - Sec[e
+ f*x])^4) - (37*a^2*Tan[e + f*x])/(105*c^5*f*(1 - Sec[e + f*x])^3) - (179*a^2*Tan[e + f*x])/(315*c^5*f*(1 -
Sec[e + f*x])^2) - (494*a^2*Tan[e + f*x])/(315*c^5*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)]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^5} \, dx &=\frac{\int \left (\frac{a^2}{(1-\sec (e+f x))^5}+\frac{2 a^2 \sec (e+f x)}{(1-\sec (e+f x))^5}+\frac{a^2 \sec ^2(e+f x)}{(1-\sec (e+f x))^5}\right ) \, dx}{c^5}\\ &=\frac{a^2 \int \frac{1}{(1-\sec (e+f x))^5} \, dx}{c^5}+\frac{a^2 \int \frac{\sec ^2(e+f x)}{(1-\sec (e+f x))^5} \, dx}{c^5}+\frac{\left (2 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^5} \, dx}{c^5}\\ &=-\frac{4 a^2 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}-\frac{a^2 \int \frac{-9-4 \sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{9 c^5}-\frac{\left (5 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{9 c^5}+\frac{\left (8 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{9 c^5}\\ &=-\frac{4 a^2 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}-\frac{16 a^2 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}+\frac{a^2 \int \frac{63+39 \sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{63 c^5}-\frac{\left (5 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{21 c^5}+\frac{\left (8 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{21 c^5}\\ &=-\frac{4 a^2 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}-\frac{16 a^2 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac{37 a^2 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac{a^2 \int \frac{-315-204 \sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{315 c^5}-\frac{\left (2 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{21 c^5}+\frac{\left (16 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{105 c^5}\\ &=-\frac{4 a^2 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}-\frac{16 a^2 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac{37 a^2 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac{179 a^2 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}+\frac{a^2 \int \frac{945+519 \sec (e+f x)}{1-\sec (e+f x)} \, dx}{945 c^5}-\frac{\left (2 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{63 c^5}+\frac{\left (16 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{315 c^5}\\ &=\frac{a^2 x}{c^5}-\frac{4 a^2 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}-\frac{16 a^2 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac{37 a^2 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac{179 a^2 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}-\frac{2 a^2 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))}+\frac{\left (488 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{315 c^5}\\ &=\frac{a^2 x}{c^5}-\frac{4 a^2 \tan (e+f x)}{9 c^5 f (1-\sec (e+f x))^5}-\frac{16 a^2 \tan (e+f x)}{63 c^5 f (1-\sec (e+f x))^4}-\frac{37 a^2 \tan (e+f x)}{105 c^5 f (1-\sec (e+f x))^3}-\frac{179 a^2 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))^2}-\frac{494 a^2 \tan (e+f x)}{315 c^5 f (1-\sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.886377, size = 283, normalized size = 1.73 \[ \frac{a^2 \csc \left (\frac{e}{2}\right ) \csc ^9\left (\frac{1}{2} (e+f x)\right ) \left (-117810 \sin \left (e+\frac{f x}{2}\right )+100002 \sin \left (e+\frac{3 f x}{2}\right )+68670 \sin \left (2 e+\frac{3 f x}{2}\right )-48978 \sin \left (2 e+\frac{5 f x}{2}\right )-23310 \sin \left (3 e+\frac{5 f x}{2}\right )+13662 \sin \left (3 e+\frac{7 f x}{2}\right )+4410 \sin \left (4 e+\frac{7 f x}{2}\right )-2008 \sin \left (4 e+\frac{9 f x}{2}\right )-39690 f x \cos \left (e+\frac{f x}{2}\right )-26460 f x \cos \left (e+\frac{3 f x}{2}\right )+26460 f x \cos \left (2 e+\frac{3 f x}{2}\right )+11340 f x \cos \left (2 e+\frac{5 f x}{2}\right )-11340 f x \cos \left (3 e+\frac{5 f x}{2}\right )-2835 f x \cos \left (3 e+\frac{7 f x}{2}\right )+2835 f x \cos \left (4 e+\frac{7 f x}{2}\right )+315 f x \cos \left (4 e+\frac{9 f x}{2}\right )-315 f x \cos \left (5 e+\frac{9 f x}{2}\right )-135198 \sin \left (\frac{f x}{2}\right )+39690 f x \cos \left (\frac{f x}{2}\right )\right )}{161280 c^5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*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] - 135198*Sin[(f*x)/2] - 117810*Sin[e + (f*x)/2] + 100002*Sin[e + (3*f*x)/2] + 68670*Sin[2*e + (3*f
*x)/2] - 48978*Sin[2*e + (5*f*x)/2] - 23310*Sin[3*e + (5*f*x)/2] + 13662*Sin[3*e + (7*f*x)/2] + 4410*Sin[4*e +
(7*f*x)/2] - 2008*Sin[4*e + (9*f*x)/2]))/(161280*c^5*f)

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Maple [A]  time = 0.119, size = 133, normalized size = 0.8 \begin{align*} 2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{5}}}+{\frac{{a}^{2}}{36\,f{c}^{5}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-9}}-{\frac{{a}^{2}}{7\,f{c}^{5}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-7}}+{\frac{7\,{a}^{2}}{20\,f{c}^{5}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}-{\frac{2\,{a}^{2}}{3\,f{c}^{5}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+2\,{\frac{{a}^{2}}{f{c}^{5}\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))^2/(c-c*sec(f*x+e))^5,x)

[Out]

2/f*a^2/c^5*arctan(tan(1/2*f*x+1/2*e))+1/36/f*a^2/c^5/tan(1/2*f*x+1/2*e)^9-1/7/f*a^2/c^5/tan(1/2*f*x+1/2*e)^7+
7/20/f*a^2/c^5/tan(1/2*f*x+1/2*e)^5-2/3/f*a^2/c^5/tan(1/2*f*x+1/2*e)^3+2/f*a^2/c^5/tan(1/2*f*x+1/2*e)

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Maxima [B]  time = 1.63463, size = 452, normalized size = 2.76 \begin{align*} \frac{a^{2}{\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{2 \, a^{2}{\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{5 \, a^{2}{\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}}}{5040 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^5,x, algorithm="maxima")

[Out]

1/5040*(a^2*(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)) - 2*a^2*(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) - 5*a^2*(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))/f

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Fricas [A]  time = 1.04967, size = 535, normalized size = 3.26 \begin{align*} \frac{1004 \, a^{2} \cos \left (f x + e\right )^{5} - 1811 \, a^{2} \cos \left (f x + e\right )^{4} + 797 \, a^{2} \cos \left (f x + e\right )^{3} + 1457 \, a^{2} \cos \left (f x + e\right )^{2} - 1661 \, a^{2} \cos \left (f x + e\right ) + 494 \, a^{2} + 315 \,{\left (a^{2} f x \cos \left (f x + e\right )^{4} - 4 \, a^{2} f x \cos \left (f x + e\right )^{3} + 6 \, a^{2} f x \cos \left (f x + e\right )^{2} - 4 \, a^{2} f x \cos \left (f x + e\right ) + a^{2} 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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^5,x, algorithm="fricas")

[Out]

1/315*(1004*a^2*cos(f*x + e)^5 - 1811*a^2*cos(f*x + e)^4 + 797*a^2*cos(f*x + e)^3 + 1457*a^2*cos(f*x + e)^2 -
1661*a^2*cos(f*x + e) + 494*a^2 + 315*(a^2*f*x*cos(f*x + e)^4 - 4*a^2*f*x*cos(f*x + e)^3 + 6*a^2*f*x*cos(f*x +
e)^2 - 4*a^2*f*x*cos(f*x + e) + a^2*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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{a^{2} \left (\int \frac{2 \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{\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{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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**5,x)

[Out]

-a**2*(Integral(2*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(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(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

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Giac [A]  time = 1.65229, size = 149, normalized size = 0.91 \begin{align*} \frac{\frac{1260 \,{\left (f x + e\right )} a^{2}}{c^{5}} + \frac{2520 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 840 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 441 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 180 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 35 \, a^{2}}{c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9}}}{1260 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^5,x, algorithm="giac")

[Out]

1/1260*(1260*(f*x + e)*a^2/c^5 + (2520*a^2*tan(1/2*f*x + 1/2*e)^8 - 840*a^2*tan(1/2*f*x + 1/2*e)^6 + 441*a^2*t
an(1/2*f*x + 1/2*e)^4 - 180*a^2*tan(1/2*f*x + 1/2*e)^2 + 35*a^2)/(c^5*tan(1/2*f*x + 1/2*e)^9))/f