3.9 \(\int \frac{(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=133 \[ -\frac{164 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))}-\frac{59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac{12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac{4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}+\frac{a^2 x}{c^4} \]

[Out]

(a^2*x)/c^4 - (4*a^2*Tan[e + f*x])/(7*c^4*f*(1 - Sec[e + f*x])^4) - (12*a^2*Tan[e + f*x])/(35*c^4*f*(1 - Sec[e
+ f*x])^3) - (59*a^2*Tan[e + f*x])/(105*c^4*f*(1 - Sec[e + f*x])^2) - (164*a^2*Tan[e + f*x])/(105*c^4*f*(1 -
Sec[e + f*x]))

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Rubi [A]  time = 0.42722, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 15, 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} \[ -\frac{164 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))}-\frac{59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac{12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac{4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}+\frac{a^2 x}{c^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^2/(c - c*Sec[e + f*x])^4,x]

[Out]

(a^2*x)/c^4 - (4*a^2*Tan[e + f*x])/(7*c^4*f*(1 - Sec[e + f*x])^4) - (12*a^2*Tan[e + f*x])/(35*c^4*f*(1 - Sec[e
+ f*x])^3) - (59*a^2*Tan[e + f*x])/(105*c^4*f*(1 - Sec[e + f*x])^2) - (164*a^2*Tan[e + f*x])/(105*c^4*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))^4} \, dx &=\frac{\int \left (\frac{a^2}{(1-\sec (e+f x))^4}+\frac{2 a^2 \sec (e+f x)}{(1-\sec (e+f x))^4}+\frac{a^2 \sec ^2(e+f x)}{(1-\sec (e+f x))^4}\right ) \, dx}{c^4}\\ &=\frac{a^2 \int \frac{1}{(1-\sec (e+f x))^4} \, dx}{c^4}+\frac{a^2 \int \frac{\sec ^2(e+f x)}{(1-\sec (e+f x))^4} \, dx}{c^4}+\frac{\left (2 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{c^4}\\ &=-\frac{4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac{a^2 \int \frac{-7-3 \sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{7 c^4}-\frac{\left (4 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{7 c^4}+\frac{\left (6 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{7 c^4}\\ &=-\frac{4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac{12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}+\frac{a^2 \int \frac{35+20 \sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{35 c^4}-\frac{\left (8 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{35 c^4}+\frac{\left (12 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{35 c^4}\\ &=-\frac{4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac{12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac{59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac{a^2 \int \frac{-105-55 \sec (e+f x)}{1-\sec (e+f x)} \, dx}{105 c^4}-\frac{\left (8 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{105 c^4}+\frac{\left (4 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{35 c^4}\\ &=\frac{a^2 x}{c^4}-\frac{4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac{12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac{59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac{4 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))}+\frac{\left (32 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{21 c^4}\\ &=\frac{a^2 x}{c^4}-\frac{4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac{12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac{59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac{164 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.615846, size = 227, normalized size = 1.71 \[ \frac{a^2 \csc \left (\frac{e}{2}\right ) \csc ^7\left (\frac{1}{2} (e+f x)\right ) \left (-10430 \sin \left (e+\frac{f x}{2}\right )+8568 \sin \left (e+\frac{3 f x}{2}\right )+4830 \sin \left (2 e+\frac{3 f x}{2}\right )-3206 \sin \left (2 e+\frac{5 f x}{2}\right )-1260 \sin \left (3 e+\frac{5 f x}{2}\right )+638 \sin \left (3 e+\frac{7 f x}{2}\right )-3675 f x \cos \left (e+\frac{f x}{2}\right )-2205 f x \cos \left (e+\frac{3 f x}{2}\right )+2205 f x \cos \left (2 e+\frac{3 f x}{2}\right )+735 f x \cos \left (2 e+\frac{5 f x}{2}\right )-735 f x \cos \left (3 e+\frac{5 f x}{2}\right )-105 f x \cos \left (3 e+\frac{7 f x}{2}\right )+105 f x \cos \left (4 e+\frac{7 f x}{2}\right )-11900 \sin \left (\frac{f x}{2}\right )+3675 f x \cos \left (\frac{f x}{2}\right )\right )}{13440 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^2/(c - c*Sec[e + f*x])^4,x]

[Out]

(a^2*Csc[e/2]*Csc[(e + f*x)/2]^7*(3675*f*x*Cos[(f*x)/2] - 3675*f*x*Cos[e + (f*x)/2] - 2205*f*x*Cos[e + (3*f*x)
/2] + 2205*f*x*Cos[2*e + (3*f*x)/2] + 735*f*x*Cos[2*e + (5*f*x)/2] - 735*f*x*Cos[3*e + (5*f*x)/2] - 105*f*x*Co
s[3*e + (7*f*x)/2] + 105*f*x*Cos[4*e + (7*f*x)/2] - 11900*Sin[(f*x)/2] - 10430*Sin[e + (f*x)/2] + 8568*Sin[e +
(3*f*x)/2] + 4830*Sin[2*e + (3*f*x)/2] - 3206*Sin[2*e + (5*f*x)/2] - 1260*Sin[3*e + (5*f*x)/2] + 638*Sin[3*e
+ (7*f*x)/2]))/(13440*c^4*f)

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Maple [A]  time = 0.109, size = 111, normalized size = 0.8 \begin{align*} 2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{4}}}-{\frac{{a}^{2}}{14\,f{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-7}}+{\frac{3\,{a}^{2}}{10\,f{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}-{\frac{2\,{a}^{2}}{3\,f{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+2\,{\frac{{a}^{2}}{f{c}^{4}\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))^4,x)

[Out]

2/f*a^2/c^4*arctan(tan(1/2*f*x+1/2*e))-1/14/f*a^2/c^4/tan(1/2*f*x+1/2*e)^7+3/10/f*a^2/c^4/tan(1/2*f*x+1/2*e)^5
-2/3/f*a^2/c^4/tan(1/2*f*x+1/2*e)^3+2/f*a^2/c^4/tan(1/2*f*x+1/2*e)

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Maxima [B]  time = 1.6022, size = 397, normalized size = 2.98 \begin{align*} \frac{5 \, a^{2}{\left (\frac{336 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{4}} + \frac{{\left (\frac{21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{77 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{315 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}\right )} + \frac{a^{2}{\left (\frac{21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 15\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}} + \frac{6 \, a^{2}{\left (\frac{21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{35 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}}{840 \, 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))^4,x, algorithm="maxima")

[Out]

1/840*(5*a^2*(336*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^4 + (21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 77*s
in(f*x + e)^4/(cos(f*x + e) + 1)^4 + 315*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 3)*(cos(f*x + e) + 1)^7/(c^4*si
n(f*x + e)^7)) + a^2*(21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 105*si
n(f*x + e)^6/(cos(f*x + e) + 1)^6 - 15)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7) + 6*a^2*(21*sin(f*x + e)^2/(
cos(f*x + e) + 1)^2 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 5)*(co
s(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7))/f

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Fricas [A]  time = 1.05406, size = 424, normalized size = 3.19 \begin{align*} \frac{319 \, a^{2} \cos \left (f x + e\right )^{4} - 327 \, a^{2} \cos \left (f x + e\right )^{3} - 95 \, a^{2} \cos \left (f x + e\right )^{2} + 387 \, a^{2} \cos \left (f x + e\right ) - 164 \, a^{2} + 105 \,{\left (a^{2} f x \cos \left (f x + e\right )^{3} - 3 \, a^{2} f x \cos \left (f x + e\right )^{2} + 3 \, a^{2} f x \cos \left (f x + e\right ) - a^{2} f x\right )} \sin \left (f x + e\right )}{105 \,{\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} 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))^4,x, algorithm="fricas")

[Out]

1/105*(319*a^2*cos(f*x + e)^4 - 327*a^2*cos(f*x + e)^3 - 95*a^2*cos(f*x + e)^2 + 387*a^2*cos(f*x + e) - 164*a^
2 + 105*(a^2*f*x*cos(f*x + e)^3 - 3*a^2*f*x*cos(f*x + e)^2 + 3*a^2*f*x*cos(f*x + e) - a^2*f*x)*sin(f*x + e))/(
(c^4*f*cos(f*x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*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 ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{2}{\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{1}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec{\left (e + f x \right )} + 1}\, dx\right )}{c^{4}} \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))**4,x)

[Out]

a**2*(Integral(2*sec(e + f*x)/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1),
x) + Integral(sec(e + f*x)**2/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1),
x) + Integral(1/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1), x))/c**4

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Giac [A]  time = 1.36476, size = 126, normalized size = 0.95 \begin{align*} \frac{\frac{210 \,{\left (f x + e\right )} a^{2}}{c^{4}} + \frac{420 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 140 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 63 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 15 \, a^{2}}{c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7}}}{210 \, 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))^4,x, algorithm="giac")

[Out]

1/210*(210*(f*x + e)*a^2/c^4 + (420*a^2*tan(1/2*f*x + 1/2*e)^6 - 140*a^2*tan(1/2*f*x + 1/2*e)^4 + 63*a^2*tan(1
/2*f*x + 1/2*e)^2 - 15*a^2)/(c^4*tan(1/2*f*x + 1/2*e)^7))/f