3.8 \(\int \frac{(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=102 \[ -\frac{23 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}-\frac{8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac{4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac{a^2 x}{c^3} \]

[Out]

(a^2*x)/c^3 - (4*a^2*Tan[e + f*x])/(5*c^3*f*(1 - Sec[e + f*x])^3) - (8*a^2*Tan[e + f*x])/(15*c^3*f*(1 - Sec[e
+ f*x])^2) - (23*a^2*Tan[e + f*x])/(15*c^3*f*(1 - Sec[e + f*x]))

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Rubi [A]  time = 0.329081, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ -\frac{23 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}-\frac{8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac{4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac{a^2 x}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^2/(c - c*Sec[e + f*x])^3,x]

[Out]

(a^2*x)/c^3 - (4*a^2*Tan[e + f*x])/(5*c^3*f*(1 - Sec[e + f*x])^3) - (8*a^2*Tan[e + f*x])/(15*c^3*f*(1 - Sec[e
+ f*x])^2) - (23*a^2*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)]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx &=\frac{\int \left (\frac{a^2}{(1-\sec (e+f x))^3}+\frac{2 a^2 \sec (e+f x)}{(1-\sec (e+f x))^3}+\frac{a^2 \sec ^2(e+f x)}{(1-\sec (e+f x))^3}\right ) \, dx}{c^3}\\ &=\frac{a^2 \int \frac{1}{(1-\sec (e+f x))^3} \, dx}{c^3}+\frac{a^2 \int \frac{\sec ^2(e+f x)}{(1-\sec (e+f x))^3} \, dx}{c^3}+\frac{\left (2 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{c^3}\\ &=-\frac{4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac{a^2 \int \frac{-5-2 \sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}-\frac{\left (3 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}+\frac{\left (4 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}\\ &=-\frac{4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac{8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}+\frac{a^2 \int \frac{15+7 \sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3}-\frac{a^2 \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{5 c^3}+\frac{\left (4 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3}\\ &=\frac{a^2 x}{c^3}-\frac{4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac{8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac{a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}+\frac{\left (22 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3}\\ &=\frac{a^2 x}{c^3}-\frac{4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac{8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac{23 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.61974, size = 171, normalized size = 1.68 \[ \frac{a^2 \csc \left (\frac{e}{2}\right ) \csc ^5\left (\frac{1}{2} (e+f x)\right ) \left (-360 \sin \left (e+\frac{f x}{2}\right )+280 \sin \left (e+\frac{3 f x}{2}\right )+150 \sin \left (2 e+\frac{3 f x}{2}\right )-86 \sin \left (2 e+\frac{5 f x}{2}\right )-150 f x \cos \left (e+\frac{f x}{2}\right )-75 f x \cos \left (e+\frac{3 f x}{2}\right )+75 f x \cos \left (2 e+\frac{3 f x}{2}\right )+15 f x \cos \left (2 e+\frac{5 f x}{2}\right )-15 f x \cos \left (3 e+\frac{5 f x}{2}\right )-500 \sin \left (\frac{f x}{2}\right )+150 f x \cos \left (\frac{f x}{2}\right )\right )}{480 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^2/(c - c*Sec[e + f*x])^3,x]

[Out]

(a^2*Csc[e/2]*Csc[(e + f*x)/2]^5*(150*f*x*Cos[(f*x)/2] - 150*f*x*Cos[e + (f*x)/2] - 75*f*x*Cos[e + (3*f*x)/2]
+ 75*f*x*Cos[2*e + (3*f*x)/2] + 15*f*x*Cos[2*e + (5*f*x)/2] - 15*f*x*Cos[3*e + (5*f*x)/2] - 500*Sin[(f*x)/2] -
360*Sin[e + (f*x)/2] + 280*Sin[e + (3*f*x)/2] + 150*Sin[2*e + (3*f*x)/2] - 86*Sin[2*e + (5*f*x)/2]))/(480*c^3
*f)

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Maple [A]  time = 0.1, size = 89, normalized size = 0.9 \begin{align*} 2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{3}}}+{\frac{{a}^{2}}{5\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}-{\frac{2\,{a}^{2}}{3\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+2\,{\frac{{a}^{2}}{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))^2/(c-c*sec(f*x+e))^3,x)

[Out]

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

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Maxima [B]  time = 1.78005, size = 290, normalized size = 2.84 \begin{align*} \frac{a^{2}{\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{2 \, a^{2}{\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^{2}{\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))^2/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/60*(a^2*(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)) - 2*a^2*(10*sin(f*x + e)^2/(c
os(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^2*(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 = 1.02706, size = 313, normalized size = 3.07 \begin{align*} \frac{43 \, a^{2} \cos \left (f x + e\right )^{3} - 11 \, a^{2} \cos \left (f x + e\right )^{2} - 31 \, a^{2} \cos \left (f x + e\right ) + 23 \, a^{2} + 15 \,{\left (a^{2} f x \cos \left (f x + e\right )^{2} - 2 \, a^{2} f x \cos \left (f x + e\right ) + a^{2} 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))^2/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(43*a^2*cos(f*x + e)^3 - 11*a^2*cos(f*x + e)^2 - 31*a^2*cos(f*x + e) + 23*a^2 + 15*(a^2*f*x*cos(f*x + e)^
2 - 2*a^2*f*x*cos(f*x + e) + a^2*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^{2} \left (\int \frac{2 \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{\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{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))**2/(c-c*sec(f*x+e))**3,x)

[Out]

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

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Giac [A]  time = 1.36649, size = 103, normalized size = 1.01 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )} a^{2}}{c^{3}} + \frac{30 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 10 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, a^{2}}{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))^2/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(15*(f*x + e)*a^2/c^3 + (30*a^2*tan(1/2*f*x + 1/2*e)^4 - 10*a^2*tan(1/2*f*x + 1/2*e)^2 + 3*a^2)/(c^3*tan(
1/2*f*x + 1/2*e)^5))/f