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

Optimal. Leaf size=71 \[ -\frac{4 a^2 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))}-\frac{4 a^2 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}+\frac{a^2 x}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.236315, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3903, 3777, 3919, 3794, 3796, 3797} \[ -\frac{4 a^2 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))}-\frac{4 a^2 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}+\frac{a^2 x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/c^2 - (4*a^2*Tan[e + f*x])/(3*c^2*f*(1 - Sec[e + f*x])^2) - (4*a^2*Tan[e + f*x])/(3*c^2*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 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))^2} \, dx &=\frac{\int \left (\frac{a^2}{(1-\sec (e+f x))^2}+\frac{2 a^2 \sec (e+f x)}{(1-\sec (e+f x))^2}+\frac{a^2 \sec ^2(e+f x)}{(1-\sec (e+f x))^2}\right ) \, dx}{c^2}\\ &=\frac{a^2 \int \frac{1}{(1-\sec (e+f x))^2} \, dx}{c^2}+\frac{a^2 \int \frac{\sec ^2(e+f x)}{(1-\sec (e+f x))^2} \, dx}{c^2}+\frac{\left (2 a^2\right ) \int \frac{\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{c^2}\\ &=-\frac{4 a^2 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}-\frac{a^2 \int \frac{-3-\sec (e+f x)}{1-\sec (e+f x)} \, dx}{3 c^2}\\ &=\frac{a^2 x}{c^2}-\frac{4 a^2 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}+\frac{\left (4 a^2\right ) \int \frac{\sec (e+f x)}{1-\sec (e+f x)} \, dx}{3 c^2}\\ &=\frac{a^2 x}{c^2}-\frac{4 a^2 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}-\frac{4 a^2 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))}\\ \end{align*}

Mathematica [C]  time = 0.0571273, size = 53, normalized size = 0.75 \[ -\frac{2 a^2 \cot ^3\left (\frac{e}{2}+\frac{f x}{2}\right ) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-\tan ^2\left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Cot[e/2 + (f*x)/2]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[e/2 + (f*x)/2]^2])/(3*c^2*f)

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Maple [A]  time = 0.091, size = 67, normalized size = 0.9 \begin{align*} 2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{2}}}-{\frac{2\,{a}^{2}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+2\,{\frac{{a}^{2}}{f{c}^{2}\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))^2,x)

[Out]

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

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Maxima [B]  time = 1.63322, size = 235, normalized size = 3.31 \begin{align*} \frac{a^{2}{\left (\frac{12 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}} + \frac{{\left (\frac{9 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}\right )} - \frac{a^{2}{\left (\frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}} + \frac{2 \, a^{2}{\left (\frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}}{6 \, 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))^2,x, algorithm="maxima")

[Out]

1/6*(a^2*(12*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2 + (9*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)*(cos(f*
x + e) + 1)^3/(c^2*sin(f*x + e)^3)) - a^2*(3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)*(cos(f*x + e) + 1)^3/(c^
2*sin(f*x + e)^3) + 2*a^2*(3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)*(cos(f*x + e) + 1)^3/(c^2*sin(f*x + e)^3
))/f

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Fricas [A]  time = 1.03363, size = 204, normalized size = 2.87 \begin{align*} \frac{8 \, a^{2} \cos \left (f x + e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} + 3 \,{\left (a^{2} f x \cos \left (f x + e\right ) - a^{2} f x\right )} \sin \left (f x + e\right )}{3 \,{\left (c^{2} f \cos \left (f x + e\right ) - c^{2} 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))^2,x, algorithm="fricas")

[Out]

1/3*(8*a^2*cos(f*x + e)^2 + 4*a^2*cos(f*x + e) - 4*a^2 + 3*(a^2*f*x*cos(f*x + e) - a^2*f*x)*sin(f*x + e))/((c^
2*f*cos(f*x + e) - c^2*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 ^{2}{\left (e + f x \right )} - 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} - 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{1}{\sec ^{2}{\left (e + f x \right )} - 2 \sec{\left (e + f x \right )} + 1}\, dx\right )}{c^{2}} \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))**2,x)

[Out]

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

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Giac [A]  time = 1.34247, size = 81, normalized size = 1.14 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )} a^{2}}{c^{2}} + \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a^{2}\right )}}{c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}}}{3 \, 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))^2,x, algorithm="giac")

[Out]

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