3.134 \(\int \frac{A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=68 \[ -\frac{2 (2 A-C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{A x}{a^2}-\frac{(A+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

(A*x)/a^2 - (2*(2*A - C)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A + C)*Tan[c + d*x])/(3*d*(a + a*Sec[c
+ d*x])^2)

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Rubi [A]  time = 0.120653, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4053, 3919, 3794} \[ -\frac{2 (2 A-C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{A x}{a^2}-\frac{(A+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^2,x]

[Out]

(A*x)/a^2 - (2*(2*A - C)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A + C)*Tan[c + d*x])/(3*d*(a + a*Sec[c
+ d*x])^2)

Rule 4053

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[
(a*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e
+ f*x])^(m + 1)*Simp[A*b*(2*m + 1) - a*(A*(m + 1) - C*m)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}
, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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]

Rubi steps

\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac{(A+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \frac{-3 a A+a (A-2 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac{A x}{a^2}-\frac{(A+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(2 (2 A-C)) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}\\ &=\frac{A x}{a^2}-\frac{(A+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{2 (2 A-C) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.495736, size = 141, normalized size = 2.07 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (12 A \sin \left (c+\frac{d x}{2}\right )-10 A \sin \left (c+\frac{3 d x}{2}\right )+9 A d x \cos \left (c+\frac{d x}{2}\right )+3 A d x \cos \left (c+\frac{3 d x}{2}\right )+3 A d x \cos \left (2 c+\frac{3 d x}{2}\right )-18 A \sin \left (\frac{d x}{2}\right )+9 A d x \cos \left (\frac{d x}{2}\right )+2 C \sin \left (c+\frac{3 d x}{2}\right )+6 C \sin \left (\frac{d x}{2}\right )\right )}{24 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^2,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(9*A*d*x*Cos[(d*x)/2] + 9*A*d*x*Cos[c + (d*x)/2] + 3*A*d*x*Cos[c + (3*d*x)/2] + 3
*A*d*x*Cos[2*c + (3*d*x)/2] - 18*A*Sin[(d*x)/2] + 6*C*Sin[(d*x)/2] + 12*A*Sin[c + (d*x)/2] - 10*A*Sin[c + (3*d
*x)/2] + 2*C*Sin[c + (3*d*x)/2]))/(24*a^2*d)

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Maple [A]  time = 0.057, size = 97, normalized size = 1.4 \begin{align*}{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3-3/2/d/a^2*A*tan(1/2*d*x+1/2*c)+1/2/d/a^2*C*t
an(1/2*d*x+1/2*c)+2/d/a^2*A*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.41635, size = 161, normalized size = 2.37 \begin{align*} -\frac{A{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - \frac{C{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*(A*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2) - C*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/
d

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Fricas [A]  time = 0.468711, size = 228, normalized size = 3.35 \begin{align*} \frac{3 \, A d x \cos \left (d x + c\right )^{2} + 6 \, A d x \cos \left (d x + c\right ) + 3 \, A d x -{\left ({\left (5 \, A - C\right )} \cos \left (d x + c\right ) + 4 \, A - 2 \, C\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(3*A*d*x*cos(d*x + c)^2 + 6*A*d*x*cos(d*x + c) + 3*A*d*x - ((5*A - C)*cos(d*x + c) + 4*A - 2*C)*sin(d*x +
c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c
+ d*x) + 1), x))/a**2

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Giac [A]  time = 1.19422, size = 113, normalized size = 1.66 \begin{align*} \frac{\frac{6 \,{\left (d x + c\right )} A}{a^{2}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*A/a^2 + (A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x +
1/2*c) + 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d