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

Optimal. Leaf size=65 \[ \frac{(A+2 B) \tan (c+d x)}{3 d \left (a^2 \sec (c+d x)+a^2\right )}+\frac{(A-B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.0797186, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {4000, 3794} \[ \frac{(A+2 B) \tan (c+d x)}{3 d \left (a^2 \sec (c+d x)+a^2\right )}+\frac{(A-B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4000

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

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

Mathematica [A]  time = 0.199152, size = 76, normalized size = 1.17 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left ((2 A+B) \sin \left (c+\frac{3 d x}{2}\right )+3 (A+B) \sin \left (\frac{d x}{2}\right )-3 A \sin \left (c+\frac{d x}{2}\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(3*(A + B)*Sin[(d*x)/2] - 3*A*Sin[c + (d*x)/2] + (2*A + B)*Sin[c + (3*d*x)/2]))/(3*
a^2*d*(1 + Cos[c + d*x])^2)

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Maple [A]  time = 0.049, size = 60, normalized size = 0.9 \begin{align*}{\frac{1}{2\,d{a}^{2}} \left ( -{\frac{A}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+A\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +B\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/a^2*(-1/3*A*tan(1/2*d*x+1/2*c)^3+1/3*B*tan(1/2*d*x+1/2*c)^3+A*tan(1/2*d*x+1/2*c)+B*tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 0.977688, size = 126, normalized size = 1.94 \begin{align*} \frac{\frac{B{\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}} + \frac{A{\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(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(B*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 + A*(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 = 0.43814, size = 144, normalized size = 2.22 \begin{align*} \frac{{\left ({\left (2 \, A + B\right )} \cos \left (d x + c\right ) + A + 2 \, B\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(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*((2*A + B)*cos(d*x + c) + A + 2*B)*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{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{B \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(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*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 = 1.23922, size = 81, normalized size = 1.25 \begin{align*} -\frac{A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{6 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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