3.121 \(\int \sec (c+d x) \sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=62 \[ \frac{2 a (3 A+B) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 B \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0944036, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {4001, 3792} \[ \frac{2 a (3 A+B) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 B \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]

[Out]

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

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, 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 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx &=\frac{2 B \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 d}+\frac{1}{3} (3 A+B) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (3 A+B) \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 B \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.158314, size = 53, normalized size = 0.85 \[ \frac{2 \tan (c+d x) \sqrt{a (\sec (c+d x)+1)} ((3 A+2 B) \cos (c+d x)+B)}{3 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]

[Out]

(2*(B + (3*A + 2*B)*Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(3*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 0.273, size = 70, normalized size = 1.1 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 3\,A\cos \left ( dx+c \right ) +2\,B\cos \left ( dx+c \right ) +B \right ) }{3\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x)

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.462124, size = 169, normalized size = 2.73 \begin{align*} \frac{2 \,{\left ({\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right ) + B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/3*((3*A + 2*B)*cos(d*x + c) + B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^2 + d*
cos(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x))*sec(c + d*x), x)

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Giac [B]  time = 4.60645, size = 174, normalized size = 2.81 \begin{align*} -\frac{2 \,{\left (3 \, \sqrt{2} A a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt{2} B a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (3 \, \sqrt{2} A a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt{2} B a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{3 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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