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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 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]

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)]

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 = 0.17, 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]))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 66, normalized size = 1.06 \[ \frac {2 \, {\left ({\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right ) + B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]

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))

________________________________________________________________________________________

giac [B]  time = 0.99, size = 129, normalized size = 2.08 \[ -\frac {2 \, {\left (3 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (3 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

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)

________________________________________________________________________________________

maple [A]  time = 1.76, size = 70, normalized size = 1.13 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (3 A \cos \left (d x +c \right )+2 B \cos \left (d x +c \right )+B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )} \]

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*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*
x+c)

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________

mupad [B]  time = 1.97, size = 159, normalized size = 2.56 \[ \frac {2\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (6\,A\,\sin \left (c+d\,x\right )+6\,B\,\sin \left (c+d\,x\right )+6\,A\,\sin \left (2\,c+2\,d\,x\right )+6\,A\,\sin \left (3\,c+3\,d\,x\right )+3\,A\,\sin \left (4\,c+4\,d\,x\right )+8\,B\,\sin \left (2\,c+2\,d\,x\right )+6\,B\,\sin \left (3\,c+3\,d\,x\right )+2\,B\,\sin \left (4\,c+4\,d\,x\right )\right )}{3\,d\,\left (12\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+4\,\cos \left (3\,c+3\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+7\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(6*A*sin(c + d*x) + 6*B*sin(c + d*x) + 6*A*sin(2*c + 2*d*x) + 6
*A*sin(3*c + 3*d*x) + 3*A*sin(4*c + 4*d*x) + 8*B*sin(2*c + 2*d*x) + 6*B*sin(3*c + 3*d*x) + 2*B*sin(4*c + 4*d*x
)))/(3*d*(12*cos(c + d*x) + 8*cos(2*c + 2*d*x) + 4*cos(3*c + 3*d*x) + cos(4*c + 4*d*x) + 7))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

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)

________________________________________________________________________________________