3.229 \(\int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=130 \[ \frac {4 a (4 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \]

[Out]

2/5*a*A*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+2/15*a*(4*A+5*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a
+a*sec(d*x+c))^(1/2)+4/15*a*(4*A+5*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {4015, 3805, 3804} \[ \frac {4 a (4 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]

[Out]

(2*a*A*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(4*A + 5*B)*Sin[c + d*x])/(15*d*
Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a*(4*A + 5*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a
+ a*Sec[c + d*x]])

Rule 3804

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

Rule 3805

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(a*Cot[
e + f*x]*(d*Csc[e + f*x])^n)/(f*n*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(a*(2*n + 1))/(2*b*d*n), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rule 4015

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {1}{5} (4 A+5 B) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {1}{15} (2 (4 A+5 B)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a (4 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.32, size = 71, normalized size = 0.55 \[ \frac {a \sin (c+d x) \sqrt {\sec (c+d x)} (2 (4 A+5 B) \cos (c+d x)+3 A \cos (2 (c+d x))+19 A+20 B)}{15 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]

[Out]

(a*(19*A + 20*B + 2*(4*A + 5*B)*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sq
rt[a*(1 + Sec[c + d*x])])

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 92, normalized size = 0.71 \[ \frac {2 \, {\left (3 \, A \cos \left (d x + c\right )^{3} + {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*A*cos(d*x + c)^3 + (4*A + 5*B)*cos(d*x + c)^2 + 2*(4*A + 5*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/
cos(d*x + c))*sin(d*x + c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*sqrt(a*sec(d*x + c) + a)/sec(d*x + c)^(5/2), x)

________________________________________________________________________________________

maple [A]  time = 2.48, size = 96, normalized size = 0.74 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (3 A \left (\cos ^{2}\left (d x +c \right )\right )+4 A \cos \left (d x +c \right )+5 B \cos \left (d x +c \right )+8 A +10 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{15 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(3*A*cos(d*x+c)^2+4*A*cos(d*x+c)+5*B*cos(d*x+c)+8*A+10*B)*(a*(1+cos(d*x+c))/cos(d*x+c)
)^(1/2)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)

________________________________________________________________________________________

maxima [B]  time = 1.26, size = 317, normalized size = 2.44 \[ \frac {\sqrt {2} {\left (30 \, \cos \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \cos \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 30 \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 5 \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 6 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} A \sqrt {a} + 10 \, \sqrt {2} {\left (3 \, \cos \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 \, \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \sin \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 2 \, \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} B \sqrt {a}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/60*(sqrt(2)*(30*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*cos(2/
5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 30*cos(5/2*d*x + 5/2*c)*sin(4/5*
arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/2*d*x + 5/
2*c), cos(5/2*d*x + 5/2*c))) + 6*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x +
5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A*sqrt(a) + 10*sqrt(2)*(3*cos(2/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*cos(3/2*d*x + 3/2*c)*sin(2/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(3/2*d*x + 3/2*c) + 3*sin(1/3*arctan2(sin(3/2*d*x + 3
/2*c), cos(3/2*d*x + 3/2*c))))*B*sqrt(a))/d

________________________________________________________________________________________

mupad [B]  time = 3.38, size = 106, normalized size = 0.82 \[ \frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (35\,A\,\sin \left (c+d\,x\right )+40\,B\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (2\,c+2\,d\,x\right )+3\,A\,\sin \left (3\,c+3\,d\,x\right )+10\,B\,\sin \left (2\,c+2\,d\,x\right )\right )}{30\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

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))/(1/cos(c + d*x))^(5/2),x)

[Out]

(cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(35*A*sin(c + d*x) + 40*B*sin
(c + d*x) + 8*A*sin(2*c + 2*d*x) + 3*A*sin(3*c + 3*d*x) + 10*B*sin(2*c + 2*d*x)))/(30*d*(cos(c + d*x) + 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))*(a+a*sec(d*x+c))**(1/2)/sec(d*x+c)**(5/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x))/sec(c + d*x)**(5/2), x)

________________________________________________________________________________________