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

Optimal. Leaf size=96 \[ \frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 \sqrt {a} B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2955, 4015, 3801, 215} \[ \frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 \sqrt {a} B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 2955

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

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 \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\left (B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\left (2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 \sqrt {a} B \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.32, size = 94, normalized size = 0.98 \[ \frac {2 \sqrt {\cos (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (A \sqrt {1-\sec (c+d x)}-B \sqrt {\sec (c+d x)} \sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right )}{d \sqrt {1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 298, normalized size = 3.10 \[ \left [\frac {4 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (B \cos \left (d x + c\right ) + B\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {2 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (B \cos \left (d x + c\right ) + B\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{d \cos \left (d x + c\right ) + d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(4*A*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (B*cos(d*x + c) + B)*sqrt(
a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x +
c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c) + d), (2*A*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (B*cos(d*x + c) + B)*sqrt(-a)*arctan(
2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d
*x + c) - 2*a)))/(d*cos(d*x + c) + d)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)*sqrt(a*sec(d*x + c) + a)*sqrt(cos(d*x + c)), x)

________________________________________________________________________________________

maple [B]  time = 1.66, size = 169, normalized size = 1.76 \[ -\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (2 A \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+B \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}-B \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{d \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(-1+cos(d*x+c))*(2*A*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+B*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x
+c)+1+sin(d*x+c))*2^(1/2))*2^(1/2)-B*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*2
^(1/2))*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/(-2/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^2

________________________________________________________________________________________

maxima [B]  time = 0.64, size = 262, normalized size = 2.73 \[ \frac {4 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B \sqrt {a} {\left (\log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) + \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + B*sqrt(a)*(log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*
c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2
*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2
*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) +
2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin
(1/2*d*x + 1/2*c) + 2)))/d

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________