3.733 \(\int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx\)

Optimal. Leaf size=155 \[ -\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\sec (c+d x)} \sqrt {\frac {a (1-\cos (c+d x))}{a+b \cos (c+d x)}} \sqrt {\frac {a (\cos (c+d x)+1)}{a+b \cos (c+d x)}} (a+b \cos (c+d x)) \Pi \left (\frac {b}{a+b};\sin ^{-1}\left (\frac {\sqrt {a+b} \sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}\right )|-\frac {a-b}{a+b}\right )}{d \sqrt {a+b}} \]

[Out]

-2*(a+b*cos(d*x+c))*csc(d*x+c)*EllipticPi((a+b)^(1/2)*cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2),b/(a+b),((-a+b)/
(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a*(1-cos(d*x+c))/(a+b*cos(d*x+c)))^(1/2)*(a*(1+cos(d*x+c))/(a+b*cos(d*x+c)))^(
1/2)*sec(d*x+c)^(1/2)/d/(a+b)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4222, 2811} \[ -\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\sec (c+d x)} \sqrt {\frac {a (1-\cos (c+d x))}{a+b \cos (c+d x)}} \sqrt {\frac {a (\cos (c+d x)+1)}{a+b \cos (c+d x)}} (a+b \cos (c+d x)) \Pi \left (\frac {b}{a+b};\sin ^{-1}\left (\frac {\sqrt {a+b} \sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}\right )|-\frac {a-b}{a+b}\right )}{d \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]],x]

[Out]

(-2*Sqrt[Cos[c + d*x]]*Sqrt[(a*(1 - Cos[c + d*x]))/(a + b*Cos[c + d*x])]*Sqrt[(a*(1 + Cos[c + d*x]))/(a + b*Co
s[c + d*x])]*(a + b*Cos[c + d*x])*Csc[c + d*x]*EllipticPi[b/(a + b), ArcSin[(Sqrt[a + b]*Sqrt[Cos[c + d*x]])/S
qrt[a + b*Cos[c + d*x]]], -((a - b)/(a + b))]*Sqrt[Sec[c + d*x]])/(Sqrt[a + b]*d)

Rule 2811

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[
(2*(a + b*Sin[e + f*x])*Sqrt[((b*c - a*d)*(1 + Sin[e + f*x]))/((c - d)*(a + b*Sin[e + f*x]))]*Sqrt[-(((b*c - a
*d)*(1 - Sin[e + f*x]))/((c + d)*(a + b*Sin[e + f*x])))]*EllipticPi[(b*(c + d))/(d*(a + b)), ArcSin[(Rt[(a + b
)/(c + d), 2]*Sqrt[c + d*Sin[e + f*x]])/Sqrt[a + b*Sin[e + f*x]]], ((a - b)*(c + d))/((a + b)*(c - d))])/(d*f*
Rt[(a + b)/(c + d), 2]*Cos[e + f*x]), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0] && PosQ[(a + b)/(c + d)]

Rule 4222

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\frac {a (1-\cos (c+d x))}{a+b \cos (c+d x)}} \sqrt {\frac {a (1+\cos (c+d x))}{a+b \cos (c+d x)}} (a+b \cos (c+d x)) \csc (c+d x) \Pi \left (\frac {b}{a+b};\sin ^{-1}\left (\frac {\sqrt {a+b} \sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}\right )|-\frac {a-b}{a+b}\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b} d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.36, size = 146, normalized size = 0.94 \[ \frac {2 \sqrt {\sec (c+d x)} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {a+b \cos (c+d x)} \left ((a-b) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+2 b \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{d (a+b) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]],x]

[Out]

(2*Sqrt[a + b*Cos[c + d*x]]*((a - b)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*b*EllipticPi[-1
, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)])*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*Sqrt[Sec[c + d*x]])/((a +
b)*d*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)])

________________________________________________________________________________________

fricas [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((a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.24, size = 199, normalized size = 1.28 \[ \frac {2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {a +b \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \left (\EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {-\frac {a -b}{a +b}}\right ) a -\EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {-\frac {a -b}{a +b}}\right ) b +2 b \EllipticPi \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, -1, \sqrt {-\frac {a -b}{a +b}}\right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \left (\sin ^{2}\left (d x +c \right )\right )}{d \sqrt {a +b \cos \left (d x +c \right )}\, \left (-1+\cos \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(EllipticF((-1+cos(d*x+c))
/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a-EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b+2*b*EllipticP
i((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2)))/(a+b*cos(d*x+c))^(1/2)*(1/cos(d*x+c))^(1/2)*sin(d*x+c)^
2/(-1+cos(d*x+c))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*cos(c + d*x))*sqrt(sec(c + d*x)), x)

________________________________________________________________________________________