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

Optimal. Leaf size=431 \[ \frac{\sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a+b \cos (c+d x)}}{d}+\frac{\sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{d \sqrt{\sec (c+d x)}}-\frac{(a-b) \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{a d \sqrt{\sec (c+d x)}}-\frac{a \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b d \sqrt{\sec (c+d x)}} \]

[Out]

-(((a - b)*Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*
Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a
- b)])/(a*d*Sqrt[Sec[c + d*x]])) + (Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Co
s[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(
a*(1 + Sec[c + d*x]))/(a - b)])/(d*Sqrt[Sec[c + d*x]]) - (a*Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*Ellipt
icPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a
*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(b*d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b*Cos
[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

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Rubi [A]  time = 0.649834, antiderivative size = 431, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4222, 2821, 3054, 2809, 12, 2801, 2816, 2994} \[ \frac{\sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a+b \cos (c+d x)}}{d}+\frac{\sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{d \sqrt{\sec (c+d x)}}-\frac{(a-b) \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{a d \sqrt{\sec (c+d x)}}-\frac{a \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - b)*Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*
Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a
- b)])/(a*d*Sqrt[Sec[c + d*x]])) + (Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Co
s[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(
a*(1 + Sec[c + d*x]))/(a - b)])/(d*Sqrt[Sec[c + d*x]]) - (a*Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*Ellipt
icPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a
*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(b*d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b*Cos
[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/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]

Rule 2821

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[1/(d*(m + n)),
Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a^2*c*d*(m + n) + b*d*(b*c*(m - 1) + a*d*n
) + (a*d*(2*b*c + a*d)*(m + n) - b*d*(a*c - b*d*(m + n - 1)))*Sin[e + f*x] + b*d*(b*c*n + a*d*(2*m + n - 1))*S
in[e + f*x]^2, x], 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] && LtQ[0, m, 2] && LtQ[-1, n, 2] && NeQ[m + n, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3054

Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.
)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[C/b^2, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x
], x] + Dist[1/b^2, Int[(A*b^2 - a^2*C - 2*a*b*C*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e +
f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0]

Rule 2809

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2801

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :
> Dist[1/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] - Dist[b/(a - b), Int[(1 +
Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), 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]

Rule 2816

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

Rule 2994

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*A*(c - d)*Tan[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c
- d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[
(c + d)/b, 2])], -((c + d)/(c - d))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]

Rubi steps

\begin{align*} \int \frac{\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 \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)} \, dx\\ &=\frac{\sqrt{a+b \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{a b}{2}+\frac{1}{2} a b \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{b}\\ &=\frac{\sqrt{a+b \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{2} \left (a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)}}{\sqrt{a+b \cos (c+d x)}} \, dx+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int -\frac{a b}{2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{b}\\ &=-\frac{a \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{b d \sqrt{\sec (c+d x)}}+\frac{\sqrt{a+b \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}-\frac{1}{2} \left (a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx\\ &=-\frac{a \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{b d \sqrt{\sec (c+d x)}}+\frac{\sqrt{a+b \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{2} \left (a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}} \, dx-\frac{1}{2} \left (a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1+\cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx\\ &=-\frac{(a-b) \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{a d \sqrt{\sec (c+d x)}}+\frac{\sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{d \sqrt{\sec (c+d x)}}-\frac{a \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{b d \sqrt{\sec (c+d x)}}+\frac{\sqrt{a+b \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 13.5347, size = 403, normalized size = 0.94 \[ \frac{\sqrt{\frac{1}{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}} \left (-4 a \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )+2 (a+b) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )-4 a \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} \Pi \left (-1;-\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )-a \tan ^5\left (\frac{1}{2} (c+d x)\right )+a \tan \left (\frac{1}{2} (c+d x)\right )+b \tan ^5\left (\frac{1}{2} (c+d x)\right )-2 b \tan ^3\left (\frac{1}{2} (c+d x)\right )+b \tan \left (\frac{1}{2} (c+d x)\right )\right )}{2 \sqrt{2} d \left (\frac{1}{\cos (c+d x)+1}\right )^{3/2} \sqrt{a+b \cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(2*(a + b)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x]
)/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2 - 4*a
*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSi
n[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2 - 4*a*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a
+ b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Se
c[(c + d*x)/2]^2 + a*Tan[(c + d*x)/2] + b*Tan[(c + d*x)/2] - 2*b*Tan[(c + d*x)/2]^3 - a*Tan[(c + d*x)/2]^5 + b
*Tan[(c + d*x)/2]^5))/(2*Sqrt[2]*d*((1 + Cos[c + d*x])^(-1))^(3/2)*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 0.753, size = 806, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

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]

1/d*(2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x
+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*sin(d*x+c)*a-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+
b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*
sin(d*x+c)*a-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+c
os(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*sin(d*x+c)*a-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)
*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*
sin(d*x+c)*b+2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1
+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*sin(d*x+c)-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*c
os(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a*sin(d*x+c)-(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin
(d*x+c),(-(a-b)/(a+b))^(1/2))*a*sin(d*x+c)-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(
d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b*sin(d*x+c)-b*cos(d*x+c)^3-a*cos(d*
x+c)^2+b*cos(d*x+c)^2+cos(d*x+c)*a)*(1/cos(d*x+c))^(1/2)/sin(d*x+c)/(a+b*cos(d*x+c))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \cos \left (d x + c\right ) + a}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \cos \left (d x + c\right ) + a}}{\sqrt{\sec \left (d x + c\right )}}, x\right ) \end{align*}

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]

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

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

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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \cos \left (d x + c\right ) + a}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

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)