3.1438 \(\int \frac {(A+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=453 \[ \frac {\sqrt {a+b} (a C+2 A b) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a b d \sqrt {\sec (c+d x)}}+\frac {a C \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{b^2 d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{b d}-\frac {C (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a b d \sqrt {\sec (c+d x)}} \]

[Out]

C*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/b/d-(a-b)*C*csc(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(
a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*
(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/b/d/sec(d*x+c)^(1/2)+(2*A*b+C*a)*csc(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/
(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)
*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/b/d/sec(d*x+c)^(1/2)+a*C*csc(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^
(1/2)/cos(d*x+c)^(1/2),(a+b)/b,((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/
2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d/sec(d*x+c)^(1/2)

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Rubi [A]  time = 0.91, antiderivative size = 453, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {4221, 3062, 3053, 2809, 2998, 2816, 2994} \[ \frac {\sqrt {a+b} (a C+2 A b) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a b d \sqrt {\sec (c+d x)}}+\frac {a C \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{b^2 d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{b d}-\frac {C (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a b d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]])/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

-(((a - b)*Sqrt[a + b]*C*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*b*d*Sqrt[Sec[c + d*x]])) + (Sqrt[a + b]*(2*A*b + a*C)*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[Ar
cSin[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*b*d*Sqrt[Sec[c + d*x]]) + (a*Sqrt[a + b]*C*Sqrt[Cos[c + d
*x]]*Csc[c + d*x]*EllipticPi[(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^2*d*Sqrt[Sec[c
+ d*x]]) + (C*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b*d)

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 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]

Rule 2998

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

Rule 3053

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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 + b*(b*B - 2*a*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, B, C}, x] && NeQ[b*c - a
*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3062

Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> -Simp[(C*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(d*f*Sqrt[a + b*Sin[e +
f*x]]), x] + Dist[1/(2*d), Int[(1*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - A*b*d)*Sin[e + f*x] - C*(b*c + a*
d)*Sin[e + f*x]^2, 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 4221

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

Rubi steps

\begin {align*} \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {C \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{b d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-a C+2 A b \cos (c+d x)-a C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{2 b}\\ &=\frac {C \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{b d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-a C+2 A b \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{2 b}-\frac {\left (a C \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}{2 b}\\ &=\frac {a \sqrt {a+b} C \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^2 d \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{b d}-\frac {\left (a C \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}{2 b}+\frac {\left ((2 A b+a C) \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}{2 b}\\ &=-\frac {(a-b) \sqrt {a+b} C \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 b d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} (2 A b+a C) \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}}}{a b d \sqrt {\sec (c+d x)}}+\frac {a \sqrt {a+b} C \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^2 d \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{b d}\\ \end {align*}

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Mathematica [A]  time = 10.86, size = 338, normalized size = 0.75 \[ \frac {\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (4 A b \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+C \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))+2 C (a+b) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-4 a C \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{b d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {a+b \cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]])/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

(Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*C*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)] + 4*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]))]*EllipticF[ArcSin[Tan[(c + d*
x)/2]], (-a + b)/(a + b)] - 4*a*C*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)] + C*Cos[c + d*x]*(a + b*Cos[c + d
*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(b*d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2])

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fricas [F]  time = 53.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {\sec \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^2 + A)*sqrt(sec(d*x + c))/sqrt(b*cos(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*sqrt(sec(d*x + c))/sqrt(b*cos(d*x + c) + a), x)

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maple [A]  time = 0.63, size = 818, normalized size = 1.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*sqrt(sec(d*x + c))/sqrt(b*cos(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*cos(c + d*x)**2)*sqrt(sec(c + d*x))/sqrt(a + b*cos(c + d*x)), x)

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