3.1415 \(\int \sqrt {a+b \cos (c+d x)} (A+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=499 \[ -\frac {(2 A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{d}-\frac {\sqrt {a+b} (2 a A-a C-2 A b) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d \sqrt {\sec (c+d x)}}+\frac {(a-b) \sqrt {a+b} (2 A-C) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{d}-\frac {a C \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{b d \sqrt {\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 1.20, antiderivative size = 499, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {4221, 3048, 3061, 3053, 2809, 2998, 2816, 2994} \[ -\frac {(2 A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{d}-\frac {\sqrt {a+b} (2 a A-a C-2 A b) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d \sqrt {\sec (c+d x)}}+\frac {(a-b) \sqrt {a+b} (2 A-C) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{d}-\frac {a C \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{b d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]

[Out]

((a - b)*Sqrt[a + b]*(2*A - 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*d*Sqrt[Sec[c + d*x]]) - (Sqrt[a + b]*(2*a*A - 2*A*b - a*C)*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*E
llipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - S
ec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*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*d*Sqrt
[Sec[c + d*x]]) + (2*A*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d - ((2*A - C)*Sqrt[a + b*Cos
[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/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 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, 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] && GtQ[m, 0] && LtQ[n, -1]

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 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(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 - d
*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - 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, B, 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 \sqrt {a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(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)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 A \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {A b}{2}-\frac {1}{2} a (A-C) \cos (c+d x)-\frac {1}{2} b (2 A-C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {2 A \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {(2 A-C) \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 {1}{2} a b (2 A-C)+A b^2 \cos (c+d x)+\frac {1}{2} a b C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{b}\\ &=\frac {2 A \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {(2 A-C) \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 {1}{2} a b (2 A-C)+A b^2 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{b}+\frac {1}{2} \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\\ &=-\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 d \sqrt {\sec (c+d x)}}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {(2 A-C) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{2} \left ((2 A b-a (2 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+\frac {1}{2} \left (a (2 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\\ &=\frac {(a-b) \sqrt {a+b} (2 A-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 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} (2 a A-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 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 d \sqrt {\sec (c+d x)}}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {(2 A-C) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 18.17, size = 699, normalized size = 1.40 \[ \frac {\sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (2 (a (A-C)+A b) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-(a+b) (2 A-C) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+2 a A \tan ^5\left (\frac {1}{2} (c+d x)\right )-2 a A \tan \left (\frac {1}{2} (c+d x)\right )+2 a C \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{a+b}} \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+2 a C \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{a+b}} \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-a C \tan ^5\left (\frac {1}{2} (c+d x)\right )+a C \tan \left (\frac {1}{2} (c+d x)\right )-2 A b \tan ^5\left (\frac {1}{2} (c+d x)\right )+4 A b \tan ^3\left (\frac {1}{2} (c+d x)\right )-2 A b \tan \left (\frac {1}{2} (c+d x)\right )+b C \tan ^5\left (\frac {1}{2} (c+d x)\right )-2 b C \tan ^3\left (\frac {1}{2} (c+d x)\right )+b C \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[a + b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]

[Out]

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

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fricas [F]  time = 1.36, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)*(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(3/2), x)

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giac [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+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)*(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.55, size = 1593, normalized size = 3.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)*(a+b*cos(d*x+c))^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)*(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\sqrt {a+b\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^(1/2),x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^(1/2), x)

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sympy [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+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2)*(a+b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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