3.756 \(\int \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=505 \[ -\frac {\sqrt {a+b} \left (3 a^2+4 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{4 b^3 d \sqrt {\sec (c+d x)}}-\frac {3 a \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{4 b^2 d}-\frac {(3 a-2 b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{4 b^2 d \sqrt {\sec (c+d x)}}+\frac {3 (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{4 b^2 d \sqrt {\sec (c+d x)}}+\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{2 b d \sqrt {\sec (c+d x)}} \]

[Out]

1/2*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d/sec(d*x+c)^(1/2)-3/4*a*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^
(1/2)/b^2/d+3/4*(a-b)*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)/b^2/d/sec(d*
x+c)^(1/2)-1/4*(3*a-2*b)*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)/b^2/d/sec
(d*x+c)^(1/2)-1/4*(3*a^2+4*b^2)*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^3/d/sec(d*x+c)^(1/2)

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Rubi [A]  time = 0.95, antiderivative size = 505, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4222, 2793, 3061, 3053, 2809, 2998, 2816, 2994} \[ -\frac {\sqrt {a+b} \left (3 a^2+4 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{4 b^3 d \sqrt {\sec (c+d x)}}-\frac {3 a \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{4 b^2 d}-\frac {(3 a-2 b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{4 b^2 d \sqrt {\sec (c+d x)}}+\frac {3 (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{4 b^2 d \sqrt {\sec (c+d x)}}+\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{2 b d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(5/2)),x]

[Out]

(3*(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)])/(4*b^2*d*Sqrt[Sec[c + d*x]]) - ((3*a - 2*b)*Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSi
n[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)])/(4*b^2*d*Sqrt[Sec[c + d*x]]) - (Sqrt[a + b]*(3*a^2 + 4*b^2)*Sqr
t[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)])/(4*b^3*
d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(2*b*d*Sqrt[Sec[c + d*x]]) - (3*a*Sqrt[a + b*C
os[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*b^2*d)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 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 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 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 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 \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {\sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {a}{2}+b \cos (c+d x)-\frac {3}{2} a \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{2 b}\\ &=\frac {\sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 b d \sqrt {\sec (c+d x)}}-\frac {3 a \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3 a^2}{2}+a b \cos (c+d x)+\frac {1}{2} \left (3 a^2+4 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{4 b^2}\\ &=\frac {\sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 b d \sqrt {\sec (c+d x)}}-\frac {3 a \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3 a^2}{2}+a b \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{4 b^2}+\frac {\left (\left (3 a^2+4 b^2\right ) \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}{8 b^2}\\ &=-\frac {\sqrt {a+b} \left (3 a^2+4 b^2\right ) \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}}}{4 b^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 b d \sqrt {\sec (c+d x)}}-\frac {3 a \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {\left (3 a^2 \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}{8 b^2}-\frac {\left (a (3 a-2 b) \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}{8 b^2}\\ &=\frac {3 (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}}}{4 b^2 d \sqrt {\sec (c+d x)}}-\frac {(3 a-2 b) \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}}}{4 b^2 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (3 a^2+4 b^2\right ) \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}}}{4 b^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 b d \sqrt {\sec (c+d x)}}-\frac {3 a \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 d}\\ \end {align*}

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Mathematica [C]  time = 18.92, size = 1153, normalized size = 2.28 \[ \frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (2 (c+d x))}{4 b d}-\frac {\sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}} \left (-3 a^2 \sqrt {\frac {a-b}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 a b \sqrt {\frac {a-b}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )-6 a b \sqrt {\frac {a-b}{a+b}} \tan ^3\left (\frac {1}{2} (c+d x)\right )+6 i a^2 \Pi \left (\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|-\frac {a+b}{a-b}\right ) \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 )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} \tan ^2\left (\frac {1}{2} (c+d x)\right )+8 i b^2 \Pi \left (\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|-\frac {a+b}{a-b}\right ) \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 )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} \tan ^2\left (\frac {1}{2} (c+d x)\right )+3 a^2 \sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+3 a b \sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+3 i a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|-\frac {a+b}{a-b}\right ) \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 )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}}-2 i \left (3 a^2-b a+2 b^2\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|-\frac {a+b}{a-b}\right ) \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 )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}}+6 i a^2 \Pi \left (\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|-\frac {a+b}{a-b}\right ) \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 )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}}+8 i b^2 \Pi \left (\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|-\frac {a+b}{a-b}\right ) \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 )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}}\right )}{4 b^2 \sqrt {\frac {a-b}{a+b}} d \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) \sqrt {\frac {\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (b \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right )-a \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(5/2)),x]

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[2*(c + d*x)])/(4*b*d) - (Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 -
b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(3*a^2*Sqrt[(a - b)/(a + b)]*Tan[(c + d*x)/2] + 3*a*b*Sqrt[(a
- b)/(a + b)]*Tan[(c + d*x)/2] - 6*a*b*Sqrt[(a - b)/(a + b)]*Tan[(c + d*x)/2]^3 - 3*a^2*Sqrt[(a - b)/(a + b)]
*Tan[(c + d*x)/2]^5 + 3*a*b*Sqrt[(a - b)/(a + b)]*Tan[(c + d*x)/2]^5 + (6*I)*a^2*EllipticPi[(a + b)/(a - b), I
*ArcSinh[Sqrt[(a - b)/(a + b)]*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)] + (8*I)*b^2*EllipticPi[(a + b)/(a - b), I*ArcSinh[Sqr
t[(a - b)/(a + b)]*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)] + (6*I)*a^2*EllipticPi[(a + b)/(a - b), I*ArcSinh[Sqrt[(a - b)/(a
+ b)]*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)] + (8*I)*b^2*EllipticPi[(a + b)/(a - b), I*ArcSinh[Sqrt[(a
- b)/(a + b)]*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)] + (3*I)*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[(a - b)/
(a + b)]*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*I)*(3*a^2 - a*b + 2*b^2)*EllipticF[I*ArcSinh[
Sqrt[(a - b)/(a + b)]*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)]))/(4*b^2*Sqrt[(a - b)/(a + b)]*d*(-1
+ Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - Tan[(c + d*x)/2]^2)]*(b*(-1 + Tan[(c + d*x)/2]^2) - a
*(1 + Tan[(c + d*x)/2]^2)))

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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(1/sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(5/2)), x)

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maple [B]  time = 0.28, size = 1248, normalized size = 2.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x)

[Out]

-1/4/d*(-3*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^2-3*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*b+6*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+8*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))*b^2+2*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*b-4*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))*b^2+2*cos(d*x+c)^4*b^2-3*(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*sin(d*x+c)-3*(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*b*sin(d*x+c)+6*(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*sin(d*x+c)+8*(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))*b^2*sin(d*x+c)+2*(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)*Elliptic
F((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b*sin(d*x+c)-4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*co
s(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^2*sin(d*x+c
)-cos(d*x+c)^3*a*b-3*cos(d*x+c)^2*a^2+3*cos(d*x+c)^2*a*b-2*cos(d*x+c)^2*b^2+3*a^2*cos(d*x+c)-2*a*b*cos(d*x+c))
*cos(d*x+c)^2*(1/cos(d*x+c))^(5/2)/sin(d*x+c)/(a+b*cos(d*x+c))^(1/2)/b^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^(1/2)),x)

[Out]

int(1/((1/cos(c + d*x))^(5/2)*(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(1/sec(d*x+c)**(5/2)/(a+b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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