3.632 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=387 \[ -\frac{2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}-\frac{2 \sqrt{a+b} \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b^2 d}-\frac{2 \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b d \sqrt{a+b}}+\frac{2 \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b d \sqrt{a+b}} \]

[Out]

(2*Cot[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)])/(b*Sqrt[a + b]*d) - (2*Cot[c + d
*x]*EllipticF[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*Sqrt[a + b]*d) - (2*Sqrt[a + b]*Cot[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) - (2*a^2*Sin[c + d*x])/(
b*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.502568, antiderivative size = 387, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2797, 2809, 2794, 2795, 2816, 2994} \[ -\frac{2 a^2 \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}-\frac{2 \sqrt{a+b} \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b^2 d}-\frac{2 \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b d \sqrt{a+b}}+\frac{2 \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cot[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)])/(b*Sqrt[a + b]*d) - (2*Cot[c + d
*x]*EllipticF[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*Sqrt[a + b]*d) - (2*Sqrt[a + b]*Cot[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) - (2*a^2*Sin[c + d*x])/(
b*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])

Rule 2797

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Dist[d/b,
Int[Sqrt[d*Sin[e + f*x]]/Sqrt[a + b*Sin[e + f*x]], x], x] - Dist[(a*d)/b, Int[Sqrt[d*Sin[e + f*x]]/(a + b*Sin
[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^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 2794

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

Rule 2795

Int[Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Dis
t[(c - d)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] - Dist[(b*c - a*d)/(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{\cos ^{\frac{3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx &=\frac{\int \frac{\sqrt{\cos (c+d x)}}{\sqrt{a+b \cos (c+d x)}} \, dx}{b}-\frac{a \int \frac{\sqrt{\cos (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx}{b}\\ &=-\frac{2 \sqrt{a+b} \cot (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}-\frac{2 a^2 \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}+\frac{a \int \frac{\sqrt{a+b \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{2 \sqrt{a+b} \cot (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}-\frac{2 a^2 \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}-\frac{a \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}} \, dx}{b (a+b)}+\frac{a^2 \int \frac{1+\cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{2 \cot (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}}}{b \sqrt{a+b} d}-\frac{2 \cot (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}}}{b \sqrt{a+b} d}-\frac{2 \sqrt{a+b} \cot (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}-\frac{2 a^2 \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 16.4351, size = 985, normalized size = 2.55 \[ \frac{2 a \sqrt{\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 a \left (\frac{i \cos \left (\frac{1}{2} (c+d x)\right ) \sqrt{a+b \cos (c+d x)} E\left (i \sinh ^{-1}\left (\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\cos (c+d x)}}\right )|-\frac{2 a}{-a-b}\right ) \sec (c+d x)}{b \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt{\frac{(a+b \cos (c+d x)) \sec (c+d x)}{a+b}}}+\frac{2 a \left (\frac{a \sqrt{\frac{(a+b) \cot ^2\left (\frac{1}{2} (c+d x)\right )}{b-a}} \sqrt{-\frac{(a+b) \cos (c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}} \sqrt{\frac{(a+b \cos (c+d x)) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}} \csc (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b \cos (c+d x)) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}}}{\sqrt{2}}\right )|-\frac{2 a}{b-a}\right ) \sin ^4\left (\frac{1}{2} (c+d x)\right )}{(a+b) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}-\frac{a \sqrt{\frac{(a+b) \cot ^2\left (\frac{1}{2} (c+d x)\right )}{b-a}} \sqrt{-\frac{(a+b) \cos (c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}} \sqrt{\frac{(a+b \cos (c+d x)) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}} \csc (c+d x) \Pi \left (-\frac{a}{b};\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b \cos (c+d x)) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}}}{\sqrt{2}}\right )|-\frac{2 a}{b-a}\right ) \sin ^4\left (\frac{1}{2} (c+d x)\right )}{b \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}\right )}{b}+\frac{\sqrt{a+b \cos (c+d x)} \sin (c+d x)}{b \sqrt{\cos (c+d x)}}\right )-4 a b \left (\frac{\sqrt{\frac{(a+b) \cot ^2\left (\frac{1}{2} (c+d x)\right )}{b-a}} \sqrt{-\frac{(a+b) \cos (c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}} \sqrt{\frac{(a+b \cos (c+d x)) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}} \csc (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b \cos (c+d x)) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}}}{\sqrt{2}}\right )|-\frac{2 a}{b-a}\right ) \sin ^4\left (\frac{1}{2} (c+d x)\right )}{(a+b) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}-\frac{\sqrt{\frac{(a+b) \cot ^2\left (\frac{1}{2} (c+d x)\right )}{b-a}} \sqrt{-\frac{(a+b) \cos (c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}} \sqrt{\frac{(a+b \cos (c+d x)) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}} \csc (c+d x) \Pi \left (-\frac{a}{b};\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b \cos (c+d x)) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}}}{\sqrt{2}}\right )|-\frac{2 a}{b-a}\right ) \sin ^4\left (\frac{1}{2} (c+d x)\right )}{b \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}}\right )}{(a-b) (a+b) d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (-4*a*b*((Sqrt[((a + b)*Cot[(
c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(
c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-
2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[
(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[
(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sq
rt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*a*((I*Cos[(c
+ d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]], (-2*a)/(-a - b)]
*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x])*Sec[c + d*x])/(a + b)]) + (
2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[
((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c +
d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d
*x]]) - (a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sq
rt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*
x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos
[c + d*x]])))/b + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*Sqrt[Cos[c + d*x]])))/((a - b)*(a + b)*d)

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Maple [B]  time = 0.515, size = 1206, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/b/(a-b)/(a+b)*(-cos(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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b-cos(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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b)
)^(1/2))*sin(d*x+c)*b^2+cos(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))*sin(d*x+c)*a^2+cos(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))*sin(d*x+c)*a*b-2*cos(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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2+2*cos(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)*EllipticPi((-1+cos(d*x+c))/sin(d*x
+c),-1,(-(a-b)/(a+b))^(1/2))*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*b*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)*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)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*El
lipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*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*b*sin
(d*x+c)-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))*a^2*sin(d*x+c)+2*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*sin(d*x+c
)+cos(d*x+c)^2*a^2-cos(d*x+c)^2*a*b-a^2*cos(d*x+c)+cos(d*x+c)*a*b)/(a+b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)/sin
(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a)^(3/2), 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} \cos \left (d x + c\right )^{\frac{3}{2}}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{\frac{3}{2}}{\left (c + d x \right )}}{\left (a + b \cos{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(3/2)/(a+b*cos(d*x+c))**(3/2),x)

[Out]

Integral(cos(c + d*x)**(3/2)/(a + b*cos(c + d*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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