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

Optimal. Leaf size=241 \[ -\frac{(8 A-14 B+9 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 \sqrt{a} d}+\frac{(8 A-2 B+7 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{8 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{(6 B-C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{12 d \sqrt{a \cos (c+d x)+a}}+\frac{C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((8*A - 14*B + 9*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*Sqrt[a]*d) + (Sqrt[2]*(A - B
+ C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) + ((8*A
- 2*B + 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(8*d*Sqrt[a + a*Cos[c + d*x]]) + ((6*B - C)*Cos[c + d*x]^(3/2)*
Sin[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]) + (C*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*
x]])

________________________________________________________________________________________

Rubi [A]  time = 0.845151, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3045, 2983, 2982, 2782, 205, 2774, 216} \[ -\frac{(8 A-14 B+9 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 \sqrt{a} d}+\frac{(8 A-2 B+7 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{8 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{(6 B-C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{12 d \sqrt{a \cos (c+d x)+a}}+\frac{C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*A - 14*B + 9*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*Sqrt[a]*d) + (Sqrt[2]*(A - B
+ C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) + ((8*A
- 2*B + 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(8*d*Sqrt[a + a*Cos[c + d*x]]) + ((6*B - C)*Cos[c + d*x]^(3/2)*
Sin[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]) + (C*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*
x]])

Rule 3045

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*
sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rule 2983

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2982

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

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx &=\frac{C \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{1}{2} a (6 A+5 C)+\frac{1}{2} a (6 B-C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{3 a}\\ &=\frac{(6 B-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{C \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{4} a^2 (6 B-C)+\frac{3}{4} a^2 (8 A-2 B+7 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac{(8 A-2 B+7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{(6 B-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{C \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\frac{3}{8} a^3 (8 A-2 B+7 C)-\frac{3}{8} a^3 (8 A-14 B+9 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{6 a^3}\\ &=\frac{(8 A-2 B+7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{(6 B-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{C \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}+(A-B+C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx-\frac{(8 A-14 B+9 C) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{16 a}\\ &=\frac{(8 A-2 B+7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{(6 B-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{C \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}-\frac{(2 a (A-B+C)) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{d}+\frac{(8 A-14 B+9 C) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 a d}\\ &=-\frac{(8 A-14 B+9 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 \sqrt{a} d}+\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{\sqrt{a} d}+\frac{(8 A-2 B+7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{(6 B-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{C \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.92385, size = 449, normalized size = 1.86 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (4 \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} (24 A+2 (6 B-C) \cos (c+d x)-6 B+4 C \cos (2 (c+d x))+25 C)+\frac{3 \sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (-16 i \sqrt{2} (A-B+C) \log \left (1+e^{i (c+d x)}\right )+i (8 A-14 B+9 C) \sinh ^{-1}\left (e^{i (c+d x)}\right )-8 i A \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )+16 i \sqrt{2} A \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-8 A d x+14 i B \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-16 i \sqrt{2} B \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+14 B d x-9 i C \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )+16 i \sqrt{2} C \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-9 C d x\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{48 d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*((3*Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(-8*A*d*x +
14*B*d*x - 9*C*d*x + I*(8*A - 14*B + 9*C)*ArcSinh[E^(I*(c + d*x))] - (16*I)*Sqrt[2]*(A - B + C)*Log[1 + E^(I*(
c + d*x))] - (8*I)*A*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] + (14*I)*B*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]]
- (9*I)*C*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] + (16*I)*Sqrt[2]*A*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 +
E^((2*I)*(c + d*x))]] - (16*I)*Sqrt[2]*B*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] + (
16*I)*Sqrt[2]*C*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/Sqrt[1 + E^((2*I)*(c + d*x)
)] + 4*Sqrt[Cos[c + d*x]]*(24*A - 6*B + 25*C + 2*(6*B - C)*Cos[c + d*x] + 4*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/
2]))/(48*d*Sqrt[a*(1 + Cos[c + d*x])])

________________________________________________________________________________________

Maple [B]  time = 0.125, size = 613, normalized size = 2.5 \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)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x)

[Out]

-1/24/d*(-1+cos(d*x+c))^4*(-24*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-48*A*sin(d*x+c)*cos
(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-12*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-24*A*
sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-6*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-8*C
*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+6*B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(3/2)+2*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+24*A*2^(1/2)*cos(d*x+c)^2*arcsin((-1+co
s(d*x+c))/sin(d*x+c))-24*B*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2+24*C*2^(1/2)*cos(d*x+c)^2*a
rcsin((-1+cos(d*x+c))/sin(d*x+c))-21*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+24*A*cos(d*x+
c)^2*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))-42*B*cos(d*x+c)^2*arctan(sin(d*x+c)*(cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+27*C*cos(d*x+c)^2*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
/cos(d*x+c)))*cos(d*x+c)^(3/2)*(a*(1+cos(d*x+c)))^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)/sin(d*x+c)^8/a

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{\sqrt{a \cos \left (d x + c\right ) + a}}\,{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)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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