3.517 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=254 \[ \frac{(3 A-11 B+35 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(3 A-43 B+115 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(2 B-5 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(A-B+C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{(A+7 B-15 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

((2*B - 5*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) + ((3*A - 43*B + 115*C)*ArcT
an[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A
- B + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((A + 7*B - 15*C)*Cos[c + d*x]^(
3/2)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + ((3*A - 11*B + 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.87189, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.178, Rules used = {3041, 2977, 2983, 2982, 2782, 205, 2774, 216} \[ \frac{(3 A-11 B+35 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(3 A-43 B+115 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(2 B-5 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(A-B+C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{(A+7 B-15 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

((2*B - 5*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) + ((3*A - 43*B + 115*C)*ArcT
an[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A
- B + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((A + 7*B - 15*C)*Cos[c + d*x]^(
3/2)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + ((3*A - 11*B + 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 3041

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

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 )}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A-B+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{1}{2} a (3 A+5 B-5 C)+a (A-B+5 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(A+7 B-15 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{4} a^2 (A+7 B-15 C)+\frac{1}{2} a^2 (3 A-11 B+35 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(A+7 B-15 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A-11 B+35 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\frac{1}{4} a^3 (3 A-11 B+35 C)+4 a^3 (2 B-5 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac{(A-B+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(A+7 B-15 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A-11 B+35 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(2 B-5 C) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{2 a^3}+\frac{(3 A-43 B+115 C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(A+7 B-15 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A-11 B+35 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(2 B-5 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 )}{a^3 d}-\frac{(3 A-43 B+115 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 )}{16 a d}\\ &=\frac{(2 B-5 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac{(3 A-43 B+115 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 )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(A+7 B-15 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A-11 B+35 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.12897, size = 434, normalized size = 1.71 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) ((7 A-15 B+55 C) \cos (c+d x)+3 A-11 B+8 C \cos (2 (c+d x))+43 C)+\frac{\sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (-i \sqrt{2} (3 A-43 B+115 C) \log \left (1+e^{i (c+d x)}\right )+3 i \sqrt{2} A \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-16 i (2 B-5 C) \sinh ^{-1}\left (e^{i (c+d x)}\right )+32 i B \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-43 i \sqrt{2} B \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+32 B d x-80 i C \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )+115 i \sqrt{2} C \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-80 C d x\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{8 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(Cos[(c + d*x)/2]^5*((Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(32*B*d*x -
80*C*d*x - (16*I)*(2*B - 5*C)*ArcSinh[E^(I*(c + d*x))] - I*Sqrt[2]*(3*A - 43*B + 115*C)*Log[1 + E^(I*(c + d*x)
)] + (32*I)*B*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] - (80*I)*C*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] + (3*I)
*Sqrt[2]*A*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] - (43*I)*Sqrt[2]*B*Log[1 - E^(I*(c
+ d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] + (115*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))] + Sqrt[Cos[c + d*x]]*(3*A - 11*B + 43*C + (7*A - 15*B
+ 55*C)*Cos[c + d*x] + 8*C*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^3*Tan[(c + d*x)/2]))/(8*d*(a*(1 + Cos[c + d*x])
)^(5/2))

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Maple [B]  time = 0.11, size = 881, normalized size = 3.5 \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))^(5/2),x)

[Out]

1/32/d*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^(3/2)*(-1+cos(d*x+c))^5*(14*A*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c
)))^(5/2)+20*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-8*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/
2)-30*B*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+3*A*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+
c)*cos(d*x+c)^3-20*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-43*B*cos(d*x+c)^3*arcsin((-1+cos(d*x+c))/sin
(d*x+c))*2^(1/2)*sin(d*x+c)-22*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+115*C*2^(1/2)*arcsin((-1+cos(d
*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3+32*C*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*A*arcsin((-1+
cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2-6*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-43*B*sin(d*x+c)*
cos(d*x+c)^2*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-64*B*sin(d*x+c)*cos(d*x+c)^3*arctan(sin(d*x+c)*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+30*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+115*C*arcsin((-1+co
s(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2+160*C*sin(d*x+c)*cos(d*x+c)^3*arctan(sin(d*x+c)*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+78*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-64*B*sin(d*x+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))+22*B*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*
x+c)))^(3/2)+160*C*sin(d*x+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))-40*
C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-70*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))/sin(d*x+
c)^11/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)/a^3

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Maxima [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}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{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)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^(5/2), x)

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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))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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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))**(5/2),x)

[Out]

Timed out

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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}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{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)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^(5/2), x)