3.2.8 \(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\) [108]

Optimal. Leaf size=216 \[ \frac {(11 A-15 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(65 A-93 B) \sin (c+d x)}{15 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d} \]

[Out]

1/2*(A-B)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/4*(11*A-15*B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1
/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/15*(65*A-93*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/10*(5*A
-9*B)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/30*(35*A-39*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^
2/d

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Rubi [A]
time = 0.39, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3056, 3062, 3047, 3102, 2830, 2728, 212} \begin {gather*} \frac {(11 A-15 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(35 A-39 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{30 a^2 d}+\frac {(A-B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(5 A-9 B) \sin (c+d x) \cos ^2(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}-\frac {(65 A-93 B) \sin (c+d x)}{15 a d \sqrt {a \cos (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((11*A - 15*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + ((A
- B)*Cos[c + d*x]^3*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((65*A - 93*B)*Sin[c + d*x])/(15*a*d*Sqr
t[a + a*Cos[c + d*x]]) - ((5*A - 9*B)*Cos[c + d*x]^2*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((35*A
- 39*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(30*a^2*d)

Rule 212

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

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

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

Rule 3047

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

Rule 3056

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 3062

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] &&
(IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx &=\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\cos ^2(c+d x) \left (3 a (A-B)-\frac {1}{2} a (5 A-9 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (-a^2 (5 A-9 B)+\frac {1}{4} a^2 (35 A-39 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {-a^2 (5 A-9 B) \cos (c+d x)+\frac {1}{4} a^2 (35 A-39 B) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}+\frac {2 \int \frac {\frac {1}{8} a^3 (35 A-39 B)-\frac {1}{4} a^3 (65 A-93 B) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{15 a^4}\\ &=\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(65 A-93 B) \sin (c+d x)}{15 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}+\frac {(11 A-15 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(65 A-93 B) \sin (c+d x)}{15 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}-\frac {(11 A-15 B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{2 a d}\\ &=\frac {(11 A-15 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(65 A-93 B) \sin (c+d x)}{15 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.94, size = 142, normalized size = 0.66 \begin {gather*} \frac {-15 (11 A-15 B) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )+\cos ^3\left (\frac {1}{2} (c+d x)\right ) (85 A-141 B+3 (20 A-39 B) \cos (c+d x)+(-10 A+6 B) \cos (2 (c+d x))-3 B \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )}{15 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*(11*A - 15*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 + Cos[(c + d*x)/2]^3*(85*A - 141*B + 3*(20*A -
39*B)*Cos[c + d*x] + (-10*A + 6*B)*Cos[2*(c + d*x)] - 3*B*Cos[3*(c + d*x)])*Sin[(c + d*x)/2])/(15*d*(a*(1 + C
os[c + d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(406\) vs. \(2(189)=378\).
time = 0.33, size = 407, normalized size = 1.88

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-96 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (5 A +6 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \sqrt {2}\, \left (33 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a -8 A \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-45 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +48 B \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+165 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a A -225 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a B -135 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+255 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(407\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/60*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-96*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^6
+16*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*(5*A+6*B)*sin(1/2*d*x+1/2*c)^4-5*2^(1/2)*(33*A*ln(4/cos(1/2
*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a-8*A*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-45*B*ln(4
/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a+48*B*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2))
*sin(1/2*d*x+1/2*c)^2+165*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a*A-225*
2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a*B-135*A*a^(1/2)*2^(1/2)*(sin(1/2
*d*x+1/2*c)^2*a)^(1/2)+255*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2))/cos(1/2*d*x+1/2*c)/a^(5/2)/sin(1/
2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]
time = 0.36, size = 224, normalized size = 1.04 \begin {gather*} -\frac {15 \, \sqrt {2} {\left ({\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right ) + 11 \, A - 15 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (12 \, B \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right ) - 95 \, A + 147 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*sqrt(2)*((11*A - 15*B)*cos(d*x + c)^2 + 2*(11*A - 15*B)*cos(d*x + c) + 11*A - 15*B)*sqrt(a)*log(-(a
*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x +
c)^2 + 2*cos(d*x + c) + 1)) - 4*(12*B*cos(d*x + c)^3 + 4*(5*A - 3*B)*cos(d*x + c)^2 - 12*(5*A - 9*B)*cos(d*x
+ c) - 95*A + 147*B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2
*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3063 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^3\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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