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

Optimal. Leaf size=201 \[ \frac {(5 A+3 B-43 C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {2 C \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac {(5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

2*C*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d-1/4*(A-B+C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a
*cos(d*x+c))^(5/2)+1/32*(5*A+3*B-43*C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))
^(1/2))/a^(5/2)/d*2^(1/2)+1/16*(5*A+3*B-11*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(3/2)

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Rubi [A]  time = 0.63, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3041, 2977, 2982, 2782, 205, 2774, 216} \[ \frac {(5 A+3 B-43 C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {2 C \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac {(5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) + ((5*A + 3*B - 43*C)*ArcTan[(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]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((5*A + 3*B - 11*C)*Sqrt[Cos[c + d*x]]*Sin
[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2))

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 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]

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 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 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 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 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)]

Rubi steps

\begin {align*} \int \frac {\sqrt {\cos (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 {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} a (5 A+3 B-3 C)+4 a C \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 A+3 B-11 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{4} a^2 (5 A+3 B-11 C)+8 a^2 C \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 A+3 B-11 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+3 B-43 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}+\frac {C \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{a^3}\\ &=-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 A+3 B-11 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(5 A+3 B-43 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 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 {2 C \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(5 A+3 B-43 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 {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 A+3 B-11 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 3.17, size = 385, normalized size = 1.92 \[ \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {\cos (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) ((A+7 B-15 C) \cos (c+d x)+5 A+3 B-11 C)+\frac {\sqrt {2} e^{\frac {1}{2} i (c+d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (-i \sqrt {2} (5 A+3 B-43 C) \log \left (1+e^{i (c+d x)}\right )+5 i \sqrt {2} A \log \left (\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+3 i \sqrt {2} B \log \left (\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+32 i C \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )-43 i \sqrt {2} C \log \left (\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-32 i C \sinh ^{-1}\left (e^{i (c+d x)}\right )+32 C d x\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right )}{8 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[Cos[c + d*x]]*(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*C*d*x -
(32*I)*C*ArcSinh[E^(I*(c + d*x))] - I*Sqrt[2]*(5*A + 3*B - 43*C)*Log[1 + E^(I*(c + d*x))] + (32*I)*C*Log[1 + S
qrt[1 + E^((2*I)*(c + d*x))]] + (5*I)*Sqrt[2]*A*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))
]] + (3*I)*Sqrt[2]*B*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] - (43*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
]]*(5*A + 3*B - 11*C + (A + 7*B - 15*C)*Cos[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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fricas [A]  time = 96.58, size = 283, normalized size = 1.41 \[ -\frac {\sqrt {2} {\left ({\left (5 \, A + 3 \, B - 43 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 3 \, B - 43 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 3 \, B - 43 \, C\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B - 43 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (A + 7 \, B - 15 \, C\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B - 11 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 64 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(sqrt(2)*((5*A + 3*B - 43*C)*cos(d*x + c)^3 + 3*(5*A + 3*B - 43*C)*cos(d*x + c)^2 + 3*(5*A + 3*B - 43*C)
*cos(d*x + c) + 5*A + 3*B - 43*C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*
sin(d*x + c))) - 2*((A + 7*B - 15*C)*cos(d*x + c) + 5*A + 3*B - 11*C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x +
c))*sin(d*x + c) + 64*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*sqrt(a)*arctan(sqrt(a*cos
(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*
a^3*d*cos(d*x + c) + a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.33, size = 747, normalized size = 3.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(1/2)*(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*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^4*(2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos
(d*x+c)^4+12*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)+14*B*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+5*A*2^(1/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c
))*cos(d*x+c)^3-12*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+3*B*cos(d*x+c)^3*2^(1/2)*sin(d*x+c)*arcsin((
-1+cos(d*x+c))/sin(d*x+c))+6*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-43*C*2^(1/2)*sin(d*x+c)*arcsin((
-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3+5*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)
-10*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+3*B*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))
-14*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-43*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(
d*x+c)*2^(1/2)-30*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-64*C*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^3-6*B*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+8*C*cos(d*x
+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-64*C*cos(d*x+c)^2*sin(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)/cos(d*x+c))+22*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))/a^3/(cos(d*x+c)/(1+cos(d*x+c)))^(5
/2)/sin(d*x+c)^9

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sqrt(cos(c + d*x))/(a*(cos(c + d*x) + 1))**(5/2), x)

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