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

Optimal. Leaf size=345 \[ -\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {a \left (15 a^4 C-a^2 b^2 (A+33 C)+b^4 (7 A+24 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d \left (a^2-b^2\right )^2}+\frac {\left (-5 a^4 C+a^2 b^2 (3 A+11 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {\left (-15 a^4 C+a^2 b^2 (A+29 C)+b^4 (5 A-8 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac {\left (15 a^6 C-a^4 b^2 (A+38 C)+5 a^2 b^4 (2 A+7 C)+3 A b^6\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d (a-b)^2 (a+b)^3} \]

[Out]

-1/4*(b^4*(5*A-8*C)-15*a^4*C+a^2*b^2*(A+29*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1
/2*d*x+1/2*c),2^(1/2))/b^3/(a^2-b^2)^2/d-1/4*a*(15*a^4*C+b^4*(7*A+24*C)-a^2*b^2*(A+33*C))*(cos(1/2*d*x+1/2*c)^
2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/b^4/(a^2-b^2)^2/d+1/4*(3*A*b^6+15*a^6*C+5*a^
2*b^4*(2*A+7*C)-a^4*b^2*(A+38*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c
),2*b/(a+b),2^(1/2))/(a-b)^2/b^4/(a+b)^3/d-1/2*(A*b^2+C*a^2)*cos(d*x+c)^(3/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*co
s(d*x+c))^2+1/4*(3*A*b^4-5*a^4*C+a^2*b^2*(3*A+11*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*
x+c))

________________________________________________________________________________________

Rubi [A]  time = 1.14, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3048, 3047, 3059, 2639, 3002, 2641, 2805} \[ -\frac {a \left (-a^2 b^2 (A+33 C)+15 a^4 C+b^4 (7 A+24 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac {\left (a^2 b^2 (A+29 C)-15 a^4 C+b^4 (5 A-8 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac {\left (-a^4 b^2 (A+38 C)+5 a^2 b^4 (2 A+7 C)+15 a^6 C+3 A b^6\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d (a-b)^2 (a+b)^3}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\left (a^2 b^2 (3 A+11 C)-5 a^4 C+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^4*(5*A - 8*C) - 15*a^4*C + a^2*b^2*(A + 29*C))*EllipticE[(c + d*x)/2, 2])/(4*b^3*(a^2 - b^2)^2*d) - (a*(1
5*a^4*C + b^4*(7*A + 24*C) - a^2*b^2*(A + 33*C))*EllipticF[(c + d*x)/2, 2])/(4*b^4*(a^2 - b^2)^2*d) + ((3*A*b^
6 + 15*a^6*C + 5*a^2*b^4*(2*A + 7*C) - a^4*b^2*(A + 38*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(4*(a -
b)^2*b^4*(a + b)^3*d) - ((A*b^2 + a^2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*
x])^2) + ((3*A*b^4 - 5*a^4*C + a^2*b^2*(3*A + 11*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(
a + b*Cos[c + d*x]))

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), 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] && GtQ[c + d, 0]

Rule 3002

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

Rule 3047

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[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3048

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

Rule 3059

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

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx &=-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{2} \left (A b^2+a^2 C\right )-2 a b (A+C) \cos (c+d x)-\frac {1}{2} \left (A b^2+5 a^2 C-4 b^2 C\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (3 A b^4-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (3 A b^4-5 a^4 C+a^2 b^2 (3 A+11 C)\right )-a b \left (3 A b^2-\left (a^2-4 b^2\right ) C\right ) \cos (c+d x)-\frac {1}{4} \left (b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (3 A b^4-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {\int \frac {-\frac {1}{4} b \left (3 A b^4-5 a^4 C+a^2 b^2 (3 A+11 C)\right )+\frac {1}{4} a \left (15 a^4 C+b^4 (7 A+24 C)-a^2 b^2 (A+33 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )^2}-\frac {\left (b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (3 A b^4-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {\left (a \left (15 a^4 C+b^4 (7 A+24 C)-a^2 b^2 (A+33 C)\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 b^4 \left (a^2-b^2\right )^2}+\frac {\left (3 A b^6+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a \left (15 a^4 C+b^4 (7 A+24 C)-a^2 b^2 (A+33 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^6+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 (a-b)^2 b^4 (a+b)^3 d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (3 A b^4-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 4.62, size = 368, normalized size = 1.07 \[ \frac {\frac {\frac {8 a \left (C \left (a^2-4 b^2\right )-3 A b^2\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac {\left (5 a^4 C+a^2 b^2 (5 A-7 C)+b^4 (A+8 C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {\left (15 a^4 C-a^2 b^2 (A+29 C)+b^4 (8 C-5 A)\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-5 a^5 C+a^3 b^2 (3 A+11 C)+\left (-7 a^4 b C+a^2 b^3 (A+13 C)+5 A b^5\right ) \cos (c+d x)+3 a A b^4\right )}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}}{8 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[Cos[c + d*x]]*(3*a*A*b^4 - 5*a^5*C + a^3*b^2*(3*A + 11*C) + (5*A*b^5 - 7*a^4*b*C + a^2*b^3*(A + 13*C)
)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + (((a^2*b^2*(5*A - 7*C) + 5*a^4*C + b^4*
(A + 8*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*a*(-3*A*b^2 + (a^2 - 4*b^2)*C)*((a + b)*Ell
ipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) + ((15*a^4*C + b^4*(-5*A + 8*C)
- a^2*b^2*(A + 29*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[C
os[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*S
qrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(8*b^2*d)

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 10.64, size = 1966, normalized size = 5.70 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*C/b^4/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(3*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))*a+EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)-4*a/b^4*(A*b^2+2*C*a^2)*(-b^2/a/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/a*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF
(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*b/a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/a/(a^2-b^2)*(
sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(
1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos
(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),
-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^
(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))-
4/b^3*(A*b^2+6*C*a^2)/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2*a^2*(A*b^2+C*a^
2)/b^4*(-1/2*b^2/a/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/
2*d*x+1/2*c)^2*b+a-b)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/
2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)-7/8/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(
1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^
(1/2))+1/4/(a+b)/(a^2-b^2)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/
2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3/8/(a+b)/(a^2-b^2)/a^2*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1
)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*b^3/a^2
/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d
*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),
2^(1/2))-3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*
x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15/4*a^2/(a^2-b^2)^2/(-2*a*b+2*b^
2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+3/2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic
Pi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-3/4/a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+
1/2*c),-2*b/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________