3.714 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=370 \[ -\frac{a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^5 d \left (a^2-b^2\right )}+\frac{\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac{a^2 \left (-3 a^2 b^2 (A-3 C)-7 a^4 C+5 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^5 d (a-b) (a+b)^2}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 b^2 d \left (a^2-b^2\right )}-\frac{a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )} \]

[Out]

((3*a^2*b^2*(5*A - 8*C) + 35*a^4*C - 2*b^4*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*b^4*(a^2 - b^2)*d) - (a*
(a^2*b^2*(9*A - 20*C) + 21*a^4*C - 4*b^4*(3*A + C))*EllipticF[(c + d*x)/2, 2])/(3*b^5*(a^2 - b^2)*d) - (a^2*(5
*A*b^4 - 3*a^2*b^2*(A - 3*C) - 7*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a - b)*b^5*(a + b)^2*d) -
(a*(3*A*b^2 + 7*a^2*C - 4*b^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^3*(a^2 - b^2)*d) + ((5*A*b^2 + 7*a^2*C
- 2*b^2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b^2*(a^2 - b^2)*d) - ((A*b^2 + a^2*C)*Cos[c + d*x]^(5/2)*Sin[c
+ d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.37583, antiderivative size = 370, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3048, 3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac{a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^5 d \left (a^2-b^2\right )}+\frac{\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac{a^2 \left (-3 a^2 b^2 (A-3 C)-7 a^4 C+5 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^5 d (a-b) (a+b)^2}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 b^2 d \left (a^2-b^2\right )}-\frac{a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2*b^2*(5*A - 8*C) + 35*a^4*C - 2*b^4*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*b^4*(a^2 - b^2)*d) - (a*
(a^2*b^2*(9*A - 20*C) + 21*a^4*C - 4*b^4*(3*A + C))*EllipticF[(c + d*x)/2, 2])/(3*b^5*(a^2 - b^2)*d) - (a^2*(5
*A*b^4 - 3*a^2*b^2*(A - 3*C) - 7*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a - b)*b^5*(a + b)^2*d) -
(a*(3*A*b^2 + 7*a^2*C - 4*b^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^3*(a^2 - b^2)*d) + ((5*A*b^2 + 7*a^2*C
- 2*b^2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b^2*(a^2 - b^2)*d) - ((A*b^2 + a^2*C)*Cos[c + d*x]^(5/2)*Sin[c
+ d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

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 3049

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

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{5}{2} \left (A b^2+a^2 C\right )-a b (A+C) \cos (c+d x)-\frac{1}{2} \left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{2 \int \frac{\sqrt{\cos (c+d x)} \left (-\frac{3}{4} a \left (5 A b^2+7 a^2 C-2 b^2 C\right )+\frac{1}{2} b \left (5 A b^2+2 a^2 C+3 b^2 C\right ) \cos (c+d x)+\frac{5}{4} a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{5 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac{\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{4 \int \frac{\frac{5}{8} a^2 \left (3 A b^2+7 a^2 C-4 b^2 C\right )-\frac{1}{4} a b \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \cos (c+d x)-\frac{3}{8} \left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^3 \left (a^2-b^2\right )}\\ &=-\frac{a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac{\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{4 \int \frac{-\frac{5}{8} a^2 b \left (3 A b^2+7 a^2 C-4 b^2 C\right )-\frac{5}{8} a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^4 \left (a^2-b^2\right )}+\frac{\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{10 b^4 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac{\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (a^2 \left (5 A b^4-3 a^2 b^2 (A-3 C)-7 a^4 C\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^5 \left (a^2-b^2\right )}-\frac{\left (a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 b^5 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^5 \left (a^2-b^2\right ) d}-\frac{a^2 \left (5 A b^4-3 a^2 b^2 (A-3 C)-7 a^4 C\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{(a-b) b^5 (a+b)^2 d}-\frac{a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac{\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.17658, size = 358, normalized size = 0.97 \[ \frac{4 \sqrt{\cos (c+d x)} \left (-\frac{15 a^2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}-20 a C \sin (c+d x)+3 b C \sin (2 (c+d x))\right )+\frac{\frac{2 \left (a^2 b^2 (15 A-32 C)+35 a^4 C-6 b^4 (5 A+3 C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{8 a \left (C \left (14 a^2+b^2\right )+15 A b^2\right ) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac{6 \left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt{\sin ^2(c+d x)}}}{(a-b) (a+b)}}{60 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*(a^2*b^2*(15*A - 32*C) + 35*a^4*C - 6*b^4*(5*A + 3*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)
+ (8*a*(15*A*b^2 + (14*a^2 + b^2)*C)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*
x)/2, 2]))/(a + b) + (6*(3*a^2*b^2*(5*A - 8*C) + 35*a^4*C - 2*b^4*(5*A + 3*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[C
os[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[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*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b)) + 4*Sqrt[Cos[c
+ d*x]]*(-20*a*C*Sin[c + d*x] - (15*a^2*(A*b^2 + a^2*C)*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])) + 3*
b*C*Sin[2*(c + d*x)]))/(60*b^3*d)

________________________________________________________________________________________

Maple [B]  time = 1.872, size = 1337, normalized size = 3.6 \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)^(5/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/5*C/b^2*(-4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2
*c)+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*EllipticE(cos(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)-6*sin(1/2*d*x+1/2*c)^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)-4/3/b^3*C*(2*a+3*b)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(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)-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))/(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2/b^4*(A*b^2+3*C*a^2+4*C*a*b+3*C*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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(2*A*a*b^2+A*b^3+4*C*a^3+3*C*a^2*b+2*C*a*b^2+C*b
^3)/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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-4*a^2/b^4*(3*A*b^2+5*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)*EllipticP
i(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*a^3*(A*b^2+C*a^2)/b^5*(-1/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*si
n(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/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*s
in(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/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)*
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))))/sin(
1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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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} + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a)^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)^(5/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^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)**(5/2)*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**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} + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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