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

Optimal. Leaf size=348 \[ \frac{\left (a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac{\left (a^2 b^2 (5 A+9 C)-3 a^4 C+A b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b^2 d \left (a^2-b^2\right )^2}+\frac{\left (-3 a^4 b^2 (A-2 C)-5 a^2 b^4 (2 A+3 C)-3 a^6 C+A b^6\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b^3 d (a-b)^2 (a+b)^3}-\frac{\left (a^2 b^2 (5 A+9 C)-3 a^4 C+A b^4\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 a b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]

[Out]

((A*b^4 - 3*a^4*C + a^2*b^2*(5*A + 9*C))*EllipticE[(c + d*x)/2, 2])/(4*a*b^2*(a^2 - b^2)^2*d) + ((a^2*b^2*(3*A
- 5*C) + 3*a^4*C + b^4*(3*A + 8*C))*EllipticF[(c + d*x)/2, 2])/(4*b^3*(a^2 - b^2)^2*d) + ((A*b^6 - 3*a^4*b^2*
(A - 2*C) - 3*a^6*C - 5*a^2*b^4*(2*A + 3*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(4*a*(a - b)^2*b^3*(a
+ b)^3*d) - ((A*b^2 + a^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((A
*b^4 - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a*b*(a^2 - b^2)^2*d*(a + b*Cos[c + d
*x]))

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Rubi [A]  time = 1.13, antiderivative size = 348, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3048, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{\left (a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac{\left (a^2 b^2 (5 A+9 C)-3 a^4 C+A b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b^2 d \left (a^2-b^2\right )^2}+\frac{\left (-3 a^4 b^2 (A-2 C)-5 a^2 b^4 (2 A+3 C)-3 a^6 C+A b^6\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b^3 d (a-b)^2 (a+b)^3}-\frac{\left (a^2 b^2 (5 A+9 C)-3 a^4 C+A b^4\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 a b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b^4 - 3*a^4*C + a^2*b^2*(5*A + 9*C))*EllipticE[(c + d*x)/2, 2])/(4*a*b^2*(a^2 - b^2)^2*d) + ((a^2*b^2*(3*A
- 5*C) + 3*a^4*C + b^4*(3*A + 8*C))*EllipticF[(c + d*x)/2, 2])/(4*b^3*(a^2 - b^2)^2*d) + ((A*b^6 - 3*a^4*b^2*
(A - 2*C) - 3*a^6*C - 5*a^2*b^4*(2*A + 3*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(4*a*(a - b)^2*b^3*(a
+ b)^3*d) - ((A*b^2 + a^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((A
*b^4 - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a*b*(a^2 - b^2)^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 3055

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*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 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{\sqrt{\cos (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 ) \sqrt{\cos (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{\frac{1}{2} \left (A b^2+a^2 C\right )-2 a b (A+C) \cos (c+d x)+\frac{1}{2} \left (A b^2-3 a^2 C+4 b^2 C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (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 ) \sqrt{\cos (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 (A b^4-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{\frac{1}{4} \left (-A b^4-a^4 C+7 a^2 b^2 (A+C)\right )-a b \left (a^2 (2 A+C)+b^2 (A+2 C)\right ) \cos (c+d x)-\frac{1}{4} \left (A b^4-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2+a^2 C\right ) \sqrt{\cos (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 (A b^4-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{4} b \left (A b^4+a^4 C-7 a^2 b^2 (A+C)\right )+\frac{1}{4} a \left (a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a b^2 \left (a^2-b^2\right )^2}+\frac{\left (A b^4-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a b^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^4-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b^2 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \sqrt{\cos (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 (A b^4-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (A b^6-3 a^4 b^2 (A-2 C)-3 a^6 C-5 a^2 b^4 (2 A+3 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 a b^3 \left (a^2-b^2\right )^2}+\frac{\left (a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^4-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b^2 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) F\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^6-3 a^4 b^2 (A-2 C)-3 a^6 C-5 a^2 b^4 (2 A+3 C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a (a-b)^2 b^3 (a+b)^3 d}-\frac{\left (A b^2+a^2 C\right ) \sqrt{\cos (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 (A b^4-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.6215, size = 368, normalized size = 1.06 \[ \frac{\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (b \left (-a^2 b^2 (5 A+9 C)+3 a^4 C-A b^4\right ) \cos (c+d x)-7 a^3 b^2 (A+C)+a^5 C+a A b^4\right )}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac{\frac{\left (a^2 b^2 (9 A+5 C)+a^4 C-3 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}-\frac{8 \left (a^3 (2 A+C)+a b^2 (A+2 C)\right ) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac{\left (-a^2 b^2 (5 A+9 C)+3 a^4 C-A b^4\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt{\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 a b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 2.326, size = 1934, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(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^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)+12/b^2*C*a/(-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*(A*b^2+C*a^2)/b^3*(-1/
2/a*b^2/(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)*EllipticF(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*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/
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)*EllipticPi(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)))+2/b^3*(A*b^2+3*C*a^2)*(-1/a*b^2/(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/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))+1/2/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(-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((A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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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((A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,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((A+C*cos(d*x+c)**2)*cos(d*x+c)**(1/2)/(a+b*cos(d*x+c))**3,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 )} \sqrt{\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

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