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

Optimal. Leaf size=364 \[ \frac{2 \left (24 a^2 C+7 b^2 (9 A+7 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b^3 d}-\frac{4 a \left (8 a^2 C+21 A b^2+18 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b^3 d}+\frac{4 a \left (a^2-b^2\right ) \left (8 a^2 C+21 A b^2+18 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (6 a^2 b^2 (7 A+4 C)+16 a^4 C-21 b^4 (9 A+7 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{4 a C \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{21 b^2 d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2}}{9 b d} \]

[Out]

(-2*(16*a^4*C + 6*a^2*b^2*(7*A + 4*C) - 21*b^4*(9*A + 7*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2
*b)/(a + b)])/(315*b^4*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (4*a*(a^2 - b^2)*(21*A*b^2 + 8*a^2*C + 18*b^2*C
)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^4*d*Sqrt[a + b*Cos[c + d*x]
]) - (4*a*(21*A*b^2 + 8*a^2*C + 18*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(315*b^3*d) + (2*(24*a^2*C +
7*b^2*(9*A + 7*C))*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b^3*d) - (4*a*C*Cos[c + d*x]*(a + b*Cos[c + d
*x])^(3/2)*Sin[c + d*x])/(21*b^2*d) + (2*C*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(9*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.805528, antiderivative size = 364, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {3050, 3049, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (24 a^2 C+7 b^2 (9 A+7 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b^3 d}-\frac{4 a \left (8 a^2 C+21 A b^2+18 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b^3 d}+\frac{4 a \left (a^2-b^2\right ) \left (8 a^2 C+21 A b^2+18 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (6 a^2 b^2 (7 A+4 C)+16 a^4 C-21 b^4 (9 A+7 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{4 a C \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{21 b^2 d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2}}{9 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(16*a^4*C + 6*a^2*b^2*(7*A + 4*C) - 21*b^4*(9*A + 7*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2
*b)/(a + b)])/(315*b^4*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (4*a*(a^2 - b^2)*(21*A*b^2 + 8*a^2*C + 18*b^2*C
)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^4*d*Sqrt[a + b*Cos[c + d*x]
]) - (4*a*(21*A*b^2 + 8*a^2*C + 18*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(315*b^3*d) + (2*(24*a^2*C +
7*b^2*(9*A + 7*C))*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b^3*d) - (4*a*C*Cos[c + d*x]*(a + b*Cos[c + d
*x])^(3/2)*Sin[c + d*x])/(21*b^2*d) + (2*C*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(9*b*d)

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 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 3023

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]

Rule 2753

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

Rule 2752

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

Rule 2663

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

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{9 b d}+\frac{2 \int \cos (c+d x) \sqrt{a+b \cos (c+d x)} \left (2 a C+\frac{1}{2} b (9 A+7 C) \cos (c+d x)-3 a C \cos ^2(c+d x)\right ) \, dx}{9 b}\\ &=-\frac{4 a C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{21 b^2 d}+\frac{2 C \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{9 b d}+\frac{4 \int \sqrt{a+b \cos (c+d x)} \left (-3 a^2 C-\frac{1}{2} a b C \cos (c+d x)+\frac{1}{4} \left (24 a^2 C+7 b^2 (9 A+7 C)\right ) \cos ^2(c+d x)\right ) \, dx}{63 b^2}\\ &=\frac{2 \left (24 a^2 C+7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^3 d}-\frac{4 a C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{21 b^2 d}+\frac{2 C \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{9 b d}+\frac{8 \int \sqrt{a+b \cos (c+d x)} \left (\frac{3}{8} b \left (63 A b^2+4 a^2 C+49 b^2 C\right )-\frac{3}{4} a \left (21 A b^2+8 a^2 C+18 b^2 C\right ) \cos (c+d x)\right ) \, dx}{315 b^3}\\ &=-\frac{4 a \left (21 A b^2+8 a^2 C+18 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}+\frac{2 \left (24 a^2 C+7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^3 d}-\frac{4 a C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{21 b^2 d}+\frac{2 C \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{9 b d}+\frac{16 \int \frac{\frac{3}{16} a b \left (147 A b^2-4 a^2 C+111 b^2 C\right )-\frac{3}{16} \left (16 a^4 C+6 a^2 b^2 (7 A+4 C)-21 b^4 (9 A+7 C)\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{945 b^3}\\ &=-\frac{4 a \left (21 A b^2+8 a^2 C+18 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}+\frac{2 \left (24 a^2 C+7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^3 d}-\frac{4 a C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{21 b^2 d}+\frac{2 C \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{9 b d}+\frac{\left (2 a \left (a^2-b^2\right ) \left (21 A b^2+8 a^2 C+18 b^2 C\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{315 b^4}-\frac{\left (16 a^4 C+6 a^2 b^2 (7 A+4 C)-21 b^4 (9 A+7 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{315 b^4}\\ &=-\frac{4 a \left (21 A b^2+8 a^2 C+18 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}+\frac{2 \left (24 a^2 C+7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^3 d}-\frac{4 a C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{21 b^2 d}+\frac{2 C \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{9 b d}-\frac{\left (\left (16 a^4 C+6 a^2 b^2 (7 A+4 C)-21 b^4 (9 A+7 C)\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{315 b^4 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (2 a \left (a^2-b^2\right ) \left (21 A b^2+8 a^2 C+18 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{315 b^4 \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{2 \left (16 a^4 C+6 a^2 b^2 (7 A+4 C)-21 b^4 (9 A+7 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{4 a \left (a^2-b^2\right ) \left (21 A b^2+8 a^2 C+18 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a \left (21 A b^2+8 a^2 C+18 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}+\frac{2 \left (24 a^2 C+7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^3 d}-\frac{4 a C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{21 b^2 d}+\frac{2 C \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{9 b d}\\ \end{align*}

Mathematica [A]  time = 1.29654, size = 269, normalized size = 0.74 \[ \frac{b (a+b \cos (c+d x)) \left (2 a \left (32 a^2 C+84 A b^2+57 b^2 C\right ) \sin (c+d x)+b \left (\left (-24 a^2 C+252 A b^2+266 b^2 C\right ) \sin (2 (c+d x))+5 b C (2 a \sin (3 (c+d x))+7 b \sin (4 (c+d x)))\right )\right )+8 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (a b^2 \left (-4 a^2 C+147 A b^2+111 b^2 C\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-\left (6 a^2 b^2 (7 A+4 C)+16 a^4 C-21 b^4 (9 A+7 C)\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{1260 b^4 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(a*b^2*(147*A*b^2 - 4*a^2*C + 111*b^2*C)*EllipticF[(c + d*x)/2, (2*b)/(a
 + b)] - (16*a^4*C + 6*a^2*b^2*(7*A + 4*C) - 21*b^4*(9*A + 7*C))*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)
] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])) + b*(a + b*Cos[c + d*x])*(2*a*(84*A*b^2 + 32*a^2*C + 57*b^2*C)*S
in[c + d*x] + b*((252*A*b^2 - 24*a^2*C + 266*b^2*C)*Sin[2*(c + d*x)] + 5*b*C*(2*a*Sin[3*(c + d*x)] + 7*b*Sin[4
*(c + d*x)]))))/(1260*b^4*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 0.457, size = 1527, normalized size = 4.2 \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)^2*(A+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2),x)

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*C*b^5*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^10+(640*C*a*b^4+2240*C*b^5)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-504*A*b^5+8*C*a^2*b^3-960*C*a*b^
4-2072*C*b^5)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(336*A*a*b^4+504*A*b^5+8*C*a^3*b^2-8*C*a^2*b^3+728*C*a*b
^4+952*C*b^5)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-42*A*a^2*b^3-168*A*a*b^4-126*A*b^5-16*C*a^4*b-4*C*a^3*
b^2-24*C*a^2*b^3-204*C*a*b^4-168*C*b^5)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+42*A*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b
^2-42*a*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d
*x+1/2*c),(-2*b/(a-b))^(1/2))*b^4-42*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-
b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b^2+42*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-
b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^3+189*A*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*
b/(a-b))^(1/2))*a*b^4-189*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*E
llipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^5+16*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+
1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^5+20*C*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*
b^2-36*a*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*
d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^4-16*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a
-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^5+16*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*
sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4*b-24*C*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b
))^(1/2))*a^3*b^2+24*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^3+147*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+
1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^4-147*C*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b
^5)/b^4/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)
^2*b+a+b)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{4} + A \cos \left (d x + c\right )^{2}\right )} \sqrt{b \cos \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^4 + A*cos(d*x + c)^2)*sqrt(b*cos(d*x + c) + a), x)

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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)**2*(A+C*cos(d*x+c)**2)*(a+b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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