3.648 \(\int \frac{\cos ^3(c+d x) (A+C \cos ^2(c+d x))}{\sqrt{a+b \cos (c+d x)}} \, dx\)
Optimal. Leaf size=378 \[ \frac{2 \left (48 a^2 C+7 b^2 (9 A+7 C)\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b^3 d}-\frac{4 a \left (32 a^2 C+42 A b^2+31 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b^4 d}-\frac{2 a \left (4 a^2 b^2 (42 A+19 C)+128 a^4 C+3 b^4 (49 A+37 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^5 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (12 a^2 b^2 (14 A+9 C)+128 a^4 C+21 b^4 (9 A+7 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^5 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{16 a C \sin (c+d x) \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)}}{63 b^2 d}+\frac{2 C \sin (c+d x) \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)}}{9 b d} \]
[Out]
(2*(128*a^4*C + 21*b^4*(9*A + 7*C) + 12*a^2*b^2*(14*A + 9*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(315*b^5*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*a*(128*a^4*C + 4*a^2*b^2*(42*A + 19*C) + 3
*b^4*(49*A + 37*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^5*d*Sqrt[
a + b*Cos[c + d*x]]) - (4*a*(42*A*b^2 + 32*a^2*C + 31*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(315*b^4*d
) + (2*(48*a^2*C + 7*b^2*(9*A + 7*C))*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(315*b^3*d) - (16*a*
C*Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(63*b^2*d) + (2*C*Cos[c + d*x]^3*Sqrt[a + b*Cos[c + d*
x]]*Sin[c + d*x])/(9*b*d)
________________________________________________________________________________________
Rubi [A] time = 0.926682, antiderivative size = 378, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used =
{3050, 3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (48 a^2 C+7 b^2 (9 A+7 C)\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b^3 d}-\frac{4 a \left (32 a^2 C+42 A b^2+31 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b^4 d}-\frac{2 a \left (4 a^2 b^2 (42 A+19 C)+128 a^4 C+3 b^4 (49 A+37 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^5 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (12 a^2 b^2 (14 A+9 C)+128 a^4 C+21 b^4 (9 A+7 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^5 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{16 a C \sin (c+d x) \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)}}{63 b^2 d}+\frac{2 C \sin (c+d x) \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)}}{9 b d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(2*(128*a^4*C + 21*b^4*(9*A + 7*C) + 12*a^2*b^2*(14*A + 9*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(315*b^5*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*a*(128*a^4*C + 4*a^2*b^2*(42*A + 19*C) + 3
*b^4*(49*A + 37*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^5*d*Sqrt[
a + b*Cos[c + d*x]]) - (4*a*(42*A*b^2 + 32*a^2*C + 31*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(315*b^4*d
) + (2*(48*a^2*C + 7*b^2*(9*A + 7*C))*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(315*b^3*d) - (16*a*
C*Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(63*b^2*d) + (2*C*Cos[c + d*x]^3*Sqrt[a + b*Cos[c + d*
x]]*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 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 \frac{\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt{a+b \cos (c+d x)}} \, dx &=\frac{2 C \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{9 b d}+\frac{2 \int \frac{\cos ^2(c+d x) \left (3 a C+\frac{1}{2} b (9 A+7 C) \cos (c+d x)-4 a C \cos ^2(c+d x)\right )}{\sqrt{a+b \cos (c+d x)}} \, dx}{9 b}\\ &=-\frac{16 a C \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{63 b^2 d}+\frac{2 C \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{9 b d}+\frac{4 \int \frac{\cos (c+d x) \left (-8 a^2 C+\frac{1}{2} a b C \cos (c+d x)+\frac{1}{4} \left (48 a^2 C+7 b^2 (9 A+7 C)\right ) \cos ^2(c+d x)\right )}{\sqrt{a+b \cos (c+d x)}} \, dx}{63 b^2}\\ &=\frac{2 \left (48 a^2 C+7 b^2 (9 A+7 C)\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}-\frac{16 a C \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{63 b^2 d}+\frac{2 C \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{9 b d}+\frac{8 \int \frac{\frac{1}{4} a \left (48 a^2 C+7 b^2 (9 A+7 C)\right )+\frac{1}{8} b \left (189 A b^2-16 a^2 C+147 b^2 C\right ) \cos (c+d x)-\frac{3}{4} a \left (42 A b^2+32 a^2 C+31 b^2 C\right ) \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{315 b^3}\\ &=-\frac{4 a \left (42 A b^2+32 a^2 C+31 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^4 d}+\frac{2 \left (48 a^2 C+7 b^2 (9 A+7 C)\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}-\frac{16 a C \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{63 b^2 d}+\frac{2 C \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{9 b d}+\frac{16 \int \frac{\frac{3}{8} a b \left (21 A b^2+16 a^2 C+18 b^2 C\right )+\frac{3}{16} \left (128 a^4 C+21 b^4 (9 A+7 C)+12 a^2 b^2 (14 A+9 C)\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{945 b^4}\\ &=-\frac{4 a \left (42 A b^2+32 a^2 C+31 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^4 d}+\frac{2 \left (48 a^2 C+7 b^2 (9 A+7 C)\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}-\frac{16 a C \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{63 b^2 d}+\frac{2 C \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{9 b d}+\frac{\left (128 a^4 C+21 b^4 (9 A+7 C)+12 a^2 b^2 (14 A+9 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{315 b^5}-\frac{\left (a \left (128 a^4 C+4 a^2 b^2 (42 A+19 C)+3 b^4 (49 A+37 C)\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{315 b^5}\\ &=-\frac{4 a \left (42 A b^2+32 a^2 C+31 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^4 d}+\frac{2 \left (48 a^2 C+7 b^2 (9 A+7 C)\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}-\frac{16 a C \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{63 b^2 d}+\frac{2 C \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{9 b d}+\frac{\left (\left (128 a^4 C+21 b^4 (9 A+7 C)+12 a^2 b^2 (14 A+9 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^5 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (a \left (128 a^4 C+4 a^2 b^2 (42 A+19 C)+3 b^4 (49 A+37 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^5 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (128 a^4 C+21 b^4 (9 A+7 C)+12 a^2 b^2 (14 A+9 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^5 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 a \left (128 a^4 C+4 a^2 b^2 (42 A+19 C)+3 b^4 (49 A+37 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^5 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a \left (42 A b^2+32 a^2 C+31 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^4 d}+\frac{2 \left (48 a^2 C+7 b^2 (9 A+7 C)\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^3 d}-\frac{16 a C \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{63 b^2 d}+\frac{2 C \cos ^3(c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{9 b d}\\ \end{align*}
Mathematica [A] time = 1.39748, size = 272, normalized size = 0.72 \[ \frac{8 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b \left (32 a^3 b C+6 a b^3 (7 A+6 C)\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (12 a^2 b^2 (14 A+9 C)+128 a^4 C+21 b^4 (9 A+7 C)\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )-b (a+b \cos (c+d x)) \left (32 a \left (2 C \left (8 a^2+9 b^2\right )+21 A b^2\right ) \sin (c+d x)-b \left (2 \left (96 a^2 C+126 A b^2+133 b^2 C\right ) \sin (2 (c+d x))+5 b C (7 b \sin (4 (c+d x))-16 a \sin (3 (c+d x)))\right )\right )}{1260 b^5 d \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(8*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b*(32*a^3*b*C + 6*a*b^3*(7*A + 6*C))*EllipticF[(c + d*x)/2, (2*b)/(a +
b)] + (128*a^4*C + 21*b^4*(9*A + 7*C) + 12*a^2*b^2*(14*A + 9*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])*(32*a*(21*A*b^2 + 2*(8*a^2 + 9*b^2)*C)*
Sin[c + d*x] - b*(2*(126*A*b^2 + 96*a^2*C + 133*b^2*C)*Sin[2*(c + d*x)] + 5*b*C*(-16*a*Sin[3*(c + d*x)] + 7*b*
Sin[4*(c + d*x)]))))/(1260*b^5*d*Sqrt[a + b*Cos[c + d*x]])
________________________________________________________________________________________
Maple [B] time = 0.422, size = 1527, normalized size = 4. \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)^3*(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+(-80*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^5-64*C*a^2*b^3+120*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)+(-84*A*a*b^4+504*A*b^5-64*C*a^3*b^2+64*C*a^2*b^3-112*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)+(168*A*a^2*b^3+42*A*a*b^4-126*A*b^5+128*C*a^4*b+32*C*
a^3*b^2+108*C*a^2*b^3+36*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)+168*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^2-168*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^5-168*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^2-147*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+128*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^5-128*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+108*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(c
os(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b^2-108*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^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^5-128*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^5-76*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*si
n(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^2-111*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)/b^5/(-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 \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt{b \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(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)*cos(d*x + c)^3/sqrt(b*cos(d*x + c) + a), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C \cos \left (d x + c\right )^{5} + A \cos \left (d x + c\right )^{3}}{\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)^3*(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)^5 + A*cos(d*x + c)^3)/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)**3*(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 \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt{b \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(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)*cos(d*x + c)^3/sqrt(b*cos(d*x + c) + a), x)