3.1015 \(\int \cos (c+d x) \sqrt{a+b \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)
Optimal. Leaf size=321 \[ \frac{2 \sin (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)}}{105 b^2 d}-\frac{2 \left (a^2-b^2\right ) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (14 a^2 b B-8 a^3 C-a b^2 (35 A+19 C)-63 b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (7 b B-4 a C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b^2 d}+\frac{2 C \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{7 b d} \]
[Out]
(-2*(14*a^2*b*B - 63*b^3*B - 8*a^3*C - a*b^2*(35*A + 19*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2
*b)/(a + b)])/(105*b^3*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(35*A*b^2 - 14*a*b*B + 8*a^2*C +
25*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(105*b^3*d*Sqrt[a + b*Cos
[c + d*x]]) + (2*(35*A*b^2 - 14*a*b*B + 8*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b^2*d)
+ (2*(7*b*B - 4*a*C)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*b^2*d) + (2*C*Cos[c + d*x]*(a + b*Cos[c + d
*x])^(3/2)*Sin[c + d*x])/(7*b*d)
________________________________________________________________________________________
Rubi [A] time = 0.574211, antiderivative size = 321, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used =
{3049, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \sin (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)}}{105 b^2 d}-\frac{2 \left (a^2-b^2\right ) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (14 a^2 b B-8 a^3 C-a b^2 (35 A+19 C)-63 b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (7 b B-4 a C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b^2 d}+\frac{2 C \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{7 b d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
[Out]
(-2*(14*a^2*b*B - 63*b^3*B - 8*a^3*C - a*b^2*(35*A + 19*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2
*b)/(a + b)])/(105*b^3*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(35*A*b^2 - 14*a*b*B + 8*a^2*C +
25*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(105*b^3*d*Sqrt[a + b*Cos
[c + d*x]]) + (2*(35*A*b^2 - 14*a*b*B + 8*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b^2*d)
+ (2*(7*b*B - 4*a*C)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*b^2*d) + (2*C*Cos[c + d*x]*(a + b*Cos[c + d
*x])^(3/2)*Sin[c + d*x])/(7*b*d)
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 (c+d x) \sqrt{a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}+\frac{2 \int \sqrt{a+b \cos (c+d x)} \left (a C+\frac{1}{2} b (7 A+5 C) \cos (c+d x)+\frac{1}{2} (7 b B-4 a C) \cos ^2(c+d x)\right ) \, dx}{7 b}\\ &=\frac{2 (7 b B-4 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}+\frac{4 \int \sqrt{a+b \cos (c+d x)} \left (\frac{1}{4} b (21 b B-2 a C)+\frac{1}{4} \left (35 A b^2-14 a b B+8 a^2 C+25 b^2 C\right ) \cos (c+d x)\right ) \, dx}{35 b^2}\\ &=\frac{2 \left (35 A b^2-14 a b B+8 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}+\frac{2 (7 b B-4 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}+\frac{8 \int \frac{\frac{1}{8} b \left (35 A b^2+49 a b B+2 a^2 C+25 b^2 C\right )-\frac{1}{8} \left (14 a^2 b B-63 b^3 B-8 a^3 C-a b^2 (35 A+19 C)\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b^2}\\ &=\frac{2 \left (35 A b^2-14 a b B+8 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}+\frac{2 (7 b B-4 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}-\frac{\left (\left (a^2-b^2\right ) \left (35 A b^2-14 a b B+8 a^2 C+25 b^2 C\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b^3}-\frac{\left (14 a^2 b B-63 b^3 B-8 a^3 C-a b^2 (35 A+19 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{105 b^3}\\ &=\frac{2 \left (35 A b^2-14 a b B+8 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}+\frac{2 (7 b B-4 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}-\frac{\left (\left (14 a^2 b B-63 b^3 B-8 a^3 C-a b^2 (35 A+19 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}{105 b^3 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (a^2-b^2\right ) \left (35 A b^2-14 a b B+8 a^2 C+25 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}{105 b^3 \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{2 \left (14 a^2 b B-63 b^3 B-8 a^3 C-a b^2 (35 A+19 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (a^2-b^2\right ) \left (35 A b^2-14 a b B+8 a^2 C+25 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 )}{105 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (35 A b^2-14 a b B+8 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}+\frac{2 (7 b B-4 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 C \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}\\ \end{align*}
Mathematica [A] time = 1.26433, size = 249, normalized size = 0.78 \[ \frac{b (a+b \cos (c+d x)) \left (\sin (c+d x) \left (-16 a^2 C+28 a b B+140 A b^2+115 b^2 C\right )+3 b (2 (a C+7 b B) \sin (2 (c+d x))+5 b C \sin (3 (c+d x)))\right )+4 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b^2 \left (2 a^2 C+49 a b B+35 A b^2+25 b^2 C\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (-14 a^2 b B+8 a^3 C+a b^2 (35 A+19 C)+63 b^3 B\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{210 b^3 d \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
[Out]
(4*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(35*A*b^2 + 49*a*b*B + 2*a^2*C + 25*b^2*C)*EllipticF[(c + d*x)/2, (
2*b)/(a + b)] + (-14*a^2*b*B + 63*b^3*B + 8*a^3*C + a*b^2*(35*A + 19*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])*((140*A*b^2 + 28*a*b*B - 16*a^2
*C + 115*b^2*C)*Sin[c + d*x] + 3*b*(2*(7*b*B + a*C)*Sin[2*(c + d*x)] + 5*b*C*Sin[3*(c + d*x)])))/(210*b^3*d*Sq
rt[a + b*Cos[c + d*x]])
________________________________________________________________________________________
Maple [B] time = 0.961, size = 1635, normalized size = 5.1 \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)*(a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
[Out]
-2/105*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-168*B*b^4-144*C*a*b^3-360*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b^4+112*B*a*b^3+168*
B*b^4-4*C*a^2*b^2+144*C*a*b^3+280*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*a*b^3-70*A*b^4-14*B*a^
2*b^2-56*B*a*b^3-42*B*b^4+8*C*a^3*b+2*C*a^2*b^2-86*C*a*b^3-80*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3
5*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^2*b^2+35*A*b^4*(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))+35*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^2-35*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^3+14*B*(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-14*B*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^3-14*B*EllipticE(cos(1/2*d*x+1/2*c)
,(-2*b/(a-b))^(1/2))*(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)*a^3*b+14
*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(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)*a^2*b^2+63*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(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)*a*b^3-63*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(
1/2))*(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)*b^4-8*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^4-17*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^2*b^2+25*C*b^4*(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))+8*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^4-8*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+19*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)*Ellip
ticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2-19*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^3)/b^3/(-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} + B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)*cos(d*x + c), 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 )^{3} + B \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\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)*(a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)^3 + B*cos(d*x + c)^2 + A*cos(d*x + c))*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)*(a+b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)
[Out]
Timed out
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Giac [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)*(a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")
[Out]
Timed out