3.297 \(\int \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\)
Optimal. Leaf size=303 \[ -\frac{2 \left (-8 a^2 B+14 a A b-25 b^2 B\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 b^2 d}+\frac{2 \left (a^2-b^2\right ) \left (-8 a^2 B+14 a A b-25 b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (14 a^2 A b-8 a^3 B-19 a b^2 B-63 A b^3\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (7 A b-4 a B) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b^2 d}+\frac{2 B \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{7 b d} \]
[Out]
(-2*(14*a^2*A*b - 63*A*b^3 - 8*a^3*B - 19*a*b^2*B)*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)*(14*a*A*b - 8*a^2*B - 25*b^2*B)*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*(14
*a*A*b - 8*a^2*B - 25*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b^2*d) + (2*(7*A*b - 4*a*B)*(a + b*Co
s[c + d*x])^(3/2)*Sin[c + d*x])/(35*b^2*d) + (2*B*Cos[c + d*x]*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(7*b*d
)
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Rubi [A] time = 0.541572, antiderivative size = 303, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used =
{2990, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (-8 a^2 B+14 a A b-25 b^2 B\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 b^2 d}+\frac{2 \left (a^2-b^2\right ) \left (-8 a^2 B+14 a A b-25 b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (14 a^2 A b-8 a^3 B-19 a b^2 B-63 A b^3\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (7 A b-4 a B) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b^2 d}+\frac{2 B \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{7 b d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]
[Out]
(-2*(14*a^2*A*b - 63*A*b^3 - 8*a^3*B - 19*a*b^2*B)*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)*(14*a*A*b - 8*a^2*B - 25*b^2*B)*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*(14
*a*A*b - 8*a^2*B - 25*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b^2*d) + (2*(7*A*b - 4*a*B)*(a + b*Co
s[c + d*x])^(3/2)*Sin[c + d*x])/(35*b^2*d) + (2*B*Cos[c + d*x]*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(7*b*d
)
Rule 2990
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] && !(IGtQ[n, 1] && ( !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)} (A+B \cos (c+d x)) \, dx &=\frac{2 B \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 B+\frac{5}{2} b B \cos (c+d x)+\frac{1}{2} (7 A b-4 a B) \cos ^2(c+d x)\right ) \, dx}{7 b}\\ &=\frac{2 (7 A b-4 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 B \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 A b-2 a B)-\frac{1}{4} \left (14 a A b-8 a^2 B-25 b^2 B\right ) \cos (c+d x)\right ) \, dx}{35 b^2}\\ &=-\frac{2 \left (14 a A b-8 a^2 B-25 b^2 B\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}+\frac{2 (7 A b-4 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 B \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 (49 a A b+2 a^2 B+25 b^2 B\right )-\frac{1}{8} \left (14 a^2 A b-63 A b^3-8 a^3 B-19 a b^2 B\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b^2}\\ &=-\frac{2 \left (14 a A b-8 a^2 B-25 b^2 B\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}+\frac{2 (7 A b-4 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 B \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 (14 a A b-8 a^2 B-25 b^2 B\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b^3}-\frac{\left (14 a^2 A b-63 A b^3-8 a^3 B-19 a b^2 B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{105 b^3}\\ &=-\frac{2 \left (14 a A b-8 a^2 B-25 b^2 B\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}+\frac{2 (7 A b-4 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 B \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 A b-63 A b^3-8 a^3 B-19 a b^2 B\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 (14 a A b-8 a^2 B-25 b^2 B\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 A b-63 A b^3-8 a^3 B-19 a b^2 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 \left (a^2-b^2\right ) \left (14 a A b-8 a^2 B-25 b^2 B\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 A b-8 a^2 B-25 b^2 B\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}+\frac{2 (7 A b-4 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac{2 B \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}\\ \end{align*}
Mathematica [A] time = 0.946124, size = 232, normalized size = 0.77 \[ \frac{b (a+b \cos (c+d x)) \left (\left (-16 a^2 B+28 a A b+115 b^2 B\right ) \sin (c+d x)+3 b (2 (a B+7 A b) \sin (2 (c+d x))+5 b B \sin (3 (c+d x)))\right )+4 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b^2 \left (2 a^2 B+49 a A b+25 b^2 B\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (-14 a^2 A b+8 a^3 B+19 a b^2 B+63 A b^3\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{210 b^3 d \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]
[Out]
(4*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(49*a*A*b + 2*a^2*B + 25*b^2*B)*EllipticF[(c + d*x)/2, (2*b)/(a + b
)] + (-14*a^2*A*b + 63*A*b^3 + 8*a^3*B + 19*a*b^2*B)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*Ellipt
icF[(c + d*x)/2, (2*b)/(a + b)])) + b*(a + b*Cos[c + d*x])*((28*a*A*b - 16*a^2*B + 115*b^2*B)*Sin[c + d*x] + 3
*b*(2*(7*A*b + a*B)*Sin[2*(c + d*x)] + 5*b*B*Sin[3*(c + d*x)])))/(210*b^3*d*Sqrt[a + b*Cos[c + d*x]])
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Maple [B] time = 4.021, size = 1305, normalized size = 4.3 \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+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x)
[Out]
-2/105*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*B*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-168*A*b^4-144*B*a*b^3-360*B*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(112*A*a*b^3+168*A*b^4-4*B*
a^2*b^2+144*B*a*b^3+280*B*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-14*A*a^2*b^2-56*A*a*b^3-42*A*b^4+8*B*
a^3*b+2*B*a^2*b^2-86*B*a*b^3-80*B*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+14*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-
14*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^3-14*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+14*A*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+63*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^3-63*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)*Elliptic
E(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^4-8*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^4-17*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^2*b^2+25*B
*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*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)*
EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4-8*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b+19*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^
2*b^2-19*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)*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 (B \cos \left (d x + c\right ) + 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+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + 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 (B \cos \left (d x + c\right )^{3} + 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+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")
[Out]
integral((B*cos(d*x + c)^3 + 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+b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)),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)^2*(a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="giac")
[Out]
Timed out