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

Optimal. Leaf size=303 \[ -\frac {2 \left (-8 a^2 C+14 a b B-25 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b^2 d}+\frac {2 \left (a^2-b^2\right ) \left (-8 a^2 C+14 a b B-25 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^3 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (7 b B-4 a C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b^2 d}+\frac {2 C \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{7 b d} \]

[Out]

2/35*(7*B*b-4*C*a)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b^2/d+2/7*C*cos(d*x+c)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/
b/d-2/105*(14*B*a*b-8*C*a^2-25*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d-2/105*(14*B*a^2*b-63*B*b^3-8*C*a
^3-19*C*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^
(1/2))*(a+b*cos(d*x+c))^(1/2)/b^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/105*(a^2-b^2)*(14*B*a*b-8*C*a^2-25*C*b^2)
*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*c
os(d*x+c))/(a+b))^(1/2)/b^3/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.60, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {3029, 2990, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (-8 a^2 C+14 a b B-25 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b^2 d}+\frac {2 \left (a^2-b^2\right ) \left (-8 a^2 C+14 a b B-25 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^3 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (14 a^2 b B-8 a^3 C-19 a b^2 C-63 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (7 b B-4 a C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b^2 d}+\frac {2 C \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*(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 - 19*a*b^2*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)*(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*(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*Co
s[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 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]

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 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 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 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 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 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 3029

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] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
 n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \cos (c+d x) \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\int \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} (B+C \cos (c+d x)) \, 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 {5}{2} b 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 (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 (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 (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-19 a b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^2}\\ &=-\frac {2 \left (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 (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-19 a b^2 C\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{105 b^3}\\ &=-\frac {2 \left (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-19 a b^2 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 (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-19 a b^2 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 (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 (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*}

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Mathematica [A]  time = 1.00, size = 232, normalized size = 0.77 \[ \frac {b (a+b \cos (c+d x)) \left (\left (-16 a^2 C+28 a b B+115 b^2 C\right ) \sin (c+d x)+3 b (2 (a C+7 b B) \sin (2 (c+d x))+5 b C \sin (3 (c+d x)))\right )+4 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 \left (2 a^2 C+49 a b B+25 b^2 C\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (8 a^3 C-14 a^2 b B+19 a b^2 C+63 b^3 B\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )\right )}{210 b^3 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*(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*(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 + 19*a*b^2*C)*((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*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*Sqrt[a + b*Cos[c + d*x]])

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{3} + B \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^(1/2)*(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)*sqrt(b*cos(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.91, size = 1305, normalized size = 4.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+b*cos(d*x+c))^(1/2)*(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)+(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)+(-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)-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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b+
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)*EllipticE(cos(1/2*d*x+1/
2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+63*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^3-63*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))*b^4+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*a*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)*Elliptic
F(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+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*(s
in(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*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^3-8*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(co
s(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))/b^3/(-2*sin(1/2*d*x+1/2*c)^4*b+(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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^(1/2)*(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))*sqrt(b*cos(d*x + c) + a)*cos(d*x + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+b\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)*(B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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