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

Optimal. Leaf size=231 \[ -\frac {2 \left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-2 a^2 C+5 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b d}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d} \]

[Out]

2/5*C*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d+2/15*(5*B*b-2*C*a)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d+2/15*(5*B
*a*b-2*C*a^2+9*C*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^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/15*(a^2-b^2)*(5*B*b-2*C*a)*(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*cos(d*x+c
))/(a+b))^(1/2)/b^2/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-2 a^2 C+5 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b d}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a*b*B - 2*a^2*C + 9*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b^2*d*Sqr
t[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(5*b*B - 2*a*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF
[(c + d*x)/2, (2*b)/(a + b)])/(15*b^2*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(5*b*B - 2*a*C)*Sqrt[a + b*Cos[c + d*x]
]*Sin[c + d*x])/(15*b*d) + (2*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*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 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]

Rubi steps

\begin {align*} \int \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac {2 \int \sqrt {a+b \cos (c+d x)} \left (\frac {3 b C}{2}+\frac {1}{2} (5 b B-2 a C) \cos (c+d x)\right ) \, dx}{5 b}\\ &=\frac {2 (5 b B-2 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac {4 \int \frac {\frac {1}{4} b (5 b B+7 a C)+\frac {1}{4} \left (5 a b B-2 a^2 C+9 b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b}\\ &=\frac {2 (5 b B-2 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}-\frac {\left (\left (a^2-b^2\right ) (5 b B-2 a C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^2}+\frac {\left (5 a b B-2 a^2 C+9 b^2 C\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b^2}\\ &=\frac {2 (5 b B-2 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac {\left (\left (5 a b B-2 a^2 C+9 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}{15 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (a^2-b^2\right ) (5 b B-2 a C) \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}{15 b^2 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (5 a b B-2 a^2 C+9 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 )}{15 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-2 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 179, normalized size = 0.77 \[ \frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (\left (-2 a^2 C+5 a b B+9 b^2 C\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+b^2 (7 a C+5 b B) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+2 b \sin (c+d x) (a+b \cos (c+d x)) (a C+5 b B+3 b C \cos (c+d x))}{15 b^2 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(5*b*B + 7*a*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (5*a*b*B -
2*a^2*C + 9*b^2*C)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))
+ 2*b*(a + b*Cos[c + d*x])*(5*b*B + a*C + 3*b*C*Cos[c + d*x])*Sin[c + d*x])/(15*b^2*d*Sqrt[a + b*Cos[c + d*x]]
)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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), x)

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maple [B]  time = 2.61, size = 993, normalized size = 4.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^6+(20*B*b^3+16*C*a*b^2+24*C*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*B*a*b^2-10*B*b^3-2*C*a^2*b-
8*C*a*b^2-6*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*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+5*B*b^3*(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))+5*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-5*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^2+2*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^3-2*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^2-2*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+2*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+9*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^2-9*C*(si
n(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^3)/b^2/(-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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((B + C*cos(c + d*x))*sqrt(a + b*cos(c + d*x))*cos(c + d*x), x)

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