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

Optimal. Leaf size=104 \[ \frac {2 a (15 A+5 B+7 C) \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (5 B-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d} \]

[Out]

2/5*C*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/15*a*(15*A+5*B+7*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/15*(5*
B-2*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.15, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3023, 2751, 2646} \[ \frac {2 a (15 A+5 B+7 C) \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (5 B-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(15*A + 5*B + 7*C)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(5*B - 2*C)*Sqrt[a + a*Cos[c + d*x]
]*Sin[c + d*x])/(15*d) + (2*C*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*a*d)

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2751

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

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

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Mathematica [A]  time = 0.18, size = 67, normalized size = 0.64 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (30 A+2 (5 B+4 C) \cos (c+d x)+20 B+3 C \cos (2 (c+d x))+19 C)}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(30*A + 20*B + 19*C + 2*(5*B + 4*C)*Cos[c + d*x] + 3*C*Cos[2*(c + d*x)])*Tan[(c +
d*x)/2])/(15*d)

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fricas [A]  time = 0.43, size = 67, normalized size = 0.64 \[ \frac {2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + {\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right ) + 15 \, A + 10 \, B + 8 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*C*cos(d*x + c)^2 + (5*B + 4*C)*cos(d*x + c) + 15*A + 10*B + 8*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)
/(d*cos(d*x + c) + d)

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giac [A]  time = 1.05, size = 139, normalized size = 1.34 \[ \frac {1}{30} \, \sqrt {2} {\left (\frac {3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {30 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} + \frac {5 \, {\left (2 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {30 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(2)*(3*C*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c)/d + 30*B*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*
x + 1/2*c)/d + 5*(2*B*sgn(cos(1/2*d*x + 1/2*c)) + C*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)/d + 30*(2*
A*sgn(cos(1/2*d*x + 1/2*c)) + C*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c)/d)*sqrt(a)

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maple [A]  time = 0.60, size = 86, normalized size = 0.83 \[ \frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 C \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-10 B -20 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 A +15 B +15 C \right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*C*sin(1/2*d*x+1/2*c)^4+(-10*B-20*C)*sin(1/2*d*x+1/2*c)^2+15*A
+15*B+15*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 0.59, size = 106, normalized size = 1.02 \[ \frac {60 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10 \, {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(60*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 10*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1
/2*c))*B*sqrt(a) + (3*sqrt(2)*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*sin(3/2*d*x + 3/2*c) + 30*sqrt(2)*sin(1/2*d*x +
1/2*c))*C*sqrt(a))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + B*cos(c + d*x) + C*cos(c + d*x)**2), x)

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