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

Optimal. Leaf size=147 \[ \frac {2 (35 A-14 B+18 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 a (35 A+49 B+27 C) \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 a d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d} \]

[Out]

2/35*(7*B+C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/105*a*(35*A+49*B+27*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2
)+2/105*(35*A-14*B+18*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+2/7*C*cos(d*x+c)^2*sin(d*x+c)*(a+a*cos(d*x+c))^(1
/2)/d

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Rubi [A]  time = 0.35, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3045, 2968, 3023, 2751, 2646} \[ \frac {2 (35 A-14 B+18 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 a (35 A+49 B+27 C) \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 a d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(35*A + 49*B + 27*C)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(35*A - 14*B + 18*C)*Sqrt[a + a*
Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*C*Cos[c + d*x]^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d) + (2*(7
*B + C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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 3045

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

Rubi steps

\begin {align*} \int \cos (c+d x) \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 \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac {2 \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (7 A+4 C)+\frac {1}{2} a (7 B+C) \cos (c+d x)\right ) \, dx}{7 a}\\ &=\frac {2 C \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac {2 \int \sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (7 A+4 C) \cos (c+d x)+\frac {1}{2} a (7 B+C) \cos ^2(c+d x)\right ) \, dx}{7 a}\\ &=\frac {2 C \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac {2 (7 B+C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d}+\frac {4 \int \sqrt {a+a \cos (c+d x)} \left (\frac {3}{4} a^2 (7 B+C)+\frac {1}{4} a^2 (35 A-14 B+18 C) \cos (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac {2 (35 A-14 B+18 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac {2 (7 B+C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d}+\frac {1}{105} (35 A+49 B+27 C) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a (35 A+49 B+27 C) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (35 A-14 B+18 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac {2 (7 B+C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 86, normalized size = 0.59 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} ((140 A+112 B+141 C) \cos (c+d x)+280 A+6 (7 B+6 C) \cos (2 (c+d x))+266 B+15 C \cos (3 (c+d x))+228 C)}{210 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*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])]*(280*A + 266*B + 228*C + (140*A + 112*B + 141*C)*Cos[c + d*x] + 6*(7*B + 6*C)*Cos[
2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(210*d)

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fricas [A]  time = 0.44, size = 87, normalized size = 0.59 \[ \frac {2 \, {\left (15 \, C \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (35 \, A + 28 \, B + 24 \, C\right )} \cos \left (d x + c\right ) + 70 \, A + 56 \, B + 48 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*C*cos(d*x + c)^3 + 3*(7*B + 6*C)*cos(d*x + c)^2 + (35*A + 28*B + 24*C)*cos(d*x + c) + 70*A + 56*B +
48*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 = 0.54, size = 182, normalized size = 1.24 \[ \frac {1}{420} \, \sqrt {2} {\left (\frac {15 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {21 \, {\left (2 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {35 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {105 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 4 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*sqrt(2)*(15*C*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c)/d + 21*(2*B*sgn(cos(1/2*d*x + 1/2*c)) + C*s
gn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c)/d + 35*(4*A*sgn(cos(1/2*d*x + 1/2*c)) + 2*B*sgn(cos(1/2*d*x + 1
/2*c)) + 3*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)/d + 105*(4*A*sgn(cos(1/2*d*x + 1/2*c)) + 4*B*sgn(
cos(1/2*d*x + 1/2*c)) + 3*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 = 0.67, size = 108, normalized size = 0.73 \[ \frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-120 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (84 B +252 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-70 A -140 B -210 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 A +105 B +105 C \right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(-120*C*sin(1/2*d*x+1/2*c)^6+(84*B+252*C)*sin(1/2*d*x+1/2*c)^4+(
-70*A-140*B-210*C)*sin(1/2*d*x+1/2*c)^2+105*A+105*B+105*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 0.62, size = 152, normalized size = 1.03 \[ \frac {140 \, {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 14 \, {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + 3 \, {\left (5 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(140*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 14*(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))*B*sqrt(a) + 3*(5*sqrt(2)*sin(7
/2*d*x + 7/2*c) + 7*sqrt(2)*sin(5/2*d*x + 5/2*c) + 35*sqrt(2)*sin(3/2*d*x + 3/2*c) + 105*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 \[ \int \cos \left (c+d\,x\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(c + d*x)*(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(-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+a*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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