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

Optimal. Leaf size=193 \[ \frac {2 (21 A+18 B+16 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 a d}-\frac {4 (21 A+18 B+16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 a (21 A+18 B+16 C) \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (9 B+C) \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d} \]

[Out]

2/105*(21*A+18*B+16*C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/45*a*(21*A+18*B+16*C)*sin(d*x+c)/d/(a+a*cos(d*x
+c))^(1/2)+2/63*a*(9*B+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-4/315*(21*A+18*B+16*C)*sin(d*x+c)*(
a+a*cos(d*x+c))^(1/2)/d+2/9*C*cos(d*x+c)^3*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.47, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3045, 2981, 2759, 2751, 2646} \[ \frac {2 (21 A+18 B+16 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 a d}-\frac {4 (21 A+18 B+16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 a (21 A+18 B+16 C) \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (9 B+C) \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(21*A + 18*B + 16*C)*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(9*B + C)*Cos[c + d*x]^3*Sin[c
+ d*x])/(63*d*Sqrt[a + a*Cos[c + d*x]]) - (4*(21*A + 18*B + 16*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*
d) + (2*C*Cos[c + d*x]^3*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(9*d) + (2*(21*A + 18*B + 16*C)*(a + a*Cos[c +
d*x])^(3/2)*Sin[c + d*x])/(105*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 2759

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(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*(b*(m + 1) - a*
Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -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 ^2(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 ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (\frac {3}{2} a (3 A+2 C)+\frac {1}{2} a (9 B+C) \cos (c+d x)\right ) \, dx}{9 a}\\ &=\frac {2 a (9 B+C) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {1}{21} (21 A+18 B+16 C) \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a (9 B+C) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (21 A+18 B+16 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}+\frac {(2 (21 A+18 B+16 C)) \int \left (\frac {3 a}{2}-a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \, dx}{105 a}\\ &=\frac {2 a (9 B+C) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 (21 A+18 B+16 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 C \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (21 A+18 B+16 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}+\frac {1}{45} (21 A+18 B+16 C) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a (21 A+18 B+16 C) \sin (c+d x)}{45 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (9 B+C) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 (21 A+18 B+16 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 C \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (21 A+18 B+16 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 114, normalized size = 0.59 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} ((672 A+94 (9 B+8 C)) \cos (c+d x)+4 (63 A+54 B+83 C) \cos (2 (c+d x))+1596 A+90 B \cos (3 (c+d x))+1368 B+80 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))+1321 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*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])]*(1596*A + 1368*B + 1321*C + (672*A + 94*(9*B + 8*C))*Cos[c + d*x] + 4*(63*A + 54*B
+ 83*C)*Cos[2*(c + d*x)] + 90*B*Cos[3*(c + d*x)] + 80*C*Cos[3*(c + d*x)] + 35*C*Cos[4*(c + d*x)])*Tan[(c + d*
x)/2])/(1260*d)

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fricas [A]  time = 0.41, size = 108, normalized size = 0.56 \[ \frac {2 \, {\left (35 \, C \cos \left (d x + c\right )^{4} + 5 \, {\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 18 \, B + 16 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (21 \, A + 18 \, B + 16 \, C\right )} \cos \left (d x + c\right ) + 168 \, A + 144 \, B + 128 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*C*cos(d*x + c)^4 + 5*(9*B + 8*C)*cos(d*x + c)^3 + 3*(21*A + 18*B + 16*C)*cos(d*x + c)^2 + 4*(21*A +
18*B + 16*C)*cos(d*x + c) + 168*A + 144*B + 128*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 = 0.76, size = 261, normalized size = 1.35 \[ \frac {1}{2520} \, \sqrt {2} {\left (\frac {35 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {45 \, {\left (2 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {126 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {210 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {1260 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} + \frac {630 \, {\left (3 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*sqrt(2)*(35*C*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c)/d + 45*(2*B*sgn(cos(1/2*d*x + 1/2*c)) + C*
sgn(cos(1/2*d*x + 1/2*c)))*sin(7/2*d*x + 7/2*c)/d + 126*(2*A*sgn(cos(1/2*d*x + 1/2*c)) + B*sgn(cos(1/2*d*x + 1
/2*c)) + 2*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c)/d + 210*(2*A*sgn(cos(1/2*d*x + 1/2*c)) + 3*B*sgn(
cos(1/2*d*x + 1/2*c)) + 2*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)/d + 1260*(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 + 630*(3*B*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 = 0.64, size = 130, normalized size = 0.67 \[ \frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (560 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-360 B -1440 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (252 A +756 B +1512 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-420 A -630 B -840 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 A +315 B +315 C \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(560*C*sin(1/2*d*x+1/2*c)^8+(-360*B-1440*C)*sin(1/2*d*x+1/2*c)^6
+(252*A+756*B+1512*C)*sin(1/2*d*x+1/2*c)^4+(-420*A-630*B-840*C)*sin(1/2*d*x+1/2*c)^2+315*A+315*B+315*C)*2^(1/2
)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 0.66, size = 194, normalized size = 1.01 \[ \frac {84 \, {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 18 \, {\left (5 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + {\left (35 \, \sqrt {2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 252 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 420 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1890 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*(84*(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))
*A*sqrt(a) + 18*(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))*B*sqrt(a) + (35*sqrt(2)*sin(9/2*d*x + 9/2*c) + 45*sqrt(2)*sin(7/2*d*x
+ 7/2*c) + 252*sqrt(2)*sin(5/2*d*x + 5/2*c) + 420*sqrt(2)*sin(3/2*d*x + 3/2*c) + 1890*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 \[ \int {\cos \left (c+d\,x\right )}^2\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(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)^2*(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)**2*(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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