3.373 \(\int \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=239 \[ \frac {2 a (99 A+88 B+80 C) \sin (c+d x) \cos ^3(c+d x)}{693 d \sqrt {a \cos (c+d x)+a}}+\frac {4 (99 A+88 B+80 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{1155 a d}-\frac {8 (99 A+88 B+80 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3465 d}+\frac {4 a (99 A+88 B+80 C) \sin (c+d x)}{495 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (11 B+C) \sin (c+d x) \cos ^4(c+d x)}{99 d \sqrt {a \cos (c+d x)+a}}+\frac {2 C \sin (c+d x) \cos ^4(c+d x) \sqrt {a \cos (c+d x)+a}}{11 d} \]

[Out]

4/1155*(99*A+88*B+80*C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+4/495*a*(99*A+88*B+80*C)*sin(d*x+c)/d/(a+a*cos(d
*x+c))^(1/2)+2/693*a*(99*A+88*B+80*C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/99*a*(11*B+C)*cos(d*x
+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-8/3465*(99*A+88*B+80*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+2/11*C*c
os(d*x+c)^4*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.55, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3045, 2981, 2770, 2759, 2751, 2646} \[ \frac {2 a (99 A+88 B+80 C) \sin (c+d x) \cos ^3(c+d x)}{693 d \sqrt {a \cos (c+d x)+a}}+\frac {4 (99 A+88 B+80 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{1155 a d}-\frac {8 (99 A+88 B+80 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3465 d}+\frac {4 a (99 A+88 B+80 C) \sin (c+d x)}{495 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (11 B+C) \sin (c+d x) \cos ^4(c+d x)}{99 d \sqrt {a \cos (c+d x)+a}}+\frac {2 C \sin (c+d x) \cos ^4(c+d x) \sqrt {a \cos (c+d x)+a}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(4*a*(99*A + 88*B + 80*C)*Sin[c + d*x])/(495*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(99*A + 88*B + 80*C)*Cos[c + d
*x]^3*Sin[c + d*x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(11*B + C)*Cos[c + d*x]^4*Sin[c + d*x])/(99*d*Sqrt
[a + a*Cos[c + d*x]]) - (8*(99*A + 88*B + 80*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3465*d) + (2*C*Cos[c +
d*x]^4*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(11*d) + (4*(99*A + 88*B + 80*C)*(a + a*Cos[c + d*x])^(3/2)*Sin
[c + d*x])/(1155*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 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

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 ^3(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 ^4(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac {2 \int \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (11 A+8 C)+\frac {1}{2} a (11 B+C) \cos (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 a (11 B+C) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac {1}{99} (99 A+88 B+80 C) \int \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a (99 A+88 B+80 C) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (11 B+C) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac {1}{231} (2 (99 A+88 B+80 C)) \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a (99 A+88 B+80 C) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (11 B+C) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac {4 (99 A+88 B+80 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 a d}+\frac {(4 (99 A+88 B+80 C)) \int \left (\frac {3 a}{2}-a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \, dx}{1155 a}\\ &=\frac {2 a (99 A+88 B+80 C) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (11 B+C) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \cos (c+d x)}}-\frac {8 (99 A+88 B+80 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac {2 C \cos ^4(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac {4 (99 A+88 B+80 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 a d}+\frac {1}{495} (2 (99 A+88 B+80 C)) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {4 a (99 A+88 B+80 C) \sin (c+d x)}{495 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (99 A+88 B+80 C) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (11 B+C) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \cos (c+d x)}}-\frac {8 (99 A+88 B+80 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac {2 C \cos ^4(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac {4 (99 A+88 B+80 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 a d}\\ \end {align*}

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Mathematica [A]  time = 1.30, size = 145, normalized size = 0.61 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (2 (9306 A+8272 B+9095 C) \cos (c+d x)+8 (594 A+913 B+830 C) \cos (2 (c+d x))+1980 A \cos (3 (c+d x))+30096 A+1760 B \cos (3 (c+d x))+770 B \cos (4 (c+d x))+29062 B+3175 C \cos (3 (c+d x))+700 C \cos (4 (c+d x))+315 C \cos (5 (c+d x))+26420 C)}{27720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*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])]*(30096*A + 29062*B + 26420*C + 2*(9306*A + 8272*B + 9095*C)*Cos[c + d*x] + 8*(594*
A + 913*B + 830*C)*Cos[2*(c + d*x)] + 1980*A*Cos[3*(c + d*x)] + 1760*B*Cos[3*(c + d*x)] + 3175*C*Cos[3*(c + d*
x)] + 770*B*Cos[4*(c + d*x)] + 700*C*Cos[4*(c + d*x)] + 315*C*Cos[5*(c + d*x)])*Tan[(c + d*x)/2])/(27720*d)

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fricas [A]  time = 0.44, size = 128, normalized size = 0.54 \[ \frac {2 \, {\left (315 \, C \cos \left (d x + c\right )^{5} + 35 \, {\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right ) + 1584 \, A + 1408 \, B + 1280 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

2/3465*(315*C*cos(d*x + c)^5 + 35*(11*B + 10*C)*cos(d*x + c)^4 + 5*(99*A + 88*B + 80*C)*cos(d*x + c)^3 + 6*(99
*A + 88*B + 80*C)*cos(d*x + c)^2 + 8*(99*A + 88*B + 80*C)*cos(d*x + c) + 1584*A + 1408*B + 1280*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 = 2.49, size = 290, normalized size = 1.21 \[ \frac {1}{55440} \, \sqrt {2} {\left (\frac {315 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )}{d} + \frac {385 \, {\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 {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {495 \, {\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 ) + 5 \, 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 {693 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 8 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, 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 {2310 \, {\left (6 \, 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 ) + 5 \, 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 {6930 \, {\left (6 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, 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)^3*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/55440*sqrt(2)*(315*C*sgn(cos(1/2*d*x + 1/2*c))*sin(11/2*d*x + 11/2*c)/d + 385*(2*B*sgn(cos(1/2*d*x + 1/2*c))
+ C*sgn(cos(1/2*d*x + 1/2*c)))*sin(9/2*d*x + 9/2*c)/d + 495*(4*A*sgn(cos(1/2*d*x + 1/2*c)) + 2*B*sgn(cos(1/2*
d*x + 1/2*c)) + 5*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(7/2*d*x + 7/2*c)/d + 693*(4*A*sgn(cos(1/2*d*x + 1/2*c)) + 8
*B*sgn(cos(1/2*d*x + 1/2*c)) + 5*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c)/d + 2310*(6*A*sgn(cos(1/2*d
*x + 1/2*c)) + 4*B*sgn(cos(1/2*d*x + 1/2*c)) + 5*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)/d + 6930*(6
*A*sgn(cos(1/2*d*x + 1/2*c)) + 6*B*sgn(cos(1/2*d*x + 1/2*c)) + 5*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 = 152, normalized size = 0.64 \[ \frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-10080 C \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6160 B +30800 C \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3960 A -15840 B -39600 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (8316 A +16632 B +27720 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-6930 A -9240 B -11550 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3465 A +3465 B +3465 C \right ) \sqrt {2}}{3465 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

2/3465*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(-10080*C*sin(1/2*d*x+1/2*c)^10+(6160*B+30800*C)*sin(1/2*d*x+1/
2*c)^8+(-3960*A-15840*B-39600*C)*sin(1/2*d*x+1/2*c)^6+(8316*A+16632*B+27720*C)*sin(1/2*d*x+1/2*c)^4+(-6930*A-9
240*B-11550*C)*sin(1/2*d*x+1/2*c)^2+3465*A+3465*B+3465*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 = 237, normalized size = 0.99 \[ \frac {396 \, {\left (5 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 22 \, {\left (35 \, \sqrt {2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 252 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 420 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1890 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + 5 \, {\left (63 \, \sqrt {2} \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 77 \, \sqrt {2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 495 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 693 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2310 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 6930 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{55440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/55440*(396*(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))*A*sqrt(a) + 22*(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))*B*sqrt(a) + 5*(63*sqrt(2)*sin(11/2*d*x + 11/2*c) + 77*sqrt(2)*sin(9/2*d*x + 9/2*c) + 495*sqrt(2)*sin(7/
2*d*x + 7/2*c) + 693*sqrt(2)*sin(5/2*d*x + 5/2*c) + 2310*sqrt(2)*sin(3/2*d*x + 3/2*c) + 6930*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 = -0.00 \[ \int {\cos \left (c+d\,x\right )}^3\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(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)^3*(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)**3*(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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