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

Optimal. Leaf size=267 \[ \frac {4 a^3 (143 A+121 B+105 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^3 (21 A+17 B+15 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (264 A+253 B+210 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{1155 d}+\frac {2 (99 A+143 B+105 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{693 d}+\frac {4 a^3 (143 A+121 B+105 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{231 d}+\frac {2 (11 B+6 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{99 a d}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{11 d} \]

[Out]

4/15*a^3*(21*A+17*B+15*C)*(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)
)/d+4/231*a^3*(143*A+121*B+105*C)*(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))/d+4/1155*a^3*(264*A+253*B+210*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/11*C*cos(d*x+c)^(3/2)*(a+a*cos(d*x+
c))^3*sin(d*x+c)/d+2/99*(11*B+6*C)*cos(d*x+c)^(3/2)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d+2/693*(99*A+143*B+10
5*C)*cos(d*x+c)^(3/2)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d+4/231*a^3*(143*A+121*B+105*C)*sin(d*x+c)*cos(d*x+c)^(1
/2)/d

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Rubi [A]  time = 0.69, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {3045, 2976, 2968, 3023, 2748, 2639, 2635, 2641} \[ \frac {4 a^3 (143 A+121 B+105 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^3 (21 A+17 B+15 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (264 A+253 B+210 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{1155 d}+\frac {2 (99 A+143 B+105 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{693 d}+\frac {4 a^3 (143 A+121 B+105 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{231 d}+\frac {2 (11 B+6 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{99 a d}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*(21*A + 17*B + 15*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^3*(143*A + 121*B + 105*C)*EllipticF[(c +
d*x)/2, 2])/(231*d) + (4*a^3*(143*A + 121*B + 105*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^3*(264*A
+ 253*B + 210*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(1155*d) + (2*C*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3*Si
n[c + d*x])/(11*d) + (2*(11*B + 6*C)*Cos[c + d*x]^(3/2)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(99*a*d) + (2
*(99*A + 143*B + 105*C)*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(693*d)

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], 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] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {2 C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2 \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \left (\frac {1}{2} a (11 A+3 C)+\frac {1}{2} a (11 B+6 C) \cos (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2 (11 B+6 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac {4 \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \left (\frac {3}{4} a^2 (33 A+11 B+15 C)+\frac {1}{4} a^2 (99 A+143 B+105 C) \cos (c+d x)\right ) \, dx}{99 a}\\ &=\frac {2 C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2 (11 B+6 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac {2 (99 A+143 B+105 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac {8 \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \left (\frac {15}{4} a^3 (33 A+22 B+21 C)+\frac {3}{4} a^3 (264 A+253 B+210 C) \cos (c+d x)\right ) \, dx}{693 a}\\ &=\frac {2 C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2 (11 B+6 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac {2 (99 A+143 B+105 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac {8 \int \sqrt {\cos (c+d x)} \left (\frac {15}{4} a^4 (33 A+22 B+21 C)+\left (\frac {15}{4} a^4 (33 A+22 B+21 C)+\frac {3}{4} a^4 (264 A+253 B+210 C)\right ) \cos (c+d x)+\frac {3}{4} a^4 (264 A+253 B+210 C) \cos ^2(c+d x)\right ) \, dx}{693 a}\\ &=\frac {4 a^3 (264 A+253 B+210 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2 (11 B+6 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac {2 (99 A+143 B+105 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac {16 \int \sqrt {\cos (c+d x)} \left (\frac {231}{8} a^4 (21 A+17 B+15 C)+\frac {45}{8} a^4 (143 A+121 B+105 C) \cos (c+d x)\right ) \, dx}{3465 a}\\ &=\frac {4 a^3 (264 A+253 B+210 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2 (11 B+6 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac {2 (99 A+143 B+105 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac {1}{15} \left (2 a^3 (21 A+17 B+15 C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{77} \left (2 a^3 (143 A+121 B+105 C)\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {4 a^3 (21 A+17 B+15 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (143 A+121 B+105 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {4 a^3 (264 A+253 B+210 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2 (11 B+6 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac {2 (99 A+143 B+105 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac {1}{231} \left (2 a^3 (143 A+121 B+105 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^3 (21 A+17 B+15 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (143 A+121 B+105 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^3 (143 A+121 B+105 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {4 a^3 (264 A+253 B+210 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2 (11 B+6 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac {2 (99 A+143 B+105 C) \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}\\ \end {align*}

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Mathematica [C]  time = 6.43, size = 1374, normalized size = 5.15 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-1/30*((21*A + 17*B + 15*C)*Cot[c])/d + ((2354
*A + 2134*B + 1953*C)*Cos[d*x]*Sin[c])/(7392*d) + ((54*A + 73*B + 75*C)*Cos[2*d*x]*Sin[2*c])/(720*d) + ((44*A
+ 132*B + 189*C)*Cos[3*d*x]*Sin[3*c])/(4928*d) + ((B + 3*C)*Cos[4*d*x]*Sin[4*c])/(288*d) + (C*Cos[5*d*x]*Sin[5
*c])/(704*d) + ((2354*A + 2134*B + 1953*C)*Cos[c]*Sin[d*x])/(7392*d) + ((54*A + 73*B + 75*C)*Cos[2*c]*Sin[2*d*
x])/(720*d) + ((44*A + 132*B + 189*C)*Cos[3*c]*Sin[3*d*x])/(4928*d) + ((B + 3*C)*Cos[4*c]*Sin[4*d*x])/(288*d)
+ (C*Cos[5*c]*Sin[5*d*x])/(704*d)) - (13*A*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4},
Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]
]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(42*d*Sqrt
[1 + Cot[c]^2]) - (11*B*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Co
t[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + C
ot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (
5*C*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 +
(d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[
d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(22*d*Sqrt[1 + Cot[c]^2]) - (7*A*(a + a*Cos[c + d
*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d
*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos
[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sq
rt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c
]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(20*d) - (17*B*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)
/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/
(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]
*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^
2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*S
qrt[1 + Tan[c]^2]]))/(60*d) - (C*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2,
-1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c
]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Ta
n[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt
[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(4*d)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{3} \cos \left (d x + c\right )^{5} + {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + {\left (A + 3 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + {\left (3 \, A + 3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{2} + {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + A a^{3}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^3*cos(d*x + c)^5 + (B + 3*C)*a^3*cos(d*x + c)^4 + (A + 3*B + 3*C)*a^3*cos(d*x + c)^3 + (3*A + 3*
B + C)*a^3*cos(d*x + c)^2 + (3*A + B)*a^3*cos(d*x + c) + A*a^3)*sqrt(cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3*sqrt(cos(d*x + c)), x)

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maple [A]  time = 1.93, size = 545, normalized size = 2.04 \[ -\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{3} \left (10080 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-6160 B -43680 C \right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (3960 A +24200 B +77280 C \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-14256 A -37532 B -72240 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (19866 A +29722 B +39270 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-6864 A -8118 B -8820 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2145 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-4851 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+1815 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3927 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+1575 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3465 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3465 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(10080*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-6160*B-43680*C)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(3960*A+24200*B+77280*C)*sin(1/2*d*x+1/2*c)
^8*cos(1/2*d*x+1/2*c)+(-14256*A-37532*B-72240*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(19866*A+29722*B+3927
0*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6864*A-8118*B-8820*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2
145*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-4851
*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+1815*B*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3927*B*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+1575*C*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3465*C*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/
2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3*sqrt(cos(d*x + c)), x)

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mupad [B]  time = 2.55, size = 507, normalized size = 1.90 \[ \frac {2\,\left (A\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+A\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}+\frac {B\,a^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {6\,A\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,B\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,C\,a^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^3\,{\cos \left (c+d\,x\right )}^{13/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {13}{4};\ \frac {17}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{13\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(A*a^3*ellipticE(c/2 + (d*x)/2, 2) + A*a^3*ellipticF(c/2 + (d*x)/2, 2) + A*a^3*cos(c + d*x)^(1/2)*sin(c + d
*x)))/d + (B*a^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d - (6*A*a^3*cos
(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (2*A*
a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2))
- (6*B*a^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^
(1/2)) - (2*B*a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(3*d*(sin(c + d
*x)^2)^(1/2)) - (2*B*a^3*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*
(sin(c + d*x)^2)^(1/2)) - (2*C*a^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2)
)/(7*d*(sin(c + d*x)^2)^(1/2)) - (2*C*a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c +
d*x)^2))/(3*d*(sin(c + d*x)^2)^(1/2)) - (6*C*a^3*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4,
cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2)) - (2*C*a^3*cos(c + d*x)^(13/2)*sin(c + d*x)*hypergeom([1/2, 13
/4], 17/4, cos(c + d*x)^2))/(13*d*(sin(c + d*x)^2)^(1/2))

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

[Out]

Timed out

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