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

Optimal. Leaf size=215 \[ \frac{4 a^2 (7 A+6 B+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^2 (12 A+9 B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^2 (21 A+27 B+19 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{4 a^2 (7 A+6 B+5 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 (9 B+4 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d}+\frac{2 C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2}{9 d} \]

[Out]

(4*a^2*(12*A + 9*B + 8*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(7*A + 6*B + 5*C)*EllipticF[(c + d*x)/2,
2])/(21*d) + (4*a^2*(7*A + 6*B + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*a^2*(21*A + 27*B + 19*C)*Co
s[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) + (2*C*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(9*d) +
(2*(9*B + 4*C)*Cos[c + d*x]^(3/2)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(63*d)

________________________________________________________________________________________

Rubi [A]  time = 0.504427, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{4 a^2 (7 A+6 B+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^2 (12 A+9 B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^2 (21 A+27 B+19 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{4 a^2 (7 A+6 B+5 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 (9 B+4 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d}+\frac{2 C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(12*A + 9*B + 8*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(7*A + 6*B + 5*C)*EllipticF[(c + d*x)/2,
2])/(21*d) + (4*a^2*(7*A + 6*B + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*a^2*(21*A + 27*B + 19*C)*Co
s[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) + (2*C*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(9*d) +
(2*(9*B + 4*C)*Cos[c + d*x]^(3/2)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(63*d)

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]

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \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))^2 \sin (c+d x)}{9 d}+\frac{2 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \left (\frac{3}{2} a (3 A+C)+\frac{1}{2} a (9 B+4 C) \cos (c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac{2 (9 B+4 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d}+\frac{4 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x)) \left (\frac{3}{4} a^2 (21 A+9 B+11 C)+\frac{3}{4} a^2 (21 A+27 B+19 C) \cos (c+d x)\right ) \, dx}{63 a}\\ &=\frac{2 C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac{2 (9 B+4 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d}+\frac{4 \int \sqrt{\cos (c+d x)} \left (\frac{3}{4} a^3 (21 A+9 B+11 C)+\left (\frac{3}{4} a^3 (21 A+9 B+11 C)+\frac{3}{4} a^3 (21 A+27 B+19 C)\right ) \cos (c+d x)+\frac{3}{4} a^3 (21 A+27 B+19 C) \cos ^2(c+d x)\right ) \, dx}{63 a}\\ &=\frac{2 a^2 (21 A+27 B+19 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac{2 (9 B+4 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d}+\frac{8 \int \sqrt{\cos (c+d x)} \left (\frac{21}{4} a^3 (12 A+9 B+8 C)+\frac{45}{4} a^3 (7 A+6 B+5 C) \cos (c+d x)\right ) \, dx}{315 a}\\ &=\frac{2 a^2 (21 A+27 B+19 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac{2 (9 B+4 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d}+\frac{1}{7} \left (2 a^2 (7 A+6 B+5 C)\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{15} \left (2 a^2 (12 A+9 B+8 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^2 (12 A+9 B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^2 (7 A+6 B+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a^2 (21 A+27 B+19 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac{2 (9 B+4 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d}+\frac{1}{21} \left (2 a^2 (7 A+6 B+5 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^2 (12 A+9 B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^2 (7 A+6 B+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^2 (7 A+6 B+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a^2 (21 A+27 B+19 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac{2 (9 B+4 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d}\\ \end{align*}

Mathematica [C]  time = 6.3569, size = 1322, normalized size = 6.15 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*(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])^2*Sec[c/2 + (d*x)/2]^4*(-((12*A + 9*B + 8*C)*Cot[c])/(15*d) + ((56*A +
51*B + 46*C)*Cos[d*x]*Sin[c])/(168*d) + ((18*A + 36*B + 37*C)*Cos[2*d*x]*Sin[2*c])/(360*d) + ((B + 2*C)*Cos[3
*d*x]*Sin[3*c])/(56*d) + (C*Cos[4*d*x]*Sin[4*c])/(144*d) + ((56*A + 51*B + 46*C)*Cos[c]*Sin[d*x])/(168*d) + ((
18*A + 36*B + 37*C)*Cos[2*c]*Sin[2*d*x])/(360*d) + ((B + 2*C)*Cos[3*c]*Sin[3*d*x])/(56*d) + (C*Cos[4*c]*Sin[4*
d*x])/(144*d)) - (A*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]
]]^2]*Sec[c/2 + (d*x)/2]^4*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]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (2*B*(
a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)
/2]^4*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]]]])/(7*d*Sqrt[1 + Cot[c]^2]) - (5*C*(a + a*Cos[c + d*x])^2
*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTa
n[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sq
rt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*Sqrt[1 + Cot[c]^2]) - (2*A*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d
*x)/2]^4*((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]]
]*Sqrt[1 + Tan[c]^2]]))/(5*d) - (3*B*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((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[T
an[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]]]*Sqrt[1 + Tan[c]^2]]))/(10*d)
- (4*C*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + A
rcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + Ar
cTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + Arc
Tan[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]]))/(15*d)

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Maple [B]  time = 0.188, size = 514, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-560*C*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1
/2*c)+(360*B+1840*C)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-252*A-1044*B-2368*C)*sin(1/2*d*x+1/2*c)^6*cos(1
/2*d*x+1/2*c)+(672*A+1134*B+1568*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-273*A-351*B-387*C)*sin(1/2*d*x+1
/2*c)^2*cos(1/2*d*x+1/2*c)+105*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))-252*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))+90*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))-189*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)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))+75*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))-168*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., size = 0, normalized size = 0. \begin{align*} \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 )}^{2} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(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)^2*sqrt(cos(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a^{2} \cos \left (d x + c\right )^{4} +{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**2*(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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{2} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(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)^2*sqrt(cos(d*x + c)), x)