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

Optimal. Leaf size=180 \[ \frac {2 (21 A+16 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 a d}-\frac {4 (21 A+16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 a (21 A+16 C) \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}+\frac {2 a C \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}} \]

[Out]

2/105*(21*A+16*C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/45*a*(21*A+16*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
+2/63*a*C*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-4/315*(21*A+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 = 0.42, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3046, 2981, 2759, 2751, 2646} \[ \frac {2 (21 A+16 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 a d}-\frac {4 (21 A+16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 a (21 A+16 C) \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}+\frac {2 a C \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(21*A + 16*C)*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*C*Cos[c + d*x]^3*Sin[c + d*x])/(63*d*S
qrt[a + a*Cos[c + d*x]]) - (4*(21*A + 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 + 16*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(1
05*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 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, 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+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 C \cos (c+d x)\right ) \, dx}{9 a}\\ &=\frac {2 a 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+16 C) \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a 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+16 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}+\frac {(2 (21 A+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 C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 (21 A+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+16 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}+\frac {1}{45} (21 A+16 C) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a (21 A+16 C) \sin (c+d x)}{45 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 (21 A+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+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 = 0.50, size = 92, normalized size = 0.51 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (16 (42 A+47 C) \cos (c+d x)+4 (63 A+83 C) \cos (2 (c+d x))+1596 A+80 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))+1321 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(1596*A + 1321*C + 16*(42*A + 47*C)*Cos[c + d*x] + 4*(63*A + 83*C)*Cos[2*(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 = 0.61, size = 93, normalized size = 0.52 \[ \frac {2 \, {\left (35 \, C \cos \left (d x + c\right )^{4} + 40 \, C \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 16 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (21 \, A + 16 \, C\right )} \cos \left (d x + c\right ) + 168 \, A + 128 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*C*cos(d*x + c)^4 + 40*C*cos(d*x + c)^3 + 3*(21*A + 16*C)*cos(d*x + c)^2 + 4*(21*A + 16*C)*cos(d*x +
c) + 168*A + 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 = 0.61, size = 202, normalized size = 1.12 \[ \frac {1}{2520} \, \sqrt {2} {\left (\frac {35 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {45 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {630 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} + \frac {252 \, {\left (A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {420 \, {\left (A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {1260 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2),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*C*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2
*d*x + 7/2*c)/d + 630*C*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)/d + 252*(A*sgn(cos(1/2*d*x + 1/2*c)) +
C*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c)/d + 420*(A*sgn(cos(1/2*d*x + 1/2*c)) + 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)*sqrt(a)

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maple [A]  time = 0.57, size = 116, normalized size = 0.64 \[ \frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (560 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1440 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (252 A +1512 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-420 A -840 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 A +315 C \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2),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)^8-1440*C*sin(1/2*d*x+1/2*c)^6+(252*A+1
512*C)*sin(1/2*d*x+1/2*c)^4+(-420*A-840*C)*sin(1/2*d*x+1/2*c)^2+315*A+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 = 0.58, size = 131, normalized size = 0.73 \[ \frac {84 \, {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (35 \, \sqrt {2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 252 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 420 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1890 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2),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) + (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 \[ \int {\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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