3.84 \(\int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=174 \[ \frac {8 a^2 (63 A+47 C) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (63 A+22 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{315 d}+\frac {2 a (63 A+47 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{21 a d} \]

[Out]

2/315*(63*A+22*C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/9*C*cos(d*x+c)^2*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2
/21*C*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/a/d+8/315*a^2*(63*A+47*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/315*a*
(63*A+47*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.36, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3046, 2968, 3023, 2751, 2647, 2646} \[ \frac {8 a^2 (63 A+47 C) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (63 A+22 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{315 d}+\frac {2 a (63 A+47 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{21 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(63*A + 47*C)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(63*A + 47*C)*Sqrt[a + a*Cos[c + d*
x]]*Sin[c + d*x])/(315*d) + (2*(63*A + 22*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*d) + (2*C*Cos[c + d
*x]^2*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(9*d) + (2*C*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(21*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 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 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 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 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 (c+d x) (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {2 \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} \left (\frac {1}{2} a (9 A+4 C)+\frac {3}{2} a C \cos (c+d x)\right ) \, dx}{9 a}\\ &=\frac {2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {2 \int (a+a \cos (c+d x))^{3/2} \left (\frac {1}{2} a (9 A+4 C) \cos (c+d x)+\frac {3}{2} a C \cos ^2(c+d x)\right ) \, dx}{9 a}\\ &=\frac {2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{21 a d}+\frac {4 \int (a+a \cos (c+d x))^{3/2} \left (\frac {15 a^2 C}{4}+\frac {1}{4} a^2 (63 A+22 C) \cos (c+d x)\right ) \, dx}{63 a^2}\\ &=\frac {2 (63 A+22 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{315 d}+\frac {2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{21 a d}+\frac {1}{105} (63 A+47 C) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac {2 a (63 A+47 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 (63 A+22 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{315 d}+\frac {2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{21 a d}+\frac {1}{315} (4 a (63 A+47 C)) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {8 a^2 (63 A+47 C) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (63 A+47 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 (63 A+22 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{315 d}+\frac {2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{21 a d}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 93, normalized size = 0.53 \[ \frac {a \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (2 (756 A+799 C) \cos (c+d x)+4 (63 A+137 C) \cos (2 (c+d x))+3276 A+170 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))+2689 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(3276*A + 2689*C + 2*(756*A + 799*C)*Cos[c + d*x] + 4*(63*A + 137*C)*Cos[2*(c +
d*x)] + 170*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 = 100, normalized size = 0.57 \[ \frac {2 \, {\left (35 \, C a \cos \left (d x + c\right )^{4} + 85 \, C a \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 34 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (189 \, A + 136 \, C\right )} a \cos \left (d x + c\right ) + 2 \, {\left (189 \, A + 136 \, C\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*C*a*cos(d*x + c)^4 + 85*C*a*cos(d*x + c)^3 + 3*(21*A + 34*C)*a*cos(d*x + c)^2 + (189*A + 136*C)*a*co
s(d*x + c) + 2*(189*A + 136*C)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)

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giac [A]  time = 0.33, size = 190, normalized size = 1.09 \[ \frac {1}{2520} \, \sqrt {2} {\left (\frac {35 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {135 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {126 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {210 \, {\left (6 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {1260 \, {\left (4 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*sqrt(2)*(35*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c)/d + 135*C*a*sgn(cos(1/2*d*x + 1/2*c))*si
n(7/2*d*x + 7/2*c)/d + 126*(2*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 3*C*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5
/2*c)/d + 210*(6*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 5*C*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)/d + 126
0*(4*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 3*C*a*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 = 118, normalized size = 0.68 \[ \frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (280 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-900 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (126 A +1134 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-315 A -735 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 A +315 C \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(280*C*sin(1/2*d*x+1/2*c)^8-900*C*sin(1/2*d*x+1/2*c)^6+(126*A+
1134*C)*sin(1/2*d*x+1/2*c)^4+(-315*A-735*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 = 0.77, size = 138, normalized size = 0.79 \[ \frac {252 \, {\left (\sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 20 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (35 \, \sqrt {2} a \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 135 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 378 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 1050 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3780 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*(252*(sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 20*sqrt(2)*a*sin(1/2*d*x + 1/
2*c))*A*sqrt(a) + (35*sqrt(2)*a*sin(9/2*d*x + 9/2*c) + 135*sqrt(2)*a*sin(7/2*d*x + 7/2*c) + 378*sqrt(2)*a*sin(
5/2*d*x + 5/2*c) + 1050*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 3780*sqrt(2)*a*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 \[ \int \cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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