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

Optimal. Leaf size=138 \[ \frac {8 a^2 (21 A+19 B) \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (7 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {2 a (21 A+19 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d} \]

[Out]

2/35*(7*A-2*B)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*B*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/a/d+8/105*a^2*(21*A
+19*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/105*a*(21*A+19*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.25, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2968, 3023, 2751, 2647, 2646} \[ \frac {8 a^2 (21 A+19 B) \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (7 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {2 a (21 A+19 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(21*A + 19*B)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(21*A + 19*B)*Sqrt[a + a*Cos[c + d*
x]]*Sin[c + d*x])/(105*d) + (2*(7*A - 2*B)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*B*(a + a*Cos[c
+ d*x])^(5/2)*Sin[c + d*x])/(7*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]

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx &=\int (a+a \cos (c+d x))^{3/2} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 B (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {2 \int (a+a \cos (c+d x))^{3/2} \left (\frac {5 a B}{2}+\frac {1}{2} a (7 A-2 B) \cos (c+d x)\right ) \, dx}{7 a}\\ &=\frac {2 (7 A-2 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {1}{35} (21 A+19 B) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac {2 a (21 A+19 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A-2 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {1}{105} (4 a (21 A+19 B)) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {8 a^2 (21 A+19 B) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (21 A+19 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A-2 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 81, normalized size = 0.59 \[ \frac {a \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} ((252 A+253 B) \cos (c+d x)+6 (7 A+13 B) \cos (2 (c+d x))+546 A+15 B \cos (3 (c+d x))+494 B)}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(546*A + 494*B + (252*A + 253*B)*Cos[c + d*x] + 6*(7*A + 13*B)*Cos[2*(c + d*x)]
+ 15*B*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(210*d)

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fricas [A]  time = 0.83, size = 88, normalized size = 0.64 \[ \frac {2 \, {\left (15 \, B a \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 13 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (63 \, A + 52 \, B\right )} a \cos \left (d x + c\right ) + 2 \, {\left (63 \, A + 52 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

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+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

2/105*(15*B*a*cos(d*x + c)^3 + 3*(7*A + 13*B)*a*cos(d*x + c)^2 + (63*A + 52*B)*a*cos(d*x + c) + 2*(63*A + 52*B
)*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 = 0.50, size = 164, normalized size = 1.19 \[ \frac {1}{420} \, \sqrt {2} {\left (\frac {15 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {21 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {35 \, {\left (6 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {105 \, {\left (8 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 7 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

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+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/420*sqrt(2)*(15*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c)/d + 21*(2*A*a*sgn(cos(1/2*d*x + 1/2*c)) +
3*B*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c)/d + 35*(6*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 5*B*a*sgn(cos
(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)/d + 105*(8*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 7*B*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 = 0.45, size = 104, normalized size = 0.75 \[ \frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-60 B \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (42 A +168 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-105 A -175 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 A +105 B \right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

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+B*cos(d*x+c)),x)

[Out]

4/105*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(-60*B*sin(1/2*d*x+1/2*c)^6+(42*A+168*B)*sin(1/2*d*x+1/2*c)^4+
(-105*A-175*B)*sin(1/2*d*x+1/2*c)^2+105*A+105*B)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 0.93, size = 123, normalized size = 0.89 \[ \frac {42 \, {\left (\sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 20 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (15 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 175 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 735 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{420 \, d} \]

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+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/420*(42*(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) + (15*sqrt(2)*a*sin(7/2*d*x + 7/2*c) + 63*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 175*sqrt(2)*a*sin(3/2
*d*x + 3/2*c) + 735*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (c+d\,x\right )\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(c + d*x)*(A + B*cos(c + d*x))*(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+B*cos(d*x+c)),x)

[Out]

Timed out

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