\(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [384]

Optimal result

Integrand size = 35, antiderivative size = 144 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {8 a^2 (35 A+21 B+19 C) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (35 A+21 B+19 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 B-2 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3102, 2830, 2726, 2725} \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {8 a^2 (35 A+21 B+19 C) \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (35 A+21 B+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 (7 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d} \]

[In]

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

[Out]

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

Rule 2725

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 2726

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] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2830

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*S
in[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 3102

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (700 A+546 B+494 C+(140 A+252 B+253 C) \cos (c+d x)+6 (7 B+13 C) \cos (2 (c+d x))+15 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \]

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(700*A + 546*B + 494*C + (140*A + 252*B + 253*C)*Cos[c + d*x] + 6*(7*B + 13*C)*C
os[2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(210*d)

Maple [A] (verified)

Time = 6.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-60 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (42 B +168 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-35 A -105 B -175 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 A +105 B +105 C \right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(110\)
parts	\(\frac {4 A \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (2 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {2}}{5 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (60 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+38\right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(217\)

[In]

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

[Out]

4/105*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(-60*C*sin(1/2*d*x+1/2*c)^6+(42*B+168*C)*sin(1/2*d*x+1/2*c)^4+
(-35*A-105*B-175*C)*sin(1/2*d*x+1/2*c)^2+105*A+105*B+105*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (15 \, C a \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, B + 13 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (35 \, A + 63 \, B + 52 \, C\right )} a \cos \left (d x + c\right ) + {\left (175 \, A + 126 \, B + 104 \, C\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/105*(15*C*a*cos(d*x + c)^3 + 3*(7*B + 13*C)*a*cos(d*x + c)^2 + (35*A + 63*B + 52*C)*a*cos(d*x + c) + (175*A
+ 126*B + 104*C)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.10 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {140 \, {\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 42 \, {\left (\sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 20 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + {\left (15 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 175 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 735 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{420 \, d} \]

[In]

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

[Out]

1/420*(140*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 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))*B*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))*C*sqrt(a))/d

Giac [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (15 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, {\left (2 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (4 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, {\left (12 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 8 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 7 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]

[In]

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

[Out]

1/420*sqrt(2)*(15*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 21*(2*B*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) + 35*(4*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 6*B*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) + 105*(12*A*a*sgn(cos(1/2*d*x + 1/2*c))
+ 8*B*a*sgn(cos(1/2*d*x + 1/2*c)) + 7*C*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int {\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

[In]

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

[Out]

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