\(\int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx\) [83]

Optimal result

Integrand size = 33, antiderivative size = 189 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {2 a^2 (39 A+34 B) \sin (c+d x)}{45 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 a (39 A+34 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d} \]

[Out]

2/105*(39*A+34*B)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/45*a^2*(39*A+34*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
+2/63*a^2*(9*A+10*B)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-4/315*a*(39*A+34*B)*sin(d*x+c)*(a+a*cos(
d*x+c))^(1/2)/d+2/9*a*B*cos(d*x+c)^3*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3055, 3060, 2838, 2830, 2725} \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {2 a^2 (9 A+10 B) \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (39 A+34 B) \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (39 A+34 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}-\frac {4 a (39 A+34 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d} \]

[In]

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

[Out]

(2*a^2*(39*A + 34*B)*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(9*A + 10*B)*Cos[c + d*x]^3*Sin[c
+ d*x])/(63*d*Sqrt[a + a*Cos[c + d*x]]) - (4*a*(39*A + 34*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*d) +
(2*a*B*Cos[c + d*x]^3*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(9*d) + (2*(39*A + 34*B)*(a + a*Cos[c + d*x])^(3/
2)*Sin[c + d*x])/(105*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 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 2838

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 3055

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)))
*Sin[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 3060

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)*Sqrt
[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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2}{9} \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (\frac {3}{2} a (3 A+2 B)+\frac {1}{2} a (9 A+10 B) \cos (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {1}{21} (a (39 A+34 B)) \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {1}{105} (2 (39 A+34 B)) \int \left (\frac {3 a}{2}-a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 a (39 A+34 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {1}{45} (a (39 A+34 B)) \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a^2 (39 A+34 B) \sin (c+d x)}{45 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 a (39 A+34 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.54 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (2964 A+2689 B+2 (759 A+799 B) \cos (c+d x)+(468 A+548 B) \cos (2 (c+d x))+90 A \cos (3 (c+d x))+170 B \cos (3 (c+d x))+35 B \cos (4 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d} \]

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(2964*A + 2689*B + 2*(759*A + 799*B)*Cos[c + d*x] + (468*A + 548*B)*Cos[2*(c + d
*x)] + 90*A*Cos[3*(c + d*x)] + 170*B*Cos[3*(c + d*x)] + 35*B*Cos[4*(c + d*x)])*Tan[(c + d*x)/2])/(1260*d)

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (280 B \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-180 A -900 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (504 A +1134 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-525 A -735 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 A +315 B \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(123\)
parts	\(\frac {4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (60 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+38\right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (280 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-220 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+114 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+47 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+94\right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(187\)

[In]

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

[Out]

4/315*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(280*B*sin(1/2*d*x+1/2*c)^8+(-180*A-900*B)*sin(1/2*d*x+1/2*c)^
6+(504*A+1134*B)*sin(1/2*d*x+1/2*c)^4+(-525*A-735*B)*sin(1/2*d*x+1/2*c)^2+315*A+315*B)*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 = 107, normalized size of antiderivative = 0.57 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (35 \, B a \cos \left (d x + c\right )^{4} + 5 \, {\left (9 \, A + 17 \, B\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{2} + 4 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right ) + 8 \, {\left (39 \, A + 34 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/315*(35*B*a*cos(d*x + c)^4 + 5*(9*A + 17*B)*a*cos(d*x + c)^3 + 3*(39*A + 34*B)*a*cos(d*x + c)^2 + 4*(39*A +
34*B)*a*cos(d*x + c) + 8*(39*A + 34*B)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {6 \, {\left (15 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 175 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 735 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (35 \, \sqrt {2} a \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 135 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 378 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 1050 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3780 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{2520 \, d} \]

[In]

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

[Out]

1/2520*(6*(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))*A*sqrt(a) + (35*sqrt(2)*a*sin(9/2*d*x + 9/2*c) + 135*sqrt(2)*a*s
in(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))*B*sqrt(a))/d

Giac [A] (verification not implemented)

none

Time = 1.06 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.01 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (35 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 378 \, {\left (A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 1050 \, {\left (A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 630 \, {\left (7 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \]

[In]

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

[Out]

1/2520*sqrt(2)*(35*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c) + 45*(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(7/2*d*x + 7/2*c) + 378*(A*a*sgn(cos(1/2*d*x + 1/2*c)) + B*a*sgn(cos(1/2*d
*x + 1/2*c)))*sin(5/2*d*x + 5/2*c) + 1050*(A*a*sgn(cos(1/2*d*x + 1/2*c)) + B*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(
3/2*d*x + 3/2*c) + 630*(7*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 6*B*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. \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^2\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

[In]

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

[Out]

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