3.393 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=184 \[ \frac{16 a^2 (21 A+15 B+13 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d}+\frac{64 a^3 (21 A+15 B+13 C) \sin (c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a (21 A+15 B+13 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}+\frac{2 (9 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{63 d}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d} \]

[Out]

(64*a^3*(21*A + 15*B + 13*C)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(21*A + 15*B + 13*C)*Sqr
t[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*d) + (2*a*(21*A + 15*B + 13*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x
])/(105*d) + (2*(9*B - 2*C)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63*d) + (2*C*(a + a*Cos[c + d*x])^(7/2)*
Sin[c + d*x])/(9*a*d)

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Rubi [A]  time = 0.245148, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {3023, 2751, 2647, 2646} \[ \frac{16 a^2 (21 A+15 B+13 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d}+\frac{64 a^3 (21 A+15 B+13 C) \sin (c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a (21 A+15 B+13 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}+\frac{2 (9 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{63 d}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(64*a^3*(21*A + 15*B + 13*C)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(21*A + 15*B + 13*C)*Sqr
t[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*d) + (2*a*(21*A + 15*B + 13*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x
])/(105*d) + (2*(9*B - 2*C)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63*d) + (2*C*(a + a*Cos[c + d*x])^(7/2)*
Sin[c + d*x])/(9*a*d)

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 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 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 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]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac{2 \int (a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (9 A+7 C)+\frac{1}{2} a (9 B-2 C) \cos (c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 (9 B-2 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac{2 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac{1}{21} (21 A+15 B+13 C) \int (a+a \cos (c+d x))^{5/2} \, dx\\ &=\frac{2 a (21 A+15 B+13 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac{2 (9 B-2 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac{2 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac{1}{105} (8 a (21 A+15 B+13 C)) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac{16 a^2 (21 A+15 B+13 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac{2 a (21 A+15 B+13 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac{2 (9 B-2 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac{2 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac{1}{315} \left (32 a^2 (21 A+15 B+13 C)\right ) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{64 a^3 (21 A+15 B+13 C) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{16 a^2 (21 A+15 B+13 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac{2 a (21 A+15 B+13 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac{2 (9 B-2 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac{2 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}\\ \end{align*}

Mathematica [A]  time = 0.658205, size = 114, normalized size = 0.62 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} ((2352 A+3030 B+3116 C) \cos (c+d x)+4 (63 A+180 B+254 C) \cos (2 (c+d x))+7476 A+90 B \cos (3 (c+d x))+6240 B+260 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))+5653 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(7476*A + 6240*B + 5653*C + (2352*A + 3030*B + 3116*C)*Cos[c + d*x] + 4*(63*A
+ 180*B + 254*C)*Cos[2*(c + d*x)] + 90*B*Cos[3*(c + d*x)] + 260*C*Cos[3*(c + d*x)] + 35*C*Cos[4*(c + d*x)])*Ta
n[(c + d*x)/2])/(1260*d)

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Maple [A]  time = 0.065, size = 132, normalized size = 0.7 \begin{align*}{\frac{8\,{a}^{3}\sqrt{2}}{315\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 140\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -90\,B-540\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}+ \left ( 63\,A+315\,B+819\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -210\,A-420\,B-630\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+315\,A+315\,B+315\,C \right ){\frac{1}{\sqrt{a \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

8/315*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(140*C*sin(1/2*d*x+1/2*c)^8+(-90*B-540*C)*sin(1/2*d*x+1/2*c)^6
+(63*A+315*B+819*C)*sin(1/2*d*x+1/2*c)^4+(-210*A-420*B-630*C)*sin(1/2*d*x+1/2*c)^2+315*A+315*B+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 = 2.30506, size = 311, normalized size = 1.69 \begin{align*} \frac{84 \,{\left (3 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 25 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 150 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} A \sqrt{a} + 30 \,{\left (3 \, \sqrt{2} a^{2} \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 21 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 77 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 315 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} B \sqrt{a} +{\left (35 \, \sqrt{2} a^{2} \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 225 \, \sqrt{2} a^{2} \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 756 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 2100 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 8190 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} C \sqrt{a}}{2520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/2520*(84*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2
*d*x + 1/2*c))*A*sqrt(a) + 30*(3*sqrt(2)*a^2*sin(7/2*d*x + 7/2*c) + 21*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 77*s
qrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 315*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*B*sqrt(a) + (35*sqrt(2)*a^2*sin(9/2*d*
x + 9/2*c) + 225*sqrt(2)*a^2*sin(7/2*d*x + 7/2*c) + 756*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 2100*sqrt(2)*a^2*si
n(3/2*d*x + 3/2*c) + 8190*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d

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Fricas [A]  time = 1.7568, size = 331, normalized size = 1.8 \begin{align*} \frac{2 \,{\left (35 \, C a^{2} \cos \left (d x + c\right )^{4} + 5 \,{\left (9 \, B + 26 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (21 \, A + 60 \, B + 73 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (294 \, A + 345 \, B + 292 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (903 \, A + 690 \, B + 584 \, C\right )} a^{2}\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

2/315*(35*C*a^2*cos(d*x + c)^4 + 5*(9*B + 26*C)*a^2*cos(d*x + c)^3 + 3*(21*A + 60*B + 73*C)*a^2*cos(d*x + c)^2
+ (294*A + 345*B + 292*C)*a^2*cos(d*x + c) + (903*A + 690*B + 584*C)*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x +
c)/(d*cos(d*x + c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(5/2), x)