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

Optimal. Leaf size=229 \[ \frac{16 a^2 (165 A+143 B+125 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{3465 d}+\frac{64 a^3 (165 A+143 B+125 C) \sin (c+d x)}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (99 A-22 B+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{693 d}+\frac{2 a (165 A+143 B+125 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{1155 d}+\frac{2 (11 B+5 C) \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{99 a d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d} \]

[Out]

(64*a^3*(165*A + 143*B + 125*C)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(165*A + 143*B + 125
*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3465*d) + (2*a*(165*A + 143*B + 125*C)*(a + a*Cos[c + d*x])^(3/2)*
Sin[c + d*x])/(1155*d) + (2*(99*A - 22*B + 26*C)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(693*d) + (2*C*Cos[c
+ d*x]^2*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(11*d) + (2*(11*B + 5*C)*(a + a*Cos[c + d*x])^(7/2)*Sin[c +
d*x])/(99*a*d)

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Rubi [A]  time = 0.493746, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3045, 2968, 3023, 2751, 2647, 2646} \[ \frac{16 a^2 (165 A+143 B+125 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{3465 d}+\frac{64 a^3 (165 A+143 B+125 C) \sin (c+d x)}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (99 A-22 B+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{693 d}+\frac{2 a (165 A+143 B+125 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{1155 d}+\frac{2 (11 B+5 C) \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{99 a d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(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*(165*A + 143*B + 125*C)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(165*A + 143*B + 125
*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3465*d) + (2*a*(165*A + 143*B + 125*C)*(a + a*Cos[c + d*x])^(3/2)*
Sin[c + d*x])/(1155*d) + (2*(99*A - 22*B + 26*C)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(693*d) + (2*C*Cos[c
+ d*x]^2*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(11*d) + (2*(11*B + 5*C)*(a + a*Cos[c + d*x])^(7/2)*Sin[c +
d*x])/(99*a*d)

Rule 3045

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(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)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 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]

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 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 \cos (c+d x) (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 \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac{2 \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (11 A+4 C)+\frac{1}{2} a (11 B+5 C) \cos (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac{2 \int (a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (11 A+4 C) \cos (c+d x)+\frac{1}{2} a (11 B+5 C) \cos ^2(c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}+\frac{4 \int (a+a \cos (c+d x))^{5/2} \left (\frac{7}{4} a^2 (11 B+5 C)+\frac{1}{4} a^2 (99 A-22 B+26 C) \cos (c+d x)\right ) \, dx}{99 a^2}\\ &=\frac{2 (99 A-22 B+26 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{693 d}+\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}+\frac{1}{231} (165 A+143 B+125 C) \int (a+a \cos (c+d x))^{5/2} \, dx\\ &=\frac{2 a (165 A+143 B+125 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac{2 (99 A-22 B+26 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{693 d}+\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}+\frac{(8 a (165 A+143 B+125 C)) \int (a+a \cos (c+d x))^{3/2} \, dx}{1155}\\ &=\frac{16 a^2 (165 A+143 B+125 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac{2 a (165 A+143 B+125 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac{2 (99 A-22 B+26 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{693 d}+\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}+\frac{\left (32 a^2 (165 A+143 B+125 C)\right ) \int \sqrt{a+a \cos (c+d x)} \, dx}{3465}\\ &=\frac{64 a^3 (165 A+143 B+125 C) \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{16 a^2 (165 A+143 B+125 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac{2 a (165 A+143 B+125 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac{2 (99 A-22 B+26 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{693 d}+\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac{2 (11 B+5 C) (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}\\ \end{align*}

Mathematica [A]  time = 1.24694, size = 147, normalized size = 0.64 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} ((66660 A+68552 B+69890 C) \cos (c+d x)+16 (990 A+1397 B+1625 C) \cos (2 (c+d x))+1980 A \cos (3 (c+d x))+137280 A+5720 B \cos (3 (c+d x))+770 B \cos (4 (c+d x))+124366 B+8675 C \cos (3 (c+d x))+2240 C \cos (4 (c+d x))+315 C \cos (5 (c+d x))+114640 C)}{27720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(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])]*(137280*A + 124366*B + 114640*C + (66660*A + 68552*B + 69890*C)*Cos[c + d*x] +
16*(990*A + 1397*B + 1625*C)*Cos[2*(c + d*x)] + 1980*A*Cos[3*(c + d*x)] + 5720*B*Cos[3*(c + d*x)] + 8675*C*Co
s[3*(c + d*x)] + 770*B*Cos[4*(c + d*x)] + 2240*C*Cos[4*(c + d*x)] + 315*C*Cos[5*(c + d*x)])*Tan[(c + d*x)/2])/
(27720*d)

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Maple [A]  time = 0.08, size = 154, normalized size = 0.7 \begin{align*}{\frac{8\,{a}^{3}\sqrt{2}}{3465\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( -2520\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}+ \left ( 1540\,B+10780\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{8}+ \left ( -990\,A-5940\,B-18810\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}+ \left ( 3465\,A+9009\,B+17325\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -4620\,A-6930\,B-9240\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+3465\,A+3465\,B+3465\,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(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

8/3465*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(-2520*C*sin(1/2*d*x+1/2*c)^10+(1540*B+10780*C)*sin(1/2*d*x+1
/2*c)^8+(-990*A-5940*B-18810*C)*sin(1/2*d*x+1/2*c)^6+(3465*A+9009*B+17325*C)*sin(1/2*d*x+1/2*c)^4+(-4620*A-693
0*B-9240*C)*sin(1/2*d*x+1/2*c)^2+3465*A+3465*B+3465*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [A]  time = 2.14184, size = 381, normalized size = 1.66 \begin{align*} \frac{660 \,{\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 )} A \sqrt{a} + 22 \,{\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 )} B \sqrt{a} + 5 \,{\left (63 \, \sqrt{2} a^{2} \sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ) + 385 \, \sqrt{2} a^{2} \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 1287 \, \sqrt{2} a^{2} \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 3465 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 8778 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 31878 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} C \sqrt{a}}{55440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/55440*(660*(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*sqrt(2)*a^2*sin(3/
2*d*x + 3/2*c) + 315*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 22*(35*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 2
25*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*sin(3/2*d*x + 3/
2*c) + 8190*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*B*sqrt(a) + 5*(63*sqrt(2)*a^2*sin(11/2*d*x + 11/2*c) + 385*sqrt(
2)*a^2*sin(9/2*d*x + 9/2*c) + 1287*sqrt(2)*a^2*sin(7/2*d*x + 7/2*c) + 3465*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) +
8778*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 31878*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d

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Fricas [A]  time = 1.93397, size = 413, normalized size = 1.8 \begin{align*} \frac{2 \,{\left (315 \, C a^{2} \cos \left (d x + c\right )^{5} + 35 \,{\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \,{\left (99 \, A + 286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (660 \, A + 803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \,{\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3465 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*C*a^2*cos(d*x + c)^5 + 35*(11*B + 32*C)*a^2*cos(d*x + c)^4 + 5*(99*A + 286*B + 355*C)*a^2*cos(d*x
+ c)^3 + 3*(660*A + 803*B + 710*C)*a^2*cos(d*x + c)^2 + (3795*A + 3212*B + 2840*C)*a^2*cos(d*x + c) + 2*(3795*
A + 3212*B + 2840*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(cos(d*x+c)*(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}} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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