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

Optimal. Leaf size=211 \[ \frac {64 a^3 (33 A+25 C) \sin (c+d x)}{693 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 (33 A+25 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{693 d}+\frac {2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{693 d}+\frac {2 a (33 A+25 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{231 d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{99 a d} \]

[Out]

2/231*a*(33*A+25*C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/693*(99*A+26*C)*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+
2/11*C*cos(d*x+c)^2*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+10/99*C*(a+a*cos(d*x+c))^(7/2)*sin(d*x+c)/a/d+64/693*a
^3*(33*A+25*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/693*a^2*(33*A+25*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.41, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3046, 2968, 3023, 2751, 2647, 2646} \[ \frac {16 a^2 (33 A+25 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{693 d}+\frac {64 a^3 (33 A+25 C) \sin (c+d x)}{693 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{693 d}+\frac {2 a (33 A+25 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{231 d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{99 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(33*A + 25*C)*Sin[c + d*x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(33*A + 25*C)*Sqrt[a + a*Cos[c
+ d*x]]*Sin[c + d*x])/(693*d) + (2*a*(33*A + 25*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(231*d) + (2*(99*A
+ 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) + (10*C*(a + a*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(99*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]

Rule 3046

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

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+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 {5}{2} a 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 {5}{2} a 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 {10 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 {35 a^2 C}{4}+\frac {1}{4} a^2 (99 A+26 C) \cos (c+d x)\right ) \, dx}{99 a^2}\\ &=\frac {2 (99 A+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 {10 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}+\frac {1}{231} (5 (33 A+25 C)) \int (a+a \cos (c+d x))^{5/2} \, dx\\ &=\frac {2 a (33 A+25 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{231 d}+\frac {2 (99 A+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 {10 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}+\frac {1}{231} (8 a (33 A+25 C)) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 (33 A+25 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{693 d}+\frac {2 a (33 A+25 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{231 d}+\frac {2 (99 A+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 {10 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}+\frac {1}{693} \left (32 a^2 (33 A+25 C)\right ) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {64 a^3 (33 A+25 C) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 (33 A+25 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{693 d}+\frac {2 a (33 A+25 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{231 d}+\frac {2 (99 A+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 {10 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 117, normalized size = 0.55 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (2 (6666 A+6989 C) \cos (c+d x)+16 (198 A+325 C) \cos (2 (c+d x))+396 A \cos (3 (c+d x))+27456 A+1735 C \cos (3 (c+d x))+448 C \cos (4 (c+d x))+63 C \cos (5 (c+d x))+22928 C)}{5544 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(27456*A + 22928*C + 2*(6666*A + 6989*C)*Cos[c + d*x] + 16*(198*A + 325*C)*Cos
[2*(c + d*x)] + 396*A*Cos[3*(c + d*x)] + 1735*C*Cos[3*(c + d*x)] + 448*C*Cos[4*(c + d*x)] + 63*C*Cos[5*(c + d*
x)])*Tan[(c + d*x)/2])/(5544*d)

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fricas [A]  time = 0.40, size = 129, normalized size = 0.61 \[ \frac {2 \, {\left (63 \, C a^{2} \cos \left (d x + c\right )^{5} + 224 \, C a^{2} \cos \left (d x + c\right )^{4} + {\left (99 \, A + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 6 \, {\left (66 \, A + 71 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (759 \, A + 568 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (759 \, A + 568 \, C\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/693*(63*C*a^2*cos(d*x + c)^5 + 224*C*a^2*cos(d*x + c)^4 + (99*A + 355*C)*a^2*cos(d*x + c)^3 + 6*(66*A + 71*C
)*a^2*cos(d*x + c)^2 + (759*A + 568*C)*a^2*cos(d*x + c) + 2*(759*A + 568*C)*a^2)*sqrt(a*cos(d*x + c) + a)*sin(
d*x + c)/(d*cos(d*x + c) + d)

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giac [A]  time = 1.44, size = 253, normalized size = 1.20 \[ \frac {1}{11088} \, \sqrt {2} {\left (\frac {63 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )}{d} + \frac {385 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {99 \, {\left (4 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 13 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {693 \, {\left (4 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {462 \, {\left (22 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 19 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {1386 \, {\left (30 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 23 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11088*sqrt(2)*(63*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(11/2*d*x + 11/2*c)/d + 385*C*a^2*sgn(cos(1/2*d*x + 1/2
*c))*sin(9/2*d*x + 9/2*c)/d + 99*(4*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 13*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(
7/2*d*x + 7/2*c)/d + 693*(4*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 5*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x +
5/2*c)/d + 462*(22*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 19*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)
/d + 1386*(30*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 23*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c)/d)*sq
rt(a)

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maple [A]  time = 0.53, size = 137, normalized size = 0.65 \[ \frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-504 C \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2156 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-198 A -3762 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (693 A +3465 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-924 A -1848 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+693 A +693 C \right ) \sqrt {2}}{693 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/693*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(-504*C*sin(1/2*d*x+1/2*c)^10+2156*C*sin(1/2*d*x+1/2*c)^8+(-19
8*A-3762*C)*sin(1/2*d*x+1/2*c)^6+(693*A+3465*C)*sin(1/2*d*x+1/2*c)^4+(-924*A-1848*C)*sin(1/2*d*x+1/2*c)^2+693*
A+693*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 0.59, size = 189, normalized size = 0.90 \[ \frac {132 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (63 \, \sqrt {2} a^{2} \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 385 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 1287 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 3465 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 8778 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 31878 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{11088 \, d} \]

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+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/11088*(132*(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) + (63*sqrt(2)*a^2*sin(11/2*d*x + 11/2*c) + 38
5*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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