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

Optimal. Leaf size=169 \[ \frac {64 a^3 (21 A+13 C) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 (21 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 a (21 A+13 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}-\frac {4 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{63 d} \]

[Out]

2/105*a*(21*A+13*C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d-4/63*C*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/9*C*(a+a*
cos(d*x+c))^(7/2)*sin(d*x+c)/a/d+64/315*a^3*(21*A+13*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/315*a^2*(21*A+1
3*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.21, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3024, 2751, 2647, 2646} \[ \frac {16 a^2 (21 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {64 a^3 (21 A+13 C) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (21 A+13 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}-\frac {4 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{63 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(21*A + 13*C)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(21*A + 13*C)*Sqrt[a + a*Cos[c
+ d*x]]*Sin[c + d*x])/(315*d) + (2*a*(21*A + 13*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(105*d) - (4*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 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 3024

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[
m, -1]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^{5/2} \left (A+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)-a C \cos (c+d x)\right ) \, dx}{9 a}\\ &=-\frac {4 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+13 C) \int (a+a \cos (c+d x))^{5/2} \, dx\\ &=\frac {2 a (21 A+13 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}-\frac {4 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+13 C)) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 (21 A+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (21 A+13 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}-\frac {4 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+13 C)\right ) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {64 a^3 (21 A+13 C) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 (21 A+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (21 A+13 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}-\frac {4 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*}

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Mathematica [A]  time = 0.46, size = 95, normalized size = 0.56 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (4 (588 A+779 C) \cos (c+d x)+4 (63 A+254 C) \cos (2 (c+d x))+7476 A+260 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))+5653 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(7476*A + 5653*C + 4*(588*A + 779*C)*Cos[c + d*x] + 4*(63*A + 254*C)*Cos[2*(c
+ d*x)] + 260*C*Cos[3*(c + d*x)] + 35*C*Cos[4*(c + d*x)])*Tan[(c + d*x)/2])/(1260*d)

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fricas [A]  time = 0.50, size = 110, normalized size = 0.65 \[ \frac {2 \, {\left (35 \, C a^{2} \cos \left (d x + c\right )^{4} + 130 \, C a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 73 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (147 \, A + 146 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (903 \, A + 584 \, C\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*C*a^2*cos(d*x + c)^4 + 130*C*a^2*cos(d*x + c)^3 + 3*(21*A + 73*C)*a^2*cos(d*x + c)^2 + 2*(147*A + 14
6*C)*a^2*cos(d*x + c) + (903*A + 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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giac [A]  time = 0.63, size = 250, normalized size = 1.48 \[ \frac {1}{2520} \, \sqrt {2} {\left (\frac {35 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {225 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {252 \, {\left (A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {2100 \, {\left (A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {630 \, {\left (12 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 7 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} + \frac {1260 \, {\left (4 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*sqrt(2)*(35*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c)/d + 225*C*a^2*sgn(cos(1/2*d*x + 1/2*c)
)*sin(7/2*d*x + 7/2*c)/d + 252*(A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 3*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d
*x + 5/2*c)/d + 2100*(A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)/
d + 630*(12*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 7*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c)/d + 1260
*(4*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 3*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c)/d)*sqrt(a)

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maple [A]  time = 0.49, size = 118, normalized size = 0.70 \[ \frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (140 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-540 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (63 A +819 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-210 A -630 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 A +315 C \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+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)^8-540*C*sin(1/2*d*x+1/2*c)^6+(63*A+8
19*C)*sin(1/2*d*x+1/2*c)^4+(-210*A-630*C)*sin(1/2*d*x+1/2*c)^2+315*A+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 = 0.56, size = 155, normalized size = 0.92 \[ \frac {84 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (35 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 225 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 756 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2100 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 8190 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+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) + (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*s
qrt(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
))*C*sqrt(a))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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