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

Optimal. Leaf size=175 \[ \frac {64 a^3 (15 B+13 C) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 (15 B+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 (9 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{63 d}+\frac {2 a (15 B+13 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d} \]

[Out]

2/105*a*(15*B+13*C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/63*(9*B-2*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*(15*B+13*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/315*a^2
*(15*B+13*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.22, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3023, 2751, 2647, 2646} \[ \frac {16 a^2 (15 B+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {64 a^3 (15 B+13 C) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (9 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{63 d}+\frac {2 a (15 B+13 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(15*B + 13*C)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(15*B + 13*C)*Sqrt[a + a*Cos[c
+ d*x]]*Sin[c + d*x])/(315*d) + (2*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 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 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]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^{5/2} \left (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 {7 a C}{2}+\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} (15 B+13 C) \int (a+a \cos (c+d x))^{5/2} \, dx\\ &=\frac {2 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 (15 B+13 C)) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 (15 B+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 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 (15 B+13 C)\right ) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {64 a^3 (15 B+13 C) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 (15 B+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 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*}

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Mathematica [A]  time = 0.71, size = 105, normalized size = 0.60 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} ((3030 B+3116 C) \cos (c+d x)+8 (90 B+127 C) \cos (2 (c+d x))+90 B \cos (3 (c+d x))+6240 B+260 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))+5653 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(6240*B + 5653*C + (3030*B + 3116*C)*Cos[c + d*x] + 8*(90*B + 127*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)])*Tan[(c + d*x)/2])/(1260*d)

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fricas [A]  time = 0.47, size = 116, normalized size = 0.66 \[ \frac {2 \, {\left (35 \, C a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (9 \, B + 26 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (60 \, B + 73 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (345 \, B + 292 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (345 \, B + 292 \, C\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(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*(60*B + 73*C)*a^2*cos(d*x + c)^2 + (345
*B + 292*C)*a^2*cos(d*x + c) + 2*(345*B + 292*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 = 1.14, size = 272, normalized size = 1.55 \[ \frac {1}{2520} \, \sqrt {2} {\left (\frac {35 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {45 \, {\left (2 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {126 \, {\left (5 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {210 \, {\left (11 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 10 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {630 \, {\left (8 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 7 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} + \frac {630 \, {\left (7 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+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 + 45*(2*B*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(7/2*d*x + 7/2*c)/d + 126*(5*B*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 6*C
*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c)/d + 210*(11*B*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 10*C*a^2*sg
n(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c)/d + 630*(8*B*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 + 630*(7*B*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 6*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 = 0.67, size = 123, normalized size = 0.70 \[ \frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (140 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-90 B -540 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (315 B +819 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-420 B -630 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 B +315 C \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(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
+(315*B+819*C)*sin(1/2*d*x+1/2*c)^4+(-420*B-630*C)*sin(1/2*d*x+1/2*c)^2+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 = 0.58, size = 172, normalized size = 0.98 \[ \frac {30 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + {\left (35 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 225 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 756 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2100 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 8190 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*(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*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))*B*sqrt(a) + (35*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 225*sq
rt(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))*C*sqrt(a))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((B*cos(c + d*x) + 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)*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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