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

Optimal. Leaf size=89 \[ \frac {64 a^3 \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

[Out]

2/5*a*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+64/15*a^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/15*a^2*sin(d*x+c)*(
a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2647, 2646} \[ \frac {64 a^3 \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(64*a^3*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(15*d)
+ (2*a*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*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]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^{5/2} \, dx &=\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{5} (8 a) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{15} \left (32 a^2\right ) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {64 a^3 \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 71, normalized size = 0.80 \[ \frac {a^2 \left (150 \sin \left (\frac {1}{2} (c+d x)\right )+25 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)}}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(150*Sin[(c + d*x)/2] + 25*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c
+ d*x))/2]))/(30*d)

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fricas [A]  time = 0.92, size = 62, normalized size = 0.70 \[ \frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + 43 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^2*cos(d*x + c)^2 + 14*a^2*cos(d*x + c) + 43*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 = 0.72, size = 90, normalized size = 1.01 \[ \frac {1}{30} \, \sqrt {2} {\left (\frac {3 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {25 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {150 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(2)*(3*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c)/d + 25*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(3/
2*d*x + 3/2*c)/d + 150*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 = 0.00, size = 73, normalized size = 0.82 \[ \frac {8 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/15*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)*(3*cos(1/2*d*x+1/2*c)^4+4*cos(1/2*d*x+1/2*c)^2+8)*2^(1/2)/(a*co
s(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 0.99, size = 60, normalized size = 0.67 \[ \frac {{\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(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))*sqrt(a)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^(5/2),x)

[Out]

int((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),x)

[Out]

Timed out

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