3.1.55 \(\int (a+a \sin (c+d x))^{5/2} \, dx\) [55]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {2726, 2725} \begin {gather*} -\frac {64 a^3 \cos (c+d x)}{15 d \sqrt {a \sin (c+d x)+a}}-\frac {16 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{15 d}-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^(5/2),x]

[Out]

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

Rule 2725

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 2726

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] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 117, normalized size = 1.31 \begin {gather*} -\frac {(a (1+\sin (c+d x)))^{5/2} \left (150 \cos \left (\frac {1}{2} (c+d x)\right )+25 \cos \left (\frac {3}{2} (c+d x)\right )-3 \cos \left (\frac {5}{2} (c+d x)\right )-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 )}{30 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.88, size = 65, normalized size = 0.73

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right ) \left (3 \left (\sin ^{2}\left (d x +c \right )\right )+14 \sin \left (d x +c \right )+43\right )}{15 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/15*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(3*sin(d*x+c)^2+14*sin(d*x+c)+43)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(d*x + c) + a)^(5/2), x)

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Fricas [A]
time = 0.40, size = 115, normalized size = 1.29 \begin {gather*} \frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{3} - 11 \, a^{2} \cos \left (d x + c\right )^{2} - 46 \, a^{2} \cos \left (d x + c\right ) - 32 \, a^{2} - {\left (3 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) - 32 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^2*cos(d*x + c)^3 - 11*a^2*cos(d*x + c)^2 - 46*a^2*cos(d*x + c) - 32*a^2 - (3*a^2*cos(d*x + c)^2 + 14
*a^2*cos(d*x + c) - 32*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \sin {\left (c + d x \right )} + a\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))**(5/2),x)

[Out]

Integral((a*sin(c + d*x) + a)**(5/2), x)

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Giac [A]
time = 0.51, size = 102, normalized size = 1.15 \begin {gather*} \frac {\sqrt {2} {\left (150 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(2)*(150*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 25*a^2*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 3*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi
+ 5/2*d*x + 5/2*c))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(5/2),x)

[Out]

int((a + a*sin(c + d*x))^(5/2), x)

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