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

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

[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)

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Rubi [A]  time = 0.0461202, antiderivative size = 89, normalized size of antiderivative = 1., 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} \[ -\frac{64 a^3 \cos (c+d x)}{15 d \sqrt{a \sin (c+d x)+a}}-\frac{16 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{15 d}-\frac{2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d} \]

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 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 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]

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*}

Mathematica [A]  time = 0.30948, size = 117, normalized size = 1.31 \[ -\frac{(a (\sin (c+d x)+1))^{5/2} \left (-150 \sin \left (\frac{1}{2} (c+d x)\right )+25 \sin \left (\frac{3}{2} (c+d x)\right )+3 \sin \left (\frac{5}{2} (c+d x)\right )+150 \cos \left (\frac{1}{2} (c+d x)\right )+25 \cos \left (\frac{3}{2} (c+d x)\right )-3 \cos \left (\frac{5}{2} (c+d x)\right )\right )}{30 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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]))/(30*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^5)

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Maple [A]  time = 0.506, size = 65, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+14\,\sin \left ( dx+c \right ) +43 \right ) }{15\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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 = 1.40427, size = 292, normalized size = 3.28 \begin{align*} \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{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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