3.46 \(\int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=86 \[ -\frac{8 a^2 \cos (c+d x)}{5 d \sqrt{a \sin (c+d x)+a}}-\frac{2 a \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{5 d}-\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d} \]

[Out]

(-8*a^2*Cos[c + d*x])/(5*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(5*d) - (2*
Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*d)

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Rubi [A]  time = 0.062433, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2751, 2647, 2646} \[ -\frac{8 a^2 \cos (c+d x)}{5 d \sqrt{a \sin (c+d x)+a}}-\frac{2 a \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{5 d}-\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-8*a^2*Cos[c + d*x])/(5*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(5*d) - (2*
Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*d)

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 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 \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac{3}{5} \int (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{2 a \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac{1}{5} (4 a) \int \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{8 a^2 \cos (c+d x)}{5 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.15917, size = 115, normalized size = 1.34 \[ -\frac{(a (\sin (c+d x)+1))^{3/2} \left (-20 \sin \left (\frac{1}{2} (c+d x)\right )+5 \sin \left (\frac{3}{2} (c+d x)\right )+\sin \left (\frac{5}{2} (c+d x)\right )+20 \cos \left (\frac{1}{2} (c+d x)\right )+5 \cos \left (\frac{3}{2} (c+d x)\right )-\cos \left (\frac{5}{2} (c+d x)\right )\right )}{10 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-((a*(1 + Sin[c + d*x]))^(3/2)*(20*Cos[(c + d*x)/2] + 5*Cos[(3*(c + d*x))/2] - Cos[(5*(c + d*x))/2] - 20*Sin[(
c + d*x)/2] + 5*Sin[(3*(c + d*x))/2] + Sin[(5*(c + d*x))/2]))/(10*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

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Maple [A]  time = 0.437, size = 63, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{2}+3\,\sin \left ( dx+c \right ) +6 \right ) }{5\,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(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/5*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)*(sin(d*x+c)^2+3*sin(d*x+c)+6)/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{3}{2}} \sin \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*sin(d*x + c), x)

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Fricas [A]  time = 1.52276, size = 259, normalized size = 3.01 \begin{align*} \frac{2 \,{\left (a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} - 7 \, a \cos \left (d x + c\right ) -{\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) - 4 \, a\right )} \sin \left (d x + c\right ) - 4 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{5 \,{\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(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/5*(a*cos(d*x + c)^3 - 2*a*cos(d*x + c)^2 - 7*a*cos(d*x + c) - (a*cos(d*x + c)^2 + 3*a*cos(d*x + c) - 4*a)*si
n(d*x + c) - 4*a)*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., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}} \sin{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral((a*(sin(c + d*x) + 1))**(3/2)*sin(c + d*x), x)

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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{3}{2}} \sin \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*sin(d*x + c), x)