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

Optimal. Leaf size=116 \[ -\frac{64 a^3 \cos (c+d x)}{21 d \sqrt{a \sin (c+d x)+a}}-\frac{16 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{21 d}-\frac{2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{7 d}-\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d} \]

[Out]

(-64*a^3*Cos[c + d*x])/(21*d*Sqrt[a + a*Sin[c + d*x]]) - (16*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(21*d)
- (2*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(7*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(7*d)

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Rubi [A]  time = 0.0829704, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2751, 2647, 2646} \[ -\frac{64 a^3 \cos (c+d x)}{21 d \sqrt{a \sin (c+d x)+a}}-\frac{16 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{21 d}-\frac{2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{7 d}-\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*Cos[c + d*x])/(21*d*Sqrt[a + a*Sin[c + d*x]]) - (16*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(21*d)
- (2*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(7*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(7*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))^{5/2} \, dx &=-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d}+\frac{5}{7} \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}}{7 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d}+\frac{1}{7} (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)}}{21 d}-\frac{2 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{7 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d}+\frac{1}{21} \left (32 a^2\right ) \int \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{64 a^3 \cos (c+d x)}{21 d \sqrt{a+a \sin (c+d x)}}-\frac{16 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{21 d}-\frac{2 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{7 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.613934, size = 141, normalized size = 1.22 \[ \frac{(a (\sin (c+d x)+1))^{5/2} \left (315 \sin \left (\frac{1}{2} (c+d x)\right )-77 \sin \left (\frac{3}{2} (c+d x)\right )-21 \sin \left (\frac{5}{2} (c+d x)\right )+3 \sin \left (\frac{7}{2} (c+d x)\right )-315 \cos \left (\frac{1}{2} (c+d x)\right )-77 \cos \left (\frac{3}{2} (c+d x)\right )+21 \cos \left (\frac{5}{2} (c+d x)\right )+3 \cos \left (\frac{7}{2} (c+d x)\right )\right )}{84 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(5/2)*(-315*Cos[(c + d*x)/2] - 77*Cos[(3*(c + d*x))/2] + 21*Cos[(5*(c + d*x))/2] + 3*C
os[(7*(c + d*x))/2] + 315*Sin[(c + d*x)/2] - 77*Sin[(3*(c + d*x))/2] - 21*Sin[(5*(c + d*x))/2] + 3*Sin[(7*(c +
d*x))/2]))/(84*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]  time = 0.506, size = 75, 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 ) ^{3}+12\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+23\,\sin \left ( dx+c \right ) +46 \right ) }{21\,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))^(5/2),x)

[Out]

2/21*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(3*sin(d*x+c)^3+12*sin(d*x+c)^2+23*sin(d*x+c)+46)/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}} \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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.38325, size = 355, normalized size = 3.06 \begin{align*} \frac{2 \,{\left (3 \, a^{2} \cos \left (d x + c\right )^{4} + 12 \, a^{2} \cos \left (d x + c\right )^{3} - 17 \, a^{2} \cos \left (d x + c\right )^{2} - 58 \, a^{2} \cos \left (d x + c\right ) - 32 \, a^{2} +{\left (3 \, a^{2} \cos \left (d x + c\right )^{3} - 9 \, a^{2} \cos \left (d x + c\right )^{2} - 26 \, 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}}{21 \,{\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))^(5/2),x, algorithm="fricas")

[Out]

2/21*(3*a^2*cos(d*x + c)^4 + 12*a^2*cos(d*x + c)^3 - 17*a^2*cos(d*x + c)^2 - 58*a^2*cos(d*x + c) - 32*a^2 + (3
*a^2*cos(d*x + c)^3 - 9*a^2*cos(d*x + c)^2 - 26*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(sin(d*x+c)*(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}} \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))^(5/2),x, algorithm="giac")

[Out]

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