3.32 \(\int \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=158 \[ -\frac{2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt{a \sin (c+d x)+a}}-\frac{16 a \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt{a \sin (c+d x)+a}}-\frac{32 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 a d}+\frac{64 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{315 d}-\frac{32 a \cos (c+d x)}{45 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-32*a*Cos[c + d*x])/(45*d*Sqrt[a + a*Sin[c + d*x]]) - (16*a*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a*Sin
[c + d*x]]) - (2*a*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (64*Cos[c + d*x]*Sqrt[a + a*S
in[c + d*x]])/(315*d) - (32*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*a*d)

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Rubi [A]  time = 0.230115, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2770, 2759, 2751, 2646} \[ -\frac{2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt{a \sin (c+d x)+a}}-\frac{16 a \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt{a \sin (c+d x)+a}}-\frac{32 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 a d}+\frac{64 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{315 d}-\frac{32 a \cos (c+d x)}{45 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-32*a*Cos[c + d*x])/(45*d*Sqrt[a + a*Sin[c + d*x]]) - (16*a*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a*Sin
[c + d*x]]) - (2*a*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (64*Cos[c + d*x]*Sqrt[a + a*S
in[c + d*x]])/(315*d) - (32*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*a*d)

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2759

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*
Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

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 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 ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx &=-\frac{2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{8}{9} \int \sin ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{16}{21} \int \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}-\frac{32 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a d}+\frac{32 \int \left (\frac{3 a}{2}-a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx}{105 a}\\ &=-\frac{16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{64 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 d}-\frac{32 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a d}+\frac{16}{45} \int \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{32 a \cos (c+d x)}{45 d \sqrt{a+a \sin (c+d x)}}-\frac{16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{64 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 d}-\frac{32 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a d}\\ \end{align*}

Mathematica [A]  time = 0.49884, size = 165, normalized size = 1.04 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (1890 \sin \left (\frac{1}{2} (c+d x)\right )-420 \sin \left (\frac{3}{2} (c+d x)\right )-252 \sin \left (\frac{5}{2} (c+d x)\right )+45 \sin \left (\frac{7}{2} (c+d x)\right )+35 \sin \left (\frac{9}{2} (c+d x)\right )-1890 \cos \left (\frac{1}{2} (c+d x)\right )-420 \cos \left (\frac{3}{2} (c+d x)\right )+252 \cos \left (\frac{5}{2} (c+d x)\right )+45 \cos \left (\frac{7}{2} (c+d x)\right )-35 \cos \left (\frac{9}{2} (c+d x)\right )\right )}{2520 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*(-1890*Cos[(c + d*x)/2] - 420*Cos[(3*(c + d*x))/2] + 252*Cos[(5*(c + d*x))/2] + 45
*Cos[(7*(c + d*x))/2] - 35*Cos[(9*(c + d*x))/2] + 1890*Sin[(c + d*x)/2] - 420*Sin[(3*(c + d*x))/2] - 252*Sin[(
5*(c + d*x))/2] + 45*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*d*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]))

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Maple [A]  time = 0.503, size = 83, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) a \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 35\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+40\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+48\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+64\,\sin \left ( dx+c \right ) +128 \right ) }{315\,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)^4*(a+a*sin(d*x+c))^(1/2),x)

[Out]

2/315*(1+sin(d*x+c))*a*(sin(d*x+c)-1)*(35*sin(d*x+c)^4+40*sin(d*x+c)^3+48*sin(d*x+c)^2+64*sin(d*x+c)+128)/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 \sqrt{a \sin \left (d x + c\right ) + a} \sin \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*sin(d*x + c)^4, x)

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Fricas [A]  time = 1.42856, size = 373, normalized size = 2.36 \begin{align*} -\frac{2 \,{\left (35 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} - 118 \, \cos \left (d x + c\right )^{3} + 26 \, \cos \left (d x + c\right )^{2} -{\left (35 \, \cos \left (d x + c\right )^{4} + 40 \, \cos \left (d x + c\right )^{3} - 78 \, \cos \left (d x + c\right )^{2} - 104 \, \cos \left (d x + c\right ) + 107\right )} \sin \left (d x + c\right ) + 211 \, \cos \left (d x + c\right ) + 107\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{315 \,{\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)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-2/315*(35*cos(d*x + c)^5 - 5*cos(d*x + c)^4 - 118*cos(d*x + c)^3 + 26*cos(d*x + c)^2 - (35*cos(d*x + c)^4 + 4
0*cos(d*x + c)^3 - 78*cos(d*x + c)^2 - 104*cos(d*x + c) + 107)*sin(d*x + c) + 211*cos(d*x + c) + 107)*sqrt(a*s
in(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 \sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )} \sin ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**4*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sin(c + d*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sin \left (d x + c\right ) + a} \sin \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*sin(d*x + c)^4, x)