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

Optimal. Leaf size=162 \[ -\frac{2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt{a \sin (c+d x)+a}}-\frac{34 a^2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt{a \sin (c+d x)+a}}-\frac{68 a^2 \cos (c+d x)}{45 d \sqrt{a \sin (c+d x)+a}}-\frac{68 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d}+\frac{136 a \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{315 d} \]

[Out]

(-68*a^2*Cos[c + d*x])/(45*d*Sqrt[a + a*Sin[c + d*x]]) - (34*a^2*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a
*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (136*a*Cos[c + d*x]*Sqr
t[a + a*Sin[c + d*x]])/(315*d) - (68*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*d)

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Rubi [A]  time = 0.241349, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2763, 21, 2770, 2759, 2751, 2646} \[ -\frac{2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt{a \sin (c+d x)+a}}-\frac{34 a^2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt{a \sin (c+d x)+a}}-\frac{68 a^2 \cos (c+d x)}{45 d \sqrt{a \sin (c+d x)+a}}-\frac{68 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d}+\frac{136 a \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{315 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-68*a^2*Cos[c + d*x])/(45*d*Sqrt[a + a*Sin[c + d*x]]) - (34*a^2*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a
*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (136*a*Cos[c + d*x]*Sqr
t[a + a*Sin[c + d*x]])/(315*d) - (68*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*d)

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{2}{9} \int \frac{\sin ^3(c+d x) \left (\frac{17 a^2}{2}+\frac{17}{2} a^2 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{1}{9} (17 a) \int \sin ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{1}{21} (34 a) \int \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}-\frac{68 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac{68}{105} \int \left (\frac{3 a}{2}-a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{136 a \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 d}-\frac{68 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac{1}{45} (34 a) \int \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{68 a^2 \cos (c+d x)}{45 d \sqrt{a+a \sin (c+d x)}}-\frac{34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{136 a \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 d}-\frac{68 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}\\ \end{align*}

Mathematica [A]  time = 0.529742, size = 165, normalized size = 1.02 \[ \frac{(a (\sin (c+d x)+1))^{3/2} \left (3780 \sin \left (\frac{1}{2} (c+d x)\right )-1050 \sin \left (\frac{3}{2} (c+d x)\right )-378 \sin \left (\frac{5}{2} (c+d x)\right )+135 \sin \left (\frac{7}{2} (c+d x)\right )+35 \sin \left (\frac{9}{2} (c+d x)\right )-3780 \cos \left (\frac{1}{2} (c+d x)\right )-1050 \cos \left (\frac{3}{2} (c+d x)\right )+378 \cos \left (\frac{5}{2} (c+d x)\right )+135 \cos \left (\frac{7}{2} (c+d x)\right )-35 \cos \left (\frac{9}{2} (c+d x)\right )\right )}{2520 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(3/2)*(-3780*Cos[(c + d*x)/2] - 1050*Cos[(3*(c + d*x))/2] + 378*Cos[(5*(c + d*x))/2] +
135*Cos[(7*(c + d*x))/2] - 35*Cos[(9*(c + d*x))/2] + 3780*Sin[(c + d*x)/2] - 1050*Sin[(3*(c + d*x))/2] - 378*
Sin[(5*(c + d*x))/2] + 135*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])^3)

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Maple [A]  time = 0.463, size = 85, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 35\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+85\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+102\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+136\,\sin \left ( dx+c \right ) +272 \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)^3*(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/315*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)*(35*sin(d*x+c)^4+85*sin(d*x+c)^3+102*sin(d*x+c)^2+136*sin(d*x+c)+272)/
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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3*(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)^3, x)

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Fricas [A]  time = 1.39264, size = 405, normalized size = 2.5 \begin{align*} -\frac{2 \,{\left (35 \, a \cos \left (d x + c\right )^{5} - 50 \, a \cos \left (d x + c\right )^{4} - 172 \, a \cos \left (d x + c\right )^{3} + 134 \, a \cos \left (d x + c\right )^{2} + 409 \, a \cos \left (d x + c\right ) -{\left (35 \, a \cos \left (d x + c\right )^{4} + 85 \, a \cos \left (d x + c\right )^{3} - 87 \, a \cos \left (d x + c\right )^{2} - 221 \, a \cos \left (d x + c\right ) + 188 \, a\right )} \sin \left (d x + c\right ) + 188 \, a\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)^3*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-2/315*(35*a*cos(d*x + c)^5 - 50*a*cos(d*x + c)^4 - 172*a*cos(d*x + c)^3 + 134*a*cos(d*x + c)^2 + 409*a*cos(d*
x + c) - (35*a*cos(d*x + c)^4 + 85*a*cos(d*x + c)^3 - 87*a*cos(d*x + c)^2 - 221*a*cos(d*x + c) + 188*a)*sin(d*
x + c) + 188*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(-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)**3*(a+a*sin(d*x+c))**(3/2),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{3}{2}} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3*(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)^3, x)