3.1.92 \(\int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx\) [92]

Optimal. Leaf size=161 \[ -\frac {63 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{220 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3}}{11 d}-\frac {67 \cos (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{2/3}}{55\ 2^{5/6} d (1+\sin (c+d x))^{7/6}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{44 a d} \]

[Out]

-63/220*cos(d*x+c)*(a+a*sin(d*x+c))^(2/3)/d-3/11*cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^(2/3)/d-67/110*cos(d
*x+c)*hypergeom([-1/6, 1/2],[3/2],1/2-1/2*sin(d*x+c))*(a+a*sin(d*x+c))^(2/3)*2^(1/6)/d/(1+sin(d*x+c))^(7/6)-3/
44*cos(d*x+c)*(a+a*sin(d*x+c))^(5/3)/a/d

________________________________________________________________________________________

Rubi [A]
time = 0.20, antiderivative size = 161, normalized size of antiderivative = 1.00, 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 = {2862, 3047, 3102, 2830, 2731, 2730} \begin {gather*} -\frac {67 \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{55\ 2^{5/6} d (\sin (c+d x)+1)^{7/6}}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{44 a d}-\frac {63 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{220 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-63*Cos[c + d*x]*(a + a*Sin[c + d*x])^(2/3))/(220*d) - (3*Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(2
/3))/(11*d) - (67*Cos[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 3/2, (1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(2
/3))/(55*2^(5/6)*d*(1 + Sin[c + d*x])^(7/6)) - (3*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/3))/(44*a*d)

Rule 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[a^IntPart[n]*((a + b*Sin[c + d*x])^FracPart
[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n]), Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2830

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*S
in[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 2862

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

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*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*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx &=-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3}}{11 d}+\frac {3 \int \sin (c+d x) \left (2 a+\frac {2}{3} a \sin (c+d x)\right ) (a+a \sin (c+d x))^{2/3} \, dx}{11 a}\\ &=-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3}}{11 d}+\frac {3 \int (a+a \sin (c+d x))^{2/3} \left (2 a \sin (c+d x)+\frac {2}{3} a \sin ^2(c+d x)\right ) \, dx}{11 a}\\ &=-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3}}{11 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{44 a d}+\frac {9 \int (a+a \sin (c+d x))^{2/3} \left (\frac {10 a^2}{9}+\frac {14}{3} a^2 \sin (c+d x)\right ) \, dx}{88 a^2}\\ &=-\frac {63 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{220 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3}}{11 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{44 a d}+\frac {67}{220} \int (a+a \sin (c+d x))^{2/3} \, dx\\ &=-\frac {63 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{220 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3}}{11 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{44 a d}+\frac {\left (67 (a+a \sin (c+d x))^{2/3}\right ) \int (1+\sin (c+d x))^{2/3} \, dx}{220 (1+\sin (c+d x))^{2/3}}\\ &=-\frac {63 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{220 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3}}{11 d}-\frac {67 \cos (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{2/3}}{55\ 2^{5/6} d (1+\sin (c+d x))^{7/6}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{44 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.47, size = 160, normalized size = 0.99 \begin {gather*} \frac {3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sin (c+d x)))^{2/3} \left (67 \sqrt {2} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {1-\sin (c+d x)} (-144+25 \cos (2 (c+d x))-92 \sin (c+d x)+10 \sin (3 (c+d x)))\right )}{440 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {1-\sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*(1 + Sin[c + d*x]))^(2/3)*(67*Sqrt[2]*Hypergeometric2F1[1/6, 1/2,
7/6, Sin[(2*c + Pi + 2*d*x)/4]^2] + Sqrt[1 - Sin[c + d*x]]*(-144 + 25*Cos[2*(c + d*x)] - 92*Sin[c + d*x] + 10*
Sin[3*(c + d*x)])))/(440*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*Sqrt[1 - Sin[c + d*x]])

________________________________________________________________________________________

Maple [F]
time = 0.33, size = 0, normalized size = 0.00 \[\int \left (\sin ^{3}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {2}{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(cos(d*x + c)^2 - 1)*(a*sin(d*x + c) + a)^(2/3)*sin(d*x + c), x)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{2/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________