\(\int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx\) [98]

Optimal result

Integrand size = 23, antiderivative size = 162 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {388\ 2^{5/6} a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [3]{a+a \sin (c+d x)}}{455 d (1+\sin (c+d x))^{5/6}}-\frac {72 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{455 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3}}{13 d}-\frac {6 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{65 a d} \]

[Out]

-388/455*2^(5/6)*a*cos(d*x+c)*hypergeom([-5/6, 1/2],[3/2],1/2-1/2*sin(d*x+c))*(a+a*sin(d*x+c))^(1/3)/d/(1+sin(
d*x+c))^(5/6)-72/455*cos(d*x+c)*(a+a*sin(d*x+c))^(4/3)/d-3/13*cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3)/d
-6/65*cos(d*x+c)*(a+a*sin(d*x+c))^(7/3)/a/d

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2862, 3047, 3102, 2830, 2731, 2730} \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {388\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{455 d (\sin (c+d x)+1)^{5/6}}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{13 d}-\frac {6 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{65 a d}-\frac {72 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{455 d} \]

[In]

Int[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^(4/3),x]

[Out]

(-388*2^(5/6)*a*Cos[c + d*x]*Hypergeometric2F1[-5/6, 1/2, 3/2, (1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(1/3
))/(455*d*(1 + Sin[c + d*x])^(5/6)) - (72*Cos[c + d*x]*(a + a*Sin[c + d*x])^(4/3))/(455*d) - (3*Cos[c + d*x]*S
in[c + d*x]^2*(a + a*Sin[c + d*x])^(4/3))/(13*d) - (6*Cos[c + d*x]*(a + a*Sin[c + d*x])^(7/3))/(65*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*} \text {integral}& = -\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3}}{13 d}+\frac {3 \int \sin (c+d x) (a+a \sin (c+d x))^{4/3} \left (2 a+\frac {4}{3} a \sin (c+d x)\right ) \, dx}{13 a} \\ & = -\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3}}{13 d}+\frac {3 \int (a+a \sin (c+d x))^{4/3} \left (2 a \sin (c+d x)+\frac {4}{3} a \sin ^2(c+d x)\right ) \, dx}{13 a} \\ & = -\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3}}{13 d}-\frac {6 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{65 a d}+\frac {9 \int (a+a \sin (c+d x))^{4/3} \left (\frac {28 a^2}{9}+\frac {16}{3} a^2 \sin (c+d x)\right ) \, dx}{130 a^2} \\ & = -\frac {72 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{455 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3}}{13 d}-\frac {6 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{65 a d}+\frac {194}{455} \int (a+a \sin (c+d x))^{4/3} \, dx \\ & = -\frac {72 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{455 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3}}{13 d}-\frac {6 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{65 a d}+\frac {\left (194 a \sqrt [3]{a+a \sin (c+d x)}\right ) \int (1+\sin (c+d x))^{4/3} \, dx}{455 \sqrt [3]{1+\sin (c+d x)}} \\ & = -\frac {388\ 2^{5/6} a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [3]{a+a \sin (c+d x)}}{455 d (1+\sin (c+d x))^{5/6}}-\frac {72 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{455 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3}}{13 d}-\frac {6 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{65 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.62 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.59 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\frac {(a (1+\sin (c+d x)))^{4/3} \left (\frac {3 (1940-790 \cos (c+d x)+98 \cos (3 (c+d x))-278 \sin (2 (c+d x))+35 \sin (4 (c+d x)))}{40 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {97 \left (-2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right )+3 \sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )\right )}{2^{2/3} \sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{8/3} \sqrt [3]{\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}}\right )}{91 d} \]

[In]

Integrate[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^(4/3),x]

[Out]

((a*(1 + Sin[c + d*x]))^(4/3)*((3*(1940 - 790*Cos[c + d*x] + 98*Cos[3*(c + d*x)] - 278*Sin[2*(c + d*x)] + 35*S
in[4*(c + d*x)]))/(40*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - (97*(-2*Cos[(2*c + Pi + 2*d*x)/4]*Hypergeomet
ricPFQ[{-1/2, -1/6}, {5/6}, Sin[(2*c + Pi + 2*d*x)/4]^2] + Sqrt[Cos[(2*c + Pi + 2*d*x)/4]^2]*(2*Cos[(2*c + Pi
+ 2*d*x)/4] + 3*Sin[(2*c + Pi + 2*d*x)/4])))/(2^(2/3)*Sqrt[Cos[(2*c + Pi + 2*d*x)/4]^2]*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^(8/3)*Sin[(2*c + Pi + 2*d*x)/4]^(1/3))))/(91*d)

Maple [F]

\[\int \left (\sin ^{3}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}d x\]

[In]

int(sin(d*x+c)^3*(a+a*sin(d*x+c))^(4/3),x)

[Out]

int(sin(d*x+c)^3*(a+a*sin(d*x+c))^(4/3),x)

Fricas [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 - (a*cos(d*x + c)^2 - a)*sin(d*x + c) + a)*(a*sin(d*x + c) + a
)^(1/3), x)

Sympy [F(-1)]

Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\text {Timed out} \]

[In]

integrate(sin(d*x+c)**3*(a+a*sin(d*x+c))**(4/3),x)

[Out]

Timed out

Maxima [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(4/3)*sin(d*x + c)^3, x)

Giac [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(4/3)*sin(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int {\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3} \,d x \]

[In]

int(sin(c + d*x)^3*(a + a*sin(c + d*x))^(4/3),x)

[Out]

int(sin(c + d*x)^3*(a + a*sin(c + d*x))^(4/3), x)