\(\int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx\) [99]

Optimal result

Integrand size = 23, antiderivative size = 127 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {37\ 2^{5/6} a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [3]{a+a \sin (c+d x)}}{35 d (1+\sin (c+d x))^{5/6}}+\frac {9 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{70 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{10 a d} \]

[Out]

-37/35*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)+9/70*cos(d*x+c)*(a+a*sin(d*x+c))^(4/3)/d-3/10*cos(d*x+c)*(a+a*sin(d*x+c))^(7/3)/a/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2838, 2830, 2731, 2730} \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {37\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{35 d (\sin (c+d x)+1)^{5/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}+\frac {9 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{70 d} \]

[In]

Int[Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(4/3),x]

[Out]

(-37*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)
)/(35*d*(1 + Sin[c + d*x])^(5/6)) + (9*Cos[c + d*x]*(a + a*Sin[c + d*x])^(4/3))/(70*d) - (3*Cos[c + d*x]*(a +
a*Sin[c + d*x])^(7/3))/(10*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 2838

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)]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{10 a d}+\frac {3 \int \left (\frac {7 a}{3}-a \sin (c+d x)\right ) (a+a \sin (c+d x))^{4/3} \, dx}{10 a} \\ & = \frac {9 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{70 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{10 a d}+\frac {37}{70} \int (a+a \sin (c+d x))^{4/3} \, dx \\ & = \frac {9 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{70 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{10 a d}+\frac {\left (37 a \sqrt [3]{a+a \sin (c+d x)}\right ) \int (1+\sin (c+d x))^{4/3} \, dx}{70 \sqrt [3]{1+\sin (c+d x)}} \\ & = -\frac {37\ 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)}}{35 d (1+\sin (c+d x))^{5/6}}+\frac {9 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{70 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{10 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.64 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.97 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\frac {(a (1+\sin (c+d x)))^{4/3} \left (-\frac {3 (-185+60 \cos (c+d x)-7 \cos (3 (c+d x))+22 \sin (2 (c+d x)))}{20 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {37 \left (-2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right )+3 \sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )\right )}{2\ 2^{2/3} \sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{8/3} \sqrt [3]{\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}}\right )}{14 d} \]

[In]

Integrate[Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(4/3),x]

[Out]

((a*(1 + Sin[c + d*x]))^(4/3)*((-3*(-185 + 60*Cos[c + d*x] - 7*Cos[3*(c + d*x)] + 22*Sin[2*(c + d*x)]))/(20*(C
os[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - (37*(-2*Cos[(2*c + Pi + 2*d*x)/4]*HypergeometricPFQ[{-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^(2/3)*Sqrt[Cos[(2*c + Pi + 2*d*x)/4]^2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^(8/3)*
Sin[(2*c + Pi + 2*d*x)/4]^(1/3))))/(14*d)

Maple [F]

\[\int \left (\sin ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}d x\]

[In]

int(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x)

[Out]

int(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x)

Fricas [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral(-(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]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {4}{3}} \sin ^{2}{\left (c + d x \right )}\, dx \]

[In]

integrate(sin(d*x+c)**2*(a+a*sin(d*x+c))**(4/3),x)

[Out]

Integral((a*(sin(c + d*x) + 1))**(4/3)*sin(c + d*x)**2, x)

Maxima [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(sin(d*x+c)^2*(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)^2, x)

Giac [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(sin(d*x+c)^2*(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)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3} \,d x \]

[In]

int(sin(c + d*x)^2*(a + a*sin(c + d*x))^(4/3),x)

[Out]

int(sin(c + d*x)^2*(a + a*sin(c + d*x))^(4/3), x)