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

Optimal. Leaf size=127 \[ -\frac {37\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{35 d (\sin (c+d x)+1)^{5/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}+\frac {9 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{70 d} \]

[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

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Rubi [A]  time = 0.14, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2759, 2751, 2652, 2651} \[ -\frac {37\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{35 d (\sin (c+d x)+1)^{5/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}+\frac {9 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{70 d} \]

Antiderivative was successfully verified.

[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 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2652

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

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 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)]

Rubi steps

\begin {align*} \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx &=-\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) \, _2F_1\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*}

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Mathematica [C]  time = 2.62, size = 363, normalized size = 2.86 \[ \frac {(a (\sin (c+d x)+1))^{4/3} \left (-\frac {3}{10} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (22 \sin (2 (c+d x))+60 \cos (c+d x)-7 \cos (3 (c+d x))-185)+\frac {111 (-1)^{3/4} e^{-\frac {3}{2} i (c+d x)} \left (e^{i (c+d x)}+i\right ) \left (-2 \left (1+i e^{-i (c+d x)}\right )^{2/3} \left (1+e^{2 i (c+d x)}\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\sin ^2\left (\frac {1}{4} (2 c+2 d x+\pi )\right )\right )+5 i \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-i e^{-i (c+d x)}\right ) \sqrt {2-2 \sin (c+d x)}+20 e^{i (c+d x)} \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {2}{3};-i e^{-i (c+d x)}\right ) \sqrt {\cos ^2\left (\frac {1}{4} (2 c+2 d x+\pi )\right )}\right )}{20 \sqrt {2} \left (1+i e^{-i (c+d x)}\right )^{2/3} \sqrt {i e^{-i (c+d x)} \left (e^{i (c+d x)}-i\right )^2}}\right )}{28 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(4/3)*((111*(-1)^(3/4)*(I + E^(I*(c + d*x)))*(20*E^(I*(c + d*x))*Sqrt[Cos[(2*c + Pi +
2*d*x)/4]^2]*Hypergeometric2F1[-1/3, 1/3, 2/3, (-I)/E^(I*(c + d*x))] - 2*(1 + I/E^(I*(c + d*x)))^(2/3)*(1 + E^
((2*I)*(c + d*x)))*Hypergeometric2F1[1/2, 5/6, 11/6, Sin[(2*c + Pi + 2*d*x)/4]^2] + (5*I)*Hypergeometric2F1[1/
3, 2/3, 5/3, (-I)/E^(I*(c + d*x))]*Sqrt[2 - 2*Sin[c + d*x]]))/(20*Sqrt[2]*E^(((3*I)/2)*(c + d*x))*(1 + I/E^(I*
(c + d*x)))^(2/3)*Sqrt[(I*(-I + E^(I*(c + d*x)))^2)/E^(I*(c + d*x))]) - (3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2
])*(-185 + 60*Cos[c + d*x] - 7*Cos[3*(c + d*x)] + 22*Sin[2*(c + d*x)]))/10))/(28*d*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^3)

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a \cos \left (d x + c\right )^{2} + {\left (a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \[ \int \left (\sin ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {4}{3}} \sin ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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