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

Optimal. Leaf size=99 \[ \frac{3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}-\frac{4 \sqrt [6]{2} \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5 a d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}} \]

[Out]

(3*Cos[c + d*x])/(5*d*(a + a*Sin[c + d*x])^(4/3)) - (4*2^(1/6)*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 3/2, (
1 - Sin[c + d*x])/2])/(5*a*d*(1 + Sin[c + d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3))

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Rubi [A]  time = 0.0789303, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2750, 2652, 2651} \[ \frac{3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}-\frac{4 \sqrt [6]{2} \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5 a d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Cos[c + d*x])/(5*d*(a + a*Sin[c + d*x])^(4/3)) - (4*2^(1/6)*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 3/2, (
1 - Sin[c + d*x])/2])/(5*a*d*(1 + Sin[c + d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3))

Rule 2750

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

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 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]

Rubi steps

\begin{align*} \int \frac{\sin (c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx &=\frac{3 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac{4 \int \frac{1}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{5 a}\\ &=\frac{3 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac{\left (4 \sqrt [3]{1+\sin (c+d x)}\right ) \int \frac{1}{\sqrt [3]{1+\sin (c+d x)}} \, dx}{5 a \sqrt [3]{a+a \sin (c+d x)}}\\ &=\frac{3 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac{4 \sqrt [6]{2} \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5 a d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.295461, size = 130, normalized size = 1.31 \[ \frac{3 \left (8 (\sin (c+d x)+1) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\sin ^2\left (\frac{1}{4} (2 c+2 d x+\pi )\right )\right )+\sqrt{2-2 \sin (c+d x)}\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{5 d \sqrt{2-2 \sin (c+d x)} (a (\sin (c+d x)+1))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(Sqrt[2 - 2*Sin[c + d*x]] + 8*H
ypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + Pi + 2*d*x)/4]^2]*(1 + Sin[c + d*x])))/(5*d*Sqrt[2 - 2*Sin[c + d*x]
]*(a*(1 + Sin[c + d*x]))^(4/3))

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Maple [F]  time = 0.082, size = 0, normalized size = 0. \begin{align*} \int{\sin \left ( dx+c \right ) \left ( a+a\sin \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sin \left (d x + c\right )}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (c + d x \right )}}{\left (a \left (\sin{\left (c + d x \right )} + 1\right )\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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