3.2.5 \(\int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx\) [105]

Optimal. Leaf size=126 \[ \frac {9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {7 \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d} \]

[Out]

9/10*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/3)-7/10*cos(d*x+c)*hypergeom([1/2, 5/6],[3/2],1/2-1/2*sin(d*x+c))*2^(1/6
)/d/(1+sin(d*x+c))^(1/6)/(a+a*sin(d*x+c))^(1/3)-3/5*cos(d*x+c)*(a+a*sin(d*x+c))^(2/3)/a/d

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Rubi [A]
time = 0.10, antiderivative size = 126, 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 = {2838, 2830, 2731, 2730} \begin {gather*} -\frac {7 \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\ 2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 a d}+\frac {9 \cos (c+d x)}{10 d \sqrt [3]{a \sin (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2/(a + a*Sin[c + d*x])^(1/3),x]

[Out]

(9*Cos[c + d*x])/(10*d*(a + a*Sin[c + d*x])^(1/3)) - (7*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sin
[c + d*x])/2])/(5*2^(5/6)*d*(1 + Sin[c + d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3)) - (3*Cos[c + d*x]*(a + a*Sin[
c + d*x])^(2/3))/(5*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*} \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx &=-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac {3 \int \frac {\frac {2 a}{3}-a \sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{5 a}\\ &=\frac {9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac {7}{10} \int \frac {1}{\sqrt [3]{a+a \sin (c+d x)}} \, dx\\ &=\frac {9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac {\left (7 \sqrt [3]{1+\sin (c+d x)}\right ) \int \frac {1}{\sqrt [3]{1+\sin (c+d x)}} \, dx}{10 \sqrt [3]{a+a \sin (c+d x)}}\\ &=\frac {9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {7 \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\ 2^{5/6} d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 95, normalized size = 0.75 \begin {gather*} -\frac {3 \cos (c+d x) \left (-14 \, _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 {2-2 \sin (c+d x)} (-1+2 \sin (c+d x))\right )}{10 d \sqrt {2-2 \sin (c+d x)} \sqrt [3]{a (1+\sin (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2/(a + a*Sin[c + d*x])^(1/3),x]

[Out]

(-3*Cos[c + d*x]*(-14*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + Pi + 2*d*x)/4]^2] + Sqrt[2 - 2*Sin[c + d*x]]
*(-1 + 2*Sin[c + d*x])))/(10*d*Sqrt[2 - 2*Sin[c + d*x]]*(a*(1 + Sin[c + d*x]))^(1/3))

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Maple [F]
time = 0.28, size = 0, normalized size = 0.00 \[\int \frac {\sin ^{2}\left (d x +c \right )}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/3),x)

[Out]

int(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/3),x)

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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)^2/(a+a*sin(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^2/(a*sin(d*x + c) + a)^(1/3), x)

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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)^2/(a+a*sin(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{2}{\left (c + d x \right )}}{\sqrt [3]{a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2/(a+a*sin(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^2/(a*sin(d*x + c) + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^2/(a + a*sin(c + d*x))^(1/3),x)

[Out]

int(sin(c + d*x)^2/(a + a*sin(c + d*x))^(1/3), x)

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