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

Optimal. Leaf size=184 \[ \frac {4 \sqrt [3]{\sin (c+d x)} \cos (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2(c+d x)\right )}{9 a^2 d \sqrt {\cos ^2(c+d x)}}-\frac {\sin ^{\frac {4}{3}}(c+d x) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\sin ^2(c+d x)\right )}{36 a^2 d \sqrt {\cos ^2(c+d x)}}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{9 a^2 d (\sin (c+d x)+1)}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2} \]

[Out]

-1/9*cos(d*x+c)*sin(d*x+c)^(1/3)/a^2/d/(1+sin(d*x+c))-1/3*cos(d*x+c)*sin(d*x+c)^(1/3)/d/(a+a*sin(d*x+c))^2+4/9
*cos(d*x+c)*hypergeom([1/6, 1/2],[7/6],sin(d*x+c)^2)*sin(d*x+c)^(1/3)/a^2/d/(cos(d*x+c)^2)^(1/2)-1/36*cos(d*x+
c)*hypergeom([1/2, 2/3],[5/3],sin(d*x+c)^2)*sin(d*x+c)^(4/3)/a^2/d/(cos(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2764, 2978, 2748, 2643} \[ -\frac {\sin ^{\frac {4}{3}}(c+d x) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\sin ^2(c+d x)\right )}{36 a^2 d \sqrt {\cos ^2(c+d x)}}+\frac {4 \sqrt [3]{\sin (c+d x)} \cos (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2(c+d x)\right )}{9 a^2 d \sqrt {\cos ^2(c+d x)}}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{9 a^2 d (\sin (c+d x)+1)}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[c + d*x]^2]*Sin[c + d*x]^(1/3))/(9*a^2*d*Sqrt[Cos[c + d*x
]^2]) - (Cos[c + d*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Sin[c + d*x]^2]*Sin[c + d*x]^(4/3))/(36*a^2*d*Sqrt[Cos[
c + d*x]^2]) - (Cos[c + d*x]*Sin[c + d*x]^(1/3))/(9*a^2*d*(1 + Sin[c + d*x])) - (Cos[c + d*x]*Sin[c + d*x]^(1/
3))/(3*d*(a + a*Sin[c + d*x])^2)

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2764

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

Rule 2978

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

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx &=-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{3 d (a+a \sin (c+d x))^2}+\frac {\int \frac {\frac {a}{3}+\frac {2}{3} a \sin (c+d x)}{\sin ^{\frac {2}{3}}(c+d x) (a+a \sin (c+d x))} \, dx}{3 a^2}\\ &=-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{9 a^2 d (1+\sin (c+d x))}-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{3 d (a+a \sin (c+d x))^2}+\frac {\int \frac {\frac {4 a^2}{9}-\frac {1}{9} a^2 \sin (c+d x)}{\sin ^{\frac {2}{3}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{9 a^2 d (1+\sin (c+d x))}-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{3 d (a+a \sin (c+d x))^2}-\frac {\int \sqrt [3]{\sin (c+d x)} \, dx}{27 a^2}+\frac {4 \int \frac {1}{\sin ^{\frac {2}{3}}(c+d x)} \, dx}{27 a^2}\\ &=\frac {4 \cos (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2(c+d x)\right ) \sqrt [3]{\sin (c+d x)}}{9 a^2 d \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\sin ^2(c+d x)\right ) \sin ^{\frac {4}{3}}(c+d x)}{36 a^2 d \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{9 a^2 d (1+\sin (c+d x))}-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{3 d (a+a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 121, normalized size = 0.66 \[ \frac {\sqrt [3]{\sin (c+d x)} \sec ^3(c+d x) \left (80 \cos ^2(c+d x)^{3/2} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2(c+d x)\right )+27 \sin (c+d x) \cos ^2(c+d x)^{3/2} \, _2F_1\left (\frac {2}{3},\frac {5}{2};\frac {5}{3};\sin ^2(c+d x)\right )+4 (27 \sin (c+d x)+5 \cos (2 (c+d x))-25)\right )}{180 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*Sin[c + d*x]^(1/3)*(80*(Cos[c + d*x]^2)^(3/2)*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[c + d*x]^2]
+ 27*(Cos[c + d*x]^2)^(3/2)*Hypergeometric2F1[2/3, 5/2, 5/3, Sin[c + d*x]^2]*Sin[c + d*x] + 4*(-25 + 5*Cos[2*
(c + d*x)] + 27*Sin[c + d*x])))/(180*a^2*d)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sin \left (d x + c\right )^{\frac {1}{3}}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.78, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{\frac {1}{3}}\left (d x +c \right )}{\left (a +a \sin \left (d x +c \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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