3.1.97 \(\int \csc ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx\) [97]

Optimal. Leaf size=77 \[ -\frac {2 \sqrt [6]{2} F_1\left (\frac {1}{2};2,-\frac {1}{6};\frac {3}{2};1-\sin (c+d x),\frac {1}{2} (1-\sin (c+d x))\right ) \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{d (1+\sin (c+d x))^{7/6}} \]

[Out]

-2*2^(1/6)*AppellF1(1/2,2,-1/6,3/2,1-sin(d*x+c),1/2-1/2*sin(d*x+c))*cos(d*x+c)*(a+a*sin(d*x+c))^(2/3)/d/(1+sin
(d*x+c))^(7/6)

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Rubi [A]
time = 0.10, antiderivative size = 77, 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 = {2866, 2864, 129, 440} \begin {gather*} -\frac {2 \sqrt [6]{2} \cos (c+d x) (a \sin (c+d x)+a)^{2/3} F_1\left (\frac {1}{2};2,-\frac {1}{6};\frac {3}{2};1-\sin (c+d x),\frac {1}{2} (1-\sin (c+d x))\right )}{d (\sin (c+d x)+1)^{7/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*2^(1/6)*AppellF1[1/2, 2, -1/6, 3/2, 1 - Sin[c + d*x], (1 - Sin[c + d*x])/2]*Cos[c + d*x]*(a + a*Sin[c + d*
x])^(2/3))/(d*(1 + Sin[c + d*x])^(7/6))

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 2864

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(-b)*(
d/b)^n*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])), Subst[Int[(a - x)^n*((2*a - x)^(m
- 1/2)/Sqrt[x]), x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !
IntegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]

Rule 2866

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

Rubi steps

\begin {align*} \int \csc ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx &=\frac {(a+a \sin (c+d x))^{2/3} \int \csc ^2(c+d x) (1+\sin (c+d x))^{2/3} \, dx}{(1+\sin (c+d x))^{2/3}}\\ &=-\frac {\left (\cos (c+d x) (a+a \sin (c+d x))^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [6]{2-x}}{(1-x)^2 \sqrt {x}} \, dx,x,1-\sin (c+d x)\right )}{d \sqrt {1-\sin (c+d x)} (1+\sin (c+d x))^{7/6}}\\ &=-\frac {\left (2 \cos (c+d x) (a+a \sin (c+d x))^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [6]{2-x^2}}{\left (1-x^2\right )^2} \, dx,x,\sqrt {1-\sin (c+d x)}\right )}{d \sqrt {1-\sin (c+d x)} (1+\sin (c+d x))^{7/6}}\\ &=-\frac {2 \sqrt [6]{2} F_1\left (\frac {1}{2};2,-\frac {1}{6};\frac {3}{2};1-\sin (c+d x),\frac {1}{2} (1-\sin (c+d x))\right ) \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{d (1+\sin (c+d x))^{7/6}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 8.79, size = 143, normalized size = 1.86 \begin {gather*} -\frac {2 e^{i (c+d x)} \left (-i-e^{i (c+d x)}+\left (1+i e^{-i (c+d x)}\right )^{2/3} \left (-i+e^{i (c+d x)}\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-i e^{-i (c+d x)}\right )\right ) (a (1+\sin (c+d x)))^{2/3}}{d \left (-i+e^{i (c+d x)}\right ) \left (i+e^{i (c+d x)}\right )^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*E^(I*(c + d*x))*(-I - E^(I*(c + d*x)) + (1 + I/E^(I*(c + d*x)))^(2/3)*(-I + E^(I*(c + d*x)))*Hypergeometri
c2F1[1/3, 2/3, 4/3, (-I)/E^(I*(c + d*x))])*(a*(1 + Sin[c + d*x]))^(2/3))/(d*(-I + E^(I*(c + d*x)))*(I + E^(I*(
c + d*x)))^2)

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \left (\csc ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {2}{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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