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

Optimal. Leaf size=78 \[ -\frac{2\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} F_1\left (\frac{1}{2};1,-\frac{5}{6};\frac{3}{2};1-\sin (c+d x),\frac{1}{2} (1-\sin (c+d x))\right )}{d (\sin (c+d x)+1)^{5/6}} \]

[Out]

(-2*2^(5/6)*a*AppellF1[1/2, 1, -5/6, 3/2, 1 - Sin[c + d*x], (1 - Sin[c + d*x])/2]*Cos[c + d*x]*(a + a*Sin[c +
d*x])^(1/3))/(d*(1 + Sin[c + d*x])^(5/6))

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Rubi [A]  time = 0.11369, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2787, 2785, 130, 429} \[ -\frac{2\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} F_1\left (\frac{1}{2};1,-\frac{5}{6};\frac{3}{2};1-\sin (c+d x),\frac{1}{2} (1-\sin (c+d x))\right )}{d (\sin (c+d x)+1)^{5/6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*2^(5/6)*a*AppellF1[1/2, 1, -5/6, 3/2, 1 - Sin[c + d*x], (1 - Sin[c + d*x])/2]*Cos[c + d*x]*(a + a*Sin[c +
d*x])^(1/3))/(d*(1 + Sin[c + d*x])^(5/6))

Rule 2787

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Sin[e + f*x])^FracPart[m])/(1 + (b*Sin[e + f*x])/a)^FracPart[m], Int[(1 + (b*Sin[e + f*x])/a)^
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]

Rule 2785

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] &&  !In
tegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]

Rule 130

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 429

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])

Rubi steps

\begin{align*} \int \csc (c+d x) (a+a \sin (c+d x))^{4/3} \, dx &=\frac{\left (a \sqrt [3]{a+a \sin (c+d x)}\right ) \int \csc (c+d x) (1+\sin (c+d x))^{4/3} \, dx}{\sqrt [3]{1+\sin (c+d x)}}\\ &=-\frac{\left (a \cos (c+d x) \sqrt [3]{a+a \sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(2-x)^{5/6}}{(1-x) \sqrt{x}} \, dx,x,1-\sin (c+d x)\right )}{d \sqrt{1-\sin (c+d x)} (1+\sin (c+d x))^{5/6}}\\ &=-\frac{\left (2 a \cos (c+d x) \sqrt [3]{a+a \sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\left (2-x^2\right )^{5/6}}{1-x^2} \, dx,x,\sqrt{1-\sin (c+d x)}\right )}{d \sqrt{1-\sin (c+d x)} (1+\sin (c+d x))^{5/6}}\\ &=-\frac{2\ 2^{5/6} a F_1\left (\frac{1}{2};1,-\frac{5}{6};\frac{3}{2};1-\sin (c+d x),\frac{1}{2} (1-\sin (c+d x))\right ) \cos (c+d x) \sqrt [3]{a+a \sin (c+d x)}}{d (1+\sin (c+d x))^{5/6}}\\ \end{align*}

Mathematica [C]  time = 9.51563, size = 2791, normalized size = 35.78 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*(a*(1 + Sin[c + d*x]))^(4/3))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - ((15 + 15*I)*AppellF1[2/3, 1/3,
1/3, 5/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]*(a*(1 + Sin[c + d*x]))^(4/3
)*(1 + Tan[(c + d*x)/2]))/(d*((5 + 5*I)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2
- I/2)*(1 + Cot[(c + d*x)/2])]*Sec[(c + d*x)/2] + AppellF1[5/3, 1/3, 4/3, 8/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/
2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]*(Csc[(c + d*x)/2] + Sec[(c + d*x)/2]) + I*AppellF1[5/3, 4/3, 1/3, 8/3
, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]*(Csc[(c + d*x)/2] + Sec[(c + d*x)/2]
))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((15/2 + (15*I)/2)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 +
Tan[(c + d*x)/2]), (1/2 - I/2)*(1 + Tan[(c + d*x)/2])]*(a*(1 + Sin[c + d*x]))^(4/3))/(d*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^2*((5 + 5*I)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Tan[(c + d*x)/2]), (1/2 - I/2)*(1
+ Tan[(c + d*x)/2])] + (AppellF1[5/3, 1/3, 4/3, 8/3, (1/2 + I/2)*(1 + Tan[(c + d*x)/2]), (1/2 - I/2)*(1 + Tan[
(c + d*x)/2])] + I*AppellF1[5/3, 4/3, 1/3, 8/3, (1/2 + I/2)*(1 + Tan[(c + d*x)/2]), (1/2 - I/2)*(1 + Tan[(c +
d*x)/2])])*(1 + Tan[(c + d*x)/2]))) - (3*Cos[(3*(c + d*x))/2]*Csc[c + d*x]*(a*(1 + Sin[c + d*x]))^(4/3)*((1 +
Tan[(c + d*x)/2])/Sqrt[Sec[(c + d*x)/2]^2])^(2/3)*(8 + (1 + I)*2^(2/3)*(((1 - I)*(I + Cot[(c + d*x)/2]))/(1 +
Cot[(c + d*x)/2]))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - I)*Tan[(c + d*x)/2])/(2 + 2*Tan[(c +
d*x)/2])]*(I + Tan[(c + d*x)/2]) - AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2 - I/
2)*(1 + Cot[(c + d*x)/2])]*((2 + 2*I) - (2 - 2*I)*Cot[(c + d*x)/2])^(1/3)*((-1 - I)*(I + Cot[(c + d*x)/2]))^(1
/3)*(1 + Tan[(c + d*x)/2])))/(4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(1 + Tan[(c + d*x)/2])*((-3*Sec[(c +
d*x)/2]^2*((1 + Tan[(c + d*x)/2])/Sqrt[Sec[(c + d*x)/2]^2])^(2/3)*(8 + (1 + I)*2^(2/3)*(((1 - I)*(I + Cot[(c
+ d*x)/2]))/(1 + Cot[(c + d*x)/2]))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - I)*Tan[(c + d*x)/2]
)/(2 + 2*Tan[(c + d*x)/2])]*(I + Tan[(c + d*x)/2]) - AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(c + d*
x)/2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]*((2 + 2*I) - (2 - 2*I)*Cot[(c + d*x)/2])^(1/3)*((-1 - I)*(I + Cot[
(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])))/(8*(1 + Tan[(c + d*x)/2])^2) + ((8 + (1 + I)*2^(2/3)*(((1 - I)*(
I + Cot[(c + d*x)/2]))/(1 + Cot[(c + d*x)/2]))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - I)*Tan[(
c + d*x)/2])/(2 + 2*Tan[(c + d*x)/2])]*(I + Tan[(c + d*x)/2]) - AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 +
Cot[(c + d*x)/2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]*((2 + 2*I) - (2 - 2*I)*Cot[(c + d*x)/2])^(1/3)*((-1 - I
)*(I + Cot[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2]))*(Sqrt[Sec[(c + d*x)/2]^2]/2 - (Tan[(c + d*x)/2]*(1 + T
an[(c + d*x)/2]))/(2*Sqrt[Sec[(c + d*x)/2]^2])))/(2*(1 + Tan[(c + d*x)/2])*((1 + Tan[(c + d*x)/2])/Sqrt[Sec[(c
+ d*x)/2]^2])^(1/3)) + (3*((1 + Tan[(c + d*x)/2])/Sqrt[Sec[(c + d*x)/2]^2])^(2/3)*(-(AppellF1[2/3, 1/3, 1/3,
5/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]*((2 + 2*I) - (2 - 2*I)*Cot[(c + d
*x)/2])^(1/3)*((-1 - I)*(I + Cot[(c + d*x)/2]))^(1/3)*Sec[(c + d*x)/2]^2)/2 + ((1 + I)*(((1 - I)*(I + Cot[(c +
d*x)/2]))/(1 + Cot[(c + d*x)/2]))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - I)*Tan[(c + d*x)/2])
/(2 + 2*Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/2^(1/3) + ((1/3 + I/3)*2^(2/3)*(((1/2 - I/2)*(I + Cot[(c + d*x)
/2])*Csc[(c + d*x)/2]^2)/(1 + Cot[(c + d*x)/2])^2 - ((1/2 - I/2)*Csc[(c + d*x)/2]^2)/(1 + Cot[(c + d*x)/2]))*H
ypergeometric2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - I)*Tan[(c + d*x)/2])/(2 + 2*Tan[(c + d*x)/2])]*(I + Tan[(c + d
*x)/2]))/(((1 - I)*(I + Cot[(c + d*x)/2]))/(1 + Cot[(c + d*x)/2]))^(2/3) - ((1/6 + I/6)*AppellF1[2/3, 1/3, 1/3
, 5/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]*((2 + 2*I) - (2 - 2*I)*Cot[(c +
d*x)/2])^(1/3)*Csc[(c + d*x)/2]^2*(1 + Tan[(c + d*x)/2]))/((-1 - I)*(I + Cot[(c + d*x)/2]))^(2/3) - ((1/3 - I
/3)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]*((-1
- I)*(I + Cot[(c + d*x)/2]))^(1/3)*Csc[(c + d*x)/2]^2*(1 + Tan[(c + d*x)/2]))/((2 + 2*I) - (2 - 2*I)*Cot[(c +
d*x)/2])^(2/3) - ((2 + 2*I) - (2 - 2*I)*Cot[(c + d*x)/2])^(1/3)*((-1 - I)*(I + Cot[(c + d*x)/2]))^(1/3)*((-1/3
0 + I/30)*AppellF1[5/3, 1/3, 4/3, 8/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2 - I/2)*(1 + Cot[(c + d*x)/2])]
*Csc[(c + d*x)/2]^2 - (1/30 + I/30)*AppellF1[5/3, 4/3, 1/3, 8/3, (1/2 + I/2)*(1 + Cot[(c + d*x)/2]), (1/2 - I/
2)*(1 + Cot[(c + d*x)/2])]*Csc[(c + d*x)/2]^2)*(1 + Tan[(c + d*x)/2]) + ((2/3 + (2*I)/3)*2^(2/3)*(((1 - I)*(I
+ Cot[(c + d*x)/2]))/(1 + Cot[(c + d*x)/2]))^(1/3)*(I + Tan[(c + d*x)/2])*(2 + 2*Tan[(c + d*x)/2])*(-((Sec[(c
+ d*x)/2]^2*((1 + I) + (1 - I)*Tan[(c + d*x)/2]))/(2 + 2*Tan[(c + d*x)/2])^2) + ((1/2 - I/2)*Sec[(c + d*x)/2]^
2)/(2 + 2*Tan[(c + d*x)/2]))*(-Hypergeometric2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - I)*Tan[(c + d*x)/2])/(2 + 2*Ta
n[(c + d*x)/2])] + (1 - ((1 + I) + (1 - I)*Tan[(c + d*x)/2])/(2 + 2*Tan[(c + d*x)/2]))^(-1/3)))/((1 + I) + (1
- I)*Tan[(c + d*x)/2])))/(4*(1 + Tan[(c + d*x)/2]))))

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Maple [F]  time = 0.096, size = 0, normalized size = 0. \begin{align*} \int \csc \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(csc(d*x+c)*(a+a*sin(d*x+c))^(4/3),x)

[Out]

int(csc(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{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{4}{3}} \csc \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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