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3.103 \(\int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx\)

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

[Out]

-2*2^(5/6)*a*AppellF1(1/2,2,-5/6,3/2,1-sin(d*x+c),1/2-1/2*sin(d*x+c))*cos(d*x+c)*(a+a*sin(d*x+c))^(1/3)/d/(1+s
in(d*x+c))^(5/6)

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Rubi [A]  time = 0.13, antiderivative size = 78, 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 = {2787, 2785, 130, 429} \[ -\frac {2\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} F_1\left (\frac {1}{2};2,-\frac {5}{6};\frac {3}{2};1-\sin (c+d x),\frac {1}{2} (1-\sin (c+d x))\right )}{d (\sin (c+d x)+1)^{5/6}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^(4/3),x]

[Out]

(-2*2^(5/6)*a*AppellF1[1/2, 2, -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 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])

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 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]

Rubi steps

\begin {align*} \int \csc ^2(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 ^2(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)^2 \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}}{\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))^{5/6}}\\ &=-\frac {2\ 2^{5/6} a F_1\left (\frac {1}{2};2,-\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*}

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Mathematica [C]  time = 10.57, size = 2800, normalized size = 35.90 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^(4/3),x]

[Out]

((-1 - Cot[c + d*x])*(a*(1 + Sin[c + d*x]))^(4/3))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - ((15/2 + (15*
I)/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])]*(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*App
ellF1[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) + ((10 + 10*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])]*(a*(1 + Sin[c + d*x]))^(4/3))/(d*(Co
s[(c + d*x)/2] + Sin[(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]))) + (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 + Tan[(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)*(-1/2*(Appe
llF1[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) + ((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 + Co
t[(c + d*x)/2]))*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]))/(((1 - I)*(I + Cot[(c + d*x)/2]))/(1 + Cot[(c + d*x)/2]))^(2/3) - ((1/6 + I/6)*Appe
llF1[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)*((-1/30 + 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*Tan[(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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \csc \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(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)^2, x)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \left (\csc ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x)

[Out]

int(csc(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \csc \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(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)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3}}{{\sin \left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(4/3)/sin(c + d*x)^2,x)

[Out]

int((a + a*sin(c + d*x))^(4/3)/sin(c + d*x)^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*(a+a*sin(d*x+c))**(4/3),x)

[Out]

Timed out

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