Integrand size = 21, antiderivative size = 78 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {2\ 2^{5/6} a \operatorname {AppellF1}\left (\frac {1}{2},1,-\frac {5}{6},\frac {3}{2},1-\sin (c+d x),\frac {1}{2} (1-\sin (c+d x))\right ) \cos (c+d x) \sqrt [3]{a+a \sin (c+d x)}}{d (1+\sin (c+d x))^{5/6}} \]
-2*2^(5/6)*a*AppellF1(1/2,1,-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+sin(d*x+c))^(5/6)
Result contains complex when optimal does not.
Time = 25.00 (sec) , antiderivative size = 2791, normalized size of antiderivative = 35.78 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\text {Result too large to show} \]
(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]))*(Co s[(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] + 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]))) - (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,...
Time = 0.35 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3266, 3042, 3264, 148, 333}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \csc (c+d x) (a \sin (c+d x)+a)^{4/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{4/3}}{\sin (c+d x)}dx\) |
\(\Big \downarrow \) 3266 |
\(\displaystyle \frac {a \sqrt [3]{a \sin (c+d x)+a} \int \csc (c+d x) (\sin (c+d x)+1)^{4/3}dx}{\sqrt [3]{\sin (c+d x)+1}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \sqrt [3]{a \sin (c+d x)+a} \int \frac {(\sin (c+d x)+1)^{4/3}}{\sin (c+d x)}dx}{\sqrt [3]{\sin (c+d x)+1}}\) |
\(\Big \downarrow \) 3264 |
\(\displaystyle -\frac {a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \int \frac {\csc (c+d x) (\sin (c+d x)+1)^{5/6}}{\sqrt {1-\sin (c+d x)}}d(1-\sin (c+d x))}{d \sqrt {1-\sin (c+d x)} (\sin (c+d x)+1)^{5/6}}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle -\frac {2 a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \int \csc (c+d x) (\sin (c+d x)+1)^{5/6}d\sqrt {1-\sin (c+d x)}}{d \sqrt {1-\sin (c+d x)} (\sin (c+d x)+1)^{5/6}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle -\frac {2\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \operatorname {AppellF1}\left (\frac {1}{2},1,-\frac {5}{6},\frac {3}{2},1-\sin (c+d x),\frac {1}{2} (1-\sin (c+d x))\right )}{d (\sin (c+d x)+1)^{5/6}}\) |
(-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))
3.2.2.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-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]
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_), x_Symbol] :> Simp[a^IntPart[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]
\[\int \csc \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}d x\]
Timed out. \[ \int \csc (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\text {Timed out} \]
Timed out. \[ \int \csc (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\text {Timed out} \]
\[ \int \csc (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \csc \left (d x + c\right ) \,d x } \]
\[ \int \csc (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \csc \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \csc (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3}}{\sin \left (c+d\,x\right )} \,d x \]