\(\int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx\) [103]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {2\ 2^{5/6} a \operatorname {AppellF1}\left (\frac {1}{2},2,-\frac {5}{6},\frac {3}{2},1-\sin (c+d x),\frac {1}{2} (1-\sin (c+d x))\right ) \cos (c+d x) \sqrt [3]{a+a \sin (c+d x)}}{d (1+\sin (c+d x))^{5/6}} \] Output:

-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+sin(d*x+c))^(5/6)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 20.24 (sec) , antiderivative size = 2800, normalized size of antiderivative = 35.90 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\text {Result too large to show} \] Input:

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

Output:

((-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*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) + ((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*(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])] + (Appe 
llF1[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 + T 
an[(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)*Hyperg...
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{4/3}}{\sin (c+d x)^2}dx\)

\(\Big \downarrow \) 3266

\(\displaystyle \frac {a \sqrt [3]{a \sin (c+d x)+a} \int \csc ^2(c+d x) (\sin (c+d x)+1)^{4/3}dx}{\sqrt [3]{\sin (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \sqrt [3]{a \sin (c+d x)+a} \int \frac {(\sin (c+d x)+1)^{4/3}}{\sin (c+d x)^2}dx}{\sqrt [3]{\sin (c+d x)+1}}\)

\(\Big \downarrow \) 3264

\(\displaystyle -\frac {a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \int \frac {\csc ^2(c+d x) (\sin (c+d x)+1)^{5/6}}{\sqrt {1-\sin (c+d x)}}d(1-\sin (c+d x))}{d \sqrt {1-\sin (c+d x)} (\sin (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 148

\(\displaystyle -\frac {2 a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \int \csc ^2(c+d x) (\sin (c+d x)+1)^{5/6}d\sqrt {1-\sin (c+d x)}}{d \sqrt {1-\sin (c+d x)} (\sin (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 333

\(\displaystyle -\frac {2\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \operatorname {AppellF1}\left (\frac {1}{2},2,-\frac {5}{6},\frac {3}{2},1-\sin (c+d x),\frac {1}{2} (1-\sin (c+d x))\right )}{d (\sin (c+d x)+1)^{5/6}}\)

Input:

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

Output:

(-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))
 

Defintions of rubi rules used

rule 148

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]
 

rule 333

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])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3264

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]
 

rule 3266

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]
 

Maple [F]

\[\int \csc \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}d x\]

Input:

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

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \csc \left (d x + c\right )^{2} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \csc \left (d x + c\right )^{2} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3}}{{\sin \left (c+d\,x\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=a^{\frac {4}{3}} \left (\int \left (\sin \left (d x +c \right )+1\right )^{\frac {1}{3}} \csc \left (d x +c \right )^{2} \sin \left (d x +c \right )d x +\int \left (\sin \left (d x +c \right )+1\right )^{\frac {1}{3}} \csc \left (d x +c \right )^{2}d x \right ) \] Input:

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

Output:

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