\(\int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx\) [122]

Optimal result

Integrand size = 23, antiderivative size = 80 \[ \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx=\frac {12 \sqrt {2} \operatorname {AppellF1}\left (\frac {13}{6},4,-\frac {3}{2},\frac {19}{6},1+\sin (e+f x),\frac {1}{2} (1+\sin (e+f x))\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{8/3}}{13 a^3 f} \] Output:

12/13*2^(1/2)*AppellF1(13/6,4,-3/2,19/6,1+sin(f*x+e),1/2+1/2*sin(f*x+e))*s 
ec(f*x+e)*(1-sin(f*x+e))^(1/2)*(a+a*sin(f*x+e))^(8/3)/a^3/f
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx=\int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx \] Input:

Integrate[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(1/3),x]
 

Output:

Integrate[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(1/3), x]
 

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.70, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3198, 149, 1013, 27, 1012}

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.

\(\displaystyle \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a \sin (e+f x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (e+f x)^4 \sqrt [3]{a \sin (e+f x)+a}}dx\)

\(\Big \downarrow \) 3198

\(\displaystyle \frac {\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \frac {\csc ^4(e+f x) (a-a \sin (e+f x))^{3/2} (\sin (e+f x) a+a)^{7/6}}{a^4}d(a \sin (e+f x))}{a f}\)

\(\Big \downarrow \) 149

\(\displaystyle \frac {6 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \frac {a^{12} \sin ^{12}(e+f x) \left (2 a-a^6 \sin ^6(e+f x)\right )^{3/2}}{\left (a-a^6 \sin ^6(e+f x)\right )^4}d\sqrt [6]{\sin (e+f x) a+a}}{a f}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {12 \sqrt {2} \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \int \frac {a^{12} \sin ^{12}(e+f x) \left (2-a^5 \sin ^6(e+f x)\right )^{3/2}}{2 \sqrt {2} \left (a-a^6 \sin ^6(e+f x)\right )^4}d\sqrt [6]{\sin (e+f x) a+a}}{f \sqrt {2-a^5 \sin ^6(e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \int \frac {a^{12} \sin ^{12}(e+f x) \left (2-a^5 \sin ^6(e+f x)\right )^{3/2}}{\left (a-a^6 \sin ^6(e+f x)\right )^4}d\sqrt [6]{\sin (e+f x) a+a}}{f \sqrt {2-a^5 \sin ^6(e+f x)}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {12 \sqrt {2} a^9 \sin ^{12}(e+f x) \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \operatorname {AppellF1}\left (\frac {13}{6},4,-\frac {3}{2},\frac {19}{6},a^5 \sin ^6(e+f x),\frac {1}{2} a^5 \sin ^6(e+f x)\right )}{13 f \sqrt {2-a^5 \sin ^6(e+f x)}}\)

Input:

Int[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(1/3),x]
 

Output:

(12*Sqrt[2]*a^9*AppellF1[13/6, 4, -3/2, 19/6, a^5*Sin[e + f*x]^6, (a^5*Sin 
[e + f*x]^6)/2]*Sin[e + f*x]^12*Sqrt[a - a*Sin[e + f*x]]*Sqrt[a + a*Sin[e 
+ f*x]]*Sqrt[2*a - a^6*Sin[e + f*x]^6]*Tan[e + f*x])/(13*f*Sqrt[2 - a^5*Si 
n[e + f*x]^6])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 149

Int[((a_) + (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 - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + 
 b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && 
IntegerQ[2*n] && IntegerQ[p]
 

rule 1012

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

rule 1013

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

rule 3042

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

rule 3198

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

Maple [F]

\[\int \frac {\cot \left (f x +e \right )^{4}}{\left (a +\sin \left (f x +e \right ) a \right )^{\frac {1}{3}}}d x\]

Input:

int(cot(f*x+e)^4/(a+sin(f*x+e)*a)^(1/3),x)
 

Output:

int(cot(f*x+e)^4/(a+sin(f*x+e)*a)^(1/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx=\int \frac {\cot ^{4}{\left (e + f x \right )}}{\sqrt [3]{a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \] Input:

integrate(cot(f*x+e)**4/(a+a*sin(f*x+e))**(1/3),x)
 

Output:

Integral(cot(e + f*x)**4/(a*(sin(e + f*x) + 1))**(1/3), x)
 

Maxima [F]

\[ \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{4}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x, algorithm="maxima")
 

Output:

integrate(cot(f*x + e)^4/(a*sin(f*x + e) + a)^(1/3), x)
 

Giac [F]

\[ \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{4}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x, algorithm="giac")
 

Output:

integrate(cot(f*x + e)^4/(a*sin(f*x + e) + a)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \] Input:

int(cot(e + f*x)^4/(a + a*sin(e + f*x))^(1/3),x)
 

Output:

int(cot(e + f*x)^4/(a + a*sin(e + f*x))^(1/3), x)
 

Reduce [F]

\[ \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx=\frac {\int \frac {\cot \left (f x +e \right )^{4}}{\left (\sin \left (f x +e \right )+1\right )^{\frac {1}{3}}}d x}{a^{\frac {1}{3}}} \] Input:

int(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x)
                                                                                    
                                                                                    
 

Output:

int(cot(e + f*x)**4/(sin(e + f*x) + 1)**(1/3),x)/a**(1/3)