\(\int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx\) [117]

Optimal result

Integrand size = 23, antiderivative size = 80 \[ \int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\frac {6 \sqrt {2} \operatorname {AppellF1}\left (\frac {11}{6},-\frac {1}{2},2,\frac {17}{6},\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{7/3}}{11 a^2 f} \] Output:

6/11*2^(1/2)*AppellF1(11/6,2,-1/2,17/6,1+sin(f*x+e),1/2+1/2*sin(f*x+e))*se 
c(f*x+e)*(1-sin(f*x+e))^(1/2)*(a+a*sin(f*x+e))^(7/3)/a^2/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 43.36 (sec) , antiderivative size = 2692, normalized size of antiderivative = 33.65 \[ \int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\text {Result too large to show} \] Input:

Integrate[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(1/3),x]
 

Output:

((15/2 + (15*I)/2)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + 
f*x)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])]*(a*(1 + Sin[e + f*x]))^(1/3) 
)/(f*((5 + 5*I)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f*x 
)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])] + (AppellF1[5/3, 1/3, 4/3, 8/3, 
 (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])] + 
 I*AppellF1[5/3, 4/3, 1/3, 8/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - 
 I/2)*(1 + Cot[(e + f*x)/2])])*(1 + Cot[(e + f*x)/2]))) + ((-4 - Cot[e + f 
*x])*(a*(1 + Sin[e + f*x]))^(1/3))/f + ((5/2 + (5*I)/2)*AppellF1[2/3, 1/3, 
 1/3, 5/3, (1/2 + I/2)*(1 + Tan[(e + f*x)/2]), (1/2 - I/2)*(1 + Tan[(e + f 
*x)/2])]*(a*(1 + Sin[e + f*x]))^(1/3))/(f*((5 + 5*I)*AppellF1[2/3, 1/3, 1/ 
3, 5/3, (1/2 + I/2)*(1 + Tan[(e + f*x)/2]), (1/2 - I/2)*(1 + Tan[(e + f*x) 
/2])] + (AppellF1[5/3, 1/3, 4/3, 8/3, (1/2 + I/2)*(1 + Tan[(e + f*x)/2]), 
(1/2 - I/2)*(1 + Tan[(e + f*x)/2])] + I*AppellF1[5/3, 4/3, 1/3, 8/3, (1/2 
+ I/2)*(1 + Tan[(e + f*x)/2]), (1/2 - I/2)*(1 + Tan[(e + f*x)/2])])*(1 + T 
an[(e + f*x)/2]))) + (Cos[(3*(e + f*x))/2]*Csc[(e + f*x)/2]*Sec[(e + f*x)/ 
2]*(a*(1 + Sin[e + f*x]))^(1/3)*((1 + Tan[(e + f*x)/2])/Sqrt[Sec[(e + f*x) 
/2]^2])^(2/3)*(8 + (1 + I)*2^(2/3)*(((1 - I)*(I + Cot[(e + f*x)/2]))/(1 + 
Cot[(e + f*x)/2]))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - 
I)*Tan[(e + f*x)/2])/(2 + 2*Tan[(e + f*x)/2])]*(I + Tan[(e + f*x)/2]) - Ap 
pellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I...
 

Rubi [A] (warning: unable to verify)

Time = 0.37 (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.

\(\displaystyle \int \cot ^2(e+f x) \sqrt [3]{a \sin (e+f x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt [3]{a \sin (e+f x)+a}}{\tan (e+f x)^2}dx\)

\(\Big \downarrow \) 3198

\(\displaystyle \frac {\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \frac {\csc ^2(e+f x) \sqrt {a-a \sin (e+f x)} (\sin (e+f x) a+a)^{5/6}}{a^2}d(a \sin (e+f x))}{a f}\)

\(\Big \downarrow \) 149

\(\displaystyle \frac {6 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \frac {a^{10} \sin ^{10}(e+f x) \sqrt {2 a-a^6 \sin ^6(e+f x)}}{\left (a-a^6 \sin ^6(e+f x)\right )^2}d\sqrt [6]{\sin (e+f x) a+a}}{a f}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {6 \sqrt {2} \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \int \frac {a^{10} \sin ^{10}(e+f x) \sqrt {2-a^5 \sin ^6(e+f x)}}{\sqrt {2} \left (a-a^6 \sin ^6(e+f x)\right )^2}d\sqrt [6]{\sin (e+f x) a+a}}{a f \sqrt {2-a^5 \sin ^6(e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \int \frac {a^{10} \sin ^{10}(e+f x) \sqrt {2-a^5 \sin ^6(e+f x)}}{\left (a-a^6 \sin ^6(e+f x)\right )^2}d\sqrt [6]{\sin (e+f x) a+a}}{a f \sqrt {2-a^5 \sin ^6(e+f x)}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {6 \sqrt {2} a^8 \sin ^{10}(e+f x) \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \operatorname {AppellF1}\left (\frac {11}{6},2,-\frac {1}{2},\frac {17}{6},a^5 \sin ^6(e+f x),\frac {1}{2} a^5 \sin ^6(e+f x)\right )}{11 f \sqrt {2-a^5 \sin ^6(e+f x)}}\)

Input:

Int[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(1/3),x]
 

Output:

(6*Sqrt[2]*a^8*AppellF1[11/6, 2, -1/2, 17/6, a^5*Sin[e + f*x]^6, (a^5*Sin[ 
e + f*x]^6)/2]*Sin[e + f*x]^10*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])/(11*f*Sqrt[2 - a^5*Sin 
[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]

\[\int \cot \left (f x +e \right )^{2} \left (a +\sin \left (f x +e \right ) a \right )^{\frac {1}{3}}d x\]

Input:

int(cot(f*x+e)^2*(a+sin(f*x+e)*a)^(1/3),x)
 

Output:

int(cot(f*x+e)^2*(a+sin(f*x+e)*a)^(1/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^(1/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\int \sqrt [3]{a \left (\sin {\left (e + f x \right )} + 1\right )} \cot ^{2}{\left (e + f x \right )}\, dx \] Input:

integrate(cot(f*x+e)**2*(a+a*sin(f*x+e))**(1/3),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**(1/3)*cot(e + f*x)**2, x)
 

Maxima [F]

\[ \int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {1}{3}} \cot \left (f x + e\right )^{2} \,d x } \] Input:

integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^(1/3),x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^(1/3)*cot(f*x + e)^2, x)
 

Giac [F]

\[ \int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {1}{3}} \cot \left (f x + e\right )^{2} \,d x } \] Input:

integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^(1/3),x, algorithm="giac")
 

Output:

integrate((a*sin(f*x + e) + a)^(1/3)*cot(f*x + e)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{1/3} \,d x \] Input:

int(cot(e + f*x)^2*(a + a*sin(e + f*x))^(1/3),x)
 

Output:

int(cot(e + f*x)^2*(a + a*sin(e + f*x))^(1/3), x)
 

Reduce [F]

\[ \int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=a^{\frac {1}{3}} \left (\int \left (\sin \left (f x +e \right )+1\right )^{\frac {1}{3}} \cot \left (f x +e \right )^{2}d x \right ) \] Input:

int(cot(f*x+e)^2*(a+a*sin(f*x+e))^(1/3),x)
                                                                                    
                                                                                    
 

Output:

a**(1/3)*int((sin(e + f*x) + 1)**(1/3)*cot(e + f*x)**2,x)