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

3.2.17.1 Optimal result
3.2.17.2 Mathematica [C] (warning: unable to verify)
3.2.17.3 Rubi [A] (warning: unable to verify)
3.2.17.4 Maple [F]
3.2.17.5 Fricas [F(-1)]
3.2.17.6 Sympy [F]
3.2.17.7 Maxima [F]
3.2.17.8 Giac [F]
3.2.17.9 Mupad [F(-1)]

3.2.17.1 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*AppellF1(11/6,2,-1/2,17/6,1+sin(f*x+e),1/2+1/2*sin(f*x+e))*sec(f*x+e) 
*(a+a*sin(f*x+e))^(7/3)*2^(1/2)*(1-sin(f*x+e))^(1/2)/a^2/f
 
3.2.17.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 35.53 (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...
 
3.2.17.3 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.

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

\(\Big \downarrow \) 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])
 

3.2.17.3.1 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]
 
3.2.17.4 Maple [F]

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

input
int(cot(f*x+e)^2*(a+a*sin(f*x+e))^(1/3),x)
 
output
int(cot(f*x+e)^2*(a+a*sin(f*x+e))^(1/3),x)
 
3.2.17.5 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
 
3.2.17.6 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)
 
3.2.17.7 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)
 
3.2.17.8 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)
 
3.2.17.9 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)