Integrand size = 23, antiderivative size = 80 \[ \int \cot ^4(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\frac {12 \sqrt {2} \operatorname {AppellF1}\left (\frac {17}{6},4,-\frac {3}{2},\frac {23}{6},1+\sin (e+f x),\frac {1}{2} (1+\sin (e+f x))\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{10/3}}{17 a^3 f} \] Output:
12/17*2^(1/2)*AppellF1(17/6,4,-3/2,23/6,1+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))^(10/3)/a^3/f
Result contains complex when optimal does not.
Time = 49.28 (sec) , antiderivative size = 2796, normalized size of antiderivative = 34.95 \[ \int \cot ^4(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\text {Result too large to show} \] Input:
Integrate[Cot[e + f*x]^4*(a + a*Sin[e + f*x])^(1/3),x]
Output:
((239/54 + (77*Cot[e + f*x])/54 - (Cot[e + f*x]*Csc[e + f*x])/18 - (Cot[e + f*x]*Csc[e + f*x]^2)/3)*(a*(1 + Sin[e + f*x]))^(1/3))/f - ((70/9 + (70*I )/9)*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)*(1 + Tan[(e + f*x)/2]))/(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])]*Sec[(e + f*x)/2] + App ellF1[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])]*(Csc[(e + f*x)/2] + Sec[(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])]*(Csc[(e + f*x)/2] + Sec[(e + f*x)/2]))*(Cos[(e + f*x)/2 ] + Sin[(e + f*x)/2])) - ((355/108 + (355*I)/108)*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 + Tan[(e + f*x)/2]))) - (239*Cos[(3*(e + f*x))/2]*Csc[e + f*x]*(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]))^(...
Time = 0.38 (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.
| \(\displaystyle \int \cot ^4(e+f x) \sqrt [3]{a \sin (e+f x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt [3]{a \sin (e+f x)+a}}{\tan (e+f x)^4}dx\) |
| \(\Big \downarrow \) 3198 |
| \(\displaystyle \frac {\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \frac {\csc ^4(e+f x) (a-a \sin (e+f x))^{3/2} (\sin (e+f x) a+a)^{11/6}}{a^4}d(a \sin (e+f x))}{a f}\) |
| \(\Big \downarrow \) 149 |
| \(\displaystyle \frac {6 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \frac {a^{16} \sin ^{16}(e+f x) \left (2 a-a^6 \sin ^6(e+f x)\right )^{3/2}}{\left (a-a^6 \sin ^6(e+f x)\right )^4}d\sqrt [6]{\sin (e+f x) a+a}}{a f}\) |
| \(\Big \downarrow \) 1013 |
| \(\displaystyle \frac {12 \sqrt {2} \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \int \frac {a^{16} \sin ^{16}(e+f x) \left (2-a^5 \sin ^6(e+f x)\right )^{3/2}}{2 \sqrt {2} \left (a-a^6 \sin ^6(e+f x)\right )^4}d\sqrt [6]{\sin (e+f x) a+a}}{f \sqrt {2-a^5 \sin ^6(e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {6 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \int \frac {a^{16} \sin ^{16}(e+f x) \left (2-a^5 \sin ^6(e+f x)\right )^{3/2}}{\left (a-a^6 \sin ^6(e+f x)\right )^4}d\sqrt [6]{\sin (e+f x) a+a}}{f \sqrt {2-a^5 \sin ^6(e+f x)}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle \frac {12 \sqrt {2} a^{13} \sin ^{16}(e+f x) \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \sqrt {2 a-a^6 \sin ^6(e+f x)} \operatorname {AppellF1}\left (\frac {17}{6},4,-\frac {3}{2},\frac {23}{6},a^5 \sin ^6(e+f x),\frac {1}{2} a^5 \sin ^6(e+f x)\right )}{17 f \sqrt {2-a^5 \sin ^6(e+f x)}}\) |
Input:
Int[Cot[e + f*x]^4*(a + a*Sin[e + f*x])^(1/3),x]
Output:
(12*Sqrt[2]*a^13*AppellF1[17/6, 4, -3/2, 23/6, a^5*Sin[e + f*x]^6, (a^5*Si n[e + f*x]^6)/2]*Sin[e + f*x]^16*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])/(17*f*Sqrt[2 - a^5*S in[e + f*x]^6])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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])
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])
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]
\[\int \cot \left (f x +e \right )^{4} \left (a +\sin \left (f x +e \right ) a \right )^{\frac {1}{3}}d x\]
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)
Timed out. \[ \int \cot ^4(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\text {Timed out} \] Input:
integrate(cot(f*x+e)^4*(a+a*sin(f*x+e))^(1/3),x, algorithm="fricas")
Output:
Timed out
\[ \int \cot ^4(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\int \sqrt [3]{a \left (\sin {\left (e + f x \right )} + 1\right )} \cot ^{4}{\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)**4*(a+a*sin(f*x+e))**(1/3),x)
Output:
Integral((a*(sin(e + f*x) + 1))**(1/3)*cot(e + f*x)**4, x)
\[ \int \cot ^4(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {1}{3}} \cot \left (f x + e\right )^{4} \,d x } \] Input:
integrate(cot(f*x+e)^4*(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)^4, x)
\[ \int \cot ^4(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {1}{3}} \cot \left (f x + e\right )^{4} \,d x } \] Input:
integrate(cot(f*x+e)^4*(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)^4, x)
Timed out. \[ \int \cot ^4(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^4\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{1/3} \,d x \] 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)
\[ \int \cot ^4(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx=a^{\frac {1}{3}} \left (\int \left (\sin \left (f x +e \right )+1\right )^{\frac {1}{3}} \cot \left (f x +e \right )^{4}d x \right ) \] Input:
int(cot(f*x+e)^4*(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)**4,x)