Integrand size = 23, antiderivative size = 151 \[ \int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx=-\frac {5 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{f}+\frac {49 a^3 \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}+\frac {31 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}+\frac {7 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac {\cot (e+f x) (a+a \sin (e+f x))^{5/2}}{f} \] Output:
-5*a^(5/2)*arctanh(a^(1/2)*cos(f*x+e)/(a+a*sin(f*x+e))^(1/2))/f+49/15*a^3* cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)+31/15*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^ (1/2)/f+7/5*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-cot(f*x+e)*(a+a*sin(f*x+ e))^(5/2)/f
Time = 6.68 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.73 \[ \int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx=-\frac {a^2 \csc ^4\left (\frac {1}{2} (e+f x)\right ) \sqrt {a (1+\sin (e+f x))} \left (125 \cos \left (\frac {1}{2} (e+f x)\right )-93 \cos \left (\frac {3}{2} (e+f x)\right )+25 \cos \left (\frac {5}{2} (e+f x)\right )+3 \cos \left (\frac {7}{2} (e+f x)\right )-125 \sin \left (\frac {1}{2} (e+f x)\right )+150 \log \left (1+\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x)-150 \log \left (1-\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x)-93 \sin \left (\frac {3}{2} (e+f x)\right )-25 \sin \left (\frac {5}{2} (e+f x)\right )+3 \sin \left (\frac {7}{2} (e+f x)\right )\right )}{30 f \left (1+\cot \left (\frac {1}{2} (e+f x)\right )\right ) \left (\csc \left (\frac {1}{4} (e+f x)\right )-\sec \left (\frac {1}{4} (e+f x)\right )\right ) \left (\csc \left (\frac {1}{4} (e+f x)\right )+\sec \left (\frac {1}{4} (e+f x)\right )\right )} \] Input:
Integrate[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2),x]
Output:
-1/30*(a^2*Csc[(e + f*x)/2]^4*Sqrt[a*(1 + Sin[e + f*x])]*(125*Cos[(e + f*x )/2] - 93*Cos[(3*(e + f*x))/2] + 25*Cos[(5*(e + f*x))/2] + 3*Cos[(7*(e + f *x))/2] - 125*Sin[(e + f*x)/2] + 150*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f *x)/2]]*Sin[e + f*x] - 150*Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Si n[e + f*x] - 93*Sin[(3*(e + f*x))/2] - 25*Sin[(5*(e + f*x))/2] + 3*Sin[(7* (e + f*x))/2]))/(f*(1 + Cot[(e + f*x)/2])*(Csc[(e + f*x)/4] - Sec[(e + f*x )/4])*(Csc[(e + f*x)/4] + Sec[(e + f*x)/4]))
Time = 1.00 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.12, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 3195, 27, 3042, 3455, 27, 3042, 3455, 27, 3042, 3460, 3042, 3252, 219}
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.
| \(\displaystyle \int \cot ^2(e+f x) (a \sin (e+f x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{\tan (e+f x)^2}dx\) |
| \(\Big \downarrow \) 3195 |
| \(\displaystyle \frac {\int \frac {1}{2} \csc (e+f x) (5 a-7 a \sin (e+f x)) (\sin (e+f x) a+a)^{5/2}dx}{a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \csc (e+f x) (5 a-7 a \sin (e+f x)) (\sin (e+f x) a+a)^{5/2}dx}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(5 a-7 a \sin (e+f x)) (\sin (e+f x) a+a)^{5/2}}{\sin (e+f x)}dx}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} \csc (e+f x) (\sin (e+f x) a+a)^{3/2} \left (25 a^2-31 a^2 \sin (e+f x)\right )dx+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int \csc (e+f x) (\sin (e+f x) a+a)^{3/2} \left (25 a^2-31 a^2 \sin (e+f x)\right )dx+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {(\sin (e+f x) a+a)^{3/2} \left (25 a^2-31 a^2 \sin (e+f x)\right )}{\sin (e+f x)}dx+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {1}{2} \csc (e+f x) \sqrt {\sin (e+f x) a+a} \left (75 a^3-49 a^3 \sin (e+f x)\right )dx+\frac {62 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \csc (e+f x) \sqrt {\sin (e+f x) a+a} \left (75 a^3-49 a^3 \sin (e+f x)\right )dx+\frac {62 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\sin (e+f x) a+a} \left (75 a^3-49 a^3 \sin (e+f x)\right )}{\sin (e+f x)}dx+\frac {62 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (75 a^3 \int \csc (e+f x) \sqrt {\sin (e+f x) a+a}dx+\frac {98 a^4 \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {62 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (75 a^3 \int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x)}dx+\frac {98 a^4 \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {62 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {98 a^4 \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {150 a^4 \int \frac {1}{a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}\right )+\frac {62 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )+\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {14 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}+\frac {1}{5} \left (\frac {62 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}+\frac {1}{3} \left (\frac {98 a^4 \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {150 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{f}\right )\right )}{2 a}-\frac {\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}\) |
Input:
Int[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2),x]
Output:
-((Cot[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/f) + ((14*a^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f) + ((62*a^3*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f) + ((-150*a^(7/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Si n[e + f*x]]])/f + (98*a^4*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]))/3)/5 )/(2*a)
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_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[-(a + b*Sin[e + f*x])^m/(f*Tan[e + f*x]), x] + Simp[1/a Int[(a + b*Sin[e + f*x])^m*((b*m - a*(m + 1)*Sin[e + f*x])/Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1 /2] && !LtQ[m, -1]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Time = 0.97 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.07
| method | result | size |
| default | \(\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (\sin \left (f x +e \right ) \left (90 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {5}{2}}-40 a^{\frac {3}{2}} \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}}+6 \sqrt {a}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {5}{2}}-75 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}}{\sqrt {a}}\right ) a^{3}\right )-15 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {5}{2}}\right )}{15 \sin \left (f x +e \right ) \sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(162\) |
Input:
int(cot(f*x+e)^2*(a+sin(f*x+e)*a)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/15*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(sin(f*x+e)*(90*(a-sin(f*x+ e)*a)^(1/2)*a^(5/2)-40*a^(3/2)*(a-sin(f*x+e)*a)^(3/2)+6*a^(1/2)*(a-sin(f*x +e)*a)^(5/2)-75*arctanh((a-sin(f*x+e)*a)^(1/2)/a^(1/2))*a^3)-15*(a-sin(f*x +e)*a)^(1/2)*a^(5/2))/sin(f*x+e)/a^(1/2)/cos(f*x+e)/(a+sin(f*x+e)*a)^(1/2) /f
Leaf count of result is larger than twice the leaf count of optimal. 363 vs. \(2 (131) = 262\).
Time = 0.10 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.40 \[ \int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx=\frac {75 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - a^{2} - {\left (a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 3\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 3\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} - 9 \, a \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right ) + 4 \, {\left (6 \, a^{2} \cos \left (f x + e\right )^{4} + 28 \, a^{2} \cos \left (f x + e\right )^{3} - 40 \, a^{2} \cos \left (f x + e\right )^{2} - 13 \, a^{2} \cos \left (f x + e\right ) + 49 \, a^{2} + {\left (6 \, a^{2} \cos \left (f x + e\right )^{3} - 22 \, a^{2} \cos \left (f x + e\right )^{2} - 62 \, a^{2} \cos \left (f x + e\right ) - 49 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{60 \, {\left (f \cos \left (f x + e\right )^{2} - {\left (f \cos \left (f x + e\right ) + f\right )} \sin \left (f x + e\right ) - f\right )}} \] Input:
integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")
Output:
1/60*(75*(a^2*cos(f*x + e)^2 - a^2 - (a^2*cos(f*x + e) + a^2)*sin(f*x + e) )*sqrt(a)*log((a*cos(f*x + e)^3 - 7*a*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + (cos(f*x + e) + 3)*sin(f*x + e) - 2*cos(f*x + e) - 3)*sqrt(a*sin(f*x + e) + a)*sqrt(a) - 9*a*cos(f*x + e) + (a*cos(f*x + e)^2 + 8*a*cos(f*x + e) - a)*sin(f*x + e) - a)/(cos(f*x + e)^3 + cos(f*x + e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) - cos(f*x + e) - 1)) + 4*(6*a^2*cos(f*x + e)^4 + 28*a^2*co s(f*x + e)^3 - 40*a^2*cos(f*x + e)^2 - 13*a^2*cos(f*x + e) + 49*a^2 + (6*a ^2*cos(f*x + e)^3 - 22*a^2*cos(f*x + e)^2 - 62*a^2*cos(f*x + e) - 49*a^2)* sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(f*cos(f*x + e)^2 - (f*cos(f*x + e ) + f)*sin(f*x + e) - f)
Timed out. \[ \int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(cot(f*x+e)**2*(a+a*sin(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \cot \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^(5/2)*cot(f*x + e)^2, x)
Time = 0.17 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.48 \[ \int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (96 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 320 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 75 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 360 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + \frac {60 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1}\right )} \sqrt {a}}{60 \, f} \] Input:
integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")
Output:
-1/60*sqrt(2)*(96*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/ 2*f*x + 1/2*e)^5 - 320*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 75*sqrt(2)*a^2*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*f*x + 1/2*e))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sgn (cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 360*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2 *e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 60*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1 /2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)/(2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1))*sqrt(a)/f
Timed out. \[ \int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \] Input:
int(cot(e + f*x)^2*(a + a*sin(e + f*x))^(5/2),x)
Output:
int(cot(e + f*x)^2*(a + a*sin(e + f*x))^(5/2), x)
\[ \int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \cot \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}d x +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \cot \left (f x +e \right )^{2} \sin \left (f x +e \right )d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \cot \left (f x +e \right )^{2}d x \right ) \] Input:
int(cot(f*x+e)^2*(a+a*sin(f*x+e))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(e + f*x) + 1)*cot(e + f*x)**2*sin(e + f*x)**2,x ) + 2*int(sqrt(sin(e + f*x) + 1)*cot(e + f*x)**2*sin(e + f*x),x) + int(sqr t(sin(e + f*x) + 1)*cot(e + f*x)**2,x))