Integrand size = 25, antiderivative size = 111 \[ \int \frac {(e \cot (c+d x))^{5/2}}{a+a \cot (c+d x)} \, dx=\frac {e^{5/2} \arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d}-\frac {e^{5/2} \arctan \left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d} \]
e^(5/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a/d-1/2*e^(5/2)*arctan(1/2*(e ^(1/2)-cot(d*x+c)*e^(1/2))*2^(1/2)/(e*cot(d*x+c))^(1/2))/a/d*2^(1/2)-2*e^2 *(e*cot(d*x+c))^(1/2)/a/d
Leaf count is larger than twice the leaf count of optimal. \(296\) vs. \(2(111)=222\).
Time = 0.92 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.67 \[ \int \frac {(e \cot (c+d x))^{5/2}}{a+a \cot (c+d x)} \, dx=\frac {e \left (8 e^{3/2} \arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )-4 \left (-e^2\right )^{3/4} \arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt [4]{-e^2}}\right )-2 \sqrt {2} e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )+2 \sqrt {2} e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )+4 \left (-e^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt [4]{-e^2}}\right )-16 e \sqrt {e \cot (c+d x)}-\sqrt {2} e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )+\sqrt {2} e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )\right )}{8 a d} \]
(e*(8*e^(3/2)*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]] - 4*(-e^2)^(3/4)*ArcTan [Sqrt[e*Cot[c + d*x]]/(-e^2)^(1/4)] - 2*Sqrt[2]*e^(3/2)*ArcTan[1 - (Sqrt[2 ]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]] + 2*Sqrt[2]*e^(3/2)*ArcTan[1 + (Sqrt[2]*S qrt[e*Cot[c + d*x]])/Sqrt[e]] + 4*(-e^2)^(3/4)*ArcTanh[Sqrt[e*Cot[c + d*x] ]/(-e^2)^(1/4)] - 16*e*Sqrt[e*Cot[c + d*x]] - Sqrt[2]*e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Cot[c + d*x]]] + Sqrt[2]*e^(3/2)*L og[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c + d*x]]]))/(8*a*d )
Time = 0.90 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4049, 27, 3042, 4136, 3042, 4015, 218, 4117, 27, 73, 216}
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.
\(\displaystyle \int \frac {(e \cot (c+d x))^{5/2}}{a \cot (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (-e \tan \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{a-a \tan \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4049 |
\(\displaystyle -\frac {2 \int \frac {a \cot ^2(c+d x) e^3+a e^3+a \cot (c+d x) e^3}{2 \sqrt {e \cot (c+d x)} (\cot (c+d x) a+a)}dx}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a \cot ^2(c+d x) e^3+a e^3+a \cot (c+d x) e^3}{\sqrt {e \cot (c+d x)} (\cot (c+d x) a+a)}dx}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {a \tan \left (c+d x+\frac {\pi }{2}\right )^2 e^3+a e^3-a \tan \left (c+d x+\frac {\pi }{2}\right ) e^3}{\sqrt {-e \tan \left (c+d x+\frac {\pi }{2}\right )} \left (a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 4136 |
\(\displaystyle -\frac {\frac {\int \frac {a^2 e^3+a^2 \cot (c+d x) e^3}{\sqrt {e \cot (c+d x)}}dx}{2 a^2}+\frac {1}{2} a e^3 \int \frac {\cot ^2(c+d x)+1}{\sqrt {e \cot (c+d x)} (\cot (c+d x) a+a)}dx}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {a^2 e^3-a^2 e^3 \tan \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {-e \tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}+\frac {1}{2} a e^3 \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-e \tan \left (c+d x+\frac {\pi }{2}\right )} \left (a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 4015 |
\(\displaystyle -\frac {\frac {1}{2} a e^3 \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-e \tan \left (c+d x+\frac {\pi }{2}\right )} \left (a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx-\frac {a^2 e^6 \int \frac {1}{-2 a^4 e^6-\left (a^2 e^3-a^2 e^3 \cot (c+d x)\right )^2 \tan (c+d x)}d\frac {a^2 e^3-a^2 e^3 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}}{d}}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {\frac {1}{2} a e^3 \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-e \tan \left (c+d x+\frac {\pi }{2}\right )} \left (a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {e^{5/2} \arctan \left (\frac {a^2 e^3-a^2 e^3 \cot (c+d x)}{\sqrt {2} a^2 e^{5/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} d}}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 4117 |
\(\displaystyle -\frac {\frac {a e^3 \int \frac {1}{a \sqrt {e \cot (c+d x)} (\cot (c+d x)+1)}d(-\cot (c+d x))}{2 d}+\frac {e^{5/2} \arctan \left (\frac {a^2 e^3-a^2 e^3 \cot (c+d x)}{\sqrt {2} a^2 e^{5/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} d}}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {e^3 \int \frac {1}{\sqrt {e \cot (c+d x)} (\cot (c+d x)+1)}d(-\cot (c+d x))}{2 d}+\frac {e^{5/2} \arctan \left (\frac {a^2 e^3-a^2 e^3 \cot (c+d x)}{\sqrt {2} a^2 e^{5/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} d}}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\frac {e^{5/2} \arctan \left (\frac {a^2 e^3-a^2 e^3 \cot (c+d x)}{\sqrt {2} a^2 e^{5/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} d}-\frac {e^2 \int \frac {1}{\frac {\cot ^2(c+d x)}{e}+1}d\sqrt {e \cot (c+d x)}}{d}}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {e^{5/2} \arctan \left (\frac {a^2 e^3-a^2 e^3 \cot (c+d x)}{\sqrt {2} a^2 e^{5/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} d}+\frac {e^{5/2} \arctan \left (\frac {\cot (c+d x)}{\sqrt {e}}\right )}{d}}{a}-\frac {2 e^2 \sqrt {e \cot (c+d x)}}{a d}\) |
-(((e^(5/2)*ArcTan[Cot[c + d*x]/Sqrt[e]])/d + (e^(5/2)*ArcTan[(a^2*e^3 - a ^2*e^3*Cot[c + d*x])/(Sqrt[2]*a^2*e^(5/2)*Sqrt[e*Cot[c + d*x]])])/(Sqrt[2] *d))/a) - (2*e^2*Sqrt[e*Cot[c + d*x]])/(a*d)
3.1.23.3.1 Defintions of rubi rules used
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_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[-2*(d^2/f) Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) )
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^ n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ (A*b^2 - a*b*B + a^2*C)/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & & !GtQ[n, 0] && !LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(311\) vs. \(2(92)=184\).
Time = 0.10 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.81
method | result | size |
derivativedivides | \(-\frac {2 e^{2} \left (\sqrt {e \cot \left (d x +c \right )}-\frac {e \left (\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{2}-\frac {\sqrt {e}\, \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2}\right )}{d a}\) | \(312\) |
default | \(-\frac {2 e^{2} \left (\sqrt {e \cot \left (d x +c \right )}-\frac {e \left (\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{2}-\frac {\sqrt {e}\, \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2}\right )}{d a}\) | \(312\) |
-2/d/a*e^2*((e*cot(d*x+c))^(1/2)-1/2*e*(1/8/e*(e^2)^(1/4)*2^(1/2)*(ln((e*c ot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x +c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^(1/2 )/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot (d*x+c))^(1/2)+1))+1/8/(e^2)^(1/4)*2^(1/2)*(ln((e*cot(d*x+c)-(e^2)^(1/4)*( e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)+(e^2)^(1/4)*(e*cot( d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d* x+c))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)))-1/2 *e^(1/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2)))
Time = 0.29 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.60 \[ \int \frac {(e \cot (c+d x))^{5/2}}{a+a \cot (c+d x)} \, dx=\left [\frac {\sqrt {2} \sqrt {-e} e^{2} \log \left ({\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) - \sqrt {2}\right )} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} - 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) + 2 \, \sqrt {-e} e^{2} \log \left (\frac {e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) + 2 \, \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right ) - 8 \, e^{2} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{4 \, a d}, -\frac {\sqrt {2} e^{\frac {5}{2}} \arctan \left (-\frac {{\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) - \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {2}\right )} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) - 2 \, e^{\frac {5}{2}} \arctan \left (\frac {\sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt {e}}\right ) + 4 \, e^{2} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \, a d}\right ] \]
[1/4*(sqrt(2)*sqrt(-e)*e^2*log((sqrt(2)*cos(2*d*x + 2*c) + sqrt(2)*sin(2*d *x + 2*c) - sqrt(2))*sqrt(-e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2* c)) - 2*e*sin(2*d*x + 2*c) + e) + 2*sqrt(-e)*e^2*log((e*cos(2*d*x + 2*c) - e*sin(2*d*x + 2*c) + 2*sqrt(-e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) + e)/(cos(2*d*x + 2*c) + sin(2*d*x + 2*c) + 1)) - 8*e^2*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(a*d), -1/2*(sqrt(2 )*e^(5/2)*arctan(-1/2*(sqrt(2)*cos(2*d*x + 2*c) - sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*sqrt(e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))/(e*cos (2*d*x + 2*c) + e)) - 2*e^(5/2)*arctan(sqrt((e*cos(2*d*x + 2*c) + e)/sin(2 *d*x + 2*c))/sqrt(e)) + 4*e^2*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2* c)))/(a*d)]
\[ \int \frac {(e \cot (c+d x))^{5/2}}{a+a \cot (c+d x)} \, dx=\frac {\int \frac {\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\cot {\left (c + d x \right )} + 1}\, dx}{a} \]
Exception generated. \[ \int \frac {(e \cot (c+d x))^{5/2}}{a+a \cot (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {(e \cot (c+d x))^{5/2}}{a+a \cot (c+d x)} \, dx=\int { \frac {\left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}}}{a \cot \left (d x + c\right ) + a} \,d x } \]
Time = 12.93 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.11 \[ \int \frac {(e \cot (c+d x))^{5/2}}{a+a \cot (c+d x)} \, dx=\frac {e^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{a\,d}-\frac {2\,e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{a\,d}+\frac {\sqrt {2}\,e^{5/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}+\frac {\sqrt {2}\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{2\,e^{3/2}}\right )\right )}{4\,a\,d} \]