\(\int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx\) [24]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 87 \[ \int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx=-\frac {e^{3/2} \arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d}+\frac {e^{3/2} \text {arctanh}\left (\frac {\sqrt {e}+\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d} \]

[Out]

-e^(3/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a/d+1/2*e^(3/2)*arctanh(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)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3654, 3613, 214, 3715, 65, 211} \[ \int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx=\frac {e^{3/2} \text {arctanh}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d} \]

[In]

Int[(e*Cot[c + d*x])^(3/2)/(a + a*Cot[c + d*x]),x]

[Out]

-((e^(3/2)*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(a*d)) + (e^(3/2)*ArcTanh[(Sqrt[e] + Sqrt[e]*Cot[c + d*x])/(S
qrt[2]*Sqrt[e*Cot[c + d*x]])])/(Sqrt[2]*a*d)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3613

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-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]

Rule 3654

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1
/(c^2 + d^2), Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e +
 f*x]], x], x] + Dist[(b*c - a*d)^2/(c^2 + d^2), Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan
[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2
, 0]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-a e^2+a e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{2 a^2}+\frac {1}{2} e^2 \int \frac {1+\cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx \\ & = \frac {e^2 \text {Subst}\left (\int \frac {1}{\sqrt {-e x} (a-a x)} \, dx,x,-\cot (c+d x)\right )}{2 d}-\frac {e^4 \text {Subst}\left (\int \frac {1}{2 a^2 e^4-e x^2} \, dx,x,\frac {-a e^2-a e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{d} \\ & = \frac {e^{3/2} \text {arctanh}\left (\frac {\sqrt {e}+\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}-\frac {e \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{e}} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{d} \\ & = -\frac {e^{3/2} \arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d}+\frac {e^{3/2} \text {arctanh}\left (\frac {\sqrt {e}+\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(87)=174\).

Time = 0.55 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.22 \[ \int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx=-\frac {8 e^{3/2} \arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )-4 \left (-e^2\right )^{3/4} \arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt [4]{-e^2}}\right )+2 \sqrt {2} e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )-2 \sqrt {2} e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )+4 \left (-e^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt [4]{-e^2}}\right )+\sqrt {2} e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )-\sqrt {2} e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{8 a d} \]

[In]

Integrate[(e*Cot[c + d*x])^(3/2)/(a + a*Cot[c + d*x]),x]

[Out]

-1/8*(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]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]] + 4*(-e^2)^(3/4)*ArcTanh[Sqrt[e*Cot[c + d*x]]/(-e^2)^(1/4)] + 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)*Log[Sqrt[e] + Sqrt[e]*C
ot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(a*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(71)=142\).

Time = 0.05 (sec) , antiderivative size = 298, normalized size of antiderivative = 3.43

method result size
derivativedivides \(-\frac {2 e^{2} \left (-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 \left (e^{2}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2 \sqrt {e}}\right )}{d a}\) \(298\)
default \(-\frac {2 e^{2} \left (-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 \left (e^{2}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2 \sqrt {e}}\right )}{d a}\) \(298\)

[In]

int((e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-2/d/a*e^2*(-1/16/e*(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))+1/16/(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)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 333, normalized size of antiderivative = 3.83 \[ \int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx=\left [-\frac {\sqrt {2} \sqrt {-e} e \arctan \left (\frac {{\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {2}\right )} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) - \sqrt {-e} e \log \left (\frac {e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) - 2 \, \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right )}{2 \, a d}, \frac {\sqrt {2} e^{\frac {3}{2}} \log \left (-{\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) - \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) - \sqrt {2}\right )} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} + 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) - 4 \, e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt {e}}\right )}{4 \, a d}\right ] \]

[In]

integrate((e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c)),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(2)*sqrt(-e)*e*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)) - sqrt(-e)*e*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)))/(a*d), 1/4*(sqrt(2)*e^(3/2)*log(-(sqrt(2)*cos(2*d*x + 2*c) - sqrt(2)*si
n(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)
- 4*e^(3/2)*arctan(sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))/sqrt(e)))/(a*d)]

Sympy [F]

\[ \int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx=\frac {\int \frac {\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\cot {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

integrate((e*cot(d*x+c))**(3/2)/(a+a*cot(d*x+c)),x)

[Out]

Integral((e*cot(c + d*x))**(3/2)/(cot(c + d*x) + 1), x)/a

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

\[ \int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx=\int { \frac {\left (e \cot \left (d x + c\right )\right )^{\frac {3}{2}}}{a \cot \left (d x + c\right ) + a} \,d x } \]

[In]

integrate((e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*cot(d*x + c))^(3/2)/(a*cot(d*x + c) + a), x)

Mupad [B] (verification not implemented)

Time = 12.78 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx=\frac {\sqrt {2}\,e^{3/2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,e^{25/2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{12\,e^{13}\,\mathrm {cot}\left (c+d\,x\right )+12\,e^{13}}\right )}{2\,a\,d}-\frac {e^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{a\,d} \]

[In]

int((e*cot(c + d*x))^(3/2)/(a + a*cot(c + d*x)),x)

[Out]

(2^(1/2)*e^(3/2)*atanh((12*2^(1/2)*e^(25/2)*(e*cot(c + d*x))^(1/2))/(12*e^13*cot(c + d*x) + 12*e^13)))/(2*a*d)
 - (e^(3/2)*atan((e*cot(c + d*x))^(1/2)/e^(1/2)))/(a*d)