3.1.55 \(\int \frac {a+b \cot (c+d x)}{(e \cot (c+d x))^{5/2}} \, dx\) [55]

Optimal. Leaf size=252 \[ -\frac {(a+b) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}+\frac {(a+b) \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}+\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {2 b}{d e^2 \sqrt {e \cot (c+d x)}}-\frac {(a-b) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}+\frac {(a-b) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}} \]

[Out]

2/3*a/d/e/(e*cot(d*x+c))^(3/2)-1/2*(a+b)*arctan(1-2^(1/2)*(e*cot(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)*2^(1/2)+1/2*
(a+b)*arctan(1+2^(1/2)*(e*cot(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)*2^(1/2)-1/4*(a-b)*ln(e^(1/2)+cot(d*x+c)*e^(1/2)
-2^(1/2)*(e*cot(d*x+c))^(1/2))/d/e^(5/2)*2^(1/2)+1/4*(a-b)*ln(e^(1/2)+cot(d*x+c)*e^(1/2)+2^(1/2)*(e*cot(d*x+c)
)^(1/2))/d/e^(5/2)*2^(1/2)+2*b/d/e^2/(e*cot(d*x+c))^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3610, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {(a+b) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}+\frac {(a+b) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{5/2}}-\frac {(a-b) \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}+\frac {(a-b) \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}+\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {2 b}{d e^2 \sqrt {e \cot (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[c + d*x])/(e*Cot[c + d*x])^(5/2),x]

[Out]

-(((a + b)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2))) + ((a + b)*ArcTan[1 + (Sqr
t[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) + (2*a)/(3*d*e*(e*Cot[c + d*x])^(3/2)) + (2*b)/(d*e^2
*Sqrt[e*Cot[c + d*x]]) - ((a - b)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(2*Sqrt[
2]*d*e^(5/2)) + ((a - b)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d*e^(5
/2))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3610

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

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rubi steps

\begin {align*} \int \frac {a+b \cot (c+d x)}{(e \cot (c+d x))^{5/2}} \, dx &=\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {\int \frac {b e-a e \cot (c+d x)}{(e \cot (c+d x))^{3/2}} \, dx}{e^2}\\ &=\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {2 b}{d e^2 \sqrt {e \cot (c+d x)}}+\frac {\int \frac {-a e^2-b e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{e^4}\\ &=\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {2 b}{d e^2 \sqrt {e \cot (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {a e^3+b e^2 x^2}{e^2+x^4} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{d e^4}\\ &=\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {2 b}{d e^2 \sqrt {e \cot (c+d x)}}+\frac {(a-b) \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{d e^2}+\frac {(a+b) \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{d e^2}\\ &=\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {2 b}{d e^2 \sqrt {e \cot (c+d x)}}-\frac {(a-b) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {(a-b) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 d e^2}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 d e^2}\\ &=\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {2 b}{d e^2 \sqrt {e \cot (c+d x)}}-\frac {(a-b) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}+\frac {(a-b) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {(a+b) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}\\ &=-\frac {(a+b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}+\frac {(a+b) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}+\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}}+\frac {2 b}{d e^2 \sqrt {e \cot (c+d x)}}-\frac {(a-b) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}+\frac {(a-b) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.71, size = 196, normalized size = 0.78 \begin {gather*} \frac {3 b \left (2 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+8 \sqrt {\tan (c+d x)}\right )-8 a \left (-1+\, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(c+d x)\right )\right ) \tan ^{\frac {3}{2}}(c+d x)}{12 d (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cot[c + d*x])/(e*Cot[c + d*x])^(5/2),x]

[Out]

(3*b*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Sq
rt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c
+ d*x]] + 8*Sqrt[Tan[c + d*x]]) - 8*a*(-1 + Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2])*Tan[c + d*x]^(3/2
))/(12*d*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))

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Maple [A]
time = 0.40, size = 311, normalized size = 1.23

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{e^{2}}+\frac {2 a}{3 e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {2 b}{e^{2} \sqrt {e \cot \left (d x +c \right )}}}{d}\)	\(311\)
default	\(\frac {-\frac {2 \left (-\frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{e^{2}}+\frac {2 a}{3 e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {2 b}{e^{2} \sqrt {e \cot \left (d x +c \right )}}}{d}\)	\(311\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cot(d*x+c))/(e*cot(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/e^2*(-1/8*a/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))-1/8*b/(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)))+2/3*a/e/(e*cot(d*x+c))^(3/2)+2*b/e^2/(e*cot(d*x+c))^(1/2))

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Maxima [A]
time = 0.55, size = 154, normalized size = 0.61 \begin {gather*} \frac {{\left (6 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 6 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 3 \, \sqrt {2} {\left (a - b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 3 \, \sqrt {2} {\left (a - b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + 8 \, {\left (a + \frac {3 \, b}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}\right )} e^{\left (-\frac {5}{2}\right )}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(d*x+c))/(e*cot(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/12*(6*sqrt(2)*(a + b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 6*sqrt(2)*(a + b)*arctan(-1/2*s
qrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + 3*sqrt(2)*(a - b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) +
1) - 3*sqrt(2)*(a - b)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + 8*(a + 3*b/tan(d*x + c))*tan(d
*x + c)^(3/2))*e^(-5/2)/d

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(d*x+c))/(e*cot(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \cot {\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(d*x+c))/(e*cot(d*x+c))**(5/2),x)

[Out]

Integral((a + b*cot(c + d*x))/(e*cot(c + d*x))**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(d*x+c))/(e*cot(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cot(d*x + c) + a)/(e*cot(d*x + c))^(5/2), x)

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Mupad [B]
time = 1.24, size = 158, normalized size = 0.63 \begin {gather*} \frac {2\,a}{3\,d\,e\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}+\frac {2\,b}{d\,e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}+\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{d\,e^{5/2}}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{d\,e^{5/2}}-\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )\,1{}\mathrm {i}}{d\,e^{5/2}}-\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )\,1{}\mathrm {i}}{d\,e^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cot(c + d*x))/(e*cot(c + d*x))^(5/2),x)

[Out]

(2*a)/(3*d*e*(e*cot(c + d*x))^(3/2)) + (2*b)/(d*e^2*(e*cot(c + d*x))^(1/2)) - ((-1)^(1/4)*a*atan(((-1)^(1/4)*(
e*cot(c + d*x))^(1/2))/e^(1/2))*1i)/(d*e^(5/2)) - ((-1)^(1/4)*a*atanh(((-1)^(1/4)*(e*cot(c + d*x))^(1/2))/e^(1
/2))*1i)/(d*e^(5/2)) + ((-1)^(1/4)*b*atan(((-1)^(1/4)*(e*cot(c + d*x))^(1/2))/e^(1/2)))/(d*e^(5/2)) - ((-1)^(1
/4)*b*atanh(((-1)^(1/4)*(e*cot(c + d*x))^(1/2))/e^(1/2)))/(d*e^(5/2))

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