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

Optimal. Leaf size=279 \[ -\frac {e^{3/2} \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d}+\frac {e^{3/2} \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d}+\frac {e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{2 \sqrt {2} a^2 d}-\frac {e \sqrt {e \cot (c+d x)}}{2 d \left (a^2 \cot (c+d x)+a^2\right )} \]

[Out]

1/2*e^(3/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^2/d+1/4*e^(3/2)*arctan(1-2^(1/2)*(e*cot(d*x+c))^(1/2)/e^(1/
2))/a^2/d*2^(1/2)-1/4*e^(3/2)*arctan(1+2^(1/2)*(e*cot(d*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)-1/8*e^(3/2)*ln(e^(1
/2)+cot(d*x+c)*e^(1/2)-2^(1/2)*(e*cot(d*x+c))^(1/2))/a^2/d*2^(1/2)+1/8*e^(3/2)*ln(e^(1/2)+cot(d*x+c)*e^(1/2)+2
^(1/2)*(e*cot(d*x+c))^(1/2))/a^2/d*2^(1/2)-1/2*e*(e*cot(d*x+c))^(1/2)/d/(a^2+a^2*cot(d*x+c))

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Rubi [A]  time = 0.56, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3567, 3653, 12, 16, 3476, 329, 297, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac {e^{3/2} \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d}+\frac {e^{3/2} \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d}+\frac {e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{2 \sqrt {2} a^2 d}-\frac {e \sqrt {e \cot (c+d x)}}{2 d \left (a^2 \cot (c+d x)+a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 63

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 204

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

Rule 205

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

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

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 628

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 1162

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 1165

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 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3567

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[
1/((m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*Tan[e
+ f*x]^2, 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] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[2*m]

Rule 3634

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]

Rule 3653

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] :> Dist[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] + Dist[(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]

Rubi steps

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

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Mathematica [A]  time = 2.83, size = 312, normalized size = 1.12 \[ -\frac {\sin ^2(c+d x) (e \cot (c+d x))^{3/2} \left (4 \cot ^{\frac {7}{2}}(c+d x)-4 \cot ^{\frac {5}{2}}(c+d x)+4 \cot ^{\frac {3}{2}}(c+d x)-4 \sqrt {\cot (c+d x)}+\sqrt {2} \cos (2 (c+d x)) \csc ^4(c+d x) \log \left (-\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}-1\right )-\sqrt {2} \cos (2 (c+d x)) \csc ^4(c+d x) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-2 \sqrt {2} \cos (2 (c+d x)) \csc ^4(c+d x) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )+2 \sqrt {2} \cos (2 (c+d x)) \csc ^4(c+d x) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-4 \cos (2 (c+d x)) \csc ^4(c+d x) \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )\right )}{8 a^2 d \cot ^{\frac {3}{2}}(c+d x) \left (\cot ^2(c+d x)-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((e*Cot[c + d*x])^(3/2)*(-4*Sqrt[Cot[c + d*x]] + 4*Cot[c + d*x]^(3/2) - 4*Cot[c + d*x]^(5/2) + 4*Cot[c +
d*x]^(7/2) - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]*Cos[2*(c + d*x)]*Csc[c + d*x]^4 + 2*Sqrt[2]*ArcT
an[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]*Cos[2*(c + d*x)]*Csc[c + d*x]^4 - 4*ArcTan[Sqrt[Cot[c + d*x]]]*Cos[2*(c + d
*x)]*Csc[c + d*x]^4 + Sqrt[2]*Cos[2*(c + d*x)]*Csc[c + d*x]^4*Log[-1 + Sqrt[2]*Sqrt[Cot[c + d*x]] - Cot[c + d*
x]] - Sqrt[2]*Cos[2*(c + d*x)]*Csc[c + d*x]^4*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])*Sin[c + d*x]
^2)/(a^2*d*Cot[c + d*x]^(3/2)*(-1 + Cot[c + d*x]^2))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cot \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \cot \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.74, size = 234, normalized size = 0.84 \[ -\frac {e^{2} \sqrt {e \cot \left (d x +c \right )}}{2 d \,a^{2} \left (e \cot \left (d x +c \right )+e \right )}+\frac {e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2 a^{2} d}-\frac {e^{2} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d \,a^{2} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {e^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d \,a^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {e^{2} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d \,a^{2} \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 232, normalized size = 0.83 \[ -\frac {e {\left (\frac {4 \, e \sqrt {\frac {e}{\tan \left (d x + c\right )}}}{a^{2} e + \frac {a^{2} e}{\tan \left (d x + c\right )}} + \frac {e {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} + \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}\right )}}{a^{2}} - \frac {4 \, \sqrt {e} \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a^{2}}\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*e*(4*e*sqrt(e/tan(d*x + c))/(a^2*e + a^2*e/tan(d*x + c)) + e*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(
e) + 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(
d*x + c)))/sqrt(e))/sqrt(e) - sqrt(2)*log(sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e) +
sqrt(2)*log(-sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e))/a^2 - 4*sqrt(e)*arctan(sqrt(
e/tan(d*x + c))/sqrt(e))/a^2)/d

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mupad [B]  time = 0.82, size = 376, normalized size = 1.35 \[ -\frac {\mathrm {atan}\left (\frac {4\,e^{16}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {e^6}{a^8\,d^4}\right )}^{1/4}}{\frac {4\,e^{18}}{a^2\,d}+4\,a^2\,d\,e^{15}\,\sqrt {-\frac {e^6}{a^8\,d^4}}}+\frac {4\,e^{13}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {e^6}{a^8\,d^4}\right )}^{3/4}}{\frac {4\,e^{18}}{a^6\,d^3}+\frac {4\,e^{15}\,\sqrt {-\frac {e^6}{a^8\,d^4}}}{a^2\,d}}\right )\,{\left (-\frac {e^6}{a^8\,d^4}\right )}^{1/4}}{2}-\mathrm {atan}\left (\frac {e^{16}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {e^6}{256\,a^8\,d^4}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {4\,e^{18}}{a^2\,d}-64\,a^2\,d\,e^{15}\,\sqrt {-\frac {e^6}{256\,a^8\,d^4}}}-\frac {e^{13}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {e^6}{256\,a^8\,d^4}\right )}^{3/4}\,256{}\mathrm {i}}{\frac {4\,e^{18}}{a^6\,d^3}-\frac {64\,e^{15}\,\sqrt {-\frac {e^6}{256\,a^8\,d^4}}}{a^2\,d}}\right )\,{\left (-\frac {e^6}{256\,a^8\,d^4}\right )}^{1/4}\,2{}\mathrm {i}-\frac {e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\left (a^2\,d\,e+a^2\,d\,e\,\mathrm {cot}\left (c+d\,x\right )\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,\sqrt {-e^3}\,1{}\mathrm {i}}{e^2}\right )\,\sqrt {-e^3}\,1{}\mathrm {i}}{2\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan(((e*cot(c + d*x))^(1/2)*(-e^3)^(1/2)*1i)/e^2)*(-e^3)^(1/2)*1i)/(2*a^2*d) - atan((e^16*(e*cot(c + d*x))^(
1/2)*(-e^6/(256*a^8*d^4))^(1/4)*16i)/((4*e^18)/(a^2*d) - 64*a^2*d*e^15*(-e^6/(256*a^8*d^4))^(1/2)) - (e^13*(e*
cot(c + d*x))^(1/2)*(-e^6/(256*a^8*d^4))^(3/4)*256i)/((4*e^18)/(a^6*d^3) - (64*e^15*(-e^6/(256*a^8*d^4))^(1/2)
)/(a^2*d)))*(-e^6/(256*a^8*d^4))^(1/4)*2i - (e^2*(e*cot(c + d*x))^(1/2))/(2*(a^2*d*e + a^2*d*e*cot(c + d*x)))
- (atan((4*e^16*(e*cot(c + d*x))^(1/2)*(-e^6/(a^8*d^4))^(1/4))/((4*e^18)/(a^2*d) + 4*a^2*d*e^15*(-e^6/(a^8*d^4
))^(1/2)) + (4*e^13*(e*cot(c + d*x))^(1/2)*(-e^6/(a^8*d^4))^(3/4))/((4*e^18)/(a^6*d^3) + (4*e^15*(-e^6/(a^8*d^
4))^(1/2))/(a^2*d)))*(-e^6/(a^8*d^4))^(1/4))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\cot ^{2}{\left (c + d x \right )} + 2 \cot {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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