3.33 \(\int \frac {1}{(e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2} \, dx\)

Optimal. Leaf size=306 \[ -\frac {\log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d e^{3/2}}+\frac {\log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d e^{3/2}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d e^{3/2}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d e^{3/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{2 \sqrt {2} a^2 d e^{3/2}}+\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \left (a^2 \cot (c+d x)+a^2\right ) \sqrt {e \cot (c+d x)}} \]

[Out]

5/2*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^2/d/e^(3/2)-1/4*arctan(1-2^(1/2)*(e*cot(d*x+c))^(1/2)/e^(1/2))/a^2/
d/e^(3/2)*2^(1/2)+1/4*arctan(1+2^(1/2)*(e*cot(d*x+c))^(1/2)/e^(1/2))/a^2/d/e^(3/2)*2^(1/2)-1/8*ln(e^(1/2)+cot(
d*x+c)*e^(1/2)-2^(1/2)*(e*cot(d*x+c))^(1/2))/a^2/d/e^(3/2)*2^(1/2)+1/8*ln(e^(1/2)+cot(d*x+c)*e^(1/2)+2^(1/2)*(
e*cot(d*x+c))^(1/2))/a^2/d/e^(3/2)*2^(1/2)+5/2/a^2/d/e/(e*cot(d*x+c))^(1/2)-1/2/d/e/(a^2+a^2*cot(d*x+c))/(e*co
t(d*x+c))^(1/2)

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Rubi [A]  time = 0.80, antiderivative size = 306, 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 = {3569, 3649, 3653, 12, 3476, 329, 211, 1165, 628, 1162, 617, 204, 3634, 63, 205} \[ -\frac {\log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d e^{3/2}}+\frac {\log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d e^{3/2}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d e^{3/2}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d e^{3/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{2 \sqrt {2} a^2 d e^{3/2}}+\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \left (a^2 \cot (c+d x)+a^2\right ) \sqrt {e \cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cot[c + d*x])^(3/2)*(a + a*Cot[c + d*x])^2),x]

[Out]

(5*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^(3/2)) + 5
/(2*a^2*d*e*Sqrt[e*Cot[c + d*x]]) - 1/(2*d*e*Sqrt[e*Cot[c + d*x]]*(a^2 + a^2*Cot[c + d*x])) - Log[Sqrt[e] + Sq
rt[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*e^(3/2))

Rule 12

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

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{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 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*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] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

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 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, 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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

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 {1}{(e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2} \, dx &=-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}-\frac {\int \frac {-\frac {5 a^2 e}{2}+a^2 e \cot (c+d x)-\frac {3}{2} a^2 e \cot ^2(c+d x)}{(e \cot (c+d x))^{3/2} (a+a \cot (c+d x))} \, dx}{2 a^3 e}\\ &=\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}-\frac {\int \frac {\frac {7 a^3 e^3}{4}+\frac {1}{2} a^3 e^3 \cot (c+d x)+\frac {5}{4} a^3 e^3 \cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{a^4 e^4}\\ &=\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}-\frac {\int \frac {a^4 e^3}{\sqrt {e \cot (c+d x)}} \, dx}{2 a^6 e^4}-\frac {5 \int \frac {1+\cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{4 a e}\\ &=\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}-\frac {\int \frac {1}{\sqrt {e \cot (c+d x)}} \, dx}{2 a^2 e}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-e x} (a-a x)} \, dx,x,-\cot (c+d x)\right )}{4 a d e}\\ &=\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \cot (c+d x)\right )}{2 a^2 d}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{a+\frac {a x^2}{e}} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 a d e^2}\\ &=\frac {5 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d e^{3/2}}+\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{a^2 d}\\ &=\frac {5 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d e^{3/2}}+\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}+\frac {\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 e}+\frac {\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 e}\\ &=\frac {5 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d e^{3/2}}+\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}-\frac {\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 e^{3/2}}-\frac {\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 e^{3/2}}+\frac {\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 e}+\frac {\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 e}\\ &=\frac {5 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d e^{3/2}}+\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}-\frac {\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 e^{3/2}}+\frac {\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 e^{3/2}}+\frac {\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 e^{3/2}}-\frac {\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 e^{3/2}}\\ &=\frac {5 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d e^{3/2}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d e^{3/2}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d e^{3/2}}+\frac {5}{2 a^2 d e \sqrt {e \cot (c+d x)}}-\frac {1}{2 d e \sqrt {e \cot (c+d x)} \left (a^2+a^2 \cot (c+d x)\right )}-\frac {\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 e^{3/2}}+\frac {\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 e^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.33, size = 203, normalized size = 0.66 \[ \frac {\cot ^{\frac {3}{2}}(c+d x) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\log \left (-\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}-1\right )}{\sqrt {2}}-\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )+\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+10 \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )+\frac {2 (4 \sin (c+d x)+5 \cos (c+d x))}{\sqrt {\cot (c+d x)} (\sin (c+d x)+\cos (c+d x))}\right )}{4 a^2 d (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cot[c + d*x])^(3/2)*(a + a*Cot[c + d*x])^2),x]

[Out]

(Cot[c + d*x]^(3/2)*(-(Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]) + Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c
+ d*x]]] + 10*ArcTan[Sqrt[Cot[c + d*x]]] + (-Log[-1 + Sqrt[2]*Sqrt[Cot[c + d*x]] - Cot[c + d*x]] + Log[1 + Sq
rt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/Sqrt[2] + (2*(5*Cos[c + d*x] + 4*Sin[c + d*x]))/(Sqrt[Cot[c + d*x]]*
(Cos[c + d*x] + Sin[c + d*x]))))/(4*a^2*d*(e*Cot[c + d*x])^(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.65, size = 255, normalized size = 0.83 \[ \frac {\sqrt {e \cot \left (d x +c \right )}}{2 d \,a^{2} e \left (e \cot \left (d x +c \right )+e \right )}+\frac {5 \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2 a^{2} d \,e^{\frac {3}{2}}}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d \,a^{2} e^{2}}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d \,a^{2} e^{2}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d \,a^{2} e^{2}}+\frac {2}{a^{2} d e \sqrt {e \cot \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.87, size = 256, normalized size = 0.84 \[ \frac {e {\left (\frac {4 \, {\left (4 \, e + \frac {5 \, e}{\tan \left (d x + c\right )}\right )}}{a^{2} e^{3} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + a^{2} e^{2} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}}} + \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{e^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{e^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{e^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{e^{\frac {3}{2}}}}{a^{2} e} + \frac {20 \, \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a^{2} e^{\frac {5}{2}}}\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*e*(4*(4*e + 5*e/tan(d*x + c))/(a^2*e^3*sqrt(e/tan(d*x + c)) + a^2*e^2*(e/tan(d*x + c))^(3/2)) + (2*sqrt(2)
*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x + c)))/sqrt(e))/e^(3/2) + 2*sqrt(2)*arctan(-1/2*sqrt(2
)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(d*x + c)))/sqrt(e))/e^(3/2) + sqrt(2)*log(sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c
)) + e + e/tan(d*x + c))/e^(3/2) - sqrt(2)*log(-sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/e^(
3/2))/(a^2*e) + 20*arctan(sqrt(e/tan(d*x + c))/sqrt(e))/(a^2*e^(5/2)))/d

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mupad [B]  time = 0.92, size = 414, normalized size = 1.35 \[ \frac {\frac {5\,\mathrm {cot}\left (c+d\,x\right )}{2}+2}{a^2\,d\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}+a^2\,d\,e\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}-\frac {\mathrm {atan}\left (\frac {2048\,a^{10}\,d^5\,e^{13}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {1}{a^8\,d^4\,e^6}\right )}^{1/4}}{51200\,a^8\,d^4\,e^{12}-2048\,a^{12}\,d^6\,e^{15}\,\sqrt {-\frac {1}{a^8\,d^4\,e^6}}}+\frac {51200\,a^{14}\,d^7\,e^{16}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {1}{a^8\,d^4\,e^6}\right )}^{3/4}}{51200\,a^8\,d^4\,e^{12}-2048\,a^{12}\,d^6\,e^{15}\,\sqrt {-\frac {1}{a^8\,d^4\,e^6}}}\right )\,{\left (-\frac {1}{a^8\,d^4\,e^6}\right )}^{1/4}}{2}-\mathrm {atan}\left (\frac {a^{10}\,d^5\,e^{13}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^4\,e^6}\right )}^{1/4}\,8192{}\mathrm {i}}{51200\,a^8\,d^4\,e^{12}+32768\,a^{12}\,d^6\,e^{15}\,\sqrt {-\frac {1}{256\,a^8\,d^4\,e^6}}}-\frac {a^{14}\,d^7\,e^{16}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^4\,e^6}\right )}^{3/4}\,3276800{}\mathrm {i}}{51200\,a^8\,d^4\,e^{12}+32768\,a^{12}\,d^6\,e^{15}\,\sqrt {-\frac {1}{256\,a^8\,d^4\,e^6}}}\right )\,{\left (-\frac {1}{256\,a^8\,d^4\,e^6}\right )}^{1/4}\,2{}\mathrm {i}+\frac {\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,\sqrt {-e^3}\,1{}\mathrm {i}}{e^2}\right )\,\sqrt {-e^3}\,5{}\mathrm {i}}{2\,a^2\,d\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*cot(c + d*x))/2 + 2)/(a^2*d*(e*cot(c + d*x))^(3/2) + a^2*d*e*(e*cot(c + d*x))^(1/2)) - (atan((2048*a^10*d^
5*e^13*(e*cot(c + d*x))^(1/2)*(-1/(a^8*d^4*e^6))^(1/4))/(51200*a^8*d^4*e^12 - 2048*a^12*d^6*e^15*(-1/(a^8*d^4*
e^6))^(1/2)) + (51200*a^14*d^7*e^16*(e*cot(c + d*x))^(1/2)*(-1/(a^8*d^4*e^6))^(3/4))/(51200*a^8*d^4*e^12 - 204
8*a^12*d^6*e^15*(-1/(a^8*d^4*e^6))^(1/2)))*(-1/(a^8*d^4*e^6))^(1/4))/2 - atan((a^10*d^5*e^13*(e*cot(c + d*x))^
(1/2)*(-1/(256*a^8*d^4*e^6))^(1/4)*8192i)/(51200*a^8*d^4*e^12 + 32768*a^12*d^6*e^15*(-1/(256*a^8*d^4*e^6))^(1/
2)) - (a^14*d^7*e^16*(e*cot(c + d*x))^(1/2)*(-1/(256*a^8*d^4*e^6))^(3/4)*3276800i)/(51200*a^8*d^4*e^12 + 32768
*a^12*d^6*e^15*(-1/(256*a^8*d^4*e^6))^(1/2)))*(-1/(256*a^8*d^4*e^6))^(1/4)*2i + (atan(((e*cot(c + d*x))^(1/2)*
(-e^3)^(1/2)*1i)/e^2)*(-e^3)^(1/2)*5i)/(2*a^2*d*e^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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