3.835 \(\int \frac{\sqrt{\cot (c+d x)}}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=396 \[ \frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{b^{3/2} \left (6 a^2 b^2+35 a^4+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 a^{5/2} d \left (a^2+b^2\right )^3}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3} \]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((a + b)*(a^2
- 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - (b^(3/2)*(35*a^4 + 6*a^2*b
^2 + 3*b^4)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(4*a^(5/2)*(a^2 + b^2)^3*d) + (b^2*Cot[c + d*x]^(3/2
))/(2*a*(a^2 + b^2)*d*(b + a*Cot[c + d*x])^2) + (b^2*(11*a^2 + 3*b^2)*Sqrt[Cot[c + d*x]])/(4*a^2*(a^2 + b^2)^2
*d*(b + a*Cot[c + d*x])) - ((a - b)*(a^2 + 4*a*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2
*Sqrt[2]*(a^2 + b^2)^3*d) + ((a - b)*(a^2 + 4*a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(
2*Sqrt[2]*(a^2 + b^2)^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.925871, antiderivative size = 396, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.609, Rules used = {3673, 3565, 3645, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{b^{3/2} \left (6 a^2 b^2+35 a^4+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 a^{5/2} d \left (a^2+b^2\right )^3}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cot[c + d*x]]/(a + b*Tan[c + d*x])^3,x]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((a + b)*(a^2
- 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - (b^(3/2)*(35*a^4 + 6*a^2*b
^2 + 3*b^4)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(4*a^(5/2)*(a^2 + b^2)^3*d) + (b^2*Cot[c + d*x]^(3/2
))/(2*a*(a^2 + b^2)*d*(b + a*Cot[c + d*x])^2) + (b^2*(11*a^2 + 3*b^2)*Sqrt[Cot[c + d*x]])/(4*a^2*(a^2 + b^2)^2
*d*(b + a*Cot[c + d*x])) - ((a - b)*(a^2 + 4*a*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2
*Sqrt[2]*(a^2 + b^2)^3*d) + ((a - b)*(a^2 + 4*a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(
2*Sqrt[2]*(a^2 + b^2)^3*d)

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*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] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3645

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*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

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]

Rule 3534

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]

Rule 1168

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 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 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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cot (c+d x)}}{(a+b \tan (c+d x))^3} \, dx &=\int \frac{\cot ^{\frac{7}{2}}(c+d x)}{(b+a \cot (c+d x))^3} \, dx\\ &=\frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{\int \frac{\sqrt{\cot (c+d x)} \left (-\frac{3 b^2}{2}+2 a b \cot (c+d x)-\frac{1}{2} \left (4 a^2+3 b^2\right ) \cot ^2(c+d x)\right )}{(b+a \cot (c+d x))^2} \, dx}{2 a \left (a^2+b^2\right )}\\ &=\frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{\int \frac{\frac{1}{4} b^2 \left (11 a^2+3 b^2\right )-4 a^3 b \cot (c+d x)+\frac{1}{4} \left (8 a^4+3 a^2 b^2+3 b^4\right ) \cot ^2(c+d x)}{\sqrt{\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{2 a^2 \left (a^2+b^2\right )^2}\\ &=\frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{\int \frac{-2 a^2 b \left (3 a^2-b^2\right )+2 a^3 \left (a^2-3 b^2\right ) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{2 a^2 \left (a^2+b^2\right )^3}+\frac{\left (b^2 \left (35 a^4+6 a^2 b^2+3 b^4\right )\right ) \int \frac{1+\cot ^2(c+d x)}{\sqrt{\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{8 a^2 \left (a^2+b^2\right )^3}\\ &=\frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{\operatorname{Subst}\left (\int \frac{2 a^2 b \left (3 a^2-b^2\right )-2 a^3 \left (a^2-3 b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 \left (a^2+b^2\right )^3 d}+\frac{\left (b^2 \left (35 a^4+6 a^2 b^2+3 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{8 a^2 \left (a^2+b^2\right )^3 d}\\ &=\frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}+\frac{\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}-\frac{\left (b^2 \left (35 a^4+6 a^2 b^2+3 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{4 a^2 \left (a^2+b^2\right )^3 d}\\ &=-\frac{b^{3/2} \left (35 a^4+6 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}+\frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}\\ &=-\frac{b^{3/2} \left (35 a^4+6 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}+\frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}\\ &=\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{b^{3/2} \left (35 a^4+6 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}+\frac{b^2 \cot ^{\frac{3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}\\ \end{align*}

Mathematica [C]  time = 6.16792, size = 499, normalized size = 1.26 \[ -\frac{\frac{4 a^2 \cot ^{\frac{9}{2}}(c+d x) \, _2F_1\left (2,\frac{9}{2};\frac{11}{2};-\frac{a \cot (c+d x)}{b}\right )}{9 b \left (a^2+b^2\right )^2}+\frac{2 a^2 \cot ^{\frac{9}{2}}(c+d x) \, _2F_1\left (3,\frac{9}{2};\frac{11}{2};-\frac{a \cot (c+d x)}{b}\right )}{9 b^3 \left (a^2+b^2\right )}-\frac{2 a \left (a^2-3 b^2\right ) \left (-7 \cot ^{\frac{3}{2}}(c+d x) \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\cot ^2(c+d x)\right )-3 \cot ^{\frac{7}{2}}(c+d x)+7 \cot ^{\frac{3}{2}}(c+d x)\right )}{21 \left (a^2+b^2\right )^3}-\frac{2 a \left (a^2-3 b^2\right ) \cot ^{\frac{7}{2}}(c+d x)}{7 \left (a^2+b^2\right )^3}+\frac{2 b \left (a^2-3 b^2\right ) \left (3 \cot ^{\frac{5}{2}}(c+d x)-5 b \left (\frac{\cot ^{\frac{3}{2}}(c+d x)}{a}-\frac{3 b \left (\frac{\sqrt{\cot (c+d x)}}{a}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{a^{3/2}}\right )}{a}\right )\right )}{15 \left (a^2+b^2\right )^3}+\frac{b \left (3 a^2-b^2\right ) \left (-8 \cot ^{\frac{5}{2}}(c+d x)+40 \sqrt{\cot (c+d x)}+\frac{5}{2} \left (2 \sqrt{2} \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-2 \sqrt{2} \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )+4 \left (\sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-\sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )\right )\right )}{20 \left (a^2+b^2\right )^3}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cot[c + d*x]]/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((-2*a*(a^2 - 3*b^2)*Cot[c + d*x]^(7/2))/(7*(a^2 + b^2)^3) + (2*b*(a^2 - 3*b^2)*(3*Cot[c + d*x]^(5/2) - 5*b*
((-3*b*(-((Sqrt[b]*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/a^(3/2)) + Sqrt[Cot[c + d*x]]/a))/a + Cot[c +
d*x]^(3/2)/a)))/(15*(a^2 + b^2)^3) - (2*a*(a^2 - 3*b^2)*(7*Cot[c + d*x]^(3/2) - 3*Cot[c + d*x]^(7/2) - 7*Cot[
c + d*x]^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2]))/(21*(a^2 + b^2)^3) + (4*a^2*Cot[c + d*x]^(9/2
)*Hypergeometric2F1[2, 9/2, 11/2, -((a*Cot[c + d*x])/b)])/(9*b*(a^2 + b^2)^2) + (2*a^2*Cot[c + d*x]^(9/2)*Hype
rgeometric2F1[3, 9/2, 11/2, -((a*Cot[c + d*x])/b)])/(9*b^3*(a^2 + b^2)) + (b*(3*a^2 - b^2)*(40*Sqrt[Cot[c + d*
x]] - 8*Cot[c + d*x]^(5/2) + (5*(4*(Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - Sqrt[2]*ArcTan[1 + Sqrt[2
]*Sqrt[Cot[c + d*x]]]) + 2*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - 2*Sqrt[2]*Log[1 + Sqrt
[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]))/2))/(20*(a^2 + b^2)^3))/d)

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Maple [C]  time = 1.323, size = 50668, normalized size = 128. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cot{\left (c + d x \right )}}}{\left (a + b \tan{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(cot(c + d*x))/(a + b*tan(c + d*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cot \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(cot(d*x + c))/(b*tan(d*x + c) + a)^3, x)