3.837 \(\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=385 \[ -\frac {b \sqrt {\cot (c+d x)}}{2 d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} d \left (a^2+b^2\right )^3} \]

[Out]

1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/2*(a+b)*(a^2-4*a*b+b^2)*
arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/4*(a-b)*(a^2+4*a*b+b^2)*ln(1+cot(d*x+c)-2^(1/2)*cot
(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)-1/4*(a-b)*(a^2+4*a*b+b^2)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+
b^2)^3/d*2^(1/2)-1/4*(3*a^4-26*a^2*b^2+3*b^4)*arctan(a^(1/2)*cot(d*x+c)^(1/2)/b^(1/2))/(a^2+b^2)^3/d/a^(1/2)/b
^(1/2)-1/2*b*cot(d*x+c)^(1/2)/(a^2+b^2)/d/(b+a*cot(d*x+c))^2+1/4*(5*a^2-3*b^2)*cot(d*x+c)^(1/2)/(a^2+b^2)^2/d/
(b+a*cot(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.89, antiderivative size = 385, normalized size of antiderivative = 1.00, 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, 3567, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac {b \sqrt {\cot (c+d x)}}{2 d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\left (-26 a^2 b^2+3 a^4+3 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(3/2)*(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) - ((3*a^4 - 26*a^2*b^2 +
3*b^4)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(4*Sqrt[a]*Sqrt[b]*(a^2 + b^2)^3*d) - (b*Sqrt[Cot[c + d*x
]])/(2*(a^2 + b^2)*d*(b + a*Cot[c + d*x])^2) + ((5*a^2 - 3*b^2)*Sqrt[Cot[c + d*x]])/(4*(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 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 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 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 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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^3} \, dx &=\int \frac {\cot ^{\frac {3}{2}}(c+d x)}{(b+a \cot (c+d x))^3} \, dx\\ &=-\frac {b \sqrt {\cot (c+d x)}}{2 \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac {\int \frac {\frac {b}{2}-2 a \cot (c+d x)-\frac {3}{2} b \cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))^2} \, dx}{2 \left (a^2+b^2\right )}\\ &=-\frac {b \sqrt {\cot (c+d x)}}{2 \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\int \frac {\frac {1}{4} b \left (3 a^2-5 b^2\right )+4 a b^2 \cot (c+d x)-\frac {1}{4} b \left (5 a^2-3 b^2\right ) \cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{2 b \left (a^2+b^2\right )^2}\\ &=-\frac {b \sqrt {\cot (c+d x)}}{2 \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\int \frac {2 b^2 \left (3 a^2-b^2\right )-2 a b \left (a^2-3 b^2\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{2 b \left (a^2+b^2\right )^3}+\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \int \frac {1+\cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{8 \left (a^2+b^2\right )^3}\\ &=-\frac {b \sqrt {\cot (c+d x)}}{2 \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\operatorname {Subst}\left (\int \frac {-2 b^2 \left (3 a^2-b^2\right )+2 a b \left (a^2-3 b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{b \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d}\\ &=-\frac {b \sqrt {\cot (c+d x)}}{2 \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 \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 (3 a^4-26 a^2 b^2+3 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 \left (a^2+b^2\right )^3 d}\\ &=-\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} \left (a^2+b^2\right )^3 d}-\frac {b \sqrt {\cot (c+d x)}}{2 \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 \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 {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} \left (a^2+b^2\right )^3 d}-\frac {b \sqrt {\cot (c+d x)}}{2 \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 \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 {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} \left (a^2+b^2\right )^3 d}-\frac {b \sqrt {\cot (c+d x)}}{2 \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {\left (5 a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{4 \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*}

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Mathematica [C]  time = 6.19, size = 492, normalized size = 1.28 \[ -\frac {\frac {4 a^2 \cot ^{\frac {5}{2}}(c+d x) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {a \cot (c+d x)}{b}\right )}{5 b \left (a^2+b^2\right )^2}+\frac {2 a \left (a^2-3 b^2\right ) \left (\cot ^{\frac {3}{2}}(c+d x)-\cot ^{\frac {3}{2}}(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )\right )}{3 \left (a^2+b^2\right )^3}-\frac {2 a \left (a^2-3 b^2\right ) \cot ^{\frac {3}{2}}(c+d x)}{3 \left (a^2+b^2\right )^3}-\frac {\frac {2 a^2 \cot ^2(c+d x)}{(a \cot (c+d x)+b)^2}+\frac {3 a \cot (c+d x)}{a \cot (c+d x)+b}-\frac {3 \sqrt {a} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b}}}{4 a \left (a^2+b^2\right ) \sqrt {\cot (c+d x)}}+\frac {2 b \left (a^2-3 b^2\right ) \left (\sqrt {\cot (c+d x)}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a}}\right )}{\left (a^2+b^2\right )^3}-\frac {b \left (3 a^2-b^2\right ) \left (8 \sqrt {\cot (c+d x)}+\sqrt {2} \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\sqrt {2} \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \left (\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )\right )}{4 \left (a^2+b^2\right )^3}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*b*(a^2 - 3*b^2)*(-((Sqrt[b]*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/Sqrt[a]) + Sqrt[Cot[c + d*x]])
)/(a^2 + b^2)^3 - (2*a*(a^2 - 3*b^2)*Cot[c + d*x]^(3/2))/(3*(a^2 + b^2)^3) - ((-3*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt
[Cot[c + d*x]])/Sqrt[b]]*Sqrt[Cot[c + d*x]])/Sqrt[b] + (2*a^2*Cot[c + d*x]^2)/(b + a*Cot[c + d*x])^2 + (3*a*Co
t[c + d*x])/(b + a*Cot[c + d*x]))/(4*a*(a^2 + b^2)*Sqrt[Cot[c + d*x]]) + (2*a*(a^2 - 3*b^2)*(Cot[c + d*x]^(3/2
) - Cot[c + d*x]^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2]))/(3*(a^2 + b^2)^3) + (4*a^2*Cot[c + d*
x]^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((a*Cot[c + d*x])/b)])/(5*b*(a^2 + b^2)^2) - (b*(3*a^2 - b^2)*(2*(Sqr
t[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]) + 8*Sqrt[Cot[c +
d*x]] + Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x
]] + Cot[c + d*x]]))/(4*(a^2 + b^2)^3))/d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.58, size = 50004, normalized size = 129.88 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [A]  time = 1.03, size = 409, normalized size = 1.06 \[ -\frac {\frac {{\left (3 \, a^{4} - 26 \, a^{2} b^{2} + 3 \, b^{4}\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {\frac {3 \, a^{2} b - 5 \, b^{3}}{\sqrt {\tan \left (d x + c\right )}} + \frac {5 \, a^{3} - 3 \, a b^{2}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6} + \frac {2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )}}{\tan \left (d x + c\right )} + \frac {a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}}{\tan \left (d x + c\right )^{2}}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((3*a^4 - 26*a^2*b^2 + 3*b^4)*arctan(a/(sqrt(a*b)*sqrt(tan(d*x + c))))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^
6)*sqrt(a*b)) - (2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))
) + 2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - sqrt(2)*
(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*(a^3 + 3*a^2*b
- 3*a*b^2 - b^3)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) -
((3*a^2*b - 5*b^3)/sqrt(tan(d*x + c)) + (5*a^3 - 3*a*b^2)/tan(d*x + c)^(3/2))/(a^4*b^2 + 2*a^2*b^4 + b^6 + 2*(
a^5*b + 2*a^3*b^3 + a*b^5)/tan(d*x + c) + (a^6 + 2*a^4*b^2 + a^2*b^4)/tan(d*x + c)^2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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