3.839 \(\int \frac{1}{\cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^3} \, dx\)

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

[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) - (a^(3/2)*(3*a^4 + 6*a^2*b^
2 + 35*b^4)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(4*b^(5/2)*(a^2 + b^2)^3*d) - (a^2*Sqrt[Cot[c + d*x]
])/(2*b*(a^2 + b^2)*d*(b + a*Cot[c + d*x])^2) - (a^2*(3*a^2 + 11*b^2)*Sqrt[Cot[c + d*x]])/(4*b^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 = 0.935857, 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, 3569, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{a^2 \left (3 a^2+11 b^2\right ) \sqrt{\cot (c+d x)}}{4 b^2 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{a^{3/2} \left (6 a^2 b^2+3 a^4+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 b^{5/2} d \left (a^2+b^2\right )^3}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(7/2)*(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) - (a^(3/2)*(3*a^4 + 6*a^2*b^
2 + 35*b^4)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(4*b^(5/2)*(a^2 + b^2)^3*d) - (a^2*Sqrt[Cot[c + d*x]
])/(2*b*(a^2 + b^2)*d*(b + a*Cot[c + d*x])^2) - (a^2*(3*a^2 + 11*b^2)*Sqrt[Cot[c + d*x]])/(4*b^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 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 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 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{1}{\cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^3} \, dx &=\int \frac{1}{\sqrt{\cot (c+d x)} (b+a \cot (c+d x))^3} \, dx\\ &=-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{\int \frac{\frac{1}{2} \left (-3 a^2-4 b^2\right )+2 a b \cot (c+d x)-\frac{3}{2} a^2 \cot ^2(c+d x)}{\sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{a^2 \left (3 a^2+11 b^2\right ) \sqrt{\cot (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{\int \frac{\frac{1}{4} \left (3 a^4+3 a^2 b^2+8 b^4\right )-4 a b^3 \cot (c+d x)+\frac{1}{4} a^2 \left (3 a^2+11 b^2\right ) \cot ^2(c+d x)}{\sqrt{\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{a^2 \left (3 a^2+11 b^2\right ) \sqrt{\cot (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{\int \frac{-2 b^3 \left (3 a^2-b^2\right )+2 a b^2 \left (a^2-3 b^2\right ) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{2 b^2 \left (a^2+b^2\right )^3}+\frac{\left (a^2 \left (3 a^4+6 a^2 b^2+35 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 b^2 \left (a^2+b^2\right )^3}\\ &=-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{a^2 \left (3 a^2+11 b^2\right ) \sqrt{\cot (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{\operatorname{Subst}\left (\int \frac{2 b^3 \left (3 a^2-b^2\right )-2 a b^2 \left (a^2-3 b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{b^2 \left (a^2+b^2\right )^3 d}+\frac{\left (a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{8 b^2 \left (a^2+b^2\right )^3 d}\\ &=-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{a^2 \left (3 a^2+11 b^2\right ) \sqrt{\cot (c+d x)}}{4 b^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 (a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{4 b^2 \left (a^2+b^2\right )^3 d}\\ &=-\frac{a^{3/2} \left (3 a^4+6 a^2 b^2+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 d}-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{a^2 \left (3 a^2+11 b^2\right ) \sqrt{\cot (c+d x)}}{4 b^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{a^{3/2} \left (3 a^4+6 a^2 b^2+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 d}-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{a^2 \left (3 a^2+11 b^2\right ) \sqrt{\cot (c+d x)}}{4 b^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{a^{3/2} \left (3 a^4+6 a^2 b^2+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 d}-\frac{a^2 \sqrt{\cot (c+d x)}}{2 b \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac{a^2 \left (3 a^2+11 b^2\right ) \sqrt{\cot (c+d x)}}{4 b^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 = 6.13294, size = 387, normalized size = 0.98 \[ -\frac{\frac{2 a \left (a^2-3 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\cot ^2(c+d x)\right )}{3 \left (a^2+b^2\right )^3}+\frac{2 a^2 \sqrt{\cot (c+d x)} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};-\frac{a \cot (c+d x)}{b}\right )}{b^3 \left (a^2+b^2\right )}+\frac{2 a^2 \sqrt{\cot (c+d x)} \left (\frac{b}{a \cot (c+d x)+b}+\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{\cot (c+d x)}}\right )}{b \left (a^2+b^2\right )^2}-\frac{2 a^{3/2} \left (a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{\sqrt{b} \left (a^2+b^2\right )^3}+\frac{b \left (3 a^2-b^2\right ) \left (2 \sqrt{2} \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-2 \sqrt{2} \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )+4 \left (\sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-\sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )\right )}{8 \left (a^2+b^2\right )^3}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^3),x]

[Out]

-(((-2*a^(3/2)*(a^2 - 3*b^2)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(Sqrt[b]*(a^2 + b^2)^3) + (2*a^2*Sq
rt[Cot[c + d*x]]*((Sqrt[b]*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(Sqrt[a]*Sqrt[Cot[c + d*x]]) + b/(b +
a*Cot[c + d*x])))/(b*(a^2 + b^2)^2) + (2*a^2*Sqrt[Cot[c + d*x]]*Hypergeometric2F1[1/2, 3, 3/2, -((a*Cot[c + d
*x])/b)])/(b^3*(a^2 + b^2)) + (2*a*(a^2 - 3*b^2)*Cot[c + d*x]^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*
x]^2])/(3*(a^2 + b^2)^3) + (b*(3*a^2 - b^2)*(4*(Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - Sqrt[2]*ArcTa
n[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]]))/(8*(a^2 + b^2)^3))/d)

________________________________________________________________________________________

Maple [C]  time = 0.809, size = 50764, normalized size = 128.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x)

[Out]

result too large to display

________________________________________________________________________________________

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(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [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(1/cot(d*x+c)**(7/2)/(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((b*tan(d*x + c) + a)^3*cot(d*x + c)^(7/2)), x)