3.619 \(\int \cot ^{\frac{11}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=422 \[ \frac{2 \left (21 a^2 A-24 a b B-A b^2\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}+\frac{2 \left (126 a^2 A b+105 a^3 B-9 a b^2 B+4 A b^3\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{315 a^2 d}-\frac{2 \left (-63 a^2 A b^2+315 a^4 A-420 a^3 b B-18 a b^3 B+8 A b^4\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{315 a^3 d}-\frac{2 (9 a B+10 A b) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}+\frac{(-b+i a)^{3/2} (-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{(b+i a)^{3/2} (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d} \]

[Out]

((I*a - b)^(3/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*
x]]*Sqrt[Tan[c + d*x]])/d - ((I*a + b)^(3/2)*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*T
an[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - (2*(315*a^4*A - 63*a^2*A*b^2 + 8*A*b^4 - 420*a^3*b*B
- 18*a*b^3*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(315*a^3*d) + (2*(126*a^2*A*b + 4*A*b^3 + 105*a^3*B
- 9*a*b^2*B)*Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(315*a^2*d) + (2*(21*a^2*A - A*b^2 - 24*a*b*B)*Cot[
c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]])/(105*a*d) - (2*(10*A*b + 9*a*B)*Cot[c + d*x]^(7/2)*Sqrt[a + b*Tan[c +
d*x]])/(63*d) - (2*a*A*Cot[c + d*x]^(9/2)*Sqrt[a + b*Tan[c + d*x]])/(9*d)

________________________________________________________________________________________

Rubi [A]  time = 2.02452, antiderivative size = 422, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4241, 3605, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (21 a^2 A-24 a b B-A b^2\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}+\frac{2 \left (126 a^2 A b+105 a^3 B-9 a b^2 B+4 A b^3\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{315 a^2 d}-\frac{2 \left (-63 a^2 A b^2+315 a^4 A-420 a^3 b B-18 a b^3 B+8 A b^4\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{315 a^3 d}-\frac{2 (9 a B+10 A b) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}+\frac{(-b+i a)^{3/2} (-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{(b+i a)^{3/2} (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(11/2)*(a + b*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

((I*a - b)^(3/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*
x]]*Sqrt[Tan[c + d*x]])/d - ((I*a + b)^(3/2)*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*T
an[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - (2*(315*a^4*A - 63*a^2*A*b^2 + 8*A*b^4 - 420*a^3*b*B
- 18*a*b^3*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(315*a^3*d) + (2*(126*a^2*A*b + 4*A*b^3 + 105*a^3*B
- 9*a*b^2*B)*Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(315*a^2*d) + (2*(21*a^2*A - A*b^2 - 24*a*b*B)*Cot[
c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]])/(105*a*d) - (2*(10*A*b + 9*a*B)*Cot[c + d*x]^(7/2)*Sqrt[a + b*Tan[c +
d*x]])/(63*d) - (2*a*A*Cot[c + d*x]^(9/2)*Sqrt[a + b*Tan[c + d*x]])/(9*d)

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 3605

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(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)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

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 3616

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]

Rule 3615

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[A^2/f, Subst[Int[((a + b*x)^m*(c + d*x)^n)/(A - B*x), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 203

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

Rule 206

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

Rubi steps

\begin{align*} \int \cot ^{\frac{11}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{11}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}+\frac{1}{9} \left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{1}{2} a (10 A b+9 a B)-\frac{9}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac{1}{2} b (8 a A-9 b B) \tan ^2(c+d x)}{\tan ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 (10 A b+9 a B) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}-\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{3}{4} a \left (21 a^2 A-A b^2-24 a b B\right )+\frac{63}{4} a \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{3}{2} a b (10 A b+9 a B) \tan ^2(c+d x)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{63 a}\\ &=\frac{2 \left (21 a^2 A-A b^2-24 a b B\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 (10 A b+9 a B) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}+\frac{\left (8 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{-\frac{3}{8} a \left (126 a^2 A b+4 A b^3+105 a^3 B-9 a b^2 B\right )+\frac{315}{8} a^2 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac{3}{2} a b \left (21 a^2 A-A b^2-24 a b B\right ) \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{315 a^2}\\ &=\frac{2 \left (126 a^2 A b+4 A b^3+105 a^3 B-9 a b^2 B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{315 a^2 d}+\frac{2 \left (21 a^2 A-A b^2-24 a b B\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 (10 A b+9 a B) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}-\frac{\left (16 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{-\frac{3}{16} a \left (315 a^4 A-63 a^2 A b^2+8 A b^4-420 a^3 b B-18 a b^3 B\right )-\frac{945}{16} a^3 \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)-\frac{3}{8} a b \left (126 a^2 A b+4 A b^3+105 a^3 B-9 a b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{945 a^3}\\ &=-\frac{2 \left (315 a^4 A-63 a^2 A b^2+8 A b^4-420 a^3 b B-18 a b^3 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{315 a^3 d}+\frac{2 \left (126 a^2 A b+4 A b^3+105 a^3 B-9 a b^2 B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{315 a^2 d}+\frac{2 \left (21 a^2 A-A b^2-24 a b B\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 (10 A b+9 a B) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}+\frac{\left (32 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{945}{32} a^4 \left (2 a A b+a^2 B-b^2 B\right )-\frac{945}{32} a^4 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{945 a^4}\\ &=-\frac{2 \left (315 a^4 A-63 a^2 A b^2+8 A b^4-420 a^3 b B-18 a b^3 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{315 a^3 d}+\frac{2 \left (126 a^2 A b+4 A b^3+105 a^3 B-9 a b^2 B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{315 a^2 d}+\frac{2 \left (21 a^2 A-A b^2-24 a b B\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 (10 A b+9 a B) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}-\frac{1}{2} \left ((a+i b)^2 (i A-B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} \left ((a-i b)^2 (i A+B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 \left (315 a^4 A-63 a^2 A b^2+8 A b^4-420 a^3 b B-18 a b^3 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{315 a^3 d}+\frac{2 \left (126 a^2 A b+4 A b^3+105 a^3 B-9 a b^2 B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{315 a^2 d}+\frac{2 \left (21 a^2 A-A b^2-24 a b B\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 (10 A b+9 a B) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}-\frac{\left ((a+i b)^2 (i A-B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left ((a-i b)^2 (i A+B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac{2 \left (315 a^4 A-63 a^2 A b^2+8 A b^4-420 a^3 b B-18 a b^3 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{315 a^3 d}+\frac{2 \left (126 a^2 A b+4 A b^3+105 a^3 B-9 a b^2 B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{315 a^2 d}+\frac{2 \left (21 a^2 A-A b^2-24 a b B\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 (10 A b+9 a B) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}-\frac{\left ((a+i b)^2 (i A-B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\left ((a-i b)^2 (i A+B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=-\frac{(a+i b)^2 (i A-B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{\sqrt{i a-b} d}-\frac{(i a+b)^{3/2} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}-\frac{2 \left (315 a^4 A-63 a^2 A b^2+8 A b^4-420 a^3 b B-18 a b^3 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{315 a^3 d}+\frac{2 \left (126 a^2 A b+4 A b^3+105 a^3 B-9 a b^2 B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{315 a^2 d}+\frac{2 \left (21 a^2 A-A b^2-24 a b B\right ) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 (10 A b+9 a B) \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{63 d}-\frac{2 a A \cot ^{\frac{9}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{9 d}\\ \end{align*}

Mathematica [A]  time = 6.59843, size = 495, normalized size = 1.17 \[ \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (-\frac{b B \sqrt{a+b \tan (c+d x)}}{4 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{1}{4} \left (-\frac{(8 a A-9 b B) \sqrt{a+b \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2 \left (-\frac{4 a (9 a B+10 A b) \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 \left (-\frac{6 a \left (21 a^2 A-24 a b B-A b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 \left (\frac{a \left (126 a^2 A b+105 a^3 B-9 a b^2 B+4 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \left (\frac{3 a \left (-63 a^2 A b^2+315 a^4 A-420 a^3 b B-18 a b^3 B+8 A b^4\right ) \sqrt{a+b \tan (c+d x)}}{2 d \sqrt{\tan (c+d x)}}+\frac{945 a^4 \left (\sqrt [4]{-1} (-a-i b)^{3/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-\sqrt [4]{-1} (a-i b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )\right )}{4 d}\right )}{3 a}\right )}{5 a}\right )}{7 a}\right )}{9 a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(11/2)*(a + b*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-(b*B*Sqrt[a + b*Tan[c + d*x]])/(4*d*Tan[c + d*x]^(9/2)) + (-((8*a*A -
9*b*B)*Sqrt[a + b*Tan[c + d*x]])/(9*d*Tan[c + d*x]^(9/2)) + (2*((-4*a*(10*A*b + 9*a*B)*Sqrt[a + b*Tan[c + d*x]
])/(7*d*Tan[c + d*x]^(7/2)) - (2*((-6*a*(21*a^2*A - A*b^2 - 24*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(5*d*Tan[c + d
*x]^(5/2)) - (2*((a*(126*a^2*A*b + 4*A*b^3 + 105*a^3*B - 9*a*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(d*Tan[c + d*x]^
(3/2)) - (2*((945*a^4*((-1)^(1/4)*(-a - I*b)^(3/2)*(A + I*B)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d
*x]])/Sqrt[a + b*Tan[c + d*x]]] - (-1)^(1/4)*(a - I*b)^(3/2)*(A - I*B)*ArcTanh[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[
Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]))/(4*d) + (3*a*(315*a^4*A - 63*a^2*A*b^2 + 8*A*b^4 - 420*a^3*b*B - 18
*a*b^3*B)*Sqrt[a + b*Tan[c + d*x]])/(2*d*Sqrt[Tan[c + d*x]])))/(3*a)))/(5*a)))/(7*a)))/(9*a))/4)

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Maple [C]  time = 2.586, size = 74462, normalized size = 176.5 \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)^(11/2)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cot \left (d x + c\right )^{\frac{11}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(11/2)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(3/2)*cot(d*x + c)^(11/2), x)

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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)^(11/2)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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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(cot(d*x+c)**(11/2)*(a+b*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [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)^(11/2)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out