Integrand size = 35, antiderivative size = 320 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\sqrt {b} \left (20 a A b+15 a^2 B-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{4 d}+\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b (4 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}} \]
(I*a-b)^(5/2)*(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c ))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+(I*a+b)^(5/2)*(I*A+B)*arctan h((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)* tan(d*x+c)^(1/2)/d+1/4*(20*A*a*b+15*B*a^2-8*B*b^2)*arctanh(b^(1/2)*tan(d*x +c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*b^(1/2)*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2 )/d+1/4*b*(4*A*b+7*B*a)*(a+b*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)+1/2*b*B* (a+b*tan(d*x+c))^(3/2)/d/cot(d*x+c)^(1/2)
Time = 3.77 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.97 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (4 \sqrt [4]{-1} (-a+i b)^{5/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-4 (-1)^{3/4} (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b (4 A b+7 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+\frac {\sqrt {a} \sqrt {b} \left (20 a A b+15 a^2 B-8 b^2 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 d} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(4*(-1)^(1/4)*(-a + I*b)^(5/2)*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan [c + d*x]]] - 4*(-1)^(3/4)*(a + I*b)^(5/2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sq rt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + b*(4*A*b + 7*a *B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + 2*b*B*Sqrt[Tan[c + d*x]] *(a + b*Tan[c + d*x])^(3/2) + (Sqrt[a]*Sqrt[b]*(20*a*A*b + 15*a^2*B - 8*b^ 2*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x ])/a])/Sqrt[a + b*Tan[c + d*x]]))/(4*d)
Time = 1.76 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.87, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4729, 3042, 4090, 27, 3042, 4130, 27, 3042, 4138, 2035, 2257, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4090 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (4 A b+7 a B) \tan ^2(c+d x)+4 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (4 a A-b B)\right )}{2 \sqrt {\tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (4 A b+7 a B) \tan ^2(c+d x)+4 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (4 a A-b B)\right )}{\sqrt {\tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (4 A b+7 a B) \tan (c+d x)^2+4 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (4 a A-b B)\right )}{\sqrt {\tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan (c+d x)^2+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {\int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {\int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {\int \left (\frac {b \left (15 B a^2+20 A b a-8 b^2 B\right )}{\sqrt {a+b \tan (c+d x)}}+\frac {8 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\sqrt {b} \left (15 a^2 B+20 a A b-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 (-b+i a)^{5/2} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 (b+i a)^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((b*B*Sqrt[Tan[c + d*x]]*(a + b*Tan[ c + d*x])^(3/2))/(2*d) + ((4*(I*a - b)^(5/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + Sqrt[b]*(20*a*A*b + 15* a^2*B - 8*b^2*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d *x]]] + 4*(I*a + b)^(5/2)*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d* x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (b*(4*A*b + 7*a*B)*Sqrt[Tan[c + d*x]]* Sqrt[a + b*Tan[c + d*x]])/d)/4)
3.7.33.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b *(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 , 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. ) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C *(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f , A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(2818\) vs. \(2(264)=528\).
Time = 0.51 (sec) , antiderivative size = 2819, normalized size of antiderivative = 8.81
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2819\) |
| default | \(\text {Expression too large to display}\) | \(2819\) |
1/4/d*((b+a*cot(d*x+c))/cot(d*x+c))^(1/2)/cot(d*x+c)^(3/2)*(2*B*(2*(a^2+b^ 2)^(1/2)-2*b)^(1/2)*(b+a*cot(d*x+c))^(1/2)*b^(5/2)*a-8*A*arctan((2*(b+a*co t(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*b)^(1/2))/(2*(a^2+b^2)^(1/2)-2*b)^(1/ 2))*(a^2+b^2)^(1/2)*b^(3/2)*a^2*cot(d*x+c)^2-4*B*arctan(((2*(a^2+b^2)^(1/2 )+2*b)^(1/2)-2*(b+a*cot(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*b)^(1/2))*(a^2 +b^2)^(1/2)*b^(5/2)*a*cot(d*x+c)^2+4*B*arctan((2*(b+a*cot(d*x+c))^(1/2)+(2 *(a^2+b^2)^(1/2)+2*b)^(1/2))/(2*(a^2+b^2)^(1/2)-2*b)^(1/2))*(a^2+b^2)^(1/2 )*b^(5/2)*a*cot(d*x+c)^2-A*ln((b+a*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2* b)^(1/2)-a*cot(d*x+c)-b-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*b)^(1/2)*(2* (a^2+b^2)^(1/2)-2*b)^(1/2)*b^(7/2)*cot(d*x+c)^2+A*ln(a*cot(d*x+c)+b+(b+a*c ot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*b)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^ 2)^(1/2)+2*b)^(1/2)*(2*(a^2+b^2)^(1/2)-2*b)^(1/2)*b^(7/2)*cot(d*x+c)^2+8*A *arctan(((2*(a^2+b^2)^(1/2)+2*b)^(1/2)-2*(b+a*cot(d*x+c))^(1/2))/(2*(a^2+b ^2)^(1/2)-2*b)^(1/2))*(a^2+b^2)^(1/2)*b^(3/2)*a^2*cot(d*x+c)^2+4*A*a*b^(5/ 2)*(2*(a^2+b^2)^(1/2)-2*b)^(1/2)*(b+a*cot(d*x+c))^(1/2)*cot(d*x+c)+9*B*a^2 *b^(3/2)*(2*(a^2+b^2)^(1/2)-2*b)^(1/2)*(b+a*cot(d*x+c))^(1/2)*cot(d*x+c)+2 0*A*(2*(a^2+b^2)^(1/2)-2*b)^(1/2)*arctanh((b+a*cot(d*x+c))^(1/2)/b^(1/2))* a^2*b^2*cot(d*x+c)^2+4*B*arctan(((2*(a^2+b^2)^(1/2)+2*b)^(1/2)-2*(b+a*cot( d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*b)^(1/2))*(a^2+b^2)^(1/2)*b^(1/2)*a^3* cot(d*x+c)^2-4*B*arctan((2*(b+a*cot(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*...
Leaf count of result is larger than twice the leaf count of optimal. 17723 vs. \(2 (258) = 516\).
Time = 7.67 (sec) , antiderivative size = 35483, normalized size of antiderivative = 110.88 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\cot \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]