Integrand size = 35, antiderivative size = 316 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {(i a-b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 \sqrt {b} d}+\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d} \]
-(I*a-b)^(5/2)*(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+ c))^(1/2))/d+(I*a+b)^(5/2)*(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/ (a+b*tan(d*x+c))^(1/2))/d+1/8*(30*A*a^2*b-16*A*b^3+5*B*a^3-40*B*a*b^2)*arc tanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/b^(1/2)+1/8*(14*A* a*b+5*B*a^2-8*B*b^2)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d+1/4*(2*A*b+ 3*B*a)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2)/d+1/3*b*B*tan(d*x+c)^(3/2)* (a+b*tan(d*x+c))^(3/2)/d
Time = 4.83 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.09 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {24 \sqrt [4]{-1} (-a+i b)^{5/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+24 \sqrt [4]{-1} (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 \left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+6 (2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}+\frac {3 \sqrt {a} \left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{24 d} \]
(24*(-1)^(1/4)*(-a + I*b)^(5/2)*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b ]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 24*(-1)^(1/4)*(a + I*b)^ (5/2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[ a + b*Tan[c + d*x]]] + 3*(14*a*A*b + 5*a^2*B - 8*b^2*B)*Sqrt[Tan[c + d*x]] *Sqrt[a + b*Tan[c + d*x]] + 6*(2*A*b + 3*a*B)*Sqrt[Tan[c + d*x]]*(a + b*Ta n[c + d*x])^(3/2) + 8*b*B*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2) + (3*Sqrt[a]*(30*a^2*A*b - 16*A*b^3 + 5*a^3*B - 40*a*b^2*B)*ArcSinh[(Sqrt[b] *Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[ a + b*Tan[c + d*x]]))/(24*d)
Time = 2.04 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4090, 27, 3042, 4130, 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.
| \(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4090 |
| \(\displaystyle \frac {1}{3} \int \frac {3}{2} \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (b (2 A b+3 a B) \tan ^2(c+d x)+2 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (2 a A-b B)\right )dx+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (b (2 A b+3 a B) \tan ^2(c+d x)+2 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (2 a A-b B)\right )dx+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (b (2 A b+3 a B) \tan (c+d x)^2+2 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (2 a A-b B)\right )dx+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {\sqrt {a+b \tan (c+d x)} \left (-b \left (5 B a^2+14 A b a-8 b^2 B\right ) \tan ^2(c+d x)-8 b \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a b (2 A b+3 a B)\right )}{2 \sqrt {\tan (c+d x)}}dx}{2 b}+\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {\int \frac {\sqrt {a+b \tan (c+d x)} \left (-b \left (5 B a^2+14 A b a-8 b^2 B\right ) \tan ^2(c+d x)-8 b \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a b (2 A b+3 a B)\right )}{\sqrt {\tan (c+d x)}}dx}{4 b}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {\int \frac {\sqrt {a+b \tan (c+d x)} \left (-b \left (5 B a^2+14 A b a-8 b^2 B\right ) \tan (c+d x)^2-8 b \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a b (2 A b+3 a B)\right )}{\sqrt {\tan (c+d x)}}dx}{4 b}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {\int \frac {-b \left (5 B a^3+30 A b a^2-40 b^2 B a-16 A b^3\right ) \tan ^2(c+d x)-16 b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a b \left (11 B a^2+18 A b a-8 b^2 B\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {b \left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}}{4 b}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {\frac {1}{2} \int \frac {-b \left (5 B a^3+30 A b a^2-40 b^2 B a-16 A b^3\right ) \tan ^2(c+d x)-16 b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a b \left (11 B a^2+18 A b a-8 b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {b \left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}}{4 b}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {\frac {1}{2} \int \frac {-b \left (5 B a^3+30 A b a^2-40 b^2 B a-16 A b^3\right ) \tan (c+d x)^2-16 b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a b \left (11 B a^2+18 A b a-8 b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {b \left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}}{4 b}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {\frac {\int \frac {-b \left (5 B a^3+30 A b a^2-40 b^2 B a-16 A b^3\right ) \tan ^2(c+d x)-16 b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a b \left (11 B a^2+18 A b a-8 b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}-\frac {b \left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}}{4 b}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {\frac {\int \frac {-b \left (5 B a^3+30 A b a^2-40 b^2 B a-16 A b^3\right ) \tan ^2(c+d x)-16 b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a b \left (11 B a^2+18 A b a-8 b^2 B\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-\frac {b \left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}}{4 b}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {\frac {\int \left (\frac {16 \left (b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right )-b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}-\frac {b \left (5 B a^3+30 A b a^2-40 b^2 B a-16 A b^3\right )}{\sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{d}-\frac {b \left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}}{4 b}\right )+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {-\frac {b \left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {-\sqrt {b} \left (5 a^3 B+30 a^2 A b-40 a b^2 B-16 A b^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+8 b (-b+i a)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-8 b (b+i a)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}}{4 b}\right )\) |
(b*B*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2))/(3*d) + (((2*A*b + 3*a *B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*d) - ((8*(I*a - b)^( 5/2)*b*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[ c + d*x]]] - Sqrt[b]*(30*a^2*A*b - 16*A*b^3 + 5*a^3*B - 40*a*b^2*B)*ArcTan h[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - 8*b*(I*a + b)^( 5/2)*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (b*(14*a*A*b + 5*a^2*B - 8*b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d)/(4*b))/2
3.5.43.3.1 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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.04 (sec) , antiderivative size = 2657119, normalized size of antiderivative = 8408.60
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 17621 vs. \(2 (260) = 520\).
Time = 7.58 (sec) , antiderivative size = 35244, normalized size of antiderivative = 111.53 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
Timed out. \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
\[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\tan \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]