Integrand size = 35, antiderivative size = 244 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a -b)^(5/2)/d+(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c) )^(1/2))/(I*a+b)^(5/2)/d+2/3*(2*A*a^2*b-4*A*b^3+B*a^3+7*B*a*b^2)*tan(d*x+c )^(1/2)/b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)+2/3*a*(A*b-B*a)*tan(d*x+c)^ (1/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)
Time = 3.35 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.26 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {-\frac {3 B \sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}}+\frac {\left (2 a A b+a^2 B+3 b^2 B\right ) \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {3 \sqrt [4]{-1} b \left (\frac {(a+i b)^2 (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {i (a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{\left (a^2+b^2\right )^2}}{3 b d} \]
((-3*B*Sqrt[Tan[c + d*x]])/(a + b*Tan[c + d*x])^(3/2) + ((2*a*A*b + a^2*B + 3*b^2*B)*Sqrt[Tan[c + d*x]])/((a^2 + b^2)*(a + b*Tan[c + d*x])^(3/2)) + (3*(-1)^(1/4)*b*(((a + I*b)^2*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]* Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + (I*(a - I* b)^2*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b]) + (2*(2*a^2*A*b - 4*A*b^3 + a^3*B + 7* a*b^2*B)*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]])/(a^2 + b^2)^2)/(3*b *d)
Time = 1.51 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.20, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4088, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}
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.
| \(\displaystyle \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{3/2} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2 \int -\frac {-\left (\left (B a^2+2 A b a+3 b^2 B\right ) \tan ^2(c+d x)\right )-3 b (A b-a B) \tan (c+d x)+a (A b-a B)}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\int \frac {-\left (\left (B a^2+2 A b a+3 b^2 B\right ) \tan ^2(c+d x)\right )-3 b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\int \frac {-\left (\left (B a^2+2 A b a+3 b^2 B\right ) \tan (c+d x)^2\right )-3 b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 \int \frac {3 \left (a b \left (A a^2+2 b B a-A b^2\right )-a b \left (-B a^2+2 A b a+b^2 B\right ) \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {3 \int \frac {a b \left (A a^2+2 b B a-A b^2\right )-a b \left (-B a^2+2 A b a+b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {3 \int \frac {a b \left (A a^2+2 b B a-A b^2\right )-a b \left (-B a^2+2 A b a+b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {1}{2} a b (a-i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a b (a+i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {1}{2} a b (a-i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a b (a+i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a b (a-i b)^2 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {a b (a+i b)^2 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a b (a-i b)^2 (A+i B) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a b (a+i b)^2 (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a b (a+i b)^2 (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a b (a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a b (a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {a b (a+i b)^2 (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
(2*a*(A*b - a*B)*Sqrt[Tan[c + d*x]])/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x ])^(3/2)) - ((3*((a*(a - I*b)^2*b*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan [c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) + (a*(a + I*b)^2* b*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d)))/(a*(a^2 + b^2)) - (2*(2*a^2*A*b - 4*A*b^3 + a^ 3*B + 7*a*b^2*B)*Sqrt[Tan[c + d*x]])/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x ]]))/(3*b*(a^2 + b^2))
3.5.64.3.1 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[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[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] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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*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] - Simp[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] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
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[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]
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[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(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]
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[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[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] + Simp[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)*Ta n[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] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 5.39 (sec) , antiderivative size = 2978130, normalized size of antiderivative = 12205.45
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 27576 vs. \(2 (204) = 408\).
Time = 14.84 (sec) , antiderivative size = 27576, normalized size of antiderivative = 113.02 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]