Integrand size = 22, antiderivative size = 193 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx=\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac {3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac {9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt {x}}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac {9 c^{5/2} (7 b B-11 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{13/2}} \]
9/28*(-11*A*c+7*B*b)/b^3/c/x^(7/2)-9/20*(-11*A*c+7*B*b)/b^4/x^(5/2)+3/4*c* (-11*A*c+7*B*b)/b^5/x^(3/2)+1/2*(A*c-B*b)/b/c/x^(7/2)/(c*x+b)^2+1/4*(11*A* c-7*B*b)/b^2/c/x^(7/2)/(c*x+b)-9/4*c^(5/2)*(-11*A*c+7*B*b)*arctan(c^(1/2)* x^(1/2)/b^(1/2))/b^(13/2)-9/4*c^2*(-11*A*c+7*B*b)/b^6/x^(1/2)
Time = 0.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx=\frac {-7 b B x \left (8 b^4-24 b^3 c x+168 b^2 c^2 x^2+525 b c^3 x^3+315 c^4 x^4\right )+A \left (-40 b^5+88 b^4 c x-264 b^3 c^2 x^2+1848 b^2 c^3 x^3+5775 b c^4 x^4+3465 c^5 x^5\right )}{140 b^6 x^{7/2} (b+c x)^2}+\frac {9 c^{5/2} (-7 b B+11 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{13/2}} \]
(-7*b*B*x*(8*b^4 - 24*b^3*c*x + 168*b^2*c^2*x^2 + 525*b*c^3*x^3 + 315*c^4* x^4) + A*(-40*b^5 + 88*b^4*c*x - 264*b^3*c^2*x^2 + 1848*b^2*c^3*x^3 + 5775 *b*c^4*x^4 + 3465*c^5*x^5))/(140*b^6*x^(7/2)*(b + c*x)^2) + (9*c^(5/2)*(-7 *b*B + 11*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(13/2))
Time = 0.24 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {9, 87, 52, 61, 61, 61, 61, 73, 218}
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.
| \(\displaystyle \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {A+B x}{x^{9/2} (b+c x)^3}dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {(7 b B-11 A c) \int \frac {1}{x^{9/2} (b+c x)^2}dx}{4 b c}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {(7 b B-11 A c) \left (\frac {9 \int \frac {1}{x^{9/2} (b+c x)}dx}{2 b}+\frac {1}{b x^{7/2} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {(7 b B-11 A c) \left (\frac {9 \left (-\frac {c \int \frac {1}{x^{7/2} (b+c x)}dx}{b}-\frac {2}{7 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{7/2} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {(7 b B-11 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {c \int \frac {1}{x^{5/2} (b+c x)}dx}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{7/2} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {(7 b B-11 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {c \left (-\frac {c \int \frac {1}{x^{3/2} (b+c x)}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{7/2} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {(7 b B-11 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {c \left (-\frac {c \left (-\frac {c \int \frac {1}{\sqrt {x} (b+c x)}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{7/2} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {(7 b B-11 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \int \frac {1}{b+c x}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{7/2} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {(7 b B-11 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {c \left (-\frac {c \left (-\frac {2 \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{7/2} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}\) |
-1/2*(b*B - A*c)/(b*c*x^(7/2)*(b + c*x)^2) - ((7*b*B - 11*A*c)*(1/(b*x^(7/ 2)*(b + c*x)) + (9*(-2/(7*b*x^(7/2)) - (c*(-2/(5*b*x^(5/2)) - (c*(-2/(3*b* x^(3/2)) - (c*(-2/(b*Sqrt[x]) - (2*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b ]])/b^(3/2)))/b))/b))/b))/(2*b)))/(4*b*c)
3.2.93.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.73
| method | result | size |
| derivativedivides | \(\frac {2 c^{3} \left (\frac {\left (\frac {19}{8} A \,c^{2}-\frac {15}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (21 A c -17 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {9 \left (11 A c -7 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{6}}-\frac {2 A}{7 b^{3} x^{\frac {7}{2}}}-\frac {2 \left (-3 A c +B b \right )}{5 b^{4} x^{\frac {5}{2}}}-\frac {2 c \left (2 A c -B b \right )}{b^{5} x^{\frac {3}{2}}}+\frac {4 c^{2} \left (5 A c -3 B b \right )}{b^{6} \sqrt {x}}\) | \(141\) |
| default | \(\frac {2 c^{3} \left (\frac {\left (\frac {19}{8} A \,c^{2}-\frac {15}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (21 A c -17 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {9 \left (11 A c -7 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{6}}-\frac {2 A}{7 b^{3} x^{\frac {7}{2}}}-\frac {2 \left (-3 A c +B b \right )}{5 b^{4} x^{\frac {5}{2}}}-\frac {2 c \left (2 A c -B b \right )}{b^{5} x^{\frac {3}{2}}}+\frac {4 c^{2} \left (5 A c -3 B b \right )}{b^{6} \sqrt {x}}\) | \(141\) |
| risch | \(-\frac {2 \left (-350 A \,c^{3} x^{3}+210 B b \,c^{2} x^{3}+70 A b \,c^{2} x^{2}-35 B \,b^{2} c \,x^{2}-21 A \,b^{2} c x +7 B \,b^{3} x +5 A \,b^{3}\right )}{35 b^{6} x^{\frac {7}{2}}}+\frac {c^{3} \left (\frac {2 \left (\frac {19}{8} A \,c^{2}-\frac {15}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (21 A c -17 B b \right ) \sqrt {x}}{4}}{\left (c x +b \right )^{2}}+\frac {9 \left (11 A c -7 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}}\right )}{b^{6}}\) | \(147\) |
2/b^6*c^3*(((19/8*A*c^2-15/8*B*b*c)*x^(3/2)+1/8*b*(21*A*c-17*B*b)*x^(1/2)) /(c*x+b)^2+9/8*(11*A*c-7*B*b)/(b*c)^(1/2)*arctan(c*x^(1/2)/(b*c)^(1/2)))-2 /7*A/b^3/x^(7/2)-2/5*(-3*A*c+B*b)/b^4/x^(5/2)-2*c*(2*A*c-B*b)/b^5/x^(3/2)+ 4*c^2*(5*A*c-3*B*b)/b^6/x^(1/2)
Time = 0.31 (sec) , antiderivative size = 490, normalized size of antiderivative = 2.54 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx=\left [-\frac {315 \, {\left ({\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{6} + 2 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{5} + {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{4}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x + 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (40 \, A b^{5} + 315 \, {\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{5} + 525 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{4} + 168 \, {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{3} - 24 \, {\left (7 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (7 \, B b^{5} - 11 \, A b^{4} c\right )} x\right )} \sqrt {x}}{280 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}, \frac {315 \, {\left ({\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{6} + 2 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{5} + {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{4}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) - {\left (40 \, A b^{5} + 315 \, {\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{5} + 525 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{4} + 168 \, {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{3} - 24 \, {\left (7 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (7 \, B b^{5} - 11 \, A b^{4} c\right )} x\right )} \sqrt {x}}{140 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}\right ] \]
[-1/280*(315*((7*B*b*c^4 - 11*A*c^5)*x^6 + 2*(7*B*b^2*c^3 - 11*A*b*c^4)*x^ 5 + (7*B*b^3*c^2 - 11*A*b^2*c^3)*x^4)*sqrt(-c/b)*log((c*x + 2*b*sqrt(x)*sq rt(-c/b) - b)/(c*x + b)) + 2*(40*A*b^5 + 315*(7*B*b*c^4 - 11*A*c^5)*x^5 + 525*(7*B*b^2*c^3 - 11*A*b*c^4)*x^4 + 168*(7*B*b^3*c^2 - 11*A*b^2*c^3)*x^3 - 24*(7*B*b^4*c - 11*A*b^3*c^2)*x^2 + 8*(7*B*b^5 - 11*A*b^4*c)*x)*sqrt(x)) /(b^6*c^2*x^6 + 2*b^7*c*x^5 + b^8*x^4), 1/140*(315*((7*B*b*c^4 - 11*A*c^5) *x^6 + 2*(7*B*b^2*c^3 - 11*A*b*c^4)*x^5 + (7*B*b^3*c^2 - 11*A*b^2*c^3)*x^4 )*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sqrt(x))) - (40*A*b^5 + 315*(7*B*b*c^4 - 11*A*c^5)*x^5 + 525*(7*B*b^2*c^3 - 11*A*b*c^4)*x^4 + 168*(7*B*b^3*c^2 - 1 1*A*b^2*c^3)*x^3 - 24*(7*B*b^4*c - 11*A*b^3*c^2)*x^2 + 8*(7*B*b^5 - 11*A*b ^4*c)*x)*sqrt(x))/(b^6*c^2*x^6 + 2*b^7*c*x^5 + b^8*x^4)]
Timed out. \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx=-\frac {40 \, A b^{5} + 315 \, {\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{5} + 525 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{4} + 168 \, {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{3} - 24 \, {\left (7 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (7 \, B b^{5} - 11 \, A b^{4} c\right )} x}{140 \, {\left (b^{6} c^{2} x^{\frac {11}{2}} + 2 \, b^{7} c x^{\frac {9}{2}} + b^{8} x^{\frac {7}{2}}\right )}} - \frac {9 \, {\left (7 \, B b c^{3} - 11 \, A c^{4}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{6}} \]
-1/140*(40*A*b^5 + 315*(7*B*b*c^4 - 11*A*c^5)*x^5 + 525*(7*B*b^2*c^3 - 11* A*b*c^4)*x^4 + 168*(7*B*b^3*c^2 - 11*A*b^2*c^3)*x^3 - 24*(7*B*b^4*c - 11*A *b^3*c^2)*x^2 + 8*(7*B*b^5 - 11*A*b^4*c)*x)/(b^6*c^2*x^(11/2) + 2*b^7*c*x^ (9/2) + b^8*x^(7/2)) - 9/4*(7*B*b*c^3 - 11*A*c^4)*arctan(c*sqrt(x)/sqrt(b* c))/(sqrt(b*c)*b^6)
Time = 0.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx=-\frac {9 \, {\left (7 \, B b c^{3} - 11 \, A c^{4}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{6}} - \frac {15 \, B b c^{4} x^{\frac {3}{2}} - 19 \, A c^{5} x^{\frac {3}{2}} + 17 \, B b^{2} c^{3} \sqrt {x} - 21 \, A b c^{4} \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{6}} - \frac {2 \, {\left (210 \, B b c^{2} x^{3} - 350 \, A c^{3} x^{3} - 35 \, B b^{2} c x^{2} + 70 \, A b c^{2} x^{2} + 7 \, B b^{3} x - 21 \, A b^{2} c x + 5 \, A b^{3}\right )}}{35 \, b^{6} x^{\frac {7}{2}}} \]
-9/4*(7*B*b*c^3 - 11*A*c^4)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^6) - 1/4*(15*B*b*c^4*x^(3/2) - 19*A*c^5*x^(3/2) + 17*B*b^2*c^3*sqrt(x) - 21*A*b *c^4*sqrt(x))/((c*x + b)^2*b^6) - 2/35*(210*B*b*c^2*x^3 - 350*A*c^3*x^3 - 35*B*b^2*c*x^2 + 70*A*b*c^2*x^2 + 7*B*b^3*x - 21*A*b^2*c*x + 5*A*b^3)/(b^6 *x^(7/2))
Time = 10.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx=\frac {\frac {2\,x\,\left (11\,A\,c-7\,B\,b\right )}{35\,b^2}-\frac {2\,A}{7\,b}+\frac {6\,c^2\,x^3\,\left (11\,A\,c-7\,B\,b\right )}{5\,b^4}+\frac {15\,c^3\,x^4\,\left (11\,A\,c-7\,B\,b\right )}{4\,b^5}+\frac {9\,c^4\,x^5\,\left (11\,A\,c-7\,B\,b\right )}{4\,b^6}-\frac {6\,c\,x^2\,\left (11\,A\,c-7\,B\,b\right )}{35\,b^3}}{b^2\,x^{7/2}+c^2\,x^{11/2}+2\,b\,c\,x^{9/2}}+\frac {9\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (11\,A\,c-7\,B\,b\right )}{4\,b^{13/2}} \]
((2*x*(11*A*c - 7*B*b))/(35*b^2) - (2*A)/(7*b) + (6*c^2*x^3*(11*A*c - 7*B* b))/(5*b^4) + (15*c^3*x^4*(11*A*c - 7*B*b))/(4*b^5) + (9*c^4*x^5*(11*A*c - 7*B*b))/(4*b^6) - (6*c*x^2*(11*A*c - 7*B*b))/(35*b^3))/(b^2*x^(7/2) + c^2 *x^(11/2) + 2*b*c*x^(9/2)) + (9*c^(5/2)*atan((c^(1/2)*x^(1/2))/b^(1/2))*(1 1*A*c - 7*B*b))/(4*b^(13/2))