Integrand size = 22, antiderivative size = 130 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^2} \, dx=\frac {5 b B-7 A c}{5 b^2 c x^{5/2}}-\frac {5 b B-7 A c}{3 b^3 x^{3/2}}+\frac {c (5 b B-7 A c)}{b^4 \sqrt {x}}-\frac {b B-A c}{b c x^{5/2} (b+c x)}+\frac {c^{3/2} (5 b B-7 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}} \]
1/5*(-7*A*c+5*B*b)/b^2/c/x^(5/2)+1/3*(7*A*c-5*B*b)/b^3/x^(3/2)+(A*c-B*b)/b /c/x^(5/2)/(c*x+b)+c^(3/2)*(-7*A*c+5*B*b)*arctan(c^(1/2)*x^(1/2)/b^(1/2))/ b^(9/2)+c*(-7*A*c+5*B*b)/b^4/x^(1/2)
Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^2} \, dx=\frac {5 b B x \left (-2 b^2+10 b c x+15 c^2 x^2\right )-A \left (6 b^3-14 b^2 c x+70 b c^2 x^2+105 c^3 x^3\right )}{15 b^4 x^{5/2} (b+c x)}+\frac {c^{3/2} (5 b B-7 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}} \]
(5*b*B*x*(-2*b^2 + 10*b*c*x + 15*c^2*x^2) - A*(6*b^3 - 14*b^2*c*x + 70*b*c ^2*x^2 + 105*c^3*x^3))/(15*b^4*x^(5/2)*(b + c*x)) + (c^(3/2)*(5*b*B - 7*A* c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/b^(9/2)
Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {9, 87, 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.
| \(\displaystyle \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {A+B x}{x^{7/2} (b+c x)^2}dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {(5 b B-7 A c) \int \frac {1}{x^{7/2} (b+c x)}dx}{2 b c}-\frac {b B-A c}{b c x^{5/2} (b+c x)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {(5 b B-7 A c) \left (-\frac {c \int \frac {1}{x^{5/2} (b+c x)}dx}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b c}-\frac {b B-A c}{b c x^{5/2} (b+c x)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {(5 b B-7 A c) \left (-\frac {c \left (-\frac {c \int \frac {1}{x^{3/2} (b+c x)}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b c}-\frac {b B-A c}{b c x^{5/2} (b+c x)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {(5 b B-7 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {c \int \frac {1}{\sqrt {x} (b+c x)}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b c}-\frac {b B-A c}{b c x^{5/2} (b+c x)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {(5 b B-7 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \int \frac {1}{b+c x}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b c}-\frac {b B-A c}{b c x^{5/2} (b+c x)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {(5 b B-7 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b c}-\frac {b B-A c}{b c x^{5/2} (b+c x)}\) |
-((b*B - A*c)/(b*c*x^(5/2)*(b + c*x))) - ((5*b*B - 7*A*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))/(2*b*c)
3.2.83.3.1 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] && 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 = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(-\frac {2 c^{2} \left (\frac {\left (\frac {A c}{2}-\frac {B b}{2}\right ) \sqrt {x}}{c x +b}+\frac {\left (7 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{4}}-\frac {2 A}{5 b^{2} x^{\frac {5}{2}}}-\frac {2 \left (-2 A c +B b \right )}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 c \left (3 A c -2 B b \right )}{b^{4} \sqrt {x}}\) | \(101\) |
| default | \(-\frac {2 c^{2} \left (\frac {\left (\frac {A c}{2}-\frac {B b}{2}\right ) \sqrt {x}}{c x +b}+\frac {\left (7 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{4}}-\frac {2 A}{5 b^{2} x^{\frac {5}{2}}}-\frac {2 \left (-2 A c +B b \right )}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 c \left (3 A c -2 B b \right )}{b^{4} \sqrt {x}}\) | \(101\) |
| risch | \(-\frac {2 \left (45 A \,c^{2} x^{2}-30 B b c \,x^{2}-10 A b c x +5 b^{2} B x +3 A \,b^{2}\right )}{15 b^{4} x^{\frac {5}{2}}}-\frac {c^{2} \left (\frac {2 \left (\frac {A c}{2}-\frac {B b}{2}\right ) \sqrt {x}}{c x +b}+\frac {\left (7 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}}\right )}{b^{4}}\) | \(103\) |
-2/b^4*c^2*((1/2*A*c-1/2*B*b)*x^(1/2)/(c*x+b)+1/2*(7*A*c-5*B*b)/(b*c)^(1/2 )*arctan(c*x^(1/2)/(b*c)^(1/2)))-2/5*A/b^2/x^(5/2)-2/3*(-2*A*c+B*b)/b^3/x^ (3/2)-2*c*(3*A*c-2*B*b)/b^4/x^(1/2)
Time = 0.42 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.45 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^2} \, dx=\left [-\frac {15 \, {\left ({\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} + {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x - 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (6 \, A b^{3} - 15 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} - 10 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 2 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x\right )} \sqrt {x}}{30 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}, -\frac {15 \, {\left ({\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} + {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) + {\left (6 \, A b^{3} - 15 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} - 10 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 2 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x\right )} \sqrt {x}}{15 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}\right ] \]
[-1/30*(15*((5*B*b*c^2 - 7*A*c^3)*x^4 + (5*B*b^2*c - 7*A*b*c^2)*x^3)*sqrt( -c/b)*log((c*x - 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(6*A*b^3 - 15* (5*B*b*c^2 - 7*A*c^3)*x^3 - 10*(5*B*b^2*c - 7*A*b*c^2)*x^2 + 2*(5*B*b^3 - 7*A*b^2*c)*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3), -1/15*(15*((5*B*b*c^2 - 7*A* c^3)*x^4 + (5*B*b^2*c - 7*A*b*c^2)*x^3)*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sq rt(x))) + (6*A*b^3 - 15*(5*B*b*c^2 - 7*A*c^3)*x^3 - 10*(5*B*b^2*c - 7*A*b* c^2)*x^2 + 2*(5*B*b^3 - 7*A*b^2*c)*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3)]
Leaf count of result is larger than twice the leaf count of optimal. 1017 vs. \(2 (119) = 238\).
Time = 44.45 (sec) , antiderivative size = 1017, normalized size of antiderivative = 7.82 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^2} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{b^{2}} & \text {for}\: c = 0 \\\frac {- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}}{c^{2}} & \text {for}\: b = 0 \\- \frac {12 A b^{3} \sqrt {- \frac {b}{c}}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} + \frac {28 A b^{2} c x \sqrt {- \frac {b}{c}}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} - \frac {105 A b c^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} + \frac {105 A b c^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} - \frac {140 A b c^{2} x^{2} \sqrt {- \frac {b}{c}}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} - \frac {105 A c^{3} x^{\frac {7}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} + \frac {105 A c^{3} x^{\frac {7}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} - \frac {210 A c^{3} x^{3} \sqrt {- \frac {b}{c}}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} - \frac {20 B b^{3} x \sqrt {- \frac {b}{c}}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} + \frac {75 B b^{2} c x^{\frac {5}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} - \frac {75 B b^{2} c x^{\frac {5}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} + \frac {100 B b^{2} c x^{2} \sqrt {- \frac {b}{c}}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} + \frac {75 B b c^{2} x^{\frac {7}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} - \frac {75 B b c^{2} x^{\frac {7}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} + \frac {150 B b c^{2} x^{3} \sqrt {- \frac {b}{c}}}{30 b^{5} x^{\frac {5}{2}} \sqrt {- \frac {b}{c}} + 30 b^{4} c x^{\frac {7}{2}} \sqrt {- \frac {b}{c}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(b, 0) & Eq(c, 0) ), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2)))/b**2, Eq(c, 0)), ((-2*A/(9*x**( 9/2)) - 2*B/(7*x**(7/2)))/c**2, Eq(b, 0)), (-12*A*b**3*sqrt(-b/c)/(30*b**5 *x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) + 28*A*b**2*c*x*sqrt (-b/c)/(30*b**5*x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) - 105 *A*b*c**2*x**(5/2)*log(sqrt(x) - sqrt(-b/c))/(30*b**5*x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) + 105*A*b*c**2*x**(5/2)*log(sqrt(x) + sqr t(-b/c))/(30*b**5*x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) - 1 40*A*b*c**2*x**2*sqrt(-b/c)/(30*b**5*x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7 /2)*sqrt(-b/c)) - 105*A*c**3*x**(7/2)*log(sqrt(x) - sqrt(-b/c))/(30*b**5*x **(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) + 105*A*c**3*x**(7/2)* log(sqrt(x) + sqrt(-b/c))/(30*b**5*x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2 )*sqrt(-b/c)) - 210*A*c**3*x**3*sqrt(-b/c)/(30*b**5*x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) - 20*B*b**3*x*sqrt(-b/c)/(30*b**5*x**(5/2)* sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) + 75*B*b**2*c*x**(5/2)*log(sqr t(x) - sqrt(-b/c))/(30*b**5*x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt( -b/c)) - 75*B*b**2*c*x**(5/2)*log(sqrt(x) + sqrt(-b/c))/(30*b**5*x**(5/2)* sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) + 100*B*b**2*c*x**2*sqrt(-b/c) /(30*b**5*x**(5/2)*sqrt(-b/c) + 30*b**4*c*x**(7/2)*sqrt(-b/c)) + 75*B*b*c* *2*x**(7/2)*log(sqrt(x) - sqrt(-b/c))/(30*b**5*x**(5/2)*sqrt(-b/c) + 30...
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^2} \, dx=-\frac {6 \, A b^{3} - 15 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} - 10 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 2 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x}{15 \, {\left (b^{4} c x^{\frac {7}{2}} + b^{5} x^{\frac {5}{2}}\right )}} + \frac {{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{4}} \]
-1/15*(6*A*b^3 - 15*(5*B*b*c^2 - 7*A*c^3)*x^3 - 10*(5*B*b^2*c - 7*A*b*c^2) *x^2 + 2*(5*B*b^3 - 7*A*b^2*c)*x)/(b^4*c*x^(7/2) + b^5*x^(5/2)) + (5*B*b*c ^2 - 7*A*c^3)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^4)
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^2} \, dx=\frac {{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{4}} + \frac {B b c^{2} \sqrt {x} - A c^{3} \sqrt {x}}{{\left (c x + b\right )} b^{4}} + \frac {2 \, {\left (30 \, B b c x^{2} - 45 \, A c^{2} x^{2} - 5 \, B b^{2} x + 10 \, A b c x - 3 \, A b^{2}\right )}}{15 \, b^{4} x^{\frac {5}{2}}} \]
(5*B*b*c^2 - 7*A*c^3)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^4) + (B*b*c ^2*sqrt(x) - A*c^3*sqrt(x))/((c*x + b)*b^4) + 2/15*(30*B*b*c*x^2 - 45*A*c^ 2*x^2 - 5*B*b^2*x + 10*A*b*c*x - 3*A*b^2)/(b^4*x^(5/2))
Time = 10.13 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^2} \, dx=-\frac {\frac {2\,A}{5\,b}-\frac {2\,x\,\left (7\,A\,c-5\,B\,b\right )}{15\,b^2}+\frac {c^2\,x^3\,\left (7\,A\,c-5\,B\,b\right )}{b^4}+\frac {2\,c\,x^2\,\left (7\,A\,c-5\,B\,b\right )}{3\,b^3}}{b\,x^{5/2}+c\,x^{7/2}}-\frac {c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (7\,A\,c-5\,B\,b\right )}{b^{9/2}} \]