Integrand size = 22, antiderivative size = 100 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {(b B-A c) \sqrt {x}}{2 b c (b+c x)^2}+\frac {(b B+3 A c) \sqrt {x}}{4 b^2 c (b+c x)}+\frac {(b B+3 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{5/2} c^{3/2}} \]
1/4*(3*A*c+B*b)*arctan(c^(1/2)*x^(1/2)/b^(1/2))/b^(5/2)/c^(3/2)-1/2*(-A*c+ B*b)*x^(1/2)/b/c/(c*x+b)^2+1/4*(3*A*c+B*b)*x^(1/2)/b^2/c/(c*x+b)
Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {\sqrt {x} \left (b^2 B-5 A b c-b B c x-3 A c^2 x\right )}{4 b^2 c (b+c x)^2}+\frac {(b B+3 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{5/2} c^{3/2}} \]
-1/4*(Sqrt[x]*(b^2*B - 5*A*b*c - b*B*c*x - 3*A*c^2*x))/(b^2*c*(b + c*x)^2) + ((b*B + 3*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(5/2)*c^(3/2))
Time = 0.19 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {9, 87, 52, 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.
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x}{\sqrt {x} (b+c x)^3}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(3 A c+b B) \int \frac {1}{\sqrt {x} (b+c x)^2}dx}{4 b c}-\frac {\sqrt {x} (b B-A c)}{2 b c (b+c x)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {(3 A c+b B) \left (\frac {\int \frac {1}{\sqrt {x} (b+c x)}dx}{2 b}+\frac {\sqrt {x}}{b (b+c x)}\right )}{4 b c}-\frac {\sqrt {x} (b B-A c)}{2 b c (b+c x)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(3 A c+b B) \left (\frac {\int \frac {1}{b+c x}d\sqrt {x}}{b}+\frac {\sqrt {x}}{b (b+c x)}\right )}{4 b c}-\frac {\sqrt {x} (b B-A c)}{2 b c (b+c x)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(3 A c+b B) \left (\frac {\arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}+\frac {\sqrt {x}}{b (b+c x)}\right )}{4 b c}-\frac {\sqrt {x} (b B-A c)}{2 b c (b+c x)^2}\) |
-1/2*((b*B - A*c)*Sqrt[x])/(b*c*(b + c*x)^2) + ((b*B + 3*A*c)*(Sqrt[x]/(b* (b + c*x)) + ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]]/(b^(3/2)*Sqrt[c])))/(4*b*c)
3.2.89.3.1 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] :> 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 = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {\frac {\left (3 A c +B b \right ) x^{\frac {3}{2}}}{4 b^{2}}+\frac {\left (5 A c -B b \right ) \sqrt {x}}{4 b c}}{\left (c x +b \right )^{2}}+\frac {\left (3 A c +B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 b^{2} c \sqrt {b c}}\) | \(80\) |
default | \(\frac {\frac {\left (3 A c +B b \right ) x^{\frac {3}{2}}}{4 b^{2}}+\frac {\left (5 A c -B b \right ) \sqrt {x}}{4 b c}}{\left (c x +b \right )^{2}}+\frac {\left (3 A c +B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 b^{2} c \sqrt {b c}}\) | \(80\) |
2*(1/8*(3*A*c+B*b)/b^2*x^(3/2)+1/8*(5*A*c-B*b)/b/c*x^(1/2))/(c*x+b)^2+1/4* (3*A*c+B*b)/b^2/c/(b*c)^(1/2)*arctan(c*x^(1/2)/(b*c)^(1/2))
Time = 0.28 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.91 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\left [-\frac {{\left (B b^{3} + 3 \, A b^{2} c + {\left (B b c^{2} + 3 \, A c^{3}\right )} x^{2} + 2 \, {\left (B b^{2} c + 3 \, A b c^{2}\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x - b - 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right ) + 2 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2} - {\left (B b^{2} c^{2} + 3 \, A b c^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (b^{3} c^{4} x^{2} + 2 \, b^{4} c^{3} x + b^{5} c^{2}\right )}}, -\frac {{\left (B b^{3} + 3 \, A b^{2} c + {\left (B b c^{2} + 3 \, A c^{3}\right )} x^{2} + 2 \, {\left (B b^{2} c + 3 \, A b c^{2}\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right ) + {\left (B b^{3} c - 5 \, A b^{2} c^{2} - {\left (B b^{2} c^{2} + 3 \, A b c^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (b^{3} c^{4} x^{2} + 2 \, b^{4} c^{3} x + b^{5} c^{2}\right )}}\right ] \]
[-1/8*((B*b^3 + 3*A*b^2*c + (B*b*c^2 + 3*A*c^3)*x^2 + 2*(B*b^2*c + 3*A*b*c ^2)*x)*sqrt(-b*c)*log((c*x - b - 2*sqrt(-b*c)*sqrt(x))/(c*x + b)) + 2*(B*b ^3*c - 5*A*b^2*c^2 - (B*b^2*c^2 + 3*A*b*c^3)*x)*sqrt(x))/(b^3*c^4*x^2 + 2* b^4*c^3*x + b^5*c^2), -1/4*((B*b^3 + 3*A*b^2*c + (B*b*c^2 + 3*A*c^3)*x^2 + 2*(B*b^2*c + 3*A*b*c^2)*x)*sqrt(b*c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (B*b ^3*c - 5*A*b^2*c^2 - (B*b^2*c^2 + 3*A*b*c^3)*x)*sqrt(x))/(b^3*c^4*x^2 + 2* b^4*c^3*x + b^5*c^2)]
Timed out. \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {{\left (B b c + 3 \, A c^{2}\right )} x^{\frac {3}{2}} - {\left (B b^{2} - 5 \, A b c\right )} \sqrt {x}}{4 \, {\left (b^{2} c^{3} x^{2} + 2 \, b^{3} c^{2} x + b^{4} c\right )}} + \frac {{\left (B b + 3 \, A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{2} c} \]
1/4*((B*b*c + 3*A*c^2)*x^(3/2) - (B*b^2 - 5*A*b*c)*sqrt(x))/(b^2*c^3*x^2 + 2*b^3*c^2*x + b^4*c) + 1/4*(B*b + 3*A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqr t(b*c)*b^2*c)
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.82 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {{\left (B b + 3 \, A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{2} c} + \frac {B b c x^{\frac {3}{2}} + 3 \, A c^{2} x^{\frac {3}{2}} - B b^{2} \sqrt {x} + 5 \, A b c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{2} c} \]
1/4*(B*b + 3*A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^2*c) + 1/4*(B*b *c*x^(3/2) + 3*A*c^2*x^(3/2) - B*b^2*sqrt(x) + 5*A*b*c*sqrt(x))/((c*x + b) ^2*b^2*c)
Time = 10.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {\frac {x^{3/2}\,\left (3\,A\,c+B\,b\right )}{4\,b^2}+\frac {\sqrt {x}\,\left (5\,A\,c-B\,b\right )}{4\,b\,c}}{b^2+2\,b\,c\,x+c^2\,x^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (3\,A\,c+B\,b\right )}{4\,b^{5/2}\,c^{3/2}} \]