Integrand size = 24, antiderivative size = 95 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{5/2}} \, dx=\frac {(2 b B+A c) \sqrt {b x+c x^2}}{b \sqrt {x}}-\frac {A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}-\frac {(2 b B+A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}} \]
-A*(c*x^2+b*x)^(3/2)/b/x^(5/2)-(A*c+2*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/ 2)/x^(1/2))/b^(1/2)+(A*c+2*B*b)*(c*x^2+b*x)^(1/2)/b/x^(1/2)
Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{5/2}} \, dx=-\frac {\sqrt {x (b+c x)} \left (\sqrt {b} (A-2 B x) \sqrt {b+c x}+(2 b B+A c) x \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{\sqrt {b} x^{3/2} \sqrt {b+c x}} \]
-((Sqrt[x*(b + c*x)]*(Sqrt[b]*(A - 2*B*x)*Sqrt[b + c*x] + (2*b*B + A*c)*x* ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(Sqrt[b]*x^(3/2)*Sqrt[b + c*x]))
Time = 0.24 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1220, 1131, 1136, 220}
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.
\(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(A c+2 b B) \int \frac {\sqrt {c x^2+b x}}{x^{3/2}}dx}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {(A c+2 b B) \left (b \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx+\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}\right )}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle \frac {(A c+2 b B) \left (2 b \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}+\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}\right )}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {(A c+2 b B) \left (\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )\right )}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}\) |
-((A*(b*x + c*x^2)^(3/2))/(b*x^(5/2))) + ((2*b*B + A*c)*((2*Sqrt[b*x + c*x ^2])/Sqrt[x] - 2*Sqrt[b]*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])]))/(2 *b)
3.2.100.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b *d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne Q[m + 2*p + 1, 0] && IntegerQ[2*p]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x _Symbol] :> Simp[2*e Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-\frac {A \left (c x +b \right )}{\sqrt {x}\, \sqrt {x \left (c x +b \right )}}+\frac {\left (2 B \sqrt {c x +b}-\frac {\left (A c +2 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{\sqrt {x \left (c x +b \right )}}\) | \(78\) |
default | \(\frac {\left (-A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c x +2 B \sqrt {c x +b}\, x \sqrt {b}-2 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b x -A \sqrt {c x +b}\, \sqrt {b}\right ) \sqrt {x \left (c x +b \right )}}{x^{\frac {3}{2}} \sqrt {c x +b}\, \sqrt {b}}\) | \(86\) |
-A*(c*x+b)/x^(1/2)/(x*(c*x+b))^(1/2)+(2*B*(c*x+b)^(1/2)-(A*c+2*B*b)/b^(1/2 )*arctanh((c*x+b)^(1/2)/b^(1/2)))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*x+b))^(1/2)
Time = 0.36 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{5/2}} \, dx=\left [\frac {{\left (2 \, B b + A c\right )} \sqrt {b} x^{2} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (2 \, B b x - A b\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{2 \, b x^{2}}, \frac {{\left (2 \, B b + A c\right )} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (2 \, B b x - A b\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{b x^{2}}\right ] \]
[1/2*((2*B*b + A*c)*sqrt(b)*x^2*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)* sqrt(b)*sqrt(x))/x^2) + 2*(2*B*b*x - A*b)*sqrt(c*x^2 + b*x)*sqrt(x))/(b*x^ 2), ((2*B*b + A*c)*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (2*B*b*x - A*b)*sqrt(c*x^2 + b*x)*sqrt(x))/(b*x^2)]
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{5/2}} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{5/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x} {\left (B x + A\right )}}{x^{\frac {5}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{5/2}} \, dx=\frac {2 \, \sqrt {c x + b} B c - \frac {\sqrt {c x + b} A c}{x} + \frac {{\left (2 \, B b c + A c^{2}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}}}{c} \]
(2*sqrt(c*x + b)*B*c - sqrt(c*x + b)*A*c/x + (2*B*b*c + A*c^2)*arctan(sqrt (c*x + b)/sqrt(-b))/sqrt(-b))/c
Timed out. \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{5/2}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^{5/2}} \,d x \]