Integrand size = 24, antiderivative size = 66 \[ \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx=-\frac {A \sqrt {b x+c x^2}}{b x^{3/2}}-\frac {(2 b B-A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}} \] Output:
-A*(c*x^2+b*x)^(1/2)/b/x^(3/2)-(-A*c+2*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1 /2)/x^(1/2))/b^(3/2)
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx=\frac {-A \sqrt {b} (b+c x)-(2 b B-A c) x \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{b^{3/2} \sqrt {x} \sqrt {x (b+c x)}} \] Input:
Integrate[(A + B*x)/(x^(3/2)*Sqrt[b*x + c*x^2]),x]
Output:
(-(A*Sqrt[b]*(b + c*x)) - (2*b*B - A*c)*x*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c *x]/Sqrt[b]])/(b^(3/2)*Sqrt[x]*Sqrt[x*(b + c*x)])
Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1220, 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.
\(\displaystyle \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(2 b B-A c) \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{2 b}-\frac {A \sqrt {b x+c x^2}}{b x^{3/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle \frac {(2 b B-A c) \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}-\frac {A \sqrt {b x+c x^2}}{b x^{3/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {(2 b B-A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {A \sqrt {b x+c x^2}}{b x^{3/2}}\) |
Input:
Int[(A + B*x)/(x^(3/2)*Sqrt[b*x + c*x^2]),x]
Output:
-((A*Sqrt[b*x + c*x^2])/(b*x^(3/2))) - ((2*b*B - A*c)*ArcTanh[Sqrt[b*x + c *x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/2)
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[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 = 0.98 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {A \left (c x +b \right )}{b \sqrt {x}\, \sqrt {x \left (c x +b \right )}}+\frac {\left (A c -2 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{b^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}\) | \(68\) |
default | \(\frac {\sqrt {x \left (c x +b \right )}\, \left (A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c x -2 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b x -A \sqrt {c x +b}\, \sqrt {b}\right )}{b^{\frac {3}{2}} x^{\frac {3}{2}} \sqrt {c x +b}}\) | \(71\) |
Input:
int((B*x+A)/x^(3/2)/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
-A/b*(c*x+b)/x^(1/2)/(x*(c*x+b))^(1/2)+(A*c-2*B*b)/b^(3/2)*arctanh((c*x+b) ^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*x+b))^(1/2)
Time = 0.08 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.23 \[ \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx=\left [-\frac {{\left (2 \, B b - A c\right )} \sqrt {b} x^{2} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, \sqrt {c x^{2} + b x} A b \sqrt {x}}{2 \, b^{2} x^{2}}, \frac {{\left (2 \, B b - A c\right )} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) - \sqrt {c x^{2} + b x} A b \sqrt {x}}{b^{2} x^{2}}\right ] \] Input:
integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")
Output:
[-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*sqrt(c*x^2 + b*x)*A*b*sqrt(x))/(b^2*x^2), ((2*B *b - A*c)*sqrt(-b)*x^2*arctan(sqrt(c*x^2 + b*x)*sqrt(-b)/(b*sqrt(x))) - sq rt(c*x^2 + b*x)*A*b*sqrt(x))/(b^2*x^2)]
\[ \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx=\int \frac {A + B x}{x^{\frac {3}{2}} \sqrt {x \left (b + c x\right )}}\, dx \] Input:
integrate((B*x+A)/x**(3/2)/(c*x**2+b*x)**(1/2),x)
Output:
Integral((A + B*x)/(x**(3/2)*sqrt(x*(b + c*x))), x)
\[ \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx=\int { \frac {B x + A}{\sqrt {c x^{2} + b x} x^{\frac {3}{2}}} \,d x } \] Input:
integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")
Output:
integrate((B*x + A)/(sqrt(c*x^2 + b*x)*x^(3/2)), x)
Time = 0.14 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx=c {\left (\frac {{\left (2 \, B b - A c\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b c} - \frac {\sqrt {c x + b} A}{b c x}\right )} \] Input:
integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")
Output:
c*((2*B*b - A*c)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b*c) - sqrt(c*x + b)*A/(b*c*x))
Timed out. \[ \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx=\int \frac {A+B\,x}{x^{3/2}\,\sqrt {c\,x^2+b\,x}} \,d x \] Input:
int((A + B*x)/(x^(3/2)*(b*x + c*x^2)^(1/2)),x)
Output:
int((A + B*x)/(x^(3/2)*(b*x + c*x^2)^(1/2)), x)
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.39 \[ \int \frac {A+B x}{x^{3/2} \sqrt {b x+c x^2}} \, dx=\frac {-2 \sqrt {c x +b}\, a b -\sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a c x +2 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} x +\sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a c x -2 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} x}{2 b^{2} x} \] Input:
int((B*x+A)/x^(3/2)/(c*x^2+b*x)^(1/2),x)
Output:
( - 2*sqrt(b + c*x)*a*b - sqrt(b)*log(sqrt(b + c*x) - sqrt(b))*a*c*x + 2*s qrt(b)*log(sqrt(b + c*x) - sqrt(b))*b**2*x + sqrt(b)*log(sqrt(b + c*x) + s qrt(b))*a*c*x - 2*sqrt(b)*log(sqrt(b + c*x) + sqrt(b))*b**2*x)/(2*b**2*x)