Integrand size = 22, antiderivative size = 83 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx=\frac {(b B+2 A c) \sqrt {b x+c x^2}}{b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{b x^2}+\frac {(b B+2 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}} \]
-2*A*(c*x^2+b*x)^(3/2)/b/x^2+(2*A*c+B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/ 2))/c^(1/2)+(2*A*c+B*b)*(c*x^2+b*x)^(1/2)/b
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (-2 A+B x+\frac {2 (b B+2 A c) \sqrt {x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{\sqrt {c} \sqrt {b+c x}}\right )}{x} \]
(Sqrt[x*(b + c*x)]*(-2*A + B*x + (2*(b*B + 2*A*c)*Sqrt[x]*ArcTanh[(Sqrt[c] *Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(Sqrt[c]*Sqrt[b + c*x])))/x
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1220, 1131, 1091, 219}
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.
\(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(2 A c+b B) \int \frac {\sqrt {c x^2+b x}}{x}dx}{b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{b x^2}\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {(2 A c+b B) \left (\frac {1}{2} b \int \frac {1}{\sqrt {c x^2+b x}}dx+\sqrt {b x+c x^2}\right )}{b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{b x^2}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {(2 A c+b B) \left (b \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}+\sqrt {b x+c x^2}\right )}{b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{b x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(2 A c+b B) \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}+\sqrt {b x+c x^2}\right )}{b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{b x^2}\) |
(-2*A*(b*x + c*x^2)^(3/2))/(b*x^2) + ((b*B + 2*A*c)*(Sqrt[b*x + c*x^2] + ( b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/Sqrt[c]))/b
3.1.72.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
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[((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 = 0.14 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-x \left (A c +\frac {B b}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\sqrt {c}\, \sqrt {x \left (c x +b \right )}\, \left (-\frac {B x}{2}+A \right )\right )}{\sqrt {c}\, x}\) | \(57\) |
risch | \(-\frac {\left (c x +b \right ) \left (-B x +2 A \right )}{\sqrt {x \left (c x +b \right )}}+\frac {\left (A c +\frac {B b}{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}\) | \(62\) |
default | \(B \left (\sqrt {c \,x^{2}+b x}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 \sqrt {c}}\right )+A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{b \,x^{2}}+\frac {2 c \left (\sqrt {c \,x^{2}+b x}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 \sqrt {c}}\right )}{b}\right )\) | \(116\) |
-2*(-x*(A*c+1/2*B*b)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+c^(1/2)*(x*(c*x+ b))^(1/2)*(-1/2*B*x+A))/c^(1/2)/x
Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx=\left [\frac {{\left (B b + 2 \, A c\right )} \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (B c x - 2 \, A c\right )} \sqrt {c x^{2} + b x}}{2 \, c x}, -\frac {{\left (B b + 2 \, A c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (B c x - 2 \, A c\right )} \sqrt {c x^{2} + b x}}{c x}\right ] \]
[1/2*((B*b + 2*A*c)*sqrt(c)*x*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(B*c*x - 2*A*c)*sqrt(c*x^2 + b*x))/(c*x), -((B*b + 2*A*c)*sqrt(-c)*x* arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (B*c*x - 2*A*c)*sqrt(c*x^2 + b* x))/(c*x)]
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{2}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx=\frac {B b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, \sqrt {c}} + A \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + \sqrt {c x^{2} + b x} B - \frac {2 \, \sqrt {c x^{2} + b x} A}{x} \]
1/2*B*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) + A*sqrt(c)*l og(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + sqrt(c*x^2 + b*x)*B - 2*sqrt (c*x^2 + b*x)*A/x
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx=\sqrt {c x^{2} + b x} B - \frac {{\left (B b + 2 \, A c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{2 \, \sqrt {c}} + \frac {2 \, A b}{\sqrt {c} x - \sqrt {c x^{2} + b x}} \]
sqrt(c*x^2 + b*x)*B - 1/2*(B*b + 2*A*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/sqrt(c) + 2*A*b/(sqrt(c)*x - sqrt(c*x^2 + b*x))
Timed out. \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^2} \,d x \]