Integrand size = 22, antiderivative size = 92 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx=-\frac {(b B-4 A c) \sqrt {b x+c x^2}}{4 c}+\frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac {b (b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}} \]
1/2*B*(c*x^2+b*x)^(3/2)/c/x-1/4*b*(-4*A*c+B*b)*arctanh(x*c^(1/2)/(c*x^2+b* x)^(1/2))/c^(3/2)-1/4*(-4*A*c+B*b)*(c*x^2+b*x)^(1/2)/c
Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} (b B+4 A c+2 B c x)+\frac {b (b B-4 A c) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{4 c^{3/2}} \]
(Sqrt[x*(b + c*x)]*(Sqrt[c]*(b*B + 4*A*c + 2*B*c*x) + (b*(b*B - 4*A*c)*Log [-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(Sqrt[x]*Sqrt[b + c*x])))/(4*c^(3/2) )
Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1221, 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.
\(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac {(b B-4 A c) \int \frac {\sqrt {c x^2+b x}}{x}dx}{4 c}\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac {(b B-4 A c) \left (\frac {1}{2} b \int \frac {1}{\sqrt {c x^2+b x}}dx+\sqrt {b x+c x^2}\right )}{4 c}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac {(b B-4 A c) \left (b \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}+\sqrt {b x+c x^2}\right )}{4 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac {(b B-4 A c) \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}+\sqrt {b x+c x^2}\right )}{4 c}\) |
(B*(b*x + c*x^2)^(3/2))/(2*c*x) - ((b*B - 4*A*c)*(Sqrt[b*x + c*x^2] + (b*A rcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/Sqrt[c]))/(4*c)
3.1.71.3.1 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[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 )/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0]
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {b \left (A c -\frac {B b}{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\sqrt {x \left (c x +b \right )}\, \left (\left (\frac {B x}{2}+A \right ) c^{\frac {3}{2}}+\frac {B \sqrt {c}\, b}{4}\right )}{c^{\frac {3}{2}}}\) | \(61\) |
risch | \(\frac {\left (2 B c x +4 A c +B b \right ) x \left (c x +b \right )}{4 c \sqrt {x \left (c x +b \right )}}+\frac {b \left (4 A c -B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\) | \(74\) |
default | \(B \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )+A \left (\sqrt {c \,x^{2}+b x}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 \sqrt {c}}\right )\) | \(103\) |
(b*(A*c-1/4*B*b)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(x*(c*x+b))^(1/2)*(( 1/2*B*x+A)*c^(3/2)+1/4*B*c^(1/2)*b))/c^(3/2)
Time = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx=\left [-\frac {{\left (B b^{2} - 4 \, A b c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{2} x + B b c + 4 \, A c^{2}\right )} \sqrt {c x^{2} + b x}}{8 \, c^{2}}, \frac {{\left (B b^{2} - 4 \, A b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (2 \, B c^{2} x + B b c + 4 \, A c^{2}\right )} \sqrt {c x^{2} + b x}}{4 \, c^{2}}\right ] \]
[-1/8*((B*b^2 - 4*A*b*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt( c)) - 2*(2*B*c^2*x + B*b*c + 4*A*c^2)*sqrt(c*x^2 + b*x))/c^2, 1/4*((B*b^2 - 4*A*b*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (2*B*c^2*x + B*b*c + 4*A*c^2)*sqrt(c*x^2 + b*x))/c^2]
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x}\, dx \]
Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx=\frac {1}{2} \, \sqrt {c x^{2} + b x} B x - \frac {B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {A b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, \sqrt {c}} + \sqrt {c x^{2} + b x} A + \frac {\sqrt {c x^{2} + b x} B b}{4 \, c} \]
1/2*sqrt(c*x^2 + b*x)*B*x - 1/8*B*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)* sqrt(c))/c^(3/2) + 1/2*A*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sq rt(c) + sqrt(c*x^2 + b*x)*A + 1/4*sqrt(c*x^2 + b*x)*B*b/c
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (2 \, B x + \frac {B b + 4 \, A c}{c}\right )} + \frac {{\left (B b^{2} - 4 \, A b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {3}{2}}} \]
1/4*sqrt(c*x^2 + b*x)*(2*B*x + (B*b + 4*A*c)/c) + 1/8*(B*b^2 - 4*A*b*c)*lo g(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(3/2)
Time = 10.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx=A\,\sqrt {c\,x^2+b\,x}+B\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {B\,b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}+\frac {A\,b\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{2\,\sqrt {c}} \]