Integrand size = 23, antiderivative size = 121 \[ \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^2} \, dx=-\frac {(A-B x) \sqrt {a+b x+c x^2}}{x}-\frac {(A b+2 a B) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {a}}+\frac {(b B+2 A c) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c}} \]
-1/2*(A*b+2*B*a)*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(1/2 )+1/2*(2*A*c+B*b)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(1/ 2)-(-B*x+A)*(c*x^2+b*x+a)^(1/2)/x
Time = 0.42 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^2} \, dx=\frac {(-A+B x) \sqrt {a+x (b+c x)}}{x}-\frac {(A b+2 a B) \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {(b B+2 A c) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{2 \sqrt {c}} \]
((-A + B*x)*Sqrt[a + x*(b + c*x)])/x - ((A*b + 2*a*B)*ArcTanh[(-(Sqrt[c]*x ) + Sqrt[a + x*(b + c*x)])/Sqrt[a]])/Sqrt[a] - ((b*B + 2*A*c)*Log[b + 2*c* x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(2*Sqrt[c])
Time = 0.31 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1230, 25, 1269, 1092, 219, 1154, 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.
\(\displaystyle \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^2} \, dx\) |
\(\Big \downarrow \) 1230 |
\(\displaystyle -\frac {1}{2} \int -\frac {A b+2 a B+(b B+2 A c) x}{x \sqrt {c x^2+b x+a}}dx-\frac {(A-B x) \sqrt {a+b x+c x^2}}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \int \frac {A b+2 a B+(b B+2 A c) x}{x \sqrt {c x^2+b x+a}}dx-\frac {(A-B x) \sqrt {a+b x+c x^2}}{x}\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {1}{2} \left ((2 A c+b B) \int \frac {1}{\sqrt {c x^2+b x+a}}dx+(2 a B+A b) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx\right )-\frac {(A-B x) \sqrt {a+b x+c x^2}}{x}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left ((2 a B+A b) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+2 (2 A c+b B) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}\right )-\frac {(A-B x) \sqrt {a+b x+c x^2}}{x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left ((2 a B+A b) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+\frac {(2 A c+b B) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )-\frac {(A-B x) \sqrt {a+b x+c x^2}}{x}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+b B) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-2 (2 a B+A b) \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}\right )-\frac {(A-B x) \sqrt {a+b x+c x^2}}{x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+b B) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-\frac {(2 a B+A b) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a}}\right )-\frac {(A-B x) \sqrt {a+b x+c x^2}}{x}\) |
-(((A - B*x)*Sqrt[a + b*x + c*x^2])/x) + (-(((A*b + 2*a*B)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/Sqrt[a]) + ((b*B + 2*A*c)*ArcTanh [(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/Sqrt[c])/2
3.10.16.3.1 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[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Simp[p/(e^2*(m + 1)*(m + 2*p + 2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, - 1] || EqQ[p, 1] || (IntegerQ[p] && !RationalQ[m])) && NeQ[m, -1] && !ILtQ [m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e 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] && !IGtQ[m, 0]
Time = 0.43 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.38
method | result | size |
risch | \(-\frac {A \sqrt {c \,x^{2}+b x +a}}{x}+A \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )+\frac {B b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}+B \sqrt {c \,x^{2}+b x +a}-\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) A b}{2 \sqrt {a}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) B\) | \(167\) |
default | \(B \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )+A \left (-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{a x}+\frac {b \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )}{2 a}+\frac {2 c \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{a}\right )\) | \(259\) |
-A*(c*x^2+b*x+a)^(1/2)/x+A*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1 /2))+1/2*B*b*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+B*(c*x^2+ b*x+a)^(1/2)-1/2/a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*A*b -a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*B
Time = 0.81 (sec) , antiderivative size = 648, normalized size of antiderivative = 5.36 \[ \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^2} \, dx=\left [\frac {{\left (2 \, B a + A b\right )} \sqrt {a} c x \log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right ) + {\left (B a b + 2 \, A a c\right )} \sqrt {c} x \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (B a c x - A a c\right )} \sqrt {c x^{2} + b x + a}}{4 \, a c x}, \frac {{\left (2 \, B a + A b\right )} \sqrt {a} c x \log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right ) - 2 \, {\left (B a b + 2 \, A a c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 4 \, {\left (B a c x - A a c\right )} \sqrt {c x^{2} + b x + a}}{4 \, a c x}, \frac {2 \, {\left (2 \, B a + A b\right )} \sqrt {-a} c x \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right ) + {\left (B a b + 2 \, A a c\right )} \sqrt {c} x \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (B a c x - A a c\right )} \sqrt {c x^{2} + b x + a}}{4 \, a c x}, \frac {{\left (2 \, B a + A b\right )} \sqrt {-a} c x \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right ) - {\left (B a b + 2 \, A a c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (B a c x - A a c\right )} \sqrt {c x^{2} + b x + a}}{2 \, a c x}\right ] \]
[1/4*((2*B*a + A*b)*sqrt(a)*c*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt (c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) + (B*a*b + 2*A*a*c)*sq rt(c)*x*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(B*a*c*x - A*a*c)*sqrt(c*x^2 + b*x + a))/(a*c*x), 1/4*((2*B*a + A*b)*sqrt(a)*c*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt( c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 2*(B*a*b + 2*A*a*c)*s qrt(-c)*x*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 4*(B*a*c*x - A*a*c)*sqrt(c*x^2 + b*x + a))/(a*c*x), 1/4*( 2*(2*B*a + A*b)*sqrt(-a)*c*x*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)* sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) + (B*a*b + 2*A*a*c)*sqrt(c)*x*log(-8*c^2 *x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c ) + 4*(B*a*c*x - A*a*c)*sqrt(c*x^2 + b*x + a))/(a*c*x), 1/2*((2*B*a + A*b) *sqrt(-a)*c*x*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x ^2 + a*b*x + a^2)) - (B*a*b + 2*A*a*c)*sqrt(-c)*x*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(B*a*c*x - A*a* c)*sqrt(c*x^2 + b*x + a))/(a*c*x)]
\[ \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^2} \, dx=\int \frac {\left (A + B x\right ) \sqrt {a + b x + c x^{2}}}{x^{2}}\, dx \]
Exception generated. \[ \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.30 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.33 \[ \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^2} \, dx=\sqrt {c x^{2} + b x + a} B + \frac {{\left (2 \, B a + A b\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {{\left (B b + 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, \sqrt {c}} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A b + 2 \, A a \sqrt {c}}{{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a} \]
sqrt(c*x^2 + b*x + a)*B + (2*B*a + A*b)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/sqrt(-a) - 1/2*(B*b + 2*A*c)*log(abs(-2*(sqrt(c)*x - s qrt(c*x^2 + b*x + a))*sqrt(c) - b))/sqrt(c) + ((sqrt(c)*x - sqrt(c*x^2 + b *x + a))*A*b + 2*A*a*sqrt(c))/((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)
Time = 10.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^2} \, dx=B\,\sqrt {c\,x^2+b\,x+a}-\frac {A\,\sqrt {c\,x^2+b\,x+a}}{x}-B\,\sqrt {a}\,\ln \left (\frac {b}{2}+\frac {a}{x}+\frac {\sqrt {a}\,\sqrt {c\,x^2+b\,x+a}}{x}\right )+A\,\sqrt {c}\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )+\frac {B\,b\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{2\,\sqrt {c}}-\frac {A\,b\,\ln \left (\frac {b}{2}+\frac {a}{x}+\frac {\sqrt {a}\,\sqrt {c\,x^2+b\,x+a}}{x}\right )}{2\,\sqrt {a}} \]
B*(a + b*x + c*x^2)^(1/2) - (A*(a + b*x + c*x^2)^(1/2))/x - B*a^(1/2)*log( b/2 + a/x + (a^(1/2)*(a + b*x + c*x^2)^(1/2))/x) + A*c^(1/2)*log((b/2 + c* x)/c^(1/2) + (a + b*x + c*x^2)^(1/2)) + (B*b*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2)))/(2*c^(1/2)) - (A*b*log(b/2 + a/x + (a^(1/2)*(a + b* x + c*x^2)^(1/2))/x))/(2*a^(1/2))