Integrand size = 29, antiderivative size = 340 \[ \int \frac {1}{(d+e x) (f+g x)^2 \sqrt {a+b x+c x^2}} \, dx=\frac {g^2 \sqrt {a+b x+c x^2}}{(e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)}+\frac {e^2 \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c d^2-b d e+a e^2} (e f-d g)^2}-\frac {g (2 c f-b g) \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{2 (e f-d g) \left (c f^2-b f g+a g^2\right )^{3/2}}-\frac {e g \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{(e f-d g)^2 \sqrt {c f^2-b f g+a g^2}} \]
-1/2*g*(-b*g+2*c*f)*arctanh(1/2*(b*f-2*a*g+(-b*g+2*c*f)*x)/(a*g^2-b*f*g+c* f^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/(-d*g+e*f)/(a*g^2-b*f*g+c*f^2)^(3/2)+e^2*a rctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x +a)^(1/2))/(-d*g+e*f)^2/(a*e^2-b*d*e+c*d^2)^(1/2)-e*g*arctanh(1/2*(b*f-2*a *g+(-b*g+2*c*f)*x)/(a*g^2-b*f*g+c*f^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/(-d*g+e* f)^2/(a*g^2-b*f*g+c*f^2)^(1/2)+g^2*(c*x^2+b*x+a)^(1/2)/(-d*g+e*f)/(a*g^2-b *f*g+c*f^2)/(g*x+f)
Time = 10.64 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(d+e x) (f+g x)^2 \sqrt {a+b x+c x^2}} \, dx=-\frac {\frac {2 g^2 (-e f+d g) \sqrt {a+x (b+c x)}}{\left (c f^2+g (-b f+a g)\right ) (f+g x)}-\frac {2 e^2 \text {arctanh}\left (\frac {-2 a e+2 c d x+b (d-e x)}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c d^2+e (-b d+a e)}}+\frac {g (2 c f (2 e f-d g)+g (-3 b e f+b d g+2 a e g)) \text {arctanh}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (-b f+a g)} \sqrt {a+x (b+c x)}}\right )}{\left (c f^2+g (-b f+a g)\right )^{3/2}}}{2 (e f-d g)^2} \]
-1/2*((2*g^2*(-(e*f) + d*g)*Sqrt[a + x*(b + c*x)])/((c*f^2 + g*(-(b*f) + a *g))*(f + g*x)) - (2*e^2*ArcTanh[(-2*a*e + 2*c*d*x + b*(d - e*x))/(2*Sqrt[ c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/Sqrt[c*d^2 + e*(-(b*d) + a*e)] + (g*(2*c*f*(2*e*f - d*g) + g*(-3*b*e*f + b*d*g + 2*a*e*g))*ArcTan h[(-2*a*g + 2*c*f*x + b*(f - g*x))/(2*Sqrt[c*f^2 + g*(-(b*f) + a*g)]*Sqrt[ a + x*(b + c*x)])])/(c*f^2 + g*(-(b*f) + a*g))^(3/2))/(e*f - d*g)^2
Time = 0.61 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1289, 2009}
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.
| \(\displaystyle \int \frac {1}{(d+e x) (f+g x)^2 \sqrt {a+b x+c x^2}} \, dx\) |
| \(\Big \downarrow \) 1289 |
| \(\displaystyle \int \left (\frac {e^2}{(d+e x) \sqrt {a+b x+c x^2} (e f-d g)^2}-\frac {e g}{(f+g x) \sqrt {a+b x+c x^2} (e f-d g)^2}-\frac {g}{(f+g x)^2 \sqrt {a+b x+c x^2} (e f-d g)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^2 \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g)^2 \sqrt {a e^2-b d e+c d^2}}-\frac {e g \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{(e f-d g)^2 \sqrt {a g^2-b f g+c f^2}}-\frac {g (2 c f-b g) \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{2 (e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}}+\frac {g^2 \sqrt {a+b x+c x^2}}{(f+g x) (e f-d g) \left (a g^2-b f g+c f^2\right )}\) |
(g^2*Sqrt[a + b*x + c*x^2])/((e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*(f + g*x) ) + (e^2*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a *e^2]*Sqrt[a + b*x + c*x^2])])/(Sqrt[c*d^2 - b*d*e + a*e^2]*(e*f - d*g)^2) - (g*(2*c*f - b*g)*ArcTanh[(b*f - 2*a*g + (2*c*f - b*g)*x)/(2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqrt[a + b*x + c*x^2])])/(2*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)^(3/2)) - (e*g*ArcTanh[(b*f - 2*a*g + (2*c*f - b*g)*x)/(2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqrt[a + b*x + c*x^2])])/((e*f - d*g)^2*Sqrt[c*f^2 - b*f *g + a*g^2])
3.9.76.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && ( IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0]))
Time = 0.88 (sec) , antiderivative size = 609, normalized size of antiderivative = 1.79
| method | result | size |
| default | \(-\frac {e \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (d g -e f \right )^{2} \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}+\frac {-\frac {g^{2} \sqrt {\left (x +\frac {f}{g}\right )^{2} c +\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}{\left (a \,g^{2}-b f g +c \,f^{2}\right ) \left (x +\frac {f}{g}\right )}+\frac {\left (b g -2 c f \right ) g \ln \left (\frac {\frac {2 a \,g^{2}-2 b f g +2 c \,f^{2}}{g^{2}}+\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+2 \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}\, \sqrt {\left (x +\frac {f}{g}\right )^{2} c +\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}{x +\frac {f}{g}}\right )}{2 \left (a \,g^{2}-b f g +c \,f^{2}\right ) \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}}{g \left (d g -e f \right )}+\frac {e \ln \left (\frac {\frac {2 a \,g^{2}-2 b f g +2 c \,f^{2}}{g^{2}}+\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+2 \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}\, \sqrt {\left (x +\frac {f}{g}\right )^{2} c +\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}{x +\frac {f}{g}}\right )}{\left (d g -e f \right )^{2} \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}\) | \(609\) |
-e/(d*g-e*f)^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e ^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b *e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))+1/g/(d*g-e*f) *(-1/(a*g^2-b*f*g+c*f^2)*g^2/(x+f/g)*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a *g^2-b*f*g+c*f^2)/g^2)^(1/2)+1/2*(b*g-2*c*f)*g/(a*g^2-b*f*g+c*f^2)/((a*g^2 -b*f*g+c*f^2)/g^2)^(1/2)*ln((2*(a*g^2-b*f*g+c*f^2)/g^2+(b*g-2*c*f)/g*(x+f/ g)+2*((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a *g^2-b*f*g+c*f^2)/g^2)^(1/2))/(x+f/g)))+e/(d*g-e*f)^2/((a*g^2-b*f*g+c*f^2) /g^2)^(1/2)*ln((2*(a*g^2-b*f*g+c*f^2)/g^2+(b*g-2*c*f)/g*(x+f/g)+2*((a*g^2- b*f*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c* f^2)/g^2)^(1/2))/(x+f/g))
Timed out. \[ \int \frac {1}{(d+e x) (f+g x)^2 \sqrt {a+b x+c x^2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(d+e x) (f+g x)^2 \sqrt {a+b x+c x^2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x) (f+g x)^2 \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )} {\left (g x + f\right )}^{2}} \,d x } \]
\[ \int \frac {1}{(d+e x) (f+g x)^2 \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )} {\left (g x + f\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{(d+e x) (f+g x)^2 \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{{\left (f+g\,x\right )}^2\,\left (d+e\,x\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \]