Integrand size = 29, antiderivative size = 352 \[ \int \frac {1}{(d+e x) (f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 e \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (e f-d g) \sqrt {a+b x+c x^2}}+\frac {2 g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right ) \sqrt {a+b x+c x^2}}+\frac {e^3 \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{\left (c d^2-b d e+a e^2\right )^{3/2} (e f-d g)}-\frac {g^3 \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{(e f-d g) \left (c f^2-b f g+a g^2\right )^{3/2}} \]
e^3*arctanh(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))/(a*e^2-b*d*e+c*d^2)^(3/2)/(-d*g+e*f)-g^3*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)-2*e*(b*c*d-b^2*e+2*a*c*e+c*(-b*e+2*c*d)*x) /(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)/(-d*g+e*f)/(c*x^2+b*x+a)^(1/2)+2*g*(b*c* f-b^2*g+2*a*c*g+c*(-b*g+2*c*f)*x)/(-4*a*c+b^2)/(-d*g+e*f)/(a*g^2-b*f*g+c*f ^2)/(c*x^2+b*x+a)^(1/2)
Time = 3.11 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(d+e x) (f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (-b^3 e g+b^2 c (d g+e (f-g x))-2 c^2 (a d g+c d f x+a e (f-g x))+b c (3 a e g+c (-d f+e f x+d g x))\right )}{\left (b^2-4 a c\right ) \left (-c d^2+e (b d-a e)\right ) \left (-c f^2+g (b f-a g)\right ) \sqrt {a+x (b+c x)}}-\frac {2 e^3 \sqrt {-c d^2+b d e-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{\left (c d^2+e (-b d+a e)\right )^2 (-e f+d g)}-\frac {2 g^3 \sqrt {-c f^2+b f g-a g^2} \arctan \left (\frac {\sqrt {c} (f+g x)-g \sqrt {a+x (b+c x)}}{\sqrt {-c f^2+g (b f-a g)}}\right )}{(e f-d g) \left (c f^2+g (-b f+a g)\right )^2} \]
(2*(-(b^3*e*g) + b^2*c*(d*g + e*(f - g*x)) - 2*c^2*(a*d*g + c*d*f*x + a*e* (f - g*x)) + b*c*(3*a*e*g + c*(-(d*f) + e*f*x + d*g*x))))/((b^2 - 4*a*c)*( -(c*d^2) + e*(b*d - a*e))*(-(c*f^2) + g*(b*f - a*g))*Sqrt[a + x*(b + c*x)] ) - (2*e^3*Sqrt[-(c*d^2) + b*d*e - a*e^2]*ArcTan[(Sqrt[c]*(d + e*x) - e*Sq rt[a + x*(b + c*x)])/Sqrt[-(c*d^2) + e*(b*d - a*e)]])/((c*d^2 + e*(-(b*d) + a*e))^2*(-(e*f) + d*g)) - (2*g^3*Sqrt[-(c*f^2) + b*f*g - a*g^2]*ArcTan[( Sqrt[c]*(f + g*x) - g*Sqrt[a + x*(b + c*x)])/Sqrt[-(c*f^2) + g*(b*f - a*g) ]])/((e*f - d*g)*(c*f^2 + g*(-(b*f) + a*g))^2)
Time = 0.67 (sec) , antiderivative size = 352, 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.
| \(\displaystyle \int \frac {1}{(d+e x) (f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1289 |
| \(\displaystyle \int \left (\frac {e}{(d+e x) \left (a+b x+c x^2\right )^{3/2} (e f-d g)}-\frac {g}{(f+g x) \left (a+b x+c x^2\right )^{3/2} (e f-d g)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^3 \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g) \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {g^3 \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{(e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}}-\frac {2 e \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (e f-d g) \left (a e^2-b d e+c d^2\right )}+\frac {2 g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right )}\) |
(-2*e*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(e*f - d*g)*Sqrt[a + b*x + c*x^2]) + (2*g*(b*c*f - b^2*g + 2*a*c*g + c*(2*c*f - b*g)*x))/((b^2 - 4*a*c)*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*Sqrt[a + b*x + c*x^2]) + (e^3*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])])/((c*d^2 - b* d*e + a*e^2)^(3/2)*(e*f - d*g)) - (g^3*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)* (c*f^2 - b*f*g + a*g^2)^(3/2))
3.9.83.3.1 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]))
Leaf count of result is larger than twice the leaf count of optimal. \(814\) vs. \(2(332)=664\).
Time = 0.77 (sec) , antiderivative size = 815, normalized size of antiderivative = 2.32
| method | result | size |
| default | \(\frac {\frac {g^{2}}{\left (a \,g^{2}-b f g +c \,f^{2}\right ) \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}}}}-\frac {\left (b g -2 c f \right ) g \left (2 c \left (x +\frac {f}{g}\right )+\frac {b g -2 c f}{g}\right )}{\left (a \,g^{2}-b f g +c \,f^{2}\right ) \left (\frac {4 c \left (a \,g^{2}-b f g +c \,f^{2}\right )}{g^{2}}-\frac {\left (b g -2 c f \right )^{2}}{g^{2}}\right ) \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}}}}-\frac {g^{2} \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 (a \,g^{2}-b f g +c \,f^{2}\right ) \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}}{d g -e f}-\frac {\frac {e^{2}}{\left (e^{2} a -b d e +c \,d^{2}\right ) \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}}}}-\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (e^{2} a -b d e +c \,d^{2}\right ) \left (\frac {4 c \left (e^{2} a -b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \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}}}}-\frac {e^{2} \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 (e^{2} a -b d e +c \,d^{2}\right ) \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}}{d g -e f}\) | \(815\) |
1/(d*g-e*f)*(1/(a*g^2-b*f*g+c*f^2)*g^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)-(b*g-2*c*f)*g/(a*g^2-b*f*g+c*f^2)*(2*c*(x+f /g)+(b*g-2*c*f)/g)/(4*c*(a*g^2-b*f*g+c*f^2)/g^2-(b*g-2*c*f)^2/g^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)-1/(a*g^2-b*f*g+ c*f^2)*g^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)))-1/(d*g-e*f)*(1/( a*e^2-b*d*e+c*d^2)*e^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)-(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(2*c*(x+d/e)+(b*e-2*c*d)/ e)/(4*c*(a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^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)-1/(a*e^2-b*d*e+c*d^2)*e^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)))
Timed out. \[ \int \frac {1}{(d+e x) (f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x) (f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (d + e x\right ) \left (f + g x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{(d+e x) (f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (e x + d\right )} {\left (g x + f\right )}} \,d x } \]
Exception generated. \[ \int \frac {1}{(d+e x) (f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {1}{(d+e x) (f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (f+g\,x\right )\,\left (d+e\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]