Integrand size = 31, antiderivative size = 553 \[ \int \frac {1}{(d+e x) (f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx=\frac {2 g^2 \sqrt {a+b x+c x^2}}{(e f-d g) \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}-\frac {\sqrt {2} \sqrt {b^2-4 a c} g \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\arcsin \left (\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{(e f-d g) \left (c f^2-b f g+a g^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {a+b x+c x^2}}-\frac {\sqrt {2} e \sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} \sqrt {1-\frac {2 c (f+g x)}{2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \operatorname {EllipticPi}\left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)},\arcsin \left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right ),\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right )}{\sqrt {c} (e f-d g)^2 \sqrt {a+b x+c x^2}} \]
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)^(1/2)-g*E llipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/ 2),(-2*g*(-4*a*c+b^2)^(1/2)/(2*c*f-g*(b+(-4*a*c+b^2)^(1/2))))^(1/2))*2^(1/ 2)*(-4*a*c+b^2)^(1/2)*(g*x+f)^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)/ (-d*g+e*f)/(a*g^2-b*f*g+c*f^2)/(c*x^2+b*x+a)^(1/2)/(c*(g*x+f)/(2*c*f-g*(b+ (-4*a*c+b^2)^(1/2))))^(1/2)-e*EllipticPi(2^(1/2)*c^(1/2)*(g*x+f)^(1/2)/(2* c*f-g*(b-(-4*a*c+b^2)^(1/2)))^(1/2),1/2*e*(2*c*f-b*g+g*(-4*a*c+b^2)^(1/2)) /c/(-d*g+e*f),((b-2*c*f/g-(-4*a*c+b^2)^(1/2))/(b-2*c*f/g+(-4*a*c+b^2)^(1/2 )))^(1/2))*2^(1/2)*(1-2*c*(g*x+f)/(2*c*f-g*(b-(-4*a*c+b^2)^(1/2))))^(1/2)* (2*c*f-g*(b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1-2*c*(g*x+f)/(2*c*f-g*(b+(-4*a*c+ b^2)^(1/2))))^(1/2)/(-d*g+e*f)^2/c^(1/2)/(c*x^2+b*x+a)^(1/2)
Result contains complex when optimal does not.
Time = 26.22 (sec) , antiderivative size = 950, normalized size of antiderivative = 1.72 \[ \int \frac {1}{(d+e x) (f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx=\frac {2 \left (\frac {g^2 (a+x (b+c x))}{e f-d g}+\frac {(f+g x)^2 \left (c+\frac {c f^2}{(f+g x)^2}-\frac {b f g}{(f+g x)^2}+\frac {a g^2}{(f+g x)^2}-\frac {2 c f}{f+g x}+\frac {b g}{f+g x}-\frac {i \sqrt {1-\frac {2 \left (c f^2+g (-b f+a g)\right )}{\left (2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}\right ) (f+g x)}} \sqrt {2+\frac {4 \left (c f^2+g (-b f+a g)\right )}{\left (-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}\right ) (f+g x)}} \left ((e f-d g) \left (2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}\right ) \left (E\left (i \text {arcsinh}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b f g+a g^2}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}}}{\sqrt {f+g x}}\right )|-\frac {-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b f g+a g^2}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}}}{\sqrt {f+g x}}\right ),-\frac {-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )\right )-2 e \left (c f^2+g (-b f+a g)\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b f g+a g^2}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}}}{\sqrt {f+g x}}\right ),-\frac {-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )+2 e \left (c f^2+g (-b f+a g)\right ) \operatorname {EllipticPi}\left (\frac {(e f-d g) \left (2 c f-b g-\sqrt {\left (b^2-4 a c\right ) g^2}\right )}{2 e \left (c f^2+g (-b f+a g)\right )},i \text {arcsinh}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b f g+a g^2}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}}}{\sqrt {f+g x}}\right ),-\frac {-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )\right )}{4 (e f-d g) \sqrt {\frac {c f^2+g (-b f+a g)}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}} \sqrt {f+g x}}\right )}{-e f+d g}\right )}{\left (c f^2+g (-b f+a g)\right ) \sqrt {f+g x} \sqrt {a+x (b+c x)}} \]
(2*((g^2*(a + x*(b + c*x)))/(e*f - d*g) + ((f + g*x)^2*(c + (c*f^2)/(f + g *x)^2 - (b*f*g)/(f + g*x)^2 + (a*g^2)/(f + g*x)^2 - (2*c*f)/(f + g*x) + (b *g)/(f + g*x) - ((I/4)*Sqrt[1 - (2*(c*f^2 + g*(-(b*f) + a*g)))/((2*c*f - b *g + Sqrt[(b^2 - 4*a*c)*g^2])*(f + g*x))]*Sqrt[2 + (4*(c*f^2 + g*(-(b*f) + a*g)))/((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(f + g*x))]*((e*f - d*g) *(2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(EllipticE[I*ArcSinh[(Sqrt[2]*Sqr t[(c*f^2 - b*f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])])/Sqrt[ f + g*x]], -((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])/(2*c*f - b*g + Sqrt[ (b^2 - 4*a*c)*g^2]))] - EllipticF[I*ArcSinh[(Sqrt[2]*Sqrt[(c*f^2 - b*f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])])/Sqrt[f + g*x]], -((-2*c *f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])/(2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2] ))]) - 2*e*(c*f^2 + g*(-(b*f) + a*g))*EllipticF[I*ArcSinh[(Sqrt[2]*Sqrt[(c *f^2 - b*f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])])/Sqrt[f + g*x]], -((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])/(2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2]))] + 2*e*(c*f^2 + g*(-(b*f) + a*g))*EllipticPi[((e*f - d*g) *(2*c*f - b*g - Sqrt[(b^2 - 4*a*c)*g^2]))/(2*e*(c*f^2 + g*(-(b*f) + a*g))) , I*ArcSinh[(Sqrt[2]*Sqrt[(c*f^2 - b*f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^ 2 - 4*a*c)*g^2])])/Sqrt[f + g*x]], -((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^ 2])/(2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2]))]))/((e*f - d*g)*Sqrt[(c*f^2 + g*(-(b*f) + a*g))/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])]*Sqrt[f + g...
Time = 1.36 (sec) , antiderivative size = 553, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1288, 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.
| \(\displaystyle \int \frac {1}{(d+e x) (f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx\) |
| \(\Big \downarrow \) 1288 |
| \(\displaystyle \int \left (\frac {e}{(d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2} (e f-d g)}-\frac {g}{(f+g x)^{3/2} \sqrt {a+b x+c x^2} (e f-d g)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt {2} g \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\arcsin \left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{\sqrt {a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {\sqrt {2} e \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} \operatorname {EllipticPi}\left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)},\arcsin \left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right ),\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right )}{\sqrt {c} \sqrt {a+b x+c x^2} (e f-d g)^2}+\frac {2 g^2 \sqrt {a+b x+c x^2}}{\sqrt {f+g x} (e f-d g) \left (a g^2-b f g+c f^2\right )}\) |
(2*g^2*Sqrt[a + b*x + c*x^2])/((e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*Sqrt[f + g*x]) - (Sqrt[2]*Sqrt[b^2 - 4*a*c]*g*Sqrt[f + g*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c *x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*g)/(2*c*f - (b + Sq rt[b^2 - 4*a*c])*g)])/((e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*Sqrt[(c*(f + g* x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[a + b*x + c*x^2]) - (Sqrt[2] *e*Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c])*g]*Sqrt[1 - (2*c*(f + g*x))/(2*c*f - (b - Sqrt[b^2 - 4*a*c])*g)]*Sqrt[1 - (2*c*(f + g*x))/(2*c*f - (b + Sqrt [b^2 - 4*a*c])*g)]*EllipticPi[(e*(2*c*f - b*g + Sqrt[b^2 - 4*a*c]*g))/(2*c *(e*f - d*g)), ArcSin[(Sqrt[2]*Sqrt[c]*Sqrt[f + g*x])/Sqrt[2*c*f - (b - Sq rt[b^2 - 4*a*c])*g]], (b - Sqrt[b^2 - 4*a*c] - (2*c*f)/g)/(b + Sqrt[b^2 - 4*a*c] - (2*c*f)/g)])/(Sqrt[c]*(e*f - d*g)^2*Sqrt[a + b*x + c*x^2])
3.10.16.3.1 Defintions of rubi rules used
Int[((f_.) + (g_.)*(x_))^(n_)/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Int[ExpandIntegrand[1/(Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2]), (f + g*x)^(n + 1/2)/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[n + 1/2]
Leaf count of result is larger than twice the leaf count of optimal. \(1263\) vs. \(2(490)=980\).
Time = 3.23 (sec) , antiderivative size = 1264, normalized size of antiderivative = 2.29
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(1264\) |
| default | \(\text {Expression too large to display}\) | \(4757\) |
((g*x+f)*(c*x^2+b*x+a))^(1/2)/(g*x+f)^(1/2)/(c*x^2+b*x+a)^(1/2)*(-2*(c*g*x ^2+b*g*x+a*g)/(a*g^2-b*f*g+c*f^2)*g/(d*g-e*f)/((x+f/g)*(c*g*x^2+b*g*x+a*g) )^(1/2)+2*(-g*(b*g-c*f)/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)+b*g^2/(a*g^2-b*f*g+c *f^2)/(d*g-e*f))*(f/g-1/2*(b+(-4*a*c+b^2)^(1/2))/c)*((x+f/g)/(f/g-1/2*(b+( -4*a*c+b^2)^(1/2))/c))^(1/2)*((x-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-f/g-1/2/ c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)*((x+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(-f/g+ 1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)/(c*g*x^3+b*g*x^2+c*f*x^2+a*g*x+b*f*x+ a*f)^(1/2)*EllipticF(((x+f/g)/(f/g-1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2),(( -f/g+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(-f/g-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^( 1/2))+2*g^2*c/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)*(f/g-1/2*(b+(-4*a*c+b^2)^(1/2) )/c)*((x+f/g)/(f/g-1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)*((x-1/2/c*(-b+(-4* a*c+b^2)^(1/2)))/(-f/g-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)*((x+1/2*(b+(- 4*a*c+b^2)^(1/2))/c)/(-f/g+1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)/(c*g*x^3+b *g*x^2+c*f*x^2+a*g*x+b*f*x+a*f)^(1/2)*((-f/g-1/2/c*(-b+(-4*a*c+b^2)^(1/2)) )*EllipticE(((x+f/g)/(f/g-1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2),((-f/g+1/2* (b+(-4*a*c+b^2)^(1/2))/c)/(-f/g-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2))+1/2 /c*(-b+(-4*a*c+b^2)^(1/2))*EllipticF(((x+f/g)/(f/g-1/2*(b+(-4*a*c+b^2)^(1/ 2))/c))^(1/2),((-f/g+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(-f/g-1/2/c*(-b+(-4*a*c +b^2)^(1/2))))^(1/2)))-2/(d*g-e*f)*(f/g-1/2*(b+(-4*a*c+b^2)^(1/2))/c)*((x+ f/g)/(f/g-1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)*((x-1/2/c*(-b+(-4*a*c+b^...
Timed out. \[ \int \frac {1}{(d+e x) (f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x) (f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{\left (d + e x\right ) \left (f + g x\right )^{\frac {3}{2}} \sqrt {a + b x + c x^{2}}}\, dx \]
\[ \int \frac {1}{(d+e x) (f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )} {\left (g x + f\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{(d+e x) (f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )} {\left (g x + f\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(d+e x) (f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{{\left (f+g\,x\right )}^{3/2}\,\left (d+e\,x\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \]