Integrand size = 33, antiderivative size = 475 \[ \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {2} \sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g} \sqrt {b-\sqrt {b^2-4 a c}+2 c x} \sqrt {\frac {(e f-d g) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g\right ) (d+e x)}} \sqrt {\frac {(e f-d g) \left (2 a+\left (b+\sqrt {b^2-4 a c}\right ) x\right )}{\left (b f+\sqrt {b^2-4 a c} f-2 a g\right ) (d+e x)}} (d+e x) \operatorname {EllipticPi}\left (\frac {e \left (2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) g},\arcsin \left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \sqrt {f+g x}}{\sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g} \sqrt {d+e x}}\right ),\frac {\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \left (2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b f+\sqrt {b^2-4 a c} f-2 a g\right )}\right )}{\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} g \sqrt {\frac {2 a c}{b+\sqrt {b^2-4 a c}}+c x} \sqrt {a+b x+c x^2}} \]
(e*x+d)*EllipticPi((g*x+f)^(1/2)*(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(e *x+d)^(1/2)/(2*c*f-g*(b+(-4*a*c+b^2)^(1/2)))^(1/2),e*(2*c*f-g*(b+(-4*a*c+b ^2)^(1/2)))/g/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))),((b*d-2*a*e+d*(-4*a*c+b^2)^ (1/2))*(2*c*f-g*(b+(-4*a*c+b^2)^(1/2)))/(b*f-2*a*g+f*(-4*a*c+b^2)^(1/2))/( 2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2))*2^(1/2)*(b+2*c*x-(-4*a*c+b^2)^(1/2 ))^(1/2)*((-d*g+e*f)*(b+2*c*x+(-4*a*c+b^2)^(1/2))/(e*x+d)/(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)*((-d*g+e*f) *(2*a+x*(b+(-4*a*c+b^2)^(1/2)))/(e*x+d)/(b*f-2*a*g+f*(-4*a*c+b^2)^(1/2)))^ (1/2)/g/(c*x^2+b*x+a)^(1/2)/(c*x+2*a*c/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(2*c* d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(1118\) vs. \(2(475)=950\).
Time = 28.87 (sec) , antiderivative size = 1118, normalized size of antiderivative = 2.35 \[ \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx=-\frac {\sqrt {2} \sqrt {-\frac {g \left (c f^2+g (-b f+a g)\right ) (d+e x)}{\left (-2 c d f g-2 a e g^2+e f \sqrt {\left (b^2-4 a c\right ) g^2}-d g \sqrt {\left (b^2-4 a c\right ) g^2}+b g (e f+d g)\right ) (f+g x)}} (f+g x)^{3/2} \left (\frac {2 e f \sqrt {\left (b^2-4 a c\right ) g^2} \sqrt {-\frac {\left (c f^2+g (-b f+a g)\right ) (a+x (b+c x))}{\left (b^2-4 a c\right ) (f+g x)^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {-2 a g^2+2 c f g x+b g (f-g x)+\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}{\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}}}{\sqrt {2}}\right ),\frac {2 \sqrt {\left (b^2-4 a c\right ) g^2} (-e f+d g)}{2 c d f g+2 a e g^2-e f \sqrt {\left (b^2-4 a c\right ) g^2}+d g \sqrt {\left (b^2-4 a c\right ) g^2}-b g (e f+d g)}\right )}{c f^2+g (-b f+a g)}+\frac {d g \left (2 a g^2-f \sqrt {\left (b^2-4 a c\right ) g^2}-2 c f g x-g \sqrt {\left (b^2-4 a c\right ) g^2} x+b g (-f+g x)\right ) \sqrt {\frac {2 a g^2-2 c f g x+b g (-f+g x)+\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}{\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {-2 a g^2+2 c f g x+b g (f-g x)+\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}{\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}}}{\sqrt {2}}\right ),\frac {2 \sqrt {\left (b^2-4 a c\right ) g^2} (-e f+d g)}{2 c d f g+2 a e g^2-e f \sqrt {\left (b^2-4 a c\right ) g^2}+d g \sqrt {\left (b^2-4 a c\right ) g^2}-b g (e f+d g)}\right )}{\left (c f^2+g (-b f+a g)\right ) (f+g x) \sqrt {\frac {-2 a g^2+2 c f g x+b g (f-g x)+\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}{\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}}}-\frac {4 e \sqrt {\left (b^2-4 a c\right ) g^2} \sqrt {-\frac {\left (c f^2+g (-b f+a g)\right ) (a+x (b+c x))}{\left (b^2-4 a c\right ) (f+g x)^2}} \operatorname {EllipticPi}\left (\frac {2 \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}},\arcsin \left (\frac {\sqrt {\frac {-2 a g^2+2 c f g x+b g (f-g x)+\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}{\sqrt {\left (b^2-4 a c\right ) g^2} (f+g x)}}}{\sqrt {2}}\right ),\frac {2 \sqrt {\left (b^2-4 a c\right ) g^2} (-e f+d g)}{2 c d f g+2 a e g^2-e f \sqrt {\left (b^2-4 a c\right ) g^2}+d g \sqrt {\left (b^2-4 a c\right ) g^2}-b g (e f+d g)}\right )}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )}{g^2 \sqrt {d+e x} \sqrt {a+x (b+c x)}} \]
-((Sqrt[2]*Sqrt[-((g*(c*f^2 + g*(-(b*f) + a*g))*(d + e*x))/((-2*c*d*f*g - 2*a*e*g^2 + e*f*Sqrt[(b^2 - 4*a*c)*g^2] - d*g*Sqrt[(b^2 - 4*a*c)*g^2] + b* g*(e*f + d*g))*(f + g*x)))]*(f + g*x)^(3/2)*((2*e*f*Sqrt[(b^2 - 4*a*c)*g^2 ]*Sqrt[-(((c*f^2 + g*(-(b*f) + a*g))*(a + x*(b + c*x)))/((b^2 - 4*a*c)*(f + g*x)^2))]*EllipticF[ArcSin[Sqrt[(-2*a*g^2 + 2*c*f*g*x + b*g*(f - g*x) + Sqrt[(b^2 - 4*a*c)*g^2]*(f + g*x))/(Sqrt[(b^2 - 4*a*c)*g^2]*(f + g*x))]/Sq rt[2]], (2*Sqrt[(b^2 - 4*a*c)*g^2]*(-(e*f) + d*g))/(2*c*d*f*g + 2*a*e*g^2 - e*f*Sqrt[(b^2 - 4*a*c)*g^2] + d*g*Sqrt[(b^2 - 4*a*c)*g^2] - b*g*(e*f + d *g))])/(c*f^2 + g*(-(b*f) + a*g)) + (d*g*(2*a*g^2 - f*Sqrt[(b^2 - 4*a*c)*g ^2] - 2*c*f*g*x - g*Sqrt[(b^2 - 4*a*c)*g^2]*x + b*g*(-f + g*x))*Sqrt[(2*a* g^2 - 2*c*f*g*x + b*g*(-f + g*x) + Sqrt[(b^2 - 4*a*c)*g^2]*(f + g*x))/(Sqr t[(b^2 - 4*a*c)*g^2]*(f + g*x))]*EllipticF[ArcSin[Sqrt[(-2*a*g^2 + 2*c*f*g *x + b*g*(f - g*x) + Sqrt[(b^2 - 4*a*c)*g^2]*(f + g*x))/(Sqrt[(b^2 - 4*a*c )*g^2]*(f + g*x))]/Sqrt[2]], (2*Sqrt[(b^2 - 4*a*c)*g^2]*(-(e*f) + d*g))/(2 *c*d*f*g + 2*a*e*g^2 - e*f*Sqrt[(b^2 - 4*a*c)*g^2] + d*g*Sqrt[(b^2 - 4*a*c )*g^2] - b*g*(e*f + d*g))])/((c*f^2 + g*(-(b*f) + a*g))*(f + g*x)*Sqrt[(-2 *a*g^2 + 2*c*f*g*x + b*g*(f - g*x) + Sqrt[(b^2 - 4*a*c)*g^2]*(f + g*x))/(S qrt[(b^2 - 4*a*c)*g^2]*(f + g*x))]) - (4*e*Sqrt[(b^2 - 4*a*c)*g^2]*Sqrt[-( ((c*f^2 + g*(-(b*f) + a*g))*(a + x*(b + c*x)))/((b^2 - 4*a*c)*(f + g*x)^2) )]*EllipticPi[(2*Sqrt[(b^2 - 4*a*c)*g^2])/(2*c*f - b*g + Sqrt[(b^2 - 4*...
Time = 0.59 (sec) , antiderivative size = 475, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {1276}
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.
| \(\displaystyle \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx\) |
| \(\Big \downarrow \) 1276 |
| \(\displaystyle \frac {\sqrt {2} (d+e x) \sqrt {-\sqrt {b^2-4 a c}+b+2 c x} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )} \sqrt {\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) (e f-d g)}{(d+e x) \left (2 c f-g \left (\sqrt {b^2-4 a c}+b\right )\right )}} \sqrt {\frac {\left (x \left (\sqrt {b^2-4 a c}+b\right )+2 a\right ) (e f-d g)}{(d+e x) \left (f \sqrt {b^2-4 a c}-2 a g+b f\right )}} \operatorname {EllipticPi}\left (\frac {e \left (2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) g},\arcsin \left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \sqrt {f+g x}}{\sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g} \sqrt {d+e x}}\right ),\frac {\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \left (2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b f+\sqrt {b^2-4 a c} f-2 a g\right )}\right )}{g \sqrt {\frac {2 a c}{\sqrt {b^2-4 a c}+b}+c x} \sqrt {a+b x+c x^2} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\) |
(Sqrt[2]*Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g]*Sqrt[b - Sqrt[b^2 - 4*a*c ] + 2*c*x]*Sqrt[((e*f - d*g)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)*(d + e*x))]*Sqrt[((e*f - d*g)*(2*a + (b + Sqrt[b^ 2 - 4*a*c])*x))/((b*f + Sqrt[b^2 - 4*a*c]*f - 2*a*g)*(d + e*x))]*(d + e*x) *EllipticPi[(e*(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g))/((2*c*d - (b + Sqrt[b^ 2 - 4*a*c])*e)*g), ArcSin[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])/(Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x])], ((b*d + Sqrt[b^2 - 4*a*c]*d - 2*a*e)*(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b*f + Sqrt[b^2 - 4*a*c]*f - 2*a*g))])/(Sqrt[ 2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*g*Sqrt[(2*a*c)/(b + Sqrt[b^2 - 4*a*c]) + c*x]*Sqrt[a + b*x + c*x^2])
3.10.18.3.1 Defintions of rubi rules used
Int[Sqrt[(d_.) + (e_.)*(x_)]/(Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_.) + (b_.)*( x_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[Sqrt [2]*Sqrt[2*c*f - g*(b + q)]*Sqrt[b - q + 2*c*x]*(d + e*x)*Sqrt[(e*f - d*g)* ((b + q + 2*c*x)/((2*c*f - g*(b + q))*(d + e*x)))]*(Sqrt[(e*f - d*g)*((2*a + (b + q)*x)/((b*f + q*f - 2*a*g)*(d + e*x)))]/(g*Sqrt[2*c*d - e*(b + q)]*S qrt[2*a*(c/(b + q)) + c*x]*Sqrt[a + b*x + c*x^2]))*EllipticPi[e*((2*c*f - g *(b + q))/(g*(2*c*d - e*(b + q)))), ArcSin[Sqrt[2*c*d - e*(b + q)]*(Sqrt[f + g*x]/(Sqrt[2*c*f - g*(b + q)]*Sqrt[d + e*x]))], (b*d + q*d - 2*a*e)*((2*c *f - g*(b + q))/((b*f + q*f - 2*a*g)*(2*c*d - e*(b + q))))], x]] /; FreeQ[{ a, b, c, d, e, f, g}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1462\) vs. \(2(420)=840\).
Time = 5.12 (sec) , antiderivative size = 1463, normalized size of antiderivative = 3.08
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(1463\) |
| default | \(\text {Expression too large to display}\) | \(10161\) |
((g*x+f)*(c*x^2+b*x+a)*(e*x+d))^(1/2)/(g*x+f)^(1/2)/(c*x^2+b*x+a)^(1/2)/(e *x+d)^(1/2)*(2*d*(-f/g+d/e)*((-d/e-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))*(x+f/g)/ (-d/e+f/g)/(x-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)*(x-1/2/c*(-b+(-4*a*c+b ^2)^(1/2)))^2*((1/2/c*(-b+(-4*a*c+b^2)^(1/2))+f/g)*(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)/(x-1/2/c*(-b+(-4*a*c+b^2)^(1/ 2))))^(1/2)*((1/2/c*(-b+(-4*a*c+b^2)^(1/2))+f/g)*(x+d/e)/(-d/e+f/g)/(x-1/2 /c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)/(-d/e-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(1 /2/c*(-b+(-4*a*c+b^2)^(1/2))+f/g)/(c*e*g*(x+f/g)*(x-1/2/c*(-b+(-4*a*c+b^2) ^(1/2)))*(x+1/2*(b+(-4*a*c+b^2)^(1/2))/c)*(x+d/e))^(1/2)*EllipticF(((-d/e- 1/2/c*(-b+(-4*a*c+b^2)^(1/2)))*(x+f/g)/(-d/e+f/g)/(x-1/2/c*(-b+(-4*a*c+b^2 )^(1/2))))^(1/2),((1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2) )/c)*(-f/g+d/e)/(-f/g+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(1/2/c*(-b+(-4*a*c+b^2 )^(1/2))+d/e))^(1/2))+2*e*(-f/g+d/e)*((-d/e-1/2/c*(-b+(-4*a*c+b^2)^(1/2))) *(x+f/g)/(-d/e+f/g)/(x-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)*(x-1/2/c*(-b+ (-4*a*c+b^2)^(1/2)))^2*((1/2/c*(-b+(-4*a*c+b^2)^(1/2))+f/g)*(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)/(x-1/2/c*(-b+(-4*a*c +b^2)^(1/2))))^(1/2)*((1/2/c*(-b+(-4*a*c+b^2)^(1/2))+f/g)*(x+d/e)/(-d/e+f/ g)/(x-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)/(-d/e-1/2/c*(-b+(-4*a*c+b^2)^( 1/2)))/(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+f/g)/(c*e*g*(x+f/g)*(x-1/2/c*(-b+(-4 *a*c+b^2)^(1/2)))*(x+1/2*(b+(-4*a*c+b^2)^(1/2))/c)*(x+d/e))^(1/2)*(1/2/...
Timed out. \[ \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx=\int \frac {\sqrt {d + e x}}{\sqrt {f + g x} \sqrt {a + b x + c x^{2}}}\, dx \]
\[ \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {\sqrt {e x + d}}{\sqrt {c x^{2} + b x + a} \sqrt {g x + f}} \,d x } \]
\[ \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {\sqrt {e x + d}}{\sqrt {c x^{2} + b x + a} \sqrt {g x + f}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx=\int \frac {\sqrt {d+e\,x}}{\sqrt {f+g\,x}\,\sqrt {c\,x^2+b\,x+a}} \,d x \]