Integrand size = 31, antiderivative size = 575 \[ \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {6+5 x+x^2}} \, dx=\frac {e \sqrt {f+g x} \sqrt {6+5 x+x^2}}{\sqrt {d+e x}}+\frac {\sqrt {d-3 e} \sqrt {f-3 g} (2+x) \sqrt {\frac {(e f-d g) (3+x)}{(f-3 g) (d+e x)}} E\left (\arcsin \left (\frac {\sqrt {d-3 e} \sqrt {f+g x}}{\sqrt {f-3 g} \sqrt {d+e x}}\right )|\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )}{\sqrt {\frac {(e f-d g) (2+x)}{(f-2 g) (d+e x)}} \sqrt {6+5 x+x^2}}-\frac {\sqrt {d-3 e} \sqrt {f-3 g} (e f-d g) (2+x) \sqrt {\frac {(e f-d g) (3+x)}{(f-3 g) (d+e x)}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d-3 e} \sqrt {f+g x}}{\sqrt {f-3 g} \sqrt {d+e x}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )}{e (f-2 g) \sqrt {\frac {(e f-d g) (2+x)}{(f-2 g) (d+e x)}} \sqrt {6+5 x+x^2}}+\frac {\sqrt {f-3 g} (e f-d g) (e (f-5 g)+d g) (2+x) \sqrt {\frac {(e f-d g) (3+x)}{(f-3 g) (d+e x)}} \operatorname {EllipticPi}\left (\frac {e (f-3 g)}{(d-3 e) g},\arcsin \left (\frac {\sqrt {d-3 e} \sqrt {f+g x}}{\sqrt {f-3 g} \sqrt {d+e x}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )}{\sqrt {d-3 e} e (f-2 g) g \sqrt {\frac {(e f-d g) (2+x)}{(f-2 g) (d+e x)}} \sqrt {6+5 x+x^2}} \] Output:
e*(g*x+f)^(1/2)*(x^2+5*x+6)^(1/2)/(e*x+d)^(1/2)+(d-3*e)^(1/2)*(f-3*g)^(1/2 )*(2+x)*((-d*g+e*f)*(3+x)/(f-3*g)/(e*x+d))^(1/2)*EllipticE((d-3*e)^(1/2)*( g*x+f)^(1/2)/(f-3*g)^(1/2)/(e*x+d)^(1/2),((d-2*e)*(f-3*g)/(d-3*e)/(f-2*g)) ^(1/2))/((-d*g+e*f)*(2+x)/(f-2*g)/(e*x+d))^(1/2)/(x^2+5*x+6)^(1/2)-(d-3*e) ^(1/2)*(f-3*g)^(1/2)*(-d*g+e*f)*(2+x)*((-d*g+e*f)*(3+x)/(f-3*g)/(e*x+d))^( 1/2)*EllipticF((d-3*e)^(1/2)*(g*x+f)^(1/2)/(f-3*g)^(1/2)/(e*x+d)^(1/2),((d -2*e)*(f-3*g)/(d-3*e)/(f-2*g))^(1/2))/e/(f-2*g)/((-d*g+e*f)*(2+x)/(f-2*g)/ (e*x+d))^(1/2)/(x^2+5*x+6)^(1/2)+(f-3*g)^(1/2)*(-d*g+e*f)*(e*(f-5*g)+d*g)* (2+x)*((-d*g+e*f)*(3+x)/(f-3*g)/(e*x+d))^(1/2)*EllipticPi((d-3*e)^(1/2)*(g *x+f)^(1/2)/(f-3*g)^(1/2)/(e*x+d)^(1/2),e*(f-3*g)/(d-3*e)/g,((d-2*e)*(f-3* g)/(d-3*e)/(f-2*g))^(1/2))/(d-3*e)^(1/2)/e/(f-2*g)/g/((-d*g+e*f)*(2+x)/(f- 2*g)/(e*x+d))^(1/2)/(x^2+5*x+6)^(1/2)
Time = 34.60 (sec) , antiderivative size = 820, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {6+5 x+x^2}} \, dx=\frac {(f-2 g) (3+x) (d+e x) (f+g x)-2 d f (f-3 g) (2+x)^2 \sqrt {-\frac {d+e x}{(d-3 e) (2+x)}} \sqrt {-\frac {(f-2 g) (3+x) (f+g x)}{(f-3 g)^2 (2+x)^2}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {(f-2 g) (3+x)}{(f-3 g) (2+x)}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )+2 e f (f-3 g) (2+x)^2 \sqrt {-\frac {d+e x}{(d-3 e) (2+x)}} \sqrt {-\frac {(f-2 g) (3+x) (f+g x)}{(f-3 g)^2 (2+x)^2}} \left (2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {(f-2 g) (3+x)}{(f-3 g) (2+x)}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )+\operatorname {EllipticPi}\left (\frac {f-3 g}{f-2 g},\arcsin \left (\sqrt {\frac {(f-2 g) (3+x)}{(f-3 g) (2+x)}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )\right )+2 d (f-3 g) g (2+x)^2 \sqrt {-\frac {d+e x}{(d-3 e) (2+x)}} \sqrt {-\frac {(f-2 g) (3+x) (f+g x)}{(f-3 g)^2 (2+x)^2}} \left (2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {(f-2 g) (3+x)}{(f-3 g) (2+x)}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )+\operatorname {EllipticPi}\left (\frac {f-3 g}{f-2 g},\arcsin \left (\sqrt {\frac {(f-2 g) (3+x)}{(f-3 g) (2+x)}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )\right )+(f-3 g) (2+x)^2 \sqrt {-\frac {d+e x}{(d-3 e) (2+x)}} \sqrt {-\frac {(f-2 g) (3+x) (f+g x)}{(f-3 g)^2 (2+x)^2}} \left ((d-3 e) (f-2 g) E\left (\arcsin \left (\sqrt {\frac {(f-2 g) (3+x)}{(f-3 g) (2+x)}}\right )|\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )+e (f-10 g) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {(f-2 g) (3+x)}{(f-3 g) (2+x)}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )-(d g+e (f+5 g)) \operatorname {EllipticPi}\left (\frac {f-3 g}{f-2 g},\arcsin \left (\sqrt {\frac {(f-2 g) (3+x)}{(f-3 g) (2+x)}}\right ),\frac {(d-2 e) (f-3 g)}{(d-3 e) (f-2 g)}\right )\right )}{(f-2 g) \sqrt {d+e x} \sqrt {f+g x} \sqrt {6+5 x+x^2}} \] Input:
Integrate[(Sqrt[d + e*x]*Sqrt[f + g*x])/Sqrt[6 + 5*x + x^2],x]
Output:
((f - 2*g)*(3 + x)*(d + e*x)*(f + g*x) - 2*d*f*(f - 3*g)*(2 + x)^2*Sqrt[-( (d + e*x)/((d - 3*e)*(2 + x)))]*Sqrt[-(((f - 2*g)*(3 + x)*(f + g*x))/((f - 3*g)^2*(2 + x)^2))]*EllipticF[ArcSin[Sqrt[((f - 2*g)*(3 + x))/((f - 3*g)* (2 + x))]], ((d - 2*e)*(f - 3*g))/((d - 3*e)*(f - 2*g))] + 2*e*f*(f - 3*g) *(2 + x)^2*Sqrt[-((d + e*x)/((d - 3*e)*(2 + x)))]*Sqrt[-(((f - 2*g)*(3 + x )*(f + g*x))/((f - 3*g)^2*(2 + x)^2))]*(2*EllipticF[ArcSin[Sqrt[((f - 2*g) *(3 + x))/((f - 3*g)*(2 + x))]], ((d - 2*e)*(f - 3*g))/((d - 3*e)*(f - 2*g ))] + EllipticPi[(f - 3*g)/(f - 2*g), ArcSin[Sqrt[((f - 2*g)*(3 + x))/((f - 3*g)*(2 + x))]], ((d - 2*e)*(f - 3*g))/((d - 3*e)*(f - 2*g))]) + 2*d*(f - 3*g)*g*(2 + x)^2*Sqrt[-((d + e*x)/((d - 3*e)*(2 + x)))]*Sqrt[-(((f - 2*g )*(3 + x)*(f + g*x))/((f - 3*g)^2*(2 + x)^2))]*(2*EllipticF[ArcSin[Sqrt[(( f - 2*g)*(3 + x))/((f - 3*g)*(2 + x))]], ((d - 2*e)*(f - 3*g))/((d - 3*e)* (f - 2*g))] + EllipticPi[(f - 3*g)/(f - 2*g), ArcSin[Sqrt[((f - 2*g)*(3 + x))/((f - 3*g)*(2 + x))]], ((d - 2*e)*(f - 3*g))/((d - 3*e)*(f - 2*g))]) + (f - 3*g)*(2 + x)^2*Sqrt[-((d + e*x)/((d - 3*e)*(2 + x)))]*Sqrt[-(((f - 2 *g)*(3 + x)*(f + g*x))/((f - 3*g)^2*(2 + x)^2))]*((d - 3*e)*(f - 2*g)*Elli pticE[ArcSin[Sqrt[((f - 2*g)*(3 + x))/((f - 3*g)*(2 + x))]], ((d - 2*e)*(f - 3*g))/((d - 3*e)*(f - 2*g))] + e*(f - 10*g)*EllipticF[ArcSin[Sqrt[((f - 2*g)*(3 + x))/((f - 3*g)*(2 + x))]], ((d - 2*e)*(f - 3*g))/((d - 3*e)*(f - 2*g))] - (d*g + e*(f + 5*g))*EllipticPi[(f - 3*g)/(f - 2*g), ArcSin[S...
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.
\(\displaystyle \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {x^2+5 x+6}} \, dx\) |
\(\Big \downarrow \) 1292 |
\(\displaystyle \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {x^2+5 x+6}}dx\) |
Input:
Int[(Sqrt[d + e*x]*Sqrt[f + g*x])/Sqrt[6 + 5*x + x^2],x]
Output:
$Aborted
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Unintegrable[(d + e*x)^m*(f + g*x)^n* (a + b*x + c*x^2)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x]
Time = 6.10 (sec) , antiderivative size = 939, normalized size of antiderivative = 1.63
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(939\) |
default | \(\text {Expression too large to display}\) | \(6596\) |
Input:
int((e*x+d)^(1/2)*(g*x+f)^(1/2)/(x^2+5*x+6)^(1/2),x,method=_RETURNVERBOSE)
Output:
((x^2+5*x+6)*(e*x+d)*(g*x+f))^(1/2)/(x^2+5*x+6)^(1/2)/(e*x+d)^(1/2)/(g*x+f )^(1/2)*(2*d*f*(f/g-3)*((-f/g+2)*(3+x)/(-f/g+3)/(2+x))^(1/2)*(2+x)^2*((x+d /e)/(-d/e+3)/(2+x))^(1/2)*((x+f/g)/(-f/g+3)/(2+x))^(1/2)/(-f/g+2)/(e*g*(3+ x)*(2+x)*(x+d/e)*(x+f/g))^(1/2)*EllipticF(((-f/g+2)*(3+x)/(-f/g+3)/(2+x))^ (1/2),((-2+d/e)*(f/g-3)/(d/e-3)/(f/g-2))^(1/2))+2*(d*g+e*f)*(f/g-3)*((-f/g +2)*(3+x)/(-f/g+3)/(2+x))^(1/2)*(2+x)^2*((x+d/e)/(-d/e+3)/(2+x))^(1/2)*((x +f/g)/(-f/g+3)/(2+x))^(1/2)/(-f/g+2)/(e*g*(3+x)*(2+x)*(x+d/e)*(x+f/g))^(1/ 2)*(-2*EllipticF(((-f/g+2)*(3+x)/(-f/g+3)/(2+x))^(1/2),((-2+d/e)*(f/g-3)/( d/e-3)/(f/g-2))^(1/2))-EllipticPi(((-f/g+2)*(3+x)/(-f/g+3)/(2+x))^(1/2),(- f/g+3)/(-f/g+2),((-2+d/e)*(f/g-3)/(d/e-3)/(f/g-2))^(1/2)))+e*g*((3+x)*(x+d /e)*(x+f/g)+(f/g-3)*((-f/g+2)*(3+x)/(-f/g+3)/(2+x))^(1/2)*(2+x)^2*((x+d/e) /(-d/e+3)/(2+x))^(1/2)*((x+f/g)/(-f/g+3)/(2+x))^(1/2)*((10-f/g)/(-f/g+2)*E llipticF(((-f/g+2)*(3+x)/(-f/g+3)/(2+x))^(1/2),((-2+d/e)*(f/g-3)/(d/e-3)/( f/g-2))^(1/2))+(d/e-3)*EllipticE(((-f/g+2)*(3+x)/(-f/g+3)/(2+x))^(1/2),((- 2+d/e)*(f/g-3)/(d/e-3)/(f/g-2))^(1/2))+(d*g+e*f+5*e*g)/e/g/(-f/g+2)*Ellipt icPi(((-f/g+2)*(3+x)/(-f/g+3)/(2+x))^(1/2),(f/g-3)/(f/g-2),((-2+d/e)*(f/g- 3)/(d/e-3)/(f/g-2))^(1/2))))/(e*g*(3+x)*(2+x)*(x+d/e)*(x+f/g))^(1/2))
\[ \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {6+5 x+x^2}} \, dx=\int { \frac {\sqrt {e x + d} \sqrt {g x + f}}{\sqrt {x^{2} + 5 \, x + 6}} \,d x } \] Input:
integrate((e*x+d)^(1/2)*(g*x+f)^(1/2)/(x^2+5*x+6)^(1/2),x, algorithm="fric as")
Output:
integral(sqrt(e*x + d)*sqrt(g*x + f)/sqrt(x^2 + 5*x + 6), x)
\[ \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {6+5 x+x^2}} \, dx=\int \frac {\sqrt {d + e x} \sqrt {f + g x}}{\sqrt {\left (x + 2\right ) \left (x + 3\right )}}\, dx \] Input:
integrate((e*x+d)**(1/2)*(g*x+f)**(1/2)/(x**2+5*x+6)**(1/2),x)
Output:
Integral(sqrt(d + e*x)*sqrt(f + g*x)/sqrt((x + 2)*(x + 3)), x)
\[ \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {6+5 x+x^2}} \, dx=\int { \frac {\sqrt {e x + d} \sqrt {g x + f}}{\sqrt {x^{2} + 5 \, x + 6}} \,d x } \] Input:
integrate((e*x+d)^(1/2)*(g*x+f)^(1/2)/(x^2+5*x+6)^(1/2),x, algorithm="maxi ma")
Output:
integrate(sqrt(e*x + d)*sqrt(g*x + f)/sqrt(x^2 + 5*x + 6), x)
\[ \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {6+5 x+x^2}} \, dx=\int { \frac {\sqrt {e x + d} \sqrt {g x + f}}{\sqrt {x^{2} + 5 \, x + 6}} \,d x } \] Input:
integrate((e*x+d)^(1/2)*(g*x+f)^(1/2)/(x^2+5*x+6)^(1/2),x, algorithm="giac ")
Output:
integrate(sqrt(e*x + d)*sqrt(g*x + f)/sqrt(x^2 + 5*x + 6), x)
Timed out. \[ \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {6+5 x+x^2}} \, dx=\int \frac {\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{\sqrt {x^2+5\,x+6}} \,d x \] Input:
int(((f + g*x)^(1/2)*(d + e*x)^(1/2))/(5*x + x^2 + 6)^(1/2),x)
Output:
int(((f + g*x)^(1/2)*(d + e*x)^(1/2))/(5*x + x^2 + 6)^(1/2), x)
\[ \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {6+5 x+x^2}} \, dx=\int \frac {\sqrt {e x +d}\, \sqrt {g x +f}}{\sqrt {x^{2}+5 x +6}}d x \] Input:
int((e*x+d)^(1/2)*(g*x+f)^(1/2)/(x^2+5*x+6)^(1/2),x)
Output:
int((e*x+d)^(1/2)*(g*x+f)^(1/2)/(x^2+5*x+6)^(1/2),x)