Integrand size = 25, antiderivative size = 220 \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {d} \sqrt {c d-b \sqrt {d} \sqrt {f}+a f}}+\frac {\text {arctanh}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {d} \sqrt {c d+b \sqrt {d} \sqrt {f}+a f}} \] Output:
1/2*arctanh(1/2*(b*d^(1/2)-2*a*f^(1/2)+(2*c*d^(1/2)-b*f^(1/2))*x)/(c*d-b*d ^(1/2)*f^(1/2)+a*f)^(1/2)/(c*x^2+b*x+a)^(1/2))/d^(1/2)/(c*d-b*d^(1/2)*f^(1 /2)+a*f)^(1/2)+1/2*arctanh(1/2*(b*d^(1/2)+2*a*f^(1/2)+(2*c*d^(1/2)+b*f^(1/ 2))*x)/(c*d+b*d^(1/2)*f^(1/2)+a*f)^(1/2)/(c*x^2+b*x+a)^(1/2))/d^(1/2)/(c*d +b*d^(1/2)*f^(1/2)+a*f)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.38 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx=-\frac {1}{2} \text {RootSum}\left [c^2 d-b^2 f+4 \sqrt {a} b f \text {$\#$1}-2 c d \text {$\#$1}^2-4 a f \text {$\#$1}^2+d \text {$\#$1}^4\&,\frac {-c \log (x)+c \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^2-\log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right ) \text {$\#$1}^2}{-\sqrt {a} b f+c d \text {$\#$1}+2 a f \text {$\#$1}-d \text {$\#$1}^3}\&\right ] \] Input:
Integrate[1/(Sqrt[a + b*x + c*x^2]*(d - f*x^2)),x]
Output:
-1/2*RootSum[c^2*d - b^2*f + 4*Sqrt[a]*b*f*#1 - 2*c*d*#1^2 - 4*a*f*#1^2 + d*#1^4 & , (-(c*Log[x]) + c*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] + Log[x]*#1^2 - Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1]*#1^2)/(-(Sqrt[ a]*b*f) + c*d*#1 + 2*a*f*#1 - d*#1^3) & ]
Time = 0.31 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1316, 27, 1154, 219}
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.
\(\displaystyle \int \frac {1}{\left (d-f x^2\right ) \sqrt {a+b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1316 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt {d} \left (\sqrt {d}-\sqrt {f} x\right ) \sqrt {c x^2+b x+a}}dx+\frac {1}{2} \int \frac {1}{\sqrt {d} \left (\sqrt {f} x+\sqrt {d}\right ) \sqrt {c x^2+b x+a}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{\left (\sqrt {d}-\sqrt {f} x\right ) \sqrt {c x^2+b x+a}}dx}{2 \sqrt {d}}+\frac {\int \frac {1}{\left (\sqrt {f} x+\sqrt {d}\right ) \sqrt {c x^2+b x+a}}dx}{2 \sqrt {d}}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle -\frac {\int \frac {1}{4 \left (-\sqrt {d} \sqrt {f} b+c d+a f\right )-\frac {\left (-2 \sqrt {f} a+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x+b \sqrt {d}\right )^2}{c x^2+b x+a}}d\left (-\frac {-2 \sqrt {f} a+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x+b \sqrt {d}}{\sqrt {c x^2+b x+a}}\right )}{\sqrt {d}}-\frac {\int \frac {1}{4 \left (\sqrt {d} \sqrt {f} b+c d+a f\right )-\frac {\left (2 \sqrt {f} a+\left (\sqrt {f} b+2 c \sqrt {d}\right ) x+b \sqrt {d}\right )^2}{c x^2+b x+a}}d\left (-\frac {2 \sqrt {f} a+\left (\sqrt {f} b+2 c \sqrt {d}\right ) x+b \sqrt {d}}{\sqrt {c x^2+b x+a}}\right )}{\sqrt {d}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 \sqrt {d} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}+\frac {\text {arctanh}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 \sqrt {d} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\) |
Input:
Int[1/(Sqrt[a + b*x + c*x^2]*(d - f*x^2)),x]
Output:
ArcTanh[(b*Sqrt[d] - 2*a*Sqrt[f] + (2*c*Sqrt[d] - b*Sqrt[f])*x)/(2*Sqrt[c* d - b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x + c*x^2])]/(2*Sqrt[d]*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]) + ArcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c*Sqrt[ d] + b*Sqrt[f])*x)/(2*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x + c *x^2])]/(2*Sqrt[d]*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[1/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Sy mbol] :> Simp[1/2 Int[1/((a - Rt[(-a)*c, 2]*x)*Sqrt[d + e*x + f*x^2]), x] , x] + Simp[1/2 Int[1/((a + Rt[(-a)*c, 2]*x)*Sqrt[d + e*x + f*x^2]), x], x] /; FreeQ[{a, c, d, e, f}, x] && NeQ[e^2 - 4*d*f, 0] && PosQ[(-a)*c]
Leaf count of result is larger than twice the leaf count of optimal. \(357\) vs. \(2(164)=328\).
Time = 2.40 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.63
method | result | size |
default | \(\frac {\ln \left (\frac {\frac {2 b \sqrt {d f}+2 a f +2 c d}{f}+\frac {\left (2 c \sqrt {d f}+f b \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {b \sqrt {d f}+a f +c d}{f}}\, \sqrt {c \left (x -\frac {\sqrt {d f}}{f}\right )^{2}+\frac {\left (2 c \sqrt {d f}+f b \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+a f +c d}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{2 \sqrt {d f}\, \sqrt {\frac {b \sqrt {d f}+a f +c d}{f}}}-\frac {\ln \left (\frac {\frac {-2 b \sqrt {d f}+2 a f +2 c d}{f}+\frac {\left (-2 c \sqrt {d f}+f b \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {-b \sqrt {d f}+a f +c d}{f}}\, \sqrt {c \left (x +\frac {\sqrt {d f}}{f}\right )^{2}+\frac {\left (-2 c \sqrt {d f}+f b \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+a f +c d}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{2 \sqrt {d f}\, \sqrt {\frac {-b \sqrt {d f}+a f +c d}{f}}}\) | \(358\) |
Input:
int(1/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x,method=_RETURNVERBOSE)
Output:
1/2/(d*f)^(1/2)/((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a*f +c*d)/f+(2*c*(d*f)^(1/2)+f*b)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+a*f+c* d)/f)^(1/2)*(c*(x-(d*f)^(1/2)/f)^2+(2*c*(d*f)^(1/2)+f*b)/f*(x-(d*f)^(1/2)/ f)+(b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))-1/2/(d*f)^(1/2)/(1 /f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+a*f+c*d)+1/f*(- 2*c*(d*f)^(1/2)+f*b)*(x+(d*f)^(1/2)/f)+2*(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1 /2)*(c*(x+(d*f)^(1/2)/f)^2+1/f*(-2*c*(d*f)^(1/2)+f*b)*(x+(d*f)^(1/2)/f)+1/ f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))/(x+(d*f)^(1/2)/f))
Leaf count of result is larger than twice the leaf count of optimal. 2641 vs. \(2 (164) = 328\).
Time = 0.35 (sec) , antiderivative size = 2641, normalized size of antiderivative = 12.00 \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx=\text {Too large to display} \] Input:
integrate(1/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x, algorithm="fricas")
Output:
1/4*sqrt((c*d + a*f + (c^2*d^3 + a^2*d*f^2 - (b^2 - 2*a*c)*d^2*f)*sqrt(b^2 *f/(c^4*d^5 + a^4*d*f^4 - 2*(b^2*c^2 - 2*a*c^3)*d^4*f + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^3*f^2 - 2*(a^2*b^2 - 2*a^3*c)*d^2*f^3)))/(c^2*d^3 + a^2*d*f^ 2 - (b^2 - 2*a*c)*d^2*f))*log((2*b*c*x + b^2 + 2*(b*c*d + a*b*f - (b*c^2*d ^3 + a^2*b*d*f^2 - (b^3 - 2*a*b*c)*d^2*f)*sqrt(b^2*f/(c^4*d^5 + a^4*d*f^4 - 2*(b^2*c^2 - 2*a*c^3)*d^4*f + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^3*f^2 - 2* (a^2*b^2 - 2*a^3*c)*d^2*f^3)))*sqrt(c*x^2 + b*x + a)*sqrt((c*d + a*f + (c^ 2*d^3 + a^2*d*f^2 - (b^2 - 2*a*c)*d^2*f)*sqrt(b^2*f/(c^4*d^5 + a^4*d*f^4 - 2*(b^2*c^2 - 2*a*c^3)*d^4*f + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^3*f^2 - 2*( a^2*b^2 - 2*a^3*c)*d^2*f^3)))/(c^2*d^3 + a^2*d*f^2 - (b^2 - 2*a*c)*d^2*f)) - (2*a*c^2*d^2 + 2*a^3*f^2 - 2*(a*b^2 - 2*a^2*c)*d*f + (b*c^2*d^2 + a^2*b *f^2 - (b^3 - 2*a*b*c)*d*f)*x)*sqrt(b^2*f/(c^4*d^5 + a^4*d*f^4 - 2*(b^2*c^ 2 - 2*a*c^3)*d^4*f + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^3*f^2 - 2*(a^2*b^2 - 2*a^3*c)*d^2*f^3)))/x) - 1/4*sqrt((c*d + a*f + (c^2*d^3 + a^2*d*f^2 - (b^2 - 2*a*c)*d^2*f)*sqrt(b^2*f/(c^4*d^5 + a^4*d*f^4 - 2*(b^2*c^2 - 2*a*c^3)*d ^4*f + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^3*f^2 - 2*(a^2*b^2 - 2*a^3*c)*d^2*f ^3)))/(c^2*d^3 + a^2*d*f^2 - (b^2 - 2*a*c)*d^2*f))*log((2*b*c*x + b^2 - 2* (b*c*d + a*b*f - (b*c^2*d^3 + a^2*b*d*f^2 - (b^3 - 2*a*b*c)*d^2*f)*sqrt(b^ 2*f/(c^4*d^5 + a^4*d*f^4 - 2*(b^2*c^2 - 2*a*c^3)*d^4*f + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^3*f^2 - 2*(a^2*b^2 - 2*a^3*c)*d^2*f^3)))*sqrt(c*x^2 + b*...
\[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx=- \int \frac {1}{- d \sqrt {a + b x + c x^{2}} + f x^{2} \sqrt {a + b x + c x^{2}}}\, dx \] Input:
integrate(1/(c*x**2+b*x+a)**(1/2)/(-f*x**2+d),x)
Output:
-Integral(1/(-d*sqrt(a + b*x + c*x**2) + f*x**2*sqrt(a + b*x + c*x**2)), x )
Exception generated. \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(((c*sqrt(4*d*f))/(2*f^2)>0)', se e `assume?
Exception generated. \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo r the root of a polynomial with parameters. This might be wrong.Non regula r value [
Timed out. \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx=\int \frac {1}{\left (d-f\,x^2\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \] Input:
int(1/((d - f*x^2)*(a + b*x + c*x^2)^(1/2)),x)
Output:
int(1/((d - f*x^2)*(a + b*x + c*x^2)^(1/2)), x)
Time = 0.87 (sec) , antiderivative size = 2178, normalized size of antiderivative = 9.90 \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx =\text {Too large to display} \] Input:
int(1/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x)
Output:
(sqrt(d)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*atan((sqrt(d)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*a*b*f - 2*sqrt(d)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*a*c*f*x + sqrt(d)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*b**2*f*x - sqrt(d)*sqrt( a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*b*c*d - 2*sqrt(d)*sq rt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*c**2*d*x + 2*sqrt (f)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*a**2*f + sq rt(f)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*a*b*f*x + 2*sqrt(f)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*a*c* d - sqrt(f)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*b** 2*d - sqrt(f)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*b *c*d*x)/(2*a**3*f**2 + 2*a**2*b*f**2*x + 4*a**2*c*d*f + 2*a**2*c*f**2*x**2 - 2*a*b**2*d*f + 4*a*b*c*d*f*x + 2*a*c**2*d**2 + 4*a*c**2*d*f*x**2 - 2*b* *3*d*f*x - 2*b**2*c*d*f*x**2 + 2*b*c**2*d**2*x + 2*c**3*d**2*x**2))*a*f + sqrt(d)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*atan((sqrt(d)*sqrt(a + b*x + c *x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*a*b*f - 2*sqrt(d)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*a*c*f*x + sqrt(d)*sqrt(a + b *x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*b**2*f*x - sqrt(d)*sqrt(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*b*c*d - 2*sqrt(d)*sqr t(a + b*x + c*x**2)*sqrt(sqrt(f)*sqrt(d)*b - a*f - c*d)*c**2*d*x + 2*sq...