Integrand size = 27, antiderivative size = 344 \[ \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} f}-\frac {\left (e^2-2 d f-e \sqrt {e^2-4 d f}\right ) \text {arctanh}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}-\frac {\left (2 d f-e \left (e+\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}} \] Output:
arctanh(c^(1/2)*x/(c*x^2+a)^(1/2))/c^(1/2)/f-1/2*(e^2-2*d*f-e*(-4*d*f+e^2) ^(1/2))*arctanh(1/2*(2*a*f-c*(e-(-4*d*f+e^2)^(1/2))*x)*2^(1/2)/(2*a*f^2+c* (e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)/(c*x^2+a)^(1/2))*2^(1/2)/f/(-4*d*f +e^2)^(1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)-1/2*(2*d*f- e*(e+(-4*d*f+e^2)^(1/2)))*arctanh(1/2*(2*a*f-c*(e+(-4*d*f+e^2)^(1/2))*x)*2 ^(1/2)/(2*a*f^2+c*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2)))^(1/2)/(c*x^2+a)^(1/2)) *2^(1/2)/f/(-4*d*f+e^2)^(1/2)/(2*a*f^2+c*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2))) ^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.58 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.76 \[ \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\frac {\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+c x^2}}\right )}{\sqrt {c}}+\text {RootSum}\left [c^2 d+2 \sqrt {a} c e \text {$\#$1}-2 c d \text {$\#$1}^2+4 a f \text {$\#$1}^2-2 \sqrt {a} e \text {$\#$1}^3+d \text {$\#$1}^4\&,\frac {-c d \log (x)+c d \log \left (-\sqrt {a}+\sqrt {a+c x^2}-x \text {$\#$1}\right )-2 \sqrt {a} e \log (x) \text {$\#$1}+2 \sqrt {a} e \log \left (-\sqrt {a}+\sqrt {a+c x^2}-x \text {$\#$1}\right ) \text {$\#$1}+d \log (x) \text {$\#$1}^2-d \log \left (-\sqrt {a}+\sqrt {a+c x^2}-x \text {$\#$1}\right ) \text {$\#$1}^2}{-\sqrt {a} c e+2 c d \text {$\#$1}-4 a f \text {$\#$1}+3 \sqrt {a} e \text {$\#$1}^2-2 d \text {$\#$1}^3}\&\right ]}{f} \] Input:
Integrate[x^2/(Sqrt[a + c*x^2]*(d + e*x + f*x^2)),x]
Output:
((2*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + c*x^2])])/Sqrt[c] + RootSum[c ^2*d + 2*Sqrt[a]*c*e*#1 - 2*c*d*#1^2 + 4*a*f*#1^2 - 2*Sqrt[a]*e*#1^3 + d*# 1^4 & , (-(c*d*Log[x]) + c*d*Log[-Sqrt[a] + Sqrt[a + c*x^2] - x*#1] - 2*Sq rt[a]*e*Log[x]*#1 + 2*Sqrt[a]*e*Log[-Sqrt[a] + Sqrt[a + c*x^2] - x*#1]*#1 + d*Log[x]*#1^2 - d*Log[-Sqrt[a] + Sqrt[a + c*x^2] - x*#1]*#1^2)/(-(Sqrt[a ]*c*e) + 2*c*d*#1 - 4*a*f*#1 + 3*Sqrt[a]*e*#1^2 - 2*d*#1^3) & ])/f
Time = 0.57 (sec) , antiderivative size = 342, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2145, 25, 224, 219, 1367, 488, 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.
\(\displaystyle \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx\) |
\(\Big \downarrow \) 2145 |
\(\displaystyle \frac {\int -\frac {d+e x}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{f}+\frac {\int \frac {1}{\sqrt {c x^2+a}}dx}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {c x^2+a}}dx}{f}-\frac {\int \frac {d+e x}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}}{f}-\frac {\int \frac {d+e x}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} f}-\frac {\int \frac {d+e x}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{f}\) |
\(\Big \downarrow \) 1367 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} f}-\frac {-\frac {\left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \int \frac {1}{\left (e+2 f x-\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+a}}dx}{\sqrt {e^2-4 d f}}-\frac {\left (2 d f-e \left (\sqrt {e^2-4 d f}+e\right )\right ) \int \frac {1}{\left (e+2 f x+\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+a}}dx}{\sqrt {e^2-4 d f}}}{f}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} f}-\frac {\frac {\left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \int \frac {1}{4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2-\frac {\left (2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x\right )^2}{c x^2+a}}d\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {c x^2+a}}}{\sqrt {e^2-4 d f}}+\frac {\left (2 d f-e \left (\sqrt {e^2-4 d f}+e\right )\right ) \int \frac {1}{4 a f^2+c \left (e+\sqrt {e^2-4 d f}\right )^2-\frac {\left (2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x\right )^2}{c x^2+a}}d\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {c x^2+a}}}{\sqrt {e^2-4 d f}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} f}-\frac {\frac {\left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \text {arctanh}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\left (2 d f-e \left (\sqrt {e^2-4 d f}+e\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}}{f}\) |
Input:
Int[x^2/(Sqrt[a + c*x^2]*(d + e*x + f*x^2)),x]
Output:
ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]]/(Sqrt[c]*f) - (((e^2 - 2*d*f - e*Sqrt [e^2 - 4*d*f])*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt [2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt [2]*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]) ]) + ((2*d*f - e*(e + Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f] )]*Sqrt[a + c*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2 *d*f + e*Sqrt[e^2 - 4*d*f])]))/f
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/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f _.)*(x_)^2]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(2*c*g - h*( b - q))/q Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Simp[(2*c*g - h*(b + q))/q Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{ a, b, c, d, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]
Int[(Px_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (f_.)*(x_)^2]), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C = Coeff[Px, x, 2]}, Simp[C/c Int[1/Sqrt[d + f*x^2], x], x] + Simp[1/c Int[(A*c - a* C + (B*c - b*C)*x)/((a + b*x + c*x^2)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a , b, c, d, f}, x] && PolyQ[Px, x, 2]
Leaf count of result is larger than twice the leaf count of optimal. \(660\) vs. \(2(300)=600\).
Time = 2.15 (sec) , antiderivative size = 661, normalized size of antiderivative = 1.92
method | result | size |
default | \(\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{f \sqrt {c}}+\frac {\left (e \sqrt {-4 d f +e^{2}}+2 d f -e^{2}\right ) \sqrt {2}\, \ln \left (\frac {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 d f c +c \,e^{2}}{f^{2}}-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}-\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 d f c +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{2 f^{2} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 d f c +c \,e^{2}}{f^{2}}}}+\frac {\left (e^{2}-2 d f +e \sqrt {-4 d f +e^{2}}\right ) \sqrt {2}\, \ln \left (\frac {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 d f c +c \,e^{2}}{f^{2}}-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}-\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 d f c +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{2 f^{2} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 d f c +c \,e^{2}}{f^{2}}}}\) | \(661\) |
Input:
int(x^2/(c*x^2+a)^(1/2)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
Output:
1/f*ln(c^(1/2)*x+(c*x^2+a)^(1/2))/c^(1/2)+1/2/f^2*(e*(-4*d*f+e^2)^(1/2)+2* d*f-e^2)/(-4*d*f+e^2)^(1/2)*2^(1/2)/((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d* f*c+c*e^2)/f^2)^(1/2)*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/ f^2-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*2^(1/ 2)*((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2)*(4*c*(x-1/2 /f*(-e+(-4*d*f+e^2)^(1/2)))^2-4*c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(- 4*d*f+e^2)^(1/2)))+2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^ (1/2))/(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2))))+1/2/f^2*(e^2-2*d*f+e*(-4*d*f+e^2 )^(1/2))/(-4*d*f+e^2)^(1/2)*2^(1/2)/(((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f *c+c*e^2)/f^2)^(1/2)*ln((((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^ 2-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)* (((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2)*(4*c*(x+1/2*(e+ (-4*d*f+e^2)^(1/2))/f)^2-4*c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^ 2)^(1/2))/f)+2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2))/ (x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))
Timed out. \[ \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(x^2/(c*x^2+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {x^{2}}{\sqrt {a + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \] Input:
integrate(x**2/(c*x**2+a)**(1/2)/(f*x**2+e*x+d),x)
Output:
Integral(x**2/(sqrt(a + c*x**2)*(d + e*x + f*x**2)), x)
Exception generated. \[ \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^2/(c*x^2+a)^(1/2)/(f*x^2+e*x+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(4*d*f-e^2>0)', see `assume?` for more deta
Exception generated. \[ \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^2/(c*x^2+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {x^2}{\sqrt {c\,x^2+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \] Input:
int(x^2/((a + c*x^2)^(1/2)*(d + e*x + f*x^2)),x)
Output:
int(x^2/((a + c*x^2)^(1/2)*(d + e*x + f*x^2)), x)
\[ \int \frac {x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\frac {-\sqrt {c}\, \mathrm {log}\left (\sqrt {c \,x^{2}+a}-\sqrt {c}\, x \right )+\sqrt {c}\, \mathrm {log}\left (\sqrt {c \,x^{2}+a}+\sqrt {c}\, x \right )-2 \left (\int \frac {\sqrt {c \,x^{2}+a}}{c f \,x^{4}+c e \,x^{3}+a f \,x^{2}+c d \,x^{2}+a e x +a d}d x \right ) c d -2 \left (\int \frac {\sqrt {c \,x^{2}+a}\, x}{c f \,x^{4}+c e \,x^{3}+a f \,x^{2}+c d \,x^{2}+a e x +a d}d x \right ) c e}{2 c f} \] Input:
int(x^2/(c*x^2+a)^(1/2)/(f*x^2+e*x+d),x)
Output:
( - sqrt(c)*log(sqrt(a + c*x**2) - sqrt(c)*x) + sqrt(c)*log(sqrt(a + c*x** 2) + sqrt(c)*x) - 2*int(sqrt(a + c*x**2)/(a*d + a*e*x + a*f*x**2 + c*d*x** 2 + c*e*x**3 + c*f*x**4),x)*c*d - 2*int((sqrt(a + c*x**2)*x)/(a*d + a*e*x + a*f*x**2 + c*d*x**2 + c*e*x**3 + c*f*x**4),x)*c*e)/(2*c*f)