Integrand size = 18, antiderivative size = 83 \[ \int x \sqrt {a+b x^2+c x^4} \, dx=\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2}} \] Output:
1/8*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c-1/16*(-4*a*c+b^2)*arctanh(1/2*(2*c *x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(3/2)
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05 \[ \int x \sqrt {a+b x^2+c x^4} \, dx=\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c}+\frac {\left (-b^2+4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2+c x^4}}\right )}{8 c^{3/2}} \] Input:
Integrate[x*Sqrt[a + b*x^2 + c*x^4],x]
Output:
((b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c) + ((-b^2 + 4*a*c)*ArcTanh[(S qrt[c]*x^2)/(-Sqrt[a] + Sqrt[a + b*x^2 + c*x^4])])/(8*c^(3/2))
Time = 0.37 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1432, 1087, 1092, 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.
\(\displaystyle \int x \sqrt {a+b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 1432 |
\(\displaystyle \frac {1}{2} \int \sqrt {c x^4+b x^2+a}dx^2\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{8 c}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{4 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{8 c^{3/2}}\right )\) |
Input:
Int[x*Sqrt[a + b*x^2 + c*x^4],x]
Output:
(((b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(8*c^(3/2)))/2
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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}\) | \(73\) |
risch | \(\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}\) | \(73\) |
elliptic | \(\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}\) | \(73\) |
pseudoelliptic | \(\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c}-\frac {\left (a c -\frac {b^{2}}{4}\right ) \left (\ln \left (2\right )-\ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}\) | \(80\) |
Input:
int(x*(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c+1/16*(4*a*c-b^2)/c^(3/2)*ln((1/2*b +c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))
Time = 0.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.37 \[ \int x \sqrt {a+b x^2+c x^4} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} + b c\right )}}{32 \, c^{2}}, \frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} + b c\right )}}{16 \, c^{2}}\right ] \] Input:
integrate(x*(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[-1/32*((b^2 - 4*a*c)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c* x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) - 4*sqrt(c*x^4 + b*x^2 + a )*(2*c^2*x^2 + b*c))/c^2, 1/16*((b^2 - 4*a*c)*sqrt(-c)*arctan(1/2*sqrt(c*x ^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*sqrt (c*x^4 + b*x^2 + a)*(2*c^2*x^2 + b*c))/c^2]
\[ \int x \sqrt {a+b x^2+c x^4} \, dx=\int x \sqrt {a + b x^{2} + c x^{4}}\, dx \] Input:
integrate(x*(c*x**4+b*x**2+a)**(1/2),x)
Output:
Integral(x*sqrt(a + b*x**2 + c*x**4), x)
Exception generated. \[ \int x \sqrt {a+b x^2+c x^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x*(c*x^4+b*x^2+a)^(1/2),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*a*c-b^2>0)', see `assume?` for more deta
Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int x \sqrt {a+b x^2+c x^4} \, dx=\frac {1}{8} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {3}{2}}} \] Input:
integrate(x*(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/8*sqrt(c*x^4 + b*x^2 + a)*(2*x^2 + b/c) + 1/16*(b^2 - 4*a*c)*log(abs(2*( sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(3/2)
Time = 19.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.87 \[ \int x \sqrt {a+b x^2+c x^4} \, dx=\frac {\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}}{2}+\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{4\,c^{3/2}} \] Input:
int(x*(a + b*x^2 + c*x^4)^(1/2),x)
Output:
((b/(4*c) + x^2/2)*(a + b*x^2 + c*x^4)^(1/2))/2 + (log((a + b*x^2 + c*x^4) ^(1/2) + (b/2 + c*x^2)/c^(1/2))*(a*c - b^2/4))/(4*c^(3/2))
Time = 0.18 (sec) , antiderivative size = 816, normalized size of antiderivative = 9.83 \[ \int x \sqrt {a+b x^2+c x^4} \, dx =\text {Too large to display} \] Input:
int(x*(c*x^4+b*x^2+a)^(1/2),x)
Output:
(16*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x **4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a*b*c + 32*sqrt(c)*sqrt(a + b*x** 2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt( 4*a*c - b**2))*a*c**2*x**2 - 4*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sq rt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**3 - 8*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x* *4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**2*c*x**2 + 8*sqrt(c)*sqrt(a + b *x**2 + c*x**4)*a*b*c + 16*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*a*c**2*x**2 + 2*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*b**3 + 20*sqrt(c)*sqrt(a + b*x**2 + c *x**4)*b**2*c*x**2 + 48*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*b*c**2*x**4 + 32 *sqrt(c)*sqrt(a + b*x**2 + c*x**4)*c**3*x**6 + 16*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*c**2 + 32*log((2 *sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a*b *c**2*x**2 + 32*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/s qrt(4*a*c - b**2))*a*c**3*x**4 - log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**4 - 8*log((2*sqrt(c)*sqrt(a + b*x** 2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**3*c*x**2 - 8*log((2*sqr t(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**2*c* *2*x**4 + 8*a*b**2*c + 32*a*b*c**2*x**2 + 32*a*c**3*x**4 + 8*b**3*c*x**2 + 40*b**2*c**2*x**4 + 64*b*c**3*x**6 + 32*c**4*x**8)/(16*c*(4*sqrt(a + b...