Integrand size = 20, antiderivative size = 104 \[ \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx=-\frac {3 b \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {a+b x^2+c x^4}}{4 c}+\frac {\left (3 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{5/2}} \] Output:
-3/8*b*(c*x^4+b*x^2+a)^(1/2)/c^2+1/4*x^2*(c*x^4+b*x^2+a)^(1/2)/c+1/16*(-4* a*c+3*b^2)*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(5/2)
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\left (-3 b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {\left (-3 b^2+4 a c\right ) \log \left (b c^2+2 c^3 x^2-2 c^{5/2} \sqrt {a+b x^2+c x^4}\right )}{16 c^{5/2}} \] Input:
Integrate[x^5/Sqrt[a + b*x^2 + c*x^4],x]
Output:
((-3*b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c^2) + ((-3*b^2 + 4*a*c)*Log [b*c^2 + 2*c^3*x^2 - 2*c^(5/2)*Sqrt[a + b*x^2 + c*x^4]])/(16*c^(5/2))
Time = 0.44 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1434, 1166, 27, 1160, 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.
\(\displaystyle \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int \frac {x^4}{\sqrt {c x^4+b x^2+a}}dx^2\) |
\(\Big \downarrow \) 1166 |
\(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {3 b x^2+2 a}{2 \sqrt {c x^4+b x^2+a}}dx^2}{2 c}+\frac {x^2 \sqrt {a+b x^2+c x^4}}{2 c}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {a+b x^2+c x^4}}{2 c}-\frac {\int \frac {3 b x^2+2 a}{\sqrt {c x^4+b x^2+a}}dx^2}{4 c}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {a+b x^2+c x^4}}{2 c}-\frac {\frac {3 b \sqrt {a+b x^2+c x^4}}{c}-\frac {\left (3 b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{2 c}}{4 c}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {a+b x^2+c x^4}}{2 c}-\frac {\frac {3 b \sqrt {a+b x^2+c x^4}}{c}-\frac {\left (3 b^2-4 a c\right ) \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{c}}{4 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {a+b x^2+c x^4}}{2 c}-\frac {\frac {3 b \sqrt {a+b x^2+c x^4}}{c}-\frac {\left (3 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 c^{3/2}}}{4 c}\right )\) |
Input:
Int[x^5/Sqrt[a + b*x^2 + c*x^4],x]
Output:
((x^2*Sqrt[a + b*x^2 + c*x^4])/(2*c) - ((3*b*Sqrt[a + b*x^2 + c*x^4])/c - ((3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]) ])/(2*c^(3/2)))/(4*c))/2
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/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[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {\left (-2 c \,x^{2}+3 b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}-\frac {\left (4 a c -3 b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}\) | \(75\) |
default | \(\frac {x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\) | \(116\) |
elliptic | \(\frac {x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\) | \(116\) |
pseudoelliptic | \(\frac {4 c^{\frac {3}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+4 a c \ln \left (2\right )-3 b^{2} \ln \left (2\right )-4 a c \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )+3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2}-6 b \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 c^{\frac {5}{2}}}\) | \(135\) |
Input:
int(x^5/(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/8*(-2*c*x^2+3*b)*(c*x^4+b*x^2+a)^(1/2)/c^2-1/16*(4*a*c-3*b^2)/c^(5/2)*l n((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))
Time = 0.08 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.95 \[ \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx=\left [-\frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} - 3 \, b c\right )}}{32 \, c^{3}}, -\frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} - 3 \, b c\right )}}{16 \, c^{3}}\right ] \] Input:
integrate(x^5/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[-1/32*((3*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 - 3*b*c))/c^3, -1/16*((3*b^2 - 4*a*c)*sqrt(-c)*arctan(1/2*s qrt(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 - 3*b*c))/c^3]
\[ \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {x^{5}}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \] Input:
integrate(x**5/(c*x**4+b*x**2+a)**(1/2),x)
Output:
Integral(x**5/sqrt(a + b*x**2 + c*x**4), x)
Exception generated. \[ \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^5/(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 = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.77 \[ \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {1}{8} \, \sqrt {c x^{4} + b x^{2} + a} {\left (\frac {2 \, x^{2}}{c} - \frac {3 \, b}{c^{2}}\right )} - \frac {{\left (3 \, b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {5}{2}}} \] Input:
integrate(x^5/(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/c - 3*b/c^2) - 1/16*(3*b^2 - 4*a*c)*log (abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(5/2)
Timed out. \[ \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {x^5}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \] Input:
int(x^5/(a + b*x^2 + c*x^4)^(1/2),x)
Output:
int(x^5/(a + b*x^2 + c*x^4)^(1/2), x)
Time = 0.19 (sec) , antiderivative size = 862, normalized size of antiderivative = 8.29 \[ \int \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx =\text {Too large to display} \] Input:
int(x^5/(c*x^4+b*x^2+a)^(1/2),x)
Output:
( - 16*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**2 - 32*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**3*x**2 + 12*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**3*c + 24*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)/sq rt(4*a*c - b**2))*b**2*c**2*x**2 - 24*sqrt(a + b*x**2 + c*x**4)*a*b*c**2 + 16*sqrt(a + b*x**2 + c*x**4)*a*c**3*x**2 - 6*sqrt(a + b*x**2 + c*x**4)*b* *3*c - 44*sqrt(a + b*x**2 + c*x**4)*b**2*c**2*x**2 - 16*sqrt(a + b*x**2 + c*x**4)*b*c**3*x**4 + 32*sqrt(a + b*x**2 + c*x**4)*c**4*x**6 - 16*sqrt(c)* 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 + 8*sqrt(c)*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**2*c - 32*sqrt(c)*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*sq rt(c)*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**3*x**4 + 3*sqrt(c)*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 + 24*sqrt(c)*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 + 24* sqrt(c)*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**2*x**4 - 24*sqrt(c)*a*b**2*c - 32*sqrt(c)*a*b*c**2*x...