Integrand size = 27, antiderivative size = 100 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=-\frac {\left (3 b B-4 A c-2 B c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {\left (3 b^2 B-4 A b c-4 a B c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{5/2}} \]
1/16*(-4*A*b*c-4*B*a*c+3*B*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)-1/8*(-2*B*c*x^2-4*A*c+3*B*b)*(c*x^4+b*x^2+a)^(1/2)/c^ 2
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\left (-3 b B+4 A c+2 B c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {\left (-3 b^2 B+4 A b c+4 a B c\right ) \log \left (b c^2+2 c^3 x^2-2 c^{5/2} \sqrt {a+b x^2+c x^4}\right )}{16 c^{5/2}} \]
((-3*b*B + 4*A*c + 2*B*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c^2) + ((-3*b^2* B + 4*A*b*c + 4*a*B*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 = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1578, 1225, 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.
\(\displaystyle \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {1}{2} \int \frac {x^2 \left (B x^2+A\right )}{\sqrt {c x^4+b x^2+a}}dx^2\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (-4 a B c-4 A b c+3 b^2 B\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{8 c^2}-\frac {\sqrt {a+b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{4 c^2}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (-4 a B c-4 A b c+3 b^2 B\right ) \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{4 c^2}-\frac {\sqrt {a+b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{4 c^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (-4 a B c-4 A b c+3 b^2 B\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{8 c^{5/2}}-\frac {\sqrt {a+b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{4 c^2}\right )\) |
(-1/4*((3*b*B - 4*A*c - 2*B*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/c^2 + ((3*b^2* B - 4*A*b*c - 4*a*B*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c *x^4])])/(8*c^(5/2)))/2
3.2.70.3.1 Defintions of rubi rules used
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_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {\left (2 B \,x^{2} c +4 A c -3 B b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}-\frac {\left (4 A b c +4 B a c -3 B \,b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}\) | \(88\) |
default | \(B \left (\frac {x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\right )+A \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 c}-\frac {b \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\right )\) | \(176\) |
elliptic | \(\frac {B \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 B b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 B \,b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {B a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}+\frac {A \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 c}-\frac {A b \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\) | \(176\) |
pseudoelliptic | \(\frac {4 B \,c^{\frac {3}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+8 A \,c^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}-4 A \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right ) b c +4 A \ln \left (2\right ) b c -6 B b \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}-4 B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right ) a c +3 B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2}+4 B \ln \left (2\right ) a c -3 B \ln \left (2\right ) b^{2}}{16 c^{\frac {5}{2}}}\) | \(205\) |
1/8*(2*B*c*x^2+4*A*c-3*B*b)*(c*x^4+b*x^2+a)^(1/2)/c^2-1/16*(4*A*b*c+4*B*a* c-3*B*b^2)/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))
Time = 0.27 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.33 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\left [-\frac {{\left (3 \, B b^{2} - 4 \, {\left (B a + A b\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (2 \, B c^{2} x^{2} - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{32 \, c^{3}}, -\frac {{\left (3 \, B b^{2} - 4 \, {\left (B a + A b\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, {\left (2 \, B c^{2} x^{2} - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{16 \, c^{3}}\right ] \]
[-1/32*((3*B*b^2 - 4*(B*a + A*b)*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*(2*B*c^2 *x^2 - 3*B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^3, -1/16*((3*B*b^2 - 4*(B*a + A*b)*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*(2*B*c^2*x^2 - 3*B*b*c + 4*A*c^2) *sqrt(c*x^4 + b*x^2 + a))/c^3]
Time = 0.92 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.33 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\begin {cases} \left (- \frac {B a}{2 c} - \frac {b \left (A - \frac {3 B b}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \left (\frac {B x^{2}}{2 c} + \frac {A - \frac {3 B b}{4 c}}{c}\right ) \sqrt {a + b x^{2} + c x^{4}} & \text {for}\: c \neq 0 \\\frac {\frac {2 A \left (- a \sqrt {a + b x^{2}} + \frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3}\right )}{b} + \frac {2 B \left (a^{2} \sqrt {a + b x^{2}} - \frac {2 a \left (a + b x^{2}\right )^{\frac {3}{2}}}{3} + \frac {\left (a + b x^{2}\right )^{\frac {5}{2}}}{5}\right )}{b^{2}}}{b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{4}}{2} + \frac {B x^{6}}{3}}{\sqrt {a}} & \text {otherwise} \end {cases}}{2} \]
Piecewise(((-B*a/(2*c) - b*(A - 3*B*b/(4*c))/(2*c))*Piecewise((log(b + 2*s qrt(c)*sqrt(a + b*x**2 + c*x**4) + 2*c*x**2)/sqrt(c), Ne(a - b**2/(4*c), 0 )), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqrt(c*(b/(2*c) + x**2)**2), Tru e)) + (B*x**2/(2*c) + (A - 3*B*b/(4*c))/c)*sqrt(a + b*x**2 + c*x**4), Ne(c , 0)), ((2*A*(-a*sqrt(a + b*x**2) + (a + b*x**2)**(3/2)/3)/b + 2*B*(a**2*s qrt(a + b*x**2) - 2*a*(a + b*x**2)**(3/2)/3 + (a + b*x**2)**(5/2)/5)/b**2) /b, Ne(b, 0)), ((A*x**4/2 + B*x**6/3)/sqrt(a), True))/2
Exception generated. \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {1}{8} \, \sqrt {c x^{4} + b x^{2} + a} {\left (\frac {2 \, B x^{2}}{c} - \frac {3 \, B b - 4 \, A c}{c^{2}}\right )} - \frac {{\left (3 \, B b^{2} - 4 \, B a c - 4 \, A b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {5}{2}}} \]
1/8*sqrt(c*x^4 + b*x^2 + a)*(2*B*x^2/c - (3*B*b - 4*A*c)/c^2) - 1/16*(3*B* b^2 - 4*B*a*c - 4*A*b*c)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a)) *sqrt(c) + b))/c^(5/2)
Timed out. \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {x^3\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \]