Integrand size = 18, antiderivative size = 124 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=-\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{128 c^2}+\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{256 c^{5/2}} \]
1/16*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(3/2)/c+3/256*(-4*a*c+b^2)^2*arctanh(1/2* (2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(5/2)-3/128*(-4*a*c+b^2)*(2*c *x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c^2
Time = 0.80 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4} \left (-3 b^2+20 a c+8 b c x^2+8 c^2 x^4\right )}{128 c^2}+\frac {3 \left (-b^2+4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2+c x^4}}\right )}{128 c^{5/2}} \]
((b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4]*(-3*b^2 + 20*a*c + 8*b*c*x^2 + 8*c^ 2*x^4))/(128*c^2) + (3*(-b^2 + 4*a*c)^2*ArcTanh[(Sqrt[c]*x^2)/(-Sqrt[a] + Sqrt[a + b*x^2 + c*x^4])])/(128*c^(5/2))
Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1432, 1087, 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.
\(\displaystyle \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1432 |
\(\displaystyle \frac {1}{2} \int \left (c x^4+b x^2+a\right )^{3/2}dx^2\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^4+b x^2+a}dx^2}{16 c}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{8 c}\right )}{16 c}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{4 c}\right )}{16 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{8 c^{3/2}}\right )}{16 c}\right )\) |
(((b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(8*c) - (3*(b^2 - 4*a*c)*(((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))))/(16*c))/2
3.10.39.3.1 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[((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 = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {\left (16 c^{3} x^{6}+24 b \,c^{2} x^{4}+40 a \,c^{2} x^{2}+2 b^{2} c \,x^{2}+20 a b c -3 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}+\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}\) | \(120\) |
pseudoelliptic | \(\frac {\frac {3 \left (a c -\frac {b^{2}}{4}\right )^{2} \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )}{16}+\frac {3 \left (\frac {5 \left (\frac {b \,x^{2}}{10}+a \right ) b \,c^{\frac {3}{2}}}{6}+\left (b \,x^{4}+\frac {5}{3} a \,x^{2}\right ) c^{\frac {5}{2}}+\frac {2 c^{\frac {7}{2}} x^{6}}{3}-\frac {\sqrt {c}\, b^{3}}{8}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}-\frac {3 \ln \left (2\right ) \left (a c -\frac {b^{2}}{4}\right )^{2}}{16}}{c^{\frac {5}{2}}}\) | \(128\) |
default | \(\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}+\frac {5 b a \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 c}+\frac {b^{2} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c}+\frac {5 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}-\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}+\frac {c \,x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}+\frac {3 b \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}+\frac {3 a^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}-\frac {3 b^{2} a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}\) | \(242\) |
elliptic | \(\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}+\frac {5 b a \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 c}+\frac {b^{2} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c}+\frac {5 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}-\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}+\frac {c \,x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}+\frac {3 b \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}+\frac {3 a^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}-\frac {3 b^{2} a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}\) | \(242\) |
1/128*(16*c^3*x^6+24*b*c^2*x^4+40*a*c^2*x^2+2*b^2*c*x^2+20*a*b*c-3*b^3)*(c *x^4+b*x^2+a)^(1/2)/c^2+3/256*(16*a^2*c^2-8*a*b^2*c+b^4)/c^(5/2)*ln((1/2*b +c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))
Time = 0.26 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.40 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{512 \, c^{3}}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{256 \, c^{3}}\right ] \]
[1/512*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*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*(16 *c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*x^ 2)*sqrt(c*x^4 + b*x^2 + a))/c^3, -1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)* 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*(16*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^3]
\[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int x \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \]
Exception generated. \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (106) = 212\).
Time = 0.32 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.51 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}}\right )} a + \frac {1}{96} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}}\right )} b + \frac {1}{768} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {7}{2}}}\right )} c \]
1/16*(2*sqrt(c*x^4 + b*x^2 + a)*(2*x^2 + b/c) + (b^2 - 4*a*c)*log(abs(2*(s qrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(3/2))*a + 1/96*(2*s qrt(c*x^4 + b*x^2 + a)*(2*(4*x^2 + b/c)*x^2 - (3*b^2 - 8*a*c)/c^2) - 3*(b^ 3 - 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))*b + 1/768*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*x^2 + b/c)*x^2 - (5*b^2*c - 12*a*c^2)/c^3)*x^2 + (15*b^3 - 52*a*b*c)/c^3) + 3*(5*b^4 - 24* a*b^2*c + 16*a^2*c^2)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sq rt(c) + b))/c^(7/2))*c
Time = 13.58 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (c\,x^2+\frac {b}{2}\right )\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{8\,c}+\frac {\left (3\,a\,c-\frac {3\,b^2}{4}\right )\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}+\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{8\,c} \]