Integrand size = 20, antiderivative size = 171 \[ \int x^7 \sqrt {a+b x^2+c x^4} \, dx=-\frac {b \left (7 b^2-12 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{256 c^4}+\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c}+\frac {\left (35 b^2-32 a c-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 c^3}+\frac {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{512 c^{9/2}} \]
1/10*x^4*(c*x^4+b*x^2+a)^(3/2)/c+1/480*(-42*b*c*x^2-32*a*c+35*b^2)*(c*x^4+ b*x^2+a)^(3/2)/c^3+1/512*b*(-12*a*c+7*b^2)*(-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^(9/2)-1/256*b*(-12*a*c+7*b^2)*(2*c* x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c^4
Time = 0.31 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97 \[ \int x^7 \sqrt {a+b x^2+c x^4} \, dx=\frac {\sqrt {a+b x^2+c x^4} \left (-105 b^4+70 b^3 c x^2+4 b^2 c \left (115 a-14 c x^4\right )+8 b c^2 x^2 \left (-29 a+6 c x^4\right )+128 c^2 \left (-2 a^2+a c x^4+3 c^2 x^8\right )\right )}{3840 c^4}-\frac {\left (7 b^5-40 a b^3 c+48 a^2 b c^2\right ) \log \left (c^4 \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )\right )}{512 c^{9/2}} \]
(Sqrt[a + b*x^2 + c*x^4]*(-105*b^4 + 70*b^3*c*x^2 + 4*b^2*c*(115*a - 14*c* x^4) + 8*b*c^2*x^2*(-29*a + 6*c*x^4) + 128*c^2*(-2*a^2 + a*c*x^4 + 3*c^2*x ^8)))/(3840*c^4) - ((7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*Log[c^4*(b + 2*c*x ^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(512*c^(9/2))
Time = 0.32 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1434, 1166, 27, 1225, 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.
| \(\displaystyle \int x^7 \sqrt {a+b x^2+c x^4} \, dx\) |
| \(\Big \downarrow \) 1434 |
| \(\displaystyle \frac {1}{2} \int x^6 \sqrt {c x^4+b x^2+a}dx^2\) |
| \(\Big \downarrow \) 1166 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {1}{2} x^2 \left (7 b x^2+4 a\right ) \sqrt {c x^4+b x^2+a}dx^2}{5 c}+\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{5 c}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{5 c}-\frac {\int x^2 \left (7 b x^2+4 a\right ) \sqrt {c x^4+b x^2+a}dx^2}{10 c}\right )\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{5 c}-\frac {\frac {5 b \left (7 b^2-12 a c\right ) \int \sqrt {c x^4+b x^2+a}dx^2}{16 c^2}-\frac {\left (-32 a c+35 b^2-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{24 c^2}}{10 c}\right )\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{5 c}-\frac {\frac {5 b \left (7 b^2-12 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{8 c}\right )}{16 c^2}-\frac {\left (-32 a c+35 b^2-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{24 c^2}}{10 c}\right )\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{5 c}-\frac {\frac {5 b \left (7 b^2-12 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{4 c}\right )}{16 c^2}-\frac {\left (-32 a c+35 b^2-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{24 c^2}}{10 c}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{5 c}-\frac {\frac {5 b \left (7 b^2-12 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{8 c^{3/2}}\right )}{16 c^2}-\frac {\left (-32 a c+35 b^2-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{24 c^2}}{10 c}\right )\) |
((x^4*(a + b*x^2 + c*x^4)^(3/2))/(5*c) - (-1/24*((35*b^2 - 32*a*c - 42*b*c *x^2)*(a + b*x^2 + c*x^4)^(3/2))/c^2 + (5*b*(7*b^2 - 12*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))/(10*c))/2
3.10.20.3.1 Defintions of rubi rules used
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[((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[((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[((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_)*((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 = 0.23 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.90
| method | result | size |
| risch | \(-\frac {\left (-384 c^{4} x^{8}-48 c^{3} x^{6} b -128 a \,c^{3} x^{4}+56 b^{2} c^{2} x^{4}+232 a b \,c^{2} x^{2}-70 x^{2} c \,b^{3}+256 a^{2} c^{2}-460 a \,b^{2} c +105 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{3840 c^{4}}+\frac {b \left (48 a^{2} c^{2}-40 a \,b^{2} c +7 b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{512 c^{\frac {9}{2}}}\) | \(154\) |
| pseudoelliptic | \(-\frac {\left (-\frac {45}{32} a^{2} b \,c^{2}+\frac {75}{64} a \,b^{3} c -\frac {105}{512} b^{5}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )+\left (\left (\frac {7}{32} b^{2} x^{4}+\frac {29}{32} a b \,x^{2}+a^{2}\right ) c^{\frac {5}{2}}-\frac {115 b^{2} \left (\frac {7 b \,x^{2}}{46}+a \right ) c^{\frac {3}{2}}}{64}-\frac {x^{4} \left (\frac {3 b \,x^{2}}{8}+a \right ) c^{\frac {7}{2}}}{2}-\frac {3 c^{\frac {9}{2}} x^{8}}{2}+\frac {105 \sqrt {c}\, b^{4}}{256}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}+\frac {45 \left (a c -\frac {7 b^{2}}{12}\right ) \ln \left (2\right ) b \left (a c -\frac {b^{2}}{4}\right )}{32}}{15 c^{\frac {9}{2}}}\) | \(172\) |
| default | \(\frac {x^{4} \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{10 c}-\frac {7 b \,x^{2} \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{80 c^{2}}+\frac {7 b^{2} \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{96 c^{3}}-\frac {7 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{128 c^{3}}-\frac {7 b^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{256 c^{4}}-\frac {5 b^{3} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a}{64 c^{\frac {7}{2}}}+\frac {7 b^{5} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{512 c^{\frac {9}{2}}}+\frac {3 b a \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{32 c^{2}}+\frac {3 b^{2} a \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c^{3}}+\frac {3 b \,a^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {5}{2}}}-\frac {a \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{15 c^{2}}\) | \(296\) |
| elliptic | \(\frac {x^{4} \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{10 c}-\frac {7 b \,x^{2} \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{80 c^{2}}+\frac {7 b^{2} \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{96 c^{3}}-\frac {7 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{128 c^{3}}-\frac {7 b^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{256 c^{4}}-\frac {5 b^{3} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a}{64 c^{\frac {7}{2}}}+\frac {7 b^{5} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{512 c^{\frac {9}{2}}}+\frac {3 b a \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{32 c^{2}}+\frac {3 b^{2} a \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c^{3}}+\frac {3 b \,a^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {5}{2}}}-\frac {a \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{15 c^{2}}\) | \(296\) |
-1/3840*(-384*c^4*x^8-48*b*c^3*x^6-128*a*c^3*x^4+56*b^2*c^2*x^4+232*a*b*c^ 2*x^2-70*b^3*c*x^2+256*a^2*c^2-460*a*b^2*c+105*b^4)*(c*x^4+b*x^2+a)^(1/2)/ c^4+1/512*b*(48*a^2*c^2-40*a*b^2*c+7*b^4)/c^(9/2)*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 = 367, normalized size of antiderivative = 2.15 \[ \int x^7 \sqrt {a+b x^2+c x^4} \, dx=\left [\frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (384 \, c^{5} x^{8} + 48 \, b c^{4} x^{6} - 105 \, b^{4} c + 460 \, a b^{2} c^{2} - 256 \, a^{2} c^{3} - 8 \, {\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} x^{4} + 2 \, {\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{15360 \, c^{5}}, -\frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, {\left (384 \, c^{5} x^{8} + 48 \, b c^{4} x^{6} - 105 \, b^{4} c + 460 \, a b^{2} c^{2} - 256 \, a^{2} c^{3} - 8 \, {\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} x^{4} + 2 \, {\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{7680 \, c^{5}}\right ] \]
[1/15360*(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*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*(384*c^5*x^8 + 48*b*c^4*x^6 - 105*b^4*c + 460*a*b^2*c^2 - 256*a^2*c^3 - 8*(7*b^2*c^3 - 16*a*c^4)*x^4 + 2*(35*b^3*c^2 - 116*a*b*c^3)*x^2)*sqrt(c *x^4 + b*x^2 + a))/c^5, -1/7680*(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*sq rt(-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*(384*c^5*x^8 + 48*b*c^4*x^6 - 105*b^4*c + 460*a*b^2* c^2 - 256*a^2*c^3 - 8*(7*b^2*c^3 - 16*a*c^4)*x^4 + 2*(35*b^3*c^2 - 116*a*b *c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^5]
\[ \int x^7 \sqrt {a+b x^2+c x^4} \, dx=\int x^{7} \sqrt {a + b x^{2} + c x^{4}}\, dx \]
Exception generated. \[ \int x^7 \sqrt {a+b x^2+c x^4} \, dx=\text {Exception raised: ValueError} \]
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 = 0.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99 \[ \int x^7 \sqrt {a+b x^2+c x^4} \, dx=\frac {1}{3840} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {7 \, b^{2} c^{2} - 16 \, a c^{3}}{c^{4}}\right )} x^{2} + \frac {35 \, b^{3} c - 116 \, a b c^{2}}{c^{4}}\right )} x^{2} - \frac {105 \, b^{4} - 460 \, a b^{2} c + 256 \, a^{2} c^{2}}{c^{4}}\right )} - \frac {{\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{512 \, c^{\frac {9}{2}}} \]
1/3840*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*(8*x^2 + b/c)*x^2 - (7*b^2*c^2 - 1 6*a*c^3)/c^4)*x^2 + (35*b^3*c - 116*a*b*c^2)/c^4)*x^2 - (105*b^4 - 460*a*b ^2*c + 256*a^2*c^2)/c^4) - 1/512*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*log(a bs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(9/2)
Time = 14.03 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.84 \[ \int x^7 \sqrt {a+b x^2+c x^4} \, dx=\frac {x^4\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{10\,c}+\frac {7\,b\,\left (\frac {a\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}+\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}-\frac {x^2\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{4\,c}+\frac {5\,b\,\left (\frac {\left (8\,c\,\left (c\,x^4+a\right )-3\,b^2+2\,b\,c\,x^2\right )\,\sqrt {c\,x^4+b\,x^2+a}}{24\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^4+b\,x^2+a}+\frac {2\,c\,x^2+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}\right )}{8\,c}\right )}{20\,c}-\frac {a\,\left (\frac {\left (8\,c\,\left (c\,x^4+a\right )-3\,b^2+2\,b\,c\,x^2\right )\,\sqrt {c\,x^4+b\,x^2+a}}{24\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^4+b\,x^2+a}+\frac {2\,c\,x^2+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}\right )}{5\,c} \]
(x^4*(a + b*x^2 + c*x^4)^(3/2))/(10*c) + (7*b*((a*((b/(4*c) + x^2/2)*(a + b*x^2 + c*x^4)^(1/2) + (log((a + b*x^2 + c*x^4)^(1/2) + (b/2 + c*x^2)/c^(1 /2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c) - (x^2*(a + b*x^2 + c*x^4)^(3/2))/ (4*c) + (5*b*(((8*c*(a + c*x^4) - 3*b^2 + 2*b*c*x^2)*(a + b*x^2 + c*x^4)^( 1/2))/(24*c^2) + (log(2*(a + b*x^2 + c*x^4)^(1/2) + (b + 2*c*x^2)/c^(1/2)) *(b^3 - 4*a*b*c))/(16*c^(5/2))))/(8*c)))/(20*c) - (a*(((8*c*(a + c*x^4) - 3*b^2 + 2*b*c*x^2)*(a + b*x^2 + c*x^4)^(1/2))/(24*c^2) + (log(2*(a + b*x^2 + c*x^4)^(1/2) + (b + 2*c*x^2)/c^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2))))/( 5*c)