3.8.23 \(\int x^7 \sqrt {a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=171 \[ \frac {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{512 c^{9/2}}-\frac {b \left (7 b^2-12 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{256 c^4}+\frac {\left (-32 a c+35 b^2-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 c^3}+\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c} \]

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Rubi [A]  time = 0.16, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1114, 742, 779, 612, 621, 206} \begin {gather*} \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}}{480 c^3}-\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 {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{512 c^{9/2}}+\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

-(b*(7*b^2 - 12*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*c^4) + (x^4*(a + b*x^2 + c*x^4)^(3/2))/(10*c)
+ ((35*b^2 - 32*a*c - 42*b*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(480*c^3) + (b*(7*b^2 - 12*a*c)*(b^2 - 4*a*c)*Ar
cTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(512*c^(9/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 612

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] - Dist[(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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

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] + Dist[(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] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int x^7 \sqrt {a+b x^2+c x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^3 \sqrt {a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c}+\frac {\operatorname {Subst}\left (\int x \left (-2 a-\frac {7 b x}{2}\right ) \sqrt {a+b x+c x^2} \, dx,x,x^2\right )}{10 c}\\ &=\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 {\left (b \left (7 b^2-12 a c\right )\right ) \operatorname {Subst}\left (\int \sqrt {a+b x+c x^2} \, dx,x,x^2\right )}{64 c^3}\\ &=-\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 {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{512 c^4}\\ &=-\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 {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^2}{\sqrt {a+b x^2+c x^4}}\right )}{256 c^4}\\ &=-\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 ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{512 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 164, normalized size = 0.96 \begin {gather*} \frac {-\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}}{48 c^2}+\frac {5 \left (12 a b c-7 b^3\right ) \left (2 \sqrt {c} \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )\right )}{256 c^{7/2}}+x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(x^4*(a + b*x^2 + c*x^4)^(3/2) - ((-35*b^2 + 32*a*c + 42*b*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(48*c^2) + (5*(-7
*b^3 + 12*a*b*c)*(2*Sqrt[c]*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4] - (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqr
t[c]*Sqrt[a + b*x^2 + c*x^4])]))/(256*c^(7/2)))/(10*c)

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IntegrateAlgebraic [A]  time = 0.46, size = 170, normalized size = 0.99 \begin {gather*} \frac {\left (-48 a^2 b c^2+40 a b^3 c-7 b^5\right ) \log \left (-2 c^{9/2} \sqrt {a+b x^2+c x^4}+b c^4+2 c^5 x^2\right )}{512 c^{9/2}}+\frac {\sqrt {a+b x^2+c x^4} \left (-256 a^2 c^2+460 a b^2 c-232 a b c^2 x^2+128 a c^3 x^4-105 b^4+70 b^3 c x^2-56 b^2 c^2 x^4+48 b c^3 x^6+384 c^4 x^8\right )}{3840 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^7*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(-105*b^4 + 460*a*b^2*c - 256*a^2*c^2 + 70*b^3*c*x^2 - 232*a*b*c^2*x^2 - 56*b^2*c^2*x
^4 + 128*a*c^3*x^4 + 48*b*c^3*x^6 + 384*c^4*x^8))/(3840*c^4) + ((-7*b^5 + 40*a*b^3*c - 48*a^2*b*c^2)*Log[b*c^4
+ 2*c^5*x^2 - 2*c^(9/2)*Sqrt[a + b*x^2 + c*x^4]])/(512*c^(9/2))

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fricas [A]  time = 0.94, size = 367, normalized size = 2.15 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[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*(1
5*(7*b^5 - 40*a*b^3*c + 48*a^2*b*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*(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]

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giac [A]  time = 0.25, size = 172, normalized size = 1.01 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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 - 16*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
(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(9/2)

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maple [A]  time = 0.02, size = 296, normalized size = 1.73 \begin {gather*} \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} x^{4}}{10 c}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a b \,x^{2}}{32 c^{2}}-\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3} x^{2}}{128 c^{3}}+\frac {3 a^{2} b \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {5}{2}}}-\frac {5 a \,b^{3} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{64 c^{\frac {7}{2}}}+\frac {7 b^{5} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{512 c^{\frac {9}{2}}}-\frac {7 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b \,x^{2}}{80 c^{2}}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,b^{2}}{64 c^{3}}-\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{4}}{256 c^{4}}-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} a}{15 c^{2}}+\frac {7 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}{96 c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(c*x^4+b*x^2+a)^(1/2),x)

[Out]

1/10*x^4*(c*x^4+b*x^2+a)^(3/2)/c-7/80*b/c^2*x^2*(c*x^4+b*x^2+a)^(3/2)+7/96*b^2/c^3*(c*x^4+b*x^2+a)^(3/2)-7/128
*b^3/c^3*(c*x^4+b*x^2+a)^(1/2)*x^2-7/256*b^4/c^4*(c*x^4+b*x^2+a)^(1/2)-5/64*b^3/c^(7/2)*ln((1/2*b+c*x^2)/c^(1/
2)+(c*x^4+b*x^2+a)^(1/2))*a+7/512*b^5/c^(9/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+3/32*b/c^2*a*(c*
x^4+b*x^2+a)^(1/2)*x^2+3/64*b^2/c^3*a*(c*x^4+b*x^2+a)^(1/2)+3/32*b/c^(5/2)*a^2*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4
+b*x^2+a)^(1/2))-1/15*a/c^2*(c*x^4+b*x^2+a)^(3/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 5.31, size = 315, normalized size = 1.84 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a + b*x^2 + c*x^4)^(1/2),x)

[Out]

(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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{7} \sqrt {a + b x^{2} + c x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral(x**7*sqrt(a + b*x**2 + c*x**4), x)

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