3.8.60 \(\int \frac {x^9}{(a+b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=190 \[ \frac {3 \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{7/2}}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c x^2 \left (5 b^2-12 a c\right )\right ) \sqrt {a+b x^2+c x^4}}{8 c^3 \left (b^2-4 a c\right )}-\frac {b x^4 \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}+\frac {x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 190, 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, 738, 832, 779, 621, 206} \begin {gather*} -\frac {\left (b \left (15 b^2-52 a c\right )-2 c x^2 \left (5 b^2-12 a c\right )\right ) \sqrt {a+b x^2+c x^4}}{8 c^3 \left (b^2-4 a c\right )}+\frac {3 \left (5 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 )}{16 c^{7/2}}+\frac {x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b x^4 \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(x^6*(2*a + b*x^2))/((b^2 - 4*a*c)*Sqrt[a + b*x^2 + c*x^4]) - (b*x^4*Sqrt[a + b*x^2 + c*x^4])/(c*(b^2 - 4*a*c)
) - ((b*(15*b^2 - 52*a*c) - 2*c*(5*b^2 - 12*a*c)*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c^3*(b^2 - 4*a*c)) + (3*(5*b
^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*c^(7/2))

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 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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && 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 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

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 \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac {x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {\operatorname {Subst}\left (\int \frac {x^2 (6 a+3 b x)}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{b^2-4 a c}\\ &=\frac {x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b x^4 \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}-\frac {\operatorname {Subst}\left (\int \frac {x \left (-6 a b-\frac {3}{2} \left (5 b^2-12 a c\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{3 c \left (b^2-4 a c\right )}\\ &=\frac {x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b x^4 \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c \left (5 b^2-12 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c^3 \left (b^2-4 a c\right )}+\frac {\left (3 \left (5 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 )}{16 c^3}\\ &=\frac {x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b x^4 \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c \left (5 b^2-12 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c^3 \left (b^2-4 a c\right )}+\frac {\left (3 \left (5 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 )}{8 c^3}\\ &=\frac {x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b x^4 \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c \left (5 b^2-12 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c^3 \left (b^2-4 a c\right )}+\frac {3 \left (5 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 )}{16 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 181, normalized size = 0.95 \begin {gather*} \frac {\frac {2 \sqrt {c} \left (4 a^2 c \left (6 c x^2-13 b\right )+a \left (15 b^3-62 b^2 c x^2-20 b c^2 x^4+8 c^3 x^6\right )+b^2 x^2 \left (15 b^2+5 b c x^2-2 c^2 x^4\right )\right )}{\sqrt {a+b x^2+c x^4}}-3 \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{7/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

((2*Sqrt[c]*(4*a^2*c*(-13*b + 6*c*x^2) + b^2*x^2*(15*b^2 + 5*b*c*x^2 - 2*c^2*x^4) + a*(15*b^3 - 62*b^2*c*x^2 -
20*b*c^2*x^4 + 8*c^3*x^6)))/Sqrt[a + b*x^2 + c*x^4] - 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*ArcTanh[(b + 2*c*x^
2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*c^(7/2)*(-b^2 + 4*a*c))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.76, size = 169, normalized size = 0.89 \begin {gather*} \frac {-52 a^2 b c+24 a^2 c^2 x^2+15 a b^3-62 a b^2 c x^2-20 a b c^2 x^4+8 a c^3 x^6+15 b^4 x^2+5 b^3 c x^4-2 b^2 c^2 x^6}{8 c^3 \left (4 a c-b^2\right ) \sqrt {a+b x^2+c x^4}}-\frac {3 \left (5 b^2-4 a c\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x^2+c x^4}+b+2 c x^2\right )}{16 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^9/(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(15*a*b^3 - 52*a^2*b*c + 15*b^4*x^2 - 62*a*b^2*c*x^2 + 24*a^2*c^2*x^2 + 5*b^3*c*x^4 - 20*a*b*c^2*x^4 - 2*b^2*c
^2*x^6 + 8*a*c^3*x^6)/(8*c^3*(-b^2 + 4*a*c)*Sqrt[a + b*x^2 + c*x^4]) - (3*(5*b^2 - 4*a*c)*Log[b + 2*c*x^2 - 2*
Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/(16*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 1.83, size = 591, normalized size = 3.11 \begin {gather*} \left [-\frac {3 \, {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{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 (2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{6} - 15 \, a b^{3} c + 52 \, a^{2} b c^{2} - 5 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{4} - {\left (15 \, b^{4} c - 62 \, a b^{2} c^{2} + 24 \, a^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{32 \, {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5} + {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{4} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x^{2}\right )}}, -\frac {3 \, {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{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 (2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{6} - 15 \, a b^{3} c + 52 \, a^{2} b c^{2} - 5 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{4} - {\left (15 \, b^{4} c - 62 \, a b^{2} c^{2} + 24 \, a^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{16 \, {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5} + {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{4} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/32*(3*(5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2 + (5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^4 + (5*b^5 - 24*a*b^3
*c + 16*a^2*b*c^2)*x^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)*sqr
t(c) - 4*a*c) - 4*(2*(b^2*c^3 - 4*a*c^4)*x^6 - 15*a*b^3*c + 52*a^2*b*c^2 - 5*(b^3*c^2 - 4*a*b*c^3)*x^4 - (15*b
^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x
^4 + (b^3*c^4 - 4*a*b*c^5)*x^2), -1/16*(3*(5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2 + (5*b^4*c - 24*a*b^2*c^2 + 16*
a^2*c^3)*x^4 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^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*(2*(b^2*c^3 - 4*a*c^4)*x^6 - 15*a*b^3*c + 52*a^2*b*c^2 - 5*(b^3*c^
2 - 4*a*b*c^3)*x^4 - (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/(a*b^2*c^4 - 4*a^2*c
^5 + (b^2*c^5 - 4*a*c^6)*x^4 + (b^3*c^4 - 4*a*b*c^5)*x^2)]

________________________________________________________________________________________

giac [A]  time = 0.27, size = 215, normalized size = 1.13 \begin {gather*} \frac {{\left ({\left (\frac {2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2}}{b^{2} c^{3} - 4 \, a c^{4}} - \frac {5 \, {\left (b^{3} c - 4 \, a b c^{2}\right )}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x^{2} - \frac {15 \, b^{4} - 62 \, a b^{2} c + 24 \, a^{2} c^{2}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x^{2} - \frac {15 \, a b^{3} - 52 \, a^{2} b c}{b^{2} c^{3} - 4 \, a c^{4}}}{8 \, \sqrt {c x^{4} + b x^{2} + a}} - \frac {3 \, {\left (5 \, 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 {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/8*(((2*(b^2*c^2 - 4*a*c^3)*x^2/(b^2*c^3 - 4*a*c^4) - 5*(b^3*c - 4*a*b*c^2)/(b^2*c^3 - 4*a*c^4))*x^2 - (15*b^
4 - 62*a*b^2*c + 24*a^2*c^2)/(b^2*c^3 - 4*a*c^4))*x^2 - (15*a*b^3 - 52*a^2*b*c)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^
4 + b*x^2 + a) - 3/16*(5*b^2 - 4*a*c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(7/2)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 354, normalized size = 1.86 \begin {gather*} \frac {x^{6}}{4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c}-\frac {13 a \,b^{2} x^{2}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}+\frac {15 b^{4} x^{2}}{16 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{3}}-\frac {5 b \,x^{4}}{8 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}-\frac {13 a \,b^{3}}{8 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{3}}+\frac {3 a \,x^{2}}{4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}+\frac {15 b^{5}}{32 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{4}}-\frac {15 b^{2} x^{2}}{16 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{3}}-\frac {3 a \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {5}{2}}}+\frac {15 b^{2} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {7}{2}}}-\frac {13 a b}{8 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{3}}+\frac {15 b^{3}}{32 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(c*x^4+b*x^2+a)^(3/2),x)

[Out]

1/4*x^6/c/(c*x^4+b*x^2+a)^(1/2)-5/8*b/c^2*x^4/(c*x^4+b*x^2+a)^(1/2)-15/16*b^2/c^3*x^2/(c*x^4+b*x^2+a)^(1/2)+15
/32*b^3/c^4/(c*x^4+b*x^2+a)^(1/2)+15/16*b^4/c^3/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)*x^2+15/32*b^5/c^4/(4*a*c-b^2
)/(c*x^4+b*x^2+a)^(1/2)+15/16*b^2/c^(7/2)*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-13/8*b/c^3*a/(c*x^4+
b*x^2+a)^(1/2)-13/4*b^2/c^2*a/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)*x^2-13/8*b^3/c^3*a/(4*a*c-b^2)/(c*x^4+b*x^2+a)
^(1/2)+3/4/c^2*a*x^2/(c*x^4+b*x^2+a)^(1/2)-3/4/c^(5/2)*a*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))

________________________________________________________________________________________

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^9/(c*x^4+b*x^2+a)^(3/2),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 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^9}{{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(a + b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^9/(a + b*x^2 + c*x^4)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{9}}{\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(x**9/(a + b*x**2 + c*x**4)**(3/2), x)

________________________________________________________________________________________