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

Optimal. Leaf size=133 \[ -\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{5/2}}+\frac {3 \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^2 x^4}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8} \]

[Out]

-1/16*(b*x^2+2*a)*(c*x^4+b*x^2+a)^(3/2)/a/x^8-3/256*(-4*a*c+b^2)^2*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^
2+a)^(1/2))/a^(5/2)+3/128*(-4*a*c+b^2)*(b*x^2+2*a)*(c*x^4+b*x^2+a)^(1/2)/a^2/x^4

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Rubi [A]  time = 0.12, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1114, 720, 724, 206} \[ \frac {3 \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^2 x^4}-\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{5/2}}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^(3/2)/x^9,x]

[Out]

(3*(b^2 - 4*a*c)*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(128*a^2*x^4) - ((2*a + b*x^2)*(a + b*x^2 + c*x^4)^(3/
2))/(16*a*x^8) - (3*(b^2 - 4*a*c)^2*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(256*a^(5/2))

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 720

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)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(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] && EqQ[m +
2*p + 2, 0] && GtQ[p, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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 {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )\\ &=-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^2\right )}{32 a}\\ &=\frac {3 \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^2 x^4}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8}+\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{256 a^2}\\ &=\frac {3 \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^2 x^4}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8}-\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}}\right )}{128 a^2}\\ &=\frac {3 \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^2 x^4}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8}-\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 138, normalized size = 1.04 \[ -\frac {\frac {3 \left (b^2-4 a c\right ) \left (x^4 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )-2 \sqrt {a} \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}\right )}{8 a^{3/2} x^4}+\frac {2 \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{x^8}}{32 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*((2*(2*a + b*x^2)*(a + b*x^2 + c*x^4)^(3/2))/x^8 + (3*(b^2 - 4*a*c)*(-2*Sqrt[a]*(2*a + b*x^2)*Sqrt[a + b
*x^2 + c*x^4] + (b^2 - 4*a*c)*x^4*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])]))/(8*a^(3/2)*x^4)
)/a

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fricas [A]  time = 1.05, size = 319, normalized size = 2.40 \[ \left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {a} x^{8} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, {\left ({\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{6} - 24 \, a^{3} b x^{2} - 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{4} - 16 \, a^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{512 \, a^{3} x^{8}}, \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-a} x^{8} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{6} - 24 \, a^{3} b x^{2} - 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{4} - 16 \, a^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{256 \, a^{3} x^{8}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/512*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(a)*x^8*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^
2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) + 4*((3*a*b^3 - 20*a^2*b*c)*x^6 - 24*a^3*b*x^2 - 2*(a^2*b^2 + 20*a^
3*c)*x^4 - 16*a^4)*sqrt(c*x^4 + b*x^2 + a))/(a^3*x^8), 1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-a)*x^8*ar
ctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 2*((3*a*b^3 - 20*a^2*b*c)
*x^6 - 24*a^3*b*x^2 - 2*(a^2*b^2 + 20*a^3*c)*x^4 - 16*a^4)*sqrt(c*x^4 + b*x^2 + a))/(a^3*x^8)]

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giac [B]  time = 0.49, size = 606, normalized size = 4.56 \[ \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{128 \, \sqrt {-a} a^{2}} - \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} b^{4} - 24 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a b^{2} c - 80 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a^{2} c^{2} - 256 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{6} a^{2} b c^{\frac {3}{2}} - 11 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a b^{4} - 168 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{2} b^{2} c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{3} c^{2} - 128 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{2} b^{3} \sqrt {c} - 11 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{2} b^{4} - 168 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{3} b^{2} c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{4} c^{2} - 256 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{4} b c^{\frac {3}{2}} + 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{3} b^{4} - 24 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{4} b^{2} c - 80 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{5} c^{2}}{128 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{4} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2)
- 1/128*(3*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*b^4 - 24*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a*b^2*
c - 80*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a^2*c^2 - 256*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^6*a^2*b
*c^(3/2) - 11*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a*b^4 - 168*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*
a^2*b^2*c - 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a^3*c^2 - 128*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))
^4*a^2*b^3*sqrt(c) - 11*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^2*b^4 - 168*(sqrt(c)*x^2 - sqrt(c*x^4 + b*
x^2 + a))^3*a^3*b^2*c - 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^4*c^2 - 256*(sqrt(c)*x^2 - sqrt(c*x^4 +
b*x^2 + a))^2*a^4*b*c^(3/2) + 3*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^3*b^4 - 24*(sqrt(c)*x^2 - sqrt(c*x^
4 + b*x^2 + a))*a^4*b^2*c - 80*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^5*c^2)/(((sqrt(c)*x^2 - sqrt(c*x^4 +
b*x^2 + a))^2 - a)^4*a^2)

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maple [B]  time = 0.02, size = 260, normalized size = 1.95 \[ -\frac {3 c^{2} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{16 \sqrt {a}}+\frac {3 b^{2} c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{32 a^{\frac {3}{2}}}-\frac {3 b^{4} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{256 a^{\frac {5}{2}}}-\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b c}{32 a \,x^{2}}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3}}{128 a^{2} x^{2}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2}}{64 a \,x^{4}}-\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c}{16 x^{4}}-\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{16 x^{6}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a}{8 x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*b^2/a/x^4*(c*x^4+b*x^2+a)^(1/2)+3/128*b^3/a^2/x^2*(c*x^4+b*x^2+a)^(1/2)-3/16*c^2/a^(1/2)*ln((b*x^2+2*a+2
*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-5/16*c/x^4*(c*x^4+b*x^2+a)^(1/2)-3/256*b^4/a^(5/2)*ln((b*x^2+2*a+2*(c*x^4
+b*x^2+a)^(1/2)*a^(1/2))/x^2)-1/8*a/x^8*(c*x^4+b*x^2+a)^(1/2)-5/32/a*c*b/x^2*(c*x^4+b*x^2+a)^(1/2)+3/32/a^(3/2
)*c*b^2*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-3/16*b/x^6*(c*x^4+b*x^2+a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(3/2)/x^9,x)

[Out]

int((a + b*x^2 + c*x^4)^(3/2)/x^9, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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