\(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^9} \, dx\) [944]

Optimal result

Integrand size = 20, antiderivative size = 133 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\frac {3 \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^2 x^4}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8}-\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{5/2}} \]

[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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1128, 734, 738, 212} \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=-\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{5/2}}+\frac {3 \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^2 x^4}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8} \]

[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 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 734

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] /; Free
Q[{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 738

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 1128

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*} \text {integral}& = \frac {1}{2} \text {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 ) \text {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 ) \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\frac {-\frac {\sqrt {a} \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4} \left (8 a^2+8 a b x^2-3 b^2 x^4+20 a c x^4\right )}{x^8}+3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{128 a^{5/2}} \]

[In]

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

[Out]

(-((Sqrt[a]*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4]*(8*a^2 + 8*a*b*x^2 - 3*b^2*x^4 + 20*a*c*x^4))/x^8) + 3*(b^2
- 4*a*c)^2*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/(128*a^(5/2))

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92

method	result	size
pseudoelliptic	\(-\frac {3 \left (x^{8} \left (a c -\frac {b^{2}}{4}\right )^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (-\frac {2 b \,x^{4} \left (10 c \,x^{2}+b \right ) a^{\frac {3}{2}}}{3}-8 x^{2} \left (\frac {5 c \,x^{2}}{3}+b \right ) a^{\frac {5}{2}}+\sqrt {a}\, b^{3} x^{6}-\frac {16 a^{\frac {7}{2}}}{3}\right )}{8}\right )}{16 a^{\frac {5}{2}} x^{8}}\)	\(123\)
risch	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (20 a b c \,x^{6}-3 b^{3} x^{6}+40 a^{2} c \,x^{4}+2 b^{2} x^{4} a +24 a^{2} b \,x^{2}+16 a^{3}\right )}{128 x^{8} a^{2}}-\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{256 a^{\frac {5}{2}}}\)	\(130\)
default	\(-\frac {b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 a \,x^{4}}+\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 a^{2} x^{2}}-\frac {3 b^{4} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{256 a^{\frac {5}{2}}}+\frac {3 b^{2} c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {3}{2}}}-\frac {5 b c \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 a \,x^{2}}-\frac {3 c^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 x^{8}}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 x^{6}}-\frac {5 c \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 x^{4}}\)	\(260\)
elliptic	\(-\frac {b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 a \,x^{4}}+\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 a^{2} x^{2}}-\frac {3 b^{4} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{256 a^{\frac {5}{2}}}+\frac {3 b^{2} c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {3}{2}}}-\frac {5 b c \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 a \,x^{2}}-\frac {3 c^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 x^{8}}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 x^{6}}-\frac {5 c \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 x^{4}}\)	\(260\)

[In]

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

[Out]

-3/16/a^(5/2)*(x^8*(a*c-1/4*b^2)^2*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)-1/8*(c*x^4+b*x^2+a)^(1/
2)*(-2/3*b*x^4*(10*c*x^2+b)*a^(3/2)-8*x^2*(5/3*c*x^2+b)*a^(5/2)+a^(1/2)*b^3*x^6-16/3*a^(7/2)))/x^8

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.40 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {a} x^{8} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, {\left ({\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{6} - 24 \, a^{3} b x^{2} - 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{4} - 16 \, a^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{512 \, a^{3} x^{8}}, \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-a} x^{8} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{6} - 24 \, a^{3} b x^{2} - 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{4} - 16 \, a^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{256 \, a^{3} x^{8}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\int \frac {\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{x^{9}}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\text {Exception raised: ValueError} \]

[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 deta

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 606 vs. \(2 (115) = 230\).

Time = 0.35 (sec) , antiderivative size = 606, normalized size of antiderivative = 4.56 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{128 \, \sqrt {-a} a^{2}} - \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} b^{4} - 24 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a b^{2} c - 80 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a^{2} c^{2} - 256 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{6} a^{2} b c^{\frac {3}{2}} - 11 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a b^{4} - 168 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{2} b^{2} c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{3} c^{2} - 128 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{2} b^{3} \sqrt {c} - 11 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{2} b^{4} - 168 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{3} b^{2} c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{4} c^{2} - 256 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{4} b c^{\frac {3}{2}} + 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{3} b^{4} - 24 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{4} b^{2} c - 80 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{5} c^{2}}{128 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{4} a^{2}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\int \frac {{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{x^9} \,d x \]

[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)