\(\int \frac {(a+b x^2+c x^4)^2}{x^{11}} \, dx\) [838]

Optimal result

Integrand size = 18, antiderivative size = 54 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx=-\frac {a^2}{10 x^{10}}-\frac {a b}{4 x^8}-\frac {b^2+2 a c}{6 x^6}-\frac {b c}{2 x^4}-\frac {c^2}{2 x^2} \]

[Out]

-1/10*a^2/x^10-1/4*a*b/x^8+1/6*(-2*a*c-b^2)/x^6-1/2*b*c/x^4-1/2*c^2/x^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1128, 712} \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx=-\frac {a^2}{10 x^{10}}-\frac {2 a c+b^2}{6 x^6}-\frac {a b}{4 x^8}-\frac {b c}{2 x^4}-\frac {c^2}{2 x^2} \]

[In]

Int[(a + b*x^2 + c*x^4)^2/x^11,x]

[Out]

-1/10*a^2/x^10 - (a*b)/(4*x^8) - (b^2 + 2*a*c)/(6*x^6) - (b*c)/(2*x^4) - c^2/(2*x^2)

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

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 )^2}{x^6} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {a^2}{x^6}+\frac {2 a b}{x^5}+\frac {b^2+2 a c}{x^4}+\frac {2 b c}{x^3}+\frac {c^2}{x^2}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a^2}{10 x^{10}}-\frac {a b}{4 x^8}-\frac {b^2+2 a c}{6 x^6}-\frac {b c}{2 x^4}-\frac {c^2}{2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx=-\frac {6 a^2+5 a \left (3 b x^2+4 c x^4\right )+10 x^4 \left (b^2+3 b c x^2+3 c^2 x^4\right )}{60 x^{10}} \]

[In]

Integrate[(a + b*x^2 + c*x^4)^2/x^11,x]

[Out]

-1/60*(6*a^2 + 5*a*(3*b*x^2 + 4*c*x^4) + 10*x^4*(b^2 + 3*b*c*x^2 + 3*c^2*x^4))/x^10

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {c^{2}}{2 x^{2}}-\frac {a b}{4 x^{8}}-\frac {2 a c +b^{2}}{6 x^{6}}-\frac {b c}{2 x^{4}}-\frac {a^{2}}{10 x^{10}}\)	\(45\)
norman	\(\frac {-\frac {c^{2} x^{8}}{2}-\frac {b c \,x^{6}}{2}+\left (-\frac {a c}{3}-\frac {b^{2}}{6}\right ) x^{4}-\frac {a b \,x^{2}}{4}-\frac {a^{2}}{10}}{x^{10}}\)	\(47\)
risch	\(\frac {-\frac {c^{2} x^{8}}{2}-\frac {b c \,x^{6}}{2}+\left (-\frac {a c}{3}-\frac {b^{2}}{6}\right ) x^{4}-\frac {a b \,x^{2}}{4}-\frac {a^{2}}{10}}{x^{10}}\)	\(47\)
gosper	\(-\frac {30 c^{2} x^{8}+30 b c \,x^{6}+20 a c \,x^{4}+10 b^{2} x^{4}+15 a b \,x^{2}+6 a^{2}}{60 x^{10}}\)	\(49\)
parallelrisch	\(\frac {-30 c^{2} x^{8}-30 b c \,x^{6}-20 a c \,x^{4}-10 b^{2} x^{4}-15 a b \,x^{2}-6 a^{2}}{60 x^{10}}\)	\(49\)

[In]

int((c*x^4+b*x^2+a)^2/x^11,x,method=_RETURNVERBOSE)

[Out]

-1/2*c^2/x^2-1/4*a*b/x^8-1/6*(2*a*c+b^2)/x^6-1/2*b*c/x^4-1/10*a^2/x^10

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx=-\frac {30 \, c^{2} x^{8} + 30 \, b c x^{6} + 10 \, {\left (b^{2} + 2 \, a c\right )} x^{4} + 15 \, a b x^{2} + 6 \, a^{2}}{60 \, x^{10}} \]

[In]

integrate((c*x^4+b*x^2+a)^2/x^11,x, algorithm="fricas")

[Out]

-1/60*(30*c^2*x^8 + 30*b*c*x^6 + 10*(b^2 + 2*a*c)*x^4 + 15*a*b*x^2 + 6*a^2)/x^10

Sympy [A] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx=\frac {- 6 a^{2} - 15 a b x^{2} - 30 b c x^{6} - 30 c^{2} x^{8} + x^{4} \left (- 20 a c - 10 b^{2}\right )}{60 x^{10}} \]

[In]

integrate((c*x**4+b*x**2+a)**2/x**11,x)

[Out]

(-6*a**2 - 15*a*b*x**2 - 30*b*c*x**6 - 30*c**2*x**8 + x**4*(-20*a*c - 10*b**2))/(60*x**10)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx=-\frac {30 \, c^{2} x^{8} + 30 \, b c x^{6} + 10 \, {\left (b^{2} + 2 \, a c\right )} x^{4} + 15 \, a b x^{2} + 6 \, a^{2}}{60 \, x^{10}} \]

[In]

integrate((c*x^4+b*x^2+a)^2/x^11,x, algorithm="maxima")

[Out]

-1/60*(30*c^2*x^8 + 30*b*c*x^6 + 10*(b^2 + 2*a*c)*x^4 + 15*a*b*x^2 + 6*a^2)/x^10

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx=-\frac {30 \, c^{2} x^{8} + 30 \, b c x^{6} + 10 \, b^{2} x^{4} + 20 \, a c x^{4} + 15 \, a b x^{2} + 6 \, a^{2}}{60 \, x^{10}} \]

[In]

integrate((c*x^4+b*x^2+a)^2/x^11,x, algorithm="giac")

[Out]

-1/60*(30*c^2*x^8 + 30*b*c*x^6 + 10*b^2*x^4 + 20*a*c*x^4 + 15*a*b*x^2 + 6*a^2)/x^10

Mupad [B] (verification not implemented)

Time = 12.90 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx=-\frac {\frac {a^2}{10}+x^4\,\left (\frac {b^2}{6}+\frac {a\,c}{3}\right )+\frac {c^2\,x^8}{2}+\frac {a\,b\,x^2}{4}+\frac {b\,c\,x^6}{2}}{x^{10}} \]

[In]

int((a + b*x^2 + c*x^4)^2/x^11,x)

[Out]

-(a^2/10 + x^4*((a*c)/3 + b^2/6) + (c^2*x^8)/2 + (a*b*x^2)/4 + (b*c*x^6)/2)/x^10