\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^2}{x^7} \, dx\) [434]

Optimal result

Integrand size = 24, antiderivative size = 49 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^7} \, dx=-\frac {a^4}{6 x^6}-\frac {a^3 b}{x^4}-\frac {3 a^2 b^2}{x^2}+\frac {b^4 x^2}{2}+4 a b^3 \log (x) \]

[Out]

-1/6*a^4/x^6-a^3*b/x^4-3*a^2*b^2/x^2+1/2*b^4*x^2+4*a*b^3*ln(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 272, 45} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^7} \, dx=-\frac {a^4}{6 x^6}-\frac {a^3 b}{x^4}-\frac {3 a^2 b^2}{x^2}+4 a b^3 \log (x)+\frac {b^4 x^2}{2} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^7,x]

[Out]

-1/6*a^4/x^6 - (a^3*b)/x^4 - (3*a^2*b^2)/x^2 + (b^4*x^2)/2 + 4*a*b^3*Log[x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\left (a b+b^2 x^2\right )^4}{x^7} \, dx}{b^4} \\ & = \frac {\text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^4}{x^4} \, dx,x,x^2\right )}{2 b^4} \\ & = \frac {\text {Subst}\left (\int \left (b^8+\frac {a^4 b^4}{x^4}+\frac {4 a^3 b^5}{x^3}+\frac {6 a^2 b^6}{x^2}+\frac {4 a b^7}{x}\right ) \, dx,x,x^2\right )}{2 b^4} \\ & = -\frac {a^4}{6 x^6}-\frac {a^3 b}{x^4}-\frac {3 a^2 b^2}{x^2}+\frac {b^4 x^2}{2}+4 a b^3 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^7} \, dx=-\frac {a^4}{6 x^6}-\frac {a^3 b}{x^4}-\frac {3 a^2 b^2}{x^2}+\frac {b^4 x^2}{2}+4 a b^3 \log (x) \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^7,x]

[Out]

-1/6*a^4/x^6 - (a^3*b)/x^4 - (3*a^2*b^2)/x^2 + (b^4*x^2)/2 + 4*a*b^3*Log[x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {a^{4}}{6 x^{6}}-\frac {a^{3} b}{x^{4}}-\frac {3 a^{2} b^{2}}{x^{2}}+\frac {b^{4} x^{2}}{2}+4 a \,b^{3} \ln \left (x \right )\)	\(46\)
norman	\(\frac {-\frac {1}{6} a^{4}+\frac {1}{2} b^{4} x^{8}-3 a^{2} b^{2} x^{4}-a^{3} b \,x^{2}}{x^{6}}+4 a \,b^{3} \ln \left (x \right )\)	\(48\)
risch	\(\frac {b^{4} x^{2}}{2}+\frac {-3 a^{2} b^{2} x^{4}-a^{3} b \,x^{2}-\frac {1}{6} a^{4}}{x^{6}}+4 a \,b^{3} \ln \left (x \right )\)	\(48\)
parallelrisch	\(\frac {3 b^{4} x^{8}+24 a \,b^{3} \ln \left (x \right ) x^{6}-18 a^{2} b^{2} x^{4}-6 a^{3} b \,x^{2}-a^{4}}{6 x^{6}}\)	\(51\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^2/x^7,x,method=_RETURNVERBOSE)

[Out]

-1/6*a^4/x^6-a^3*b/x^4-3*a^2*b^2/x^2+1/2*b^4*x^2+4*a*b^3*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^7} \, dx=\frac {3 \, b^{4} x^{8} + 24 \, a b^{3} x^{6} \log \left (x\right ) - 18 \, a^{2} b^{2} x^{4} - 6 \, a^{3} b x^{2} - a^{4}}{6 \, x^{6}} \]

[In]

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

[Out]

1/6*(3*b^4*x^8 + 24*a*b^3*x^6*log(x) - 18*a^2*b^2*x^4 - 6*a^3*b*x^2 - a^4)/x^6

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^7} \, dx=4 a b^{3} \log {\left (x \right )} + \frac {b^{4} x^{2}}{2} + \frac {- a^{4} - 6 a^{3} b x^{2} - 18 a^{2} b^{2} x^{4}}{6 x^{6}} \]

[In]

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

[Out]

4*a*b**3*log(x) + b**4*x**2/2 + (-a**4 - 6*a**3*b*x**2 - 18*a**2*b**2*x**4)/(6*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^7} \, dx=\frac {1}{2} \, b^{4} x^{2} + 2 \, a b^{3} \log \left (x^{2}\right ) - \frac {18 \, a^{2} b^{2} x^{4} + 6 \, a^{3} b x^{2} + a^{4}}{6 \, x^{6}} \]

[In]

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

[Out]

1/2*b^4*x^2 + 2*a*b^3*log(x^2) - 1/6*(18*a^2*b^2*x^4 + 6*a^3*b*x^2 + a^4)/x^6

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^7} \, dx=\frac {1}{2} \, b^{4} x^{2} + 2 \, a b^{3} \log \left (x^{2}\right ) - \frac {22 \, a b^{3} x^{6} + 18 \, a^{2} b^{2} x^{4} + 6 \, a^{3} b x^{2} + a^{4}}{6 \, x^{6}} \]

[In]

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

[Out]

1/2*b^4*x^2 + 2*a*b^3*log(x^2) - 1/6*(22*a*b^3*x^6 + 18*a^2*b^2*x^4 + 6*a^3*b*x^2 + a^4)/x^6

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^7} \, dx=\frac {b^4\,x^2}{2}-\frac {\frac {a^4}{6}+a^3\,b\,x^2+3\,a^2\,b^2\,x^4}{x^6}+4\,a\,b^3\,\ln \left (x\right ) \]

[In]

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

[Out]

(b^4*x^2)/2 - (a^4/6 + a^3*b*x^2 + 3*a^2*b^2*x^4)/x^6 + 4*a*b^3*log(x)