\(\int (a+\frac {b}{x^2})^2 x^3 \, dx\) [1820]

Optimal result

Integrand size = 13, antiderivative size = 23 \[ \int \left (a+\frac {b}{x^2}\right )^2 x^3 \, dx=a b x^2+\frac {a^2 x^4}{4}+b^2 \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {269, 272, 45} \[ \int \left (a+\frac {b}{x^2}\right )^2 x^3 \, dx=\frac {a^2 x^4}{4}+a b x^2+b^2 \log (x) \]

[In]

Int[(a + b/x^2)^2*x^3,x]

[Out]

a*b*x^2 + (a^2*x^4)/4 + b^2*Log[x]

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 269

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

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}& = \int \frac {\left (b+a x^2\right )^2}{x} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {(b+a x)^2}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (2 a b+\frac {b^2}{x}+a^2 x\right ) \, dx,x,x^2\right ) \\ & = a b x^2+\frac {a^2 x^4}{4}+b^2 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x^2}\right )^2 x^3 \, dx=a b x^2+\frac {a^2 x^4}{4}+b^2 \log (x) \]

[In]

Integrate[(a + b/x^2)^2*x^3,x]

[Out]

a*b*x^2 + (a^2*x^4)/4 + b^2*Log[x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
default	\(a b \,x^{2}+\frac {a^{2} x^{4}}{4}+b^{2} \ln \left (x \right )\)	\(22\)
parallelrisch	\(a b \,x^{2}+\frac {a^{2} x^{4}}{4}+b^{2} \ln \left (x \right )\)	\(22\)
risch	\(\frac {a^{2} x^{4}}{4}+a b \,x^{2}+b^{2}+b^{2} \ln \left (x \right )\)	\(25\)
norman	\(\frac {a b \,x^{5}+\frac {1}{4} a^{2} x^{7}}{x^{3}}+b^{2} \ln \left (x \right )\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \left (a+\frac {b}{x^2}\right )^2 x^3 \, dx=\frac {1}{4} \, a^{2} x^{4} + a b x^{2} + b^{2} \log \left (x\right ) \]

[In]

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

[Out]

1/4*a^2*x^4 + a*b*x^2 + b^2*log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \left (a+\frac {b}{x^2}\right )^2 x^3 \, dx=\frac {a^{2} x^{4}}{4} + a b x^{2} + b^{2} \log {\left (x \right )} \]

[In]

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

[Out]

a**2*x**4/4 + a*b*x**2 + b**2*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \left (a+\frac {b}{x^2}\right )^2 x^3 \, dx=\frac {1}{4} \, a^{2} x^{4} + a b x^{2} + \frac {1}{2} \, b^{2} \log \left (x^{2}\right ) \]

[In]

integrate((a+b/x^2)^2*x^3,x, algorithm="maxima")

[Out]

1/4*a^2*x^4 + a*b*x^2 + 1/2*b^2*log(x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \left (a+\frac {b}{x^2}\right )^2 x^3 \, dx=\frac {1}{4} \, a^{2} x^{4} + a b x^{2} + b^{2} \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

1/4*a^2*x^4 + a*b*x^2 + b^2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \left (a+\frac {b}{x^2}\right )^2 x^3 \, dx=b^2\,\ln \left (x\right )+\frac {a^2\,x^4}{4}+a\,b\,x^2 \]

[In]

int(x^3*(a + b/x^2)^2,x)

[Out]

b^2*log(x) + (a^2*x^4)/4 + a*b*x^2