\(\int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx\) [479]

Optimal result

Integrand size = 24, antiderivative size = 33 \[ \int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a}{2 b^2 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, 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} \[ \int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a}{2 b^2 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^2} \]

[In]

Int[x^3/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

a/(2*b^2*(a + b*x^2)) + Log[a + b*x^2]/(2*b^2)

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}& = b^2 \int \frac {x^3}{\left (a b+b^2 x^2\right )^2} \, dx \\ & = \frac {1}{2} b^2 \text {Subst}\left (\int \frac {x}{\left (a b+b^2 x\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} b^2 \text {Subst}\left (\int \left (-\frac {a}{b^3 (a+b x)^2}+\frac {1}{b^3 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {a}{2 b^2 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\frac {a}{a+b x^2}+\log \left (a+b x^2\right )}{2 b^2} \]

[In]

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

[Out]

(a/(a + b*x^2) + Log[a + b*x^2])/(2*b^2)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {a}{2 b^{2} \left (b \,x^{2}+a \right )}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(30\)
norman	\(\frac {a}{2 b^{2} \left (b \,x^{2}+a \right )}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(30\)
risch	\(\frac {a}{2 b^{2} \left (b \,x^{2}+a \right )}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(30\)
parallelrisch	\(\frac {\ln \left (b \,x^{2}+a \right ) x^{2} b +\ln \left (b \,x^{2}+a \right ) a +a}{2 b^{2} \left (b \,x^{2}+a \right )}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {{\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \]

[In]

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

[Out]

1/2*((b*x^2 + a)*log(b*x^2 + a) + a)/(b^3*x^2 + a*b^2)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {\log {\left (a + b x^{2} \right )}}{2 b^{2}} \]

[In]

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

[Out]

a/(2*a*b**2 + 2*b**3*x**2) + log(a + b*x**2)/(2*b**2)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

1/2*a/(b^3*x^2 + a*b^2) + 1/2*log(b*x^2 + a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{2}} + \frac {a}{2 \, {\left (b x^{2} + a\right )} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\ln \left (b\,x^2+a\right )}{2\,b^2}+\frac {a}{2\,b^2\,\left (b\,x^2+a\right )} \]

[In]

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

[Out]

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