\(\int \frac {x^3}{(b x^2+c x^4)^2} \, dx\) [199]

Optimal result

Integrand size = 17, antiderivative size = 38 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^2} \, dx=\frac {1}{2 b \left (b+c x^2\right )}+\frac {\log (x)}{b^2}-\frac {\log \left (b+c x^2\right )}{2 b^2} \]

[Out]

1/2/b/(c*x^2+b)+ln(x)/b^2-1/2*ln(c*x^2+b)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1598, 272, 46} \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {\log \left (b+c x^2\right )}{2 b^2}+\frac {\log (x)}{b^2}+\frac {1}{2 b \left (b+c x^2\right )} \]

[In]

Int[x^3/(b*x^2 + c*x^4)^2,x]

[Out]

1/(2*b*(b + c*x^2)) + Log[x]/b^2 - Log[b + c*x^2]/(2*b^2)

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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]]

Rule 1598

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x \left (b+c x^2\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x (b+c x)^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{b^2 x}-\frac {c}{b (b+c x)^2}-\frac {c}{b^2 (b+c x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{2 b \left (b+c x^2\right )}+\frac {\log (x)}{b^2}-\frac {\log \left (b+c x^2\right )}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {b}{b+c x^2}+2 \log (x)-\log \left (b+c x^2\right )}{2 b^2} \]

[In]

Integrate[x^3/(b*x^2 + c*x^4)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {1}{2 b \left (c \,x^{2}+b \right )}+\frac {\ln \left (x \right )}{b^{2}}-\frac {\ln \left (c \,x^{2}+b \right )}{2 b^{2}}\)	\(35\)
norman	\(-\frac {c \,x^{2}}{2 b^{2} \left (c \,x^{2}+b \right )}+\frac {\ln \left (x \right )}{b^{2}}-\frac {\ln \left (c \,x^{2}+b \right )}{2 b^{2}}\)	\(39\)
default	\(\frac {\ln \left (x \right )}{b^{2}}-\frac {c \left (\frac {\ln \left (c \,x^{2}+b \right )}{c}-\frac {b}{c \left (c \,x^{2}+b \right )}\right )}{2 b^{2}}\)	\(42\)
parallelrisch	\(\frac {2 c \ln \left (x \right ) x^{2}-c \ln \left (c \,x^{2}+b \right ) x^{2}-c \,x^{2}+2 b \ln \left (x \right )-b \ln \left (c \,x^{2}+b \right )}{2 b^{2} \left (c \,x^{2}+b \right )}\)	\(60\)

[In]

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

[Out]

1/2/b/(c*x^2+b)+ln(x)/b^2-1/2*ln(c*x^2+b)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {{\left (c x^{2} + b\right )} \log \left (c x^{2} + b\right ) - 2 \, {\left (c x^{2} + b\right )} \log \left (x\right ) - b}{2 \, {\left (b^{2} c x^{2} + b^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^2} \, dx=\frac {1}{2 b^{2} + 2 b c x^{2}} + \frac {\log {\left (x \right )}}{b^{2}} - \frac {\log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{2}} \]

[In]

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

[Out]

1/(2*b**2 + 2*b*c*x**2) + log(x)/b**2 - log(b/c + x**2)/(2*b**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^2} \, dx=\frac {1}{2 \, {\left (b c x^{2} + b^{2}\right )}} - \frac {\log \left (c x^{2} + b\right )}{2 \, b^{2}} + \frac {\log \left (x^{2}\right )}{2 \, b^{2}} \]

[In]

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

[Out]

1/2/(b*c*x^2 + b^2) - 1/2*log(c*x^2 + b)/b^2 + 1/2*log(x^2)/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {\log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{2}} + \frac {\log \left ({\left | x \right |}\right )}{b^{2}} + \frac {1}{2 \, {\left (c x^{2} + b\right )} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.82 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\ln \left (x\right )}{b^2}+\frac {1}{2\,b\,\left (c\,x^2+b\right )}-\frac {\ln \left (c\,x^2+b\right )}{2\,b^2} \]

[In]

int(x^3/(b*x^2 + c*x^4)^2,x)

[Out]

log(x)/b^2 + 1/(2*b*(b + c*x^2)) - log(b + c*x^2)/(2*b^2)