\(\int \frac {x^5}{b x^2+c x^4} \, dx\) [179]

Optimal result

Integrand size = 17, antiderivative size = 27 \[ \int \frac {x^5}{b x^2+c x^4} \, dx=\frac {x^2}{2 c}-\frac {b \log \left (b+c x^2\right )}{2 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, 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, 45} \[ \int \frac {x^5}{b x^2+c x^4} \, dx=\frac {x^2}{2 c}-\frac {b \log \left (b+c x^2\right )}{2 c^2} \]

[In]

Int[x^5/(b*x^2 + c*x^4),x]

[Out]

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

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]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{b x^2+c x^4} \, dx=\frac {x^2}{2 c}-\frac {b \log \left (b+c x^2\right )}{2 c^2} \]

[In]

Integrate[x^5/(b*x^2 + c*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
parallelrisch	\(-\frac {-c \,x^{2}+b \ln \left (c \,x^{2}+b \right )}{2 c^{2}}\)	\(23\)
default	\(\frac {x^{2}}{2 c}-\frac {b \ln \left (c \,x^{2}+b \right )}{2 c^{2}}\)	\(24\)
norman	\(\frac {x^{2}}{2 c}-\frac {b \ln \left (c \,x^{2}+b \right )}{2 c^{2}}\)	\(24\)
risch	\(\frac {x^{2}}{2 c}-\frac {b \ln \left (c \,x^{2}+b \right )}{2 c^{2}}\)	\(24\)

[In]

int(x^5/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {x^5}{b x^2+c x^4} \, dx=\frac {c x^{2} - b \log \left (c x^{2} + b\right )}{2 \, c^{2}} \]

[In]

integrate(x^5/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/2*(c*x^2 - b*log(c*x^2 + b))/c^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {x^5}{b x^2+c x^4} \, dx=- \frac {b \log {\left (b + c x^{2} \right )}}{2 c^{2}} + \frac {x^{2}}{2 c} \]

[In]

integrate(x**5/(c*x**4+b*x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {x^5}{b x^2+c x^4} \, dx=\frac {x^{2}}{2 \, c} - \frac {b \log \left (c x^{2} + b\right )}{2 \, c^{2}} \]

[In]

integrate(x^5/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/2*x^2/c - 1/2*b*log(c*x^2 + b)/c^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {x^5}{b x^2+c x^4} \, dx=\frac {x^{2}}{2 \, c} - \frac {b \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{2}} \]

[In]

integrate(x^5/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {x^5}{b x^2+c x^4} \, dx=-\frac {b\,\ln \left (c\,x^2+b\right )-c\,x^2}{2\,c^2} \]

[In]

int(x^5/(b*x^2 + c*x^4),x)

[Out]

-(b*log(b + c*x^2) - c*x^2)/(2*c^2)