3.3.14 \(\int \frac {x^5}{(b x^2+c x^4)^3} \, dx\) [214]

Optimal. Leaf size=54 \[ \frac {1}{4 b \left (b+c x^2\right )^2}+\frac {1}{2 b^2 \left (b+c x^2\right )}+\frac {\log (x)}{b^3}-\frac {\log \left (b+c x^2\right )}{2 b^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1598, 272, 46} \begin {gather*} -\frac {\log \left (b+c x^2\right )}{2 b^3}+\frac {\log (x)}{b^3}+\frac {1}{2 b^2 \left (b+c x^2\right )}+\frac {1}{4 b \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(b*x^2 + c*x^4)^3,x]

[Out]

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

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*} \int \frac {x^5}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {1}{x \left (b+c x^2\right )^3} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (b+c x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{b^3 x}-\frac {c}{b (b+c x)^3}-\frac {c}{b^2 (b+c x)^2}-\frac {c}{b^3 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{4 b \left (b+c x^2\right )^2}+\frac {1}{2 b^2 \left (b+c x^2\right )}+\frac {\log (x)}{b^3}-\frac {\log \left (b+c x^2\right )}{2 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 43, normalized size = 0.80 \begin {gather*} \frac {\frac {b \left (3 b+2 c x^2\right )}{\left (b+c x^2\right )^2}+4 \log (x)-2 \log \left (b+c x^2\right )}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(b*x^2 + c*x^4)^3,x]

[Out]

((b*(3*b + 2*c*x^2))/(b + c*x^2)^2 + 4*Log[x] - 2*Log[b + c*x^2])/(4*b^3)

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Maple [A]
time = 0.10, size = 59, normalized size = 1.09

method	result	size
risch	\(\frac {\frac {c \,x^{2}}{2 b^{2}}+\frac {3}{4 b}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\ln \left (x \right )}{b^{3}}-\frac {\ln \left (c \,x^{2}+b \right )}{2 b^{3}}\)	\(46\)
norman	\(\frac {-\frac {c \,x^{7}}{b^{2}}-\frac {3 c^{2} x^{9}}{4 b^{3}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}+\frac {\ln \left (x \right )}{b^{3}}-\frac {\ln \left (c \,x^{2}+b \right )}{2 b^{3}}\)	\(55\)
default	\(-\frac {c \left (-\frac {b}{c \left (c \,x^{2}+b \right )}+\frac {\ln \left (c \,x^{2}+b \right )}{c}-\frac {b^{2}}{2 c \left (c \,x^{2}+b \right )^{2}}\right )}{2 b^{3}}+\frac {\ln \left (x \right )}{b^{3}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 60, normalized size = 1.11 \begin {gather*} \frac {2 \, c x^{2} + 3 \, b}{4 \, {\left (b^{2} c^{2} x^{4} + 2 \, b^{3} c x^{2} + b^{4}\right )}} - \frac {\log \left (c x^{2} + b\right )}{2 \, b^{3}} + \frac {\log \left (x^{2}\right )}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/4*(2*c*x^2 + 3*b)/(b^2*c^2*x^4 + 2*b^3*c*x^2 + b^4) - 1/2*log(c*x^2 + b)/b^3 + 1/2*log(x^2)/b^3

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Fricas [A]
time = 0.37, size = 90, normalized size = 1.67 \begin {gather*} \frac {2 \, b c x^{2} + 3 \, b^{2} - 2 \, {\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \log \left (c x^{2} + b\right ) + 4 \, {\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \log \left (x\right )}{4 \, {\left (b^{3} c^{2} x^{4} + 2 \, b^{4} c x^{2} + b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/4*(2*b*c*x^2 + 3*b^2 - 2*(c^2*x^4 + 2*b*c*x^2 + b^2)*log(c*x^2 + b) + 4*(c^2*x^4 + 2*b*c*x^2 + b^2)*log(x))/
(b^3*c^2*x^4 + 2*b^4*c*x^2 + b^5)

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Sympy [A]
time = 0.21, size = 56, normalized size = 1.04 \begin {gather*} \frac {3 b + 2 c x^{2}}{4 b^{4} + 8 b^{3} c x^{2} + 4 b^{2} c^{2} x^{4}} + \frac {\log {\left (x \right )}}{b^{3}} - \frac {\log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**4+b*x**2)**3,x)

[Out]

(3*b + 2*c*x**2)/(4*b**4 + 8*b**3*c*x**2 + 4*b**2*c**2*x**4) + log(x)/b**3 - log(b/c + x**2)/(2*b**3)

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Giac [A]
time = 4.55, size = 59, normalized size = 1.09 \begin {gather*} \frac {\log \left (x^{2}\right )}{2 \, b^{3}} - \frac {\log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{3}} + \frac {3 \, c^{2} x^{4} + 8 \, b c x^{2} + 6 \, b^{2}}{4 \, {\left (c x^{2} + b\right )}^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

1/2*log(x^2)/b^3 - 1/2*log(abs(c*x^2 + b))/b^3 + 1/4*(3*c^2*x^4 + 8*b*c*x^2 + 6*b^2)/((c*x^2 + b)^2*b^3)

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Mupad [B]
time = 0.06, size = 56, normalized size = 1.04 \begin {gather*} \frac {\ln \left (x\right )}{b^3}+\frac {\frac {3}{4\,b}+\frac {c\,x^2}{2\,b^2}}{b^2+2\,b\,c\,x^2+c^2\,x^4}-\frac {\ln \left (c\,x^2+b\right )}{2\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^2 + c*x^4)^3,x)

[Out]

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

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