3.1.75 \(\int \frac {x^9}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=44 \[ -\frac {b^2}{2 c^3 \left (b+c x^2\right )}-\frac {b \log \left (b+c x^2\right )}{c^3}+\frac {x^2}{2 c^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 44, 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 = {1584, 266, 43} \begin {gather*} -\frac {b^2}{2 c^3 \left (b+c x^2\right )}-\frac {b \log \left (b+c x^2\right )}{c^3}+\frac {x^2}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(b*x^2 + c*x^4)^2,x]

[Out]

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

Rule 43

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 266

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 1584

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^9}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^5}{\left (b+c x^2\right )^2} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{(b+c x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{c^2}+\frac {b^2}{c^2 (b+c x)^2}-\frac {2 b}{c^2 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {x^2}{2 c^2}-\frac {b^2}{2 c^3 \left (b+c x^2\right )}-\frac {b \log \left (b+c x^2\right )}{c^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 38, normalized size = 0.86 \begin {gather*} \frac {-\frac {b^2}{b+c x^2}-2 b \log \left (b+c x^2\right )+c x^2}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(b*x^2 + c*x^4)^2,x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^9}{\left (b x^2+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^9/(b*x^2 + c*x^4)^2,x]

[Out]

IntegrateAlgebraic[x^9/(b*x^2 + c*x^4)^2, x]

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 56, normalized size = 1.27 \begin {gather*} \frac {c^{2} x^{4} + b c x^{2} - b^{2} - 2 \, {\left (b c x^{2} + b^{2}\right )} \log \left (c x^{2} + b\right )}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.17, size = 49, normalized size = 1.11 \begin {gather*} \frac {x^{2}}{2 \, c^{2}} - \frac {b \log \left ({\left | c x^{2} + b \right |}\right )}{c^{3}} + \frac {2 \, b c x^{2} + b^{2}}{2 \, {\left (c x^{2} + b\right )} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 41, normalized size = 0.93 \begin {gather*} \frac {x^{2}}{2 c^{2}}-\frac {b^{2}}{2 \left (c \,x^{2}+b \right ) c^{3}}-\frac {b \ln \left (c \,x^{2}+b \right )}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(c*x^4+b*x^2)^2,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.32, size = 43, normalized size = 0.98 \begin {gather*} -\frac {b^{2}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {x^{2}}{2 \, c^{2}} - \frac {b \log \left (c x^{2} + b\right )}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 45, normalized size = 1.02 \begin {gather*} \frac {x^2}{2\,c^2}-\frac {b^2}{2\,\left (c^4\,x^2+b\,c^3\right )}-\frac {b\,\ln \left (c\,x^2+b\right )}{c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^2 + c*x^4)^2,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.28, size = 39, normalized size = 0.89 \begin {gather*} - \frac {b^{2}}{2 b c^{3} + 2 c^{4} x^{2}} - \frac {b \log {\left (b + c x^{2} \right )}}{c^{3}} + \frac {x^{2}}{2 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(c*x**4+b*x**2)**2,x)

[Out]

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

________________________________________________________________________________________