3.191 \(\int \frac {x^{11}}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=57 \[ \frac {b^3}{2 c^4 \left (b+c x^2\right )}+\frac {3 b^2 \log \left (b+c x^2\right )}{2 c^4}-\frac {b x^2}{c^3}+\frac {x^4}{4 c^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 57, 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} \[ \frac {b^3}{2 c^4 \left (b+c x^2\right )}+\frac {3 b^2 \log \left (b+c x^2\right )}{2 c^4}-\frac {b x^2}{c^3}+\frac {x^4}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^11/(b*x^2 + c*x^4)^2,x]

[Out]

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

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^{11}}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^7}{\left (b+c x^2\right )^2} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3}{(b+c x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {2 b}{c^3}+\frac {x}{c^2}-\frac {b^3}{c^3 (b+c x)^2}+\frac {3 b^2}{c^3 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {b x^2}{c^3}+\frac {x^4}{4 c^2}+\frac {b^3}{2 c^4 \left (b+c x^2\right )}+\frac {3 b^2 \log \left (b+c x^2\right )}{2 c^4}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.86 \[ \frac {\frac {2 b^3}{b+c x^2}+6 b^2 \log \left (b+c x^2\right )-4 b c x^2+c^2 x^4}{4 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/(b*x^2 + c*x^4)^2,x]

[Out]

(-4*b*c*x^2 + c^2*x^4 + (2*b^3)/(b + c*x^2) + 6*b^2*Log[b + c*x^2])/(4*c^4)

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fricas [A]  time = 0.80, size = 70, normalized size = 1.23 \[ \frac {c^{3} x^{6} - 3 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} + 2 \, b^{3} + 6 \, {\left (b^{2} c x^{2} + b^{3}\right )} \log \left (c x^{2} + b\right )}{4 \, {\left (c^{5} x^{2} + b c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/4*(c^3*x^6 - 3*b*c^2*x^4 - 4*b^2*c*x^2 + 2*b^3 + 6*(b^2*c*x^2 + b^3)*log(c*x^2 + b))/(c^5*x^2 + b*c^4)

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giac [A]  time = 0.18, size = 67, normalized size = 1.18 \[ \frac {3 \, b^{2} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{4}} + \frac {c^{2} x^{4} - 4 \, b c x^{2}}{4 \, c^{4}} - \frac {3 \, b^{2} c x^{2} + 2 \, b^{3}}{2 \, {\left (c x^{2} + b\right )} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 52, normalized size = 0.91 \[ \frac {x^{4}}{4 c^{2}}-\frac {b \,x^{2}}{c^{3}}+\frac {b^{3}}{2 \left (c \,x^{2}+b \right ) c^{4}}+\frac {3 b^{2} \ln \left (c \,x^{2}+b \right )}{2 c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(c*x^4+b*x^2)^2,x)

[Out]

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

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maxima [A]  time = 1.32, size = 54, normalized size = 0.95 \[ \frac {b^{3}}{2 \, {\left (c^{5} x^{2} + b c^{4}\right )}} + \frac {3 \, b^{2} \log \left (c x^{2} + b\right )}{2 \, c^{4}} + \frac {c x^{4} - 4 \, b x^{2}}{4 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.14, size = 57, normalized size = 1.00 \[ \frac {x^4}{4\,c^2}+\frac {b^3}{2\,c\,\left (c^4\,x^2+b\,c^3\right )}-\frac {b\,x^2}{c^3}+\frac {3\,b^2\,\ln \left (c\,x^2+b\right )}{2\,c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(b*x^2 + c*x^4)^2,x)

[Out]

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

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sympy [A]  time = 0.30, size = 53, normalized size = 0.93 \[ \frac {b^{3}}{2 b c^{4} + 2 c^{5} x^{2}} + \frac {3 b^{2} \log {\left (b + c x^{2} \right )}}{2 c^{4}} - \frac {b x^{2}}{c^{3}} + \frac {x^{4}}{4 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(c*x**4+b*x**2)**2,x)

[Out]

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

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