\(\int \frac {x^{11}}{(b x^2+c x^4)^3} \, dx\) [208]

Optimal result

Integrand size = 17, antiderivative size = 49 \[ \int \frac {x^{11}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {b^2}{4 c^3 \left (b+c x^2\right )^2}+\frac {b}{c^3 \left (b+c x^2\right )}+\frac {\log \left (b+c x^2\right )}{2 c^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, 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} \[ \int \frac {x^{11}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {b^2}{4 c^3 \left (b+c x^2\right )^2}+\frac {b}{c^3 \left (b+c x^2\right )}+\frac {\log \left (b+c x^2\right )}{2 c^3} \]

[In]

Int[x^11/(b*x^2 + c*x^4)^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int \frac {x^{11}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {b \left (3 b+4 c x^2\right )}{\left (b+c x^2\right )^2}+2 \log \left (b+c x^2\right )}{4 c^3} \]

[In]

Integrate[x^11/(b*x^2 + c*x^4)^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {\frac {b \,x^{2}}{c^{2}}+\frac {3 b^{2}}{4 c^{3}}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\ln \left (c \,x^{2}+b \right )}{2 c^{3}}\)	\(42\)
default	\(-\frac {b^{2}}{4 c^{3} \left (c \,x^{2}+b \right )^{2}}+\frac {b}{c^{3} \left (c \,x^{2}+b \right )}+\frac {\ln \left (c \,x^{2}+b \right )}{2 c^{3}}\)	\(46\)
norman	\(\frac {\frac {b \,x^{7}}{c^{2}}+\frac {3 b^{2} x^{5}}{4 c^{3}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}+\frac {\ln \left (c \,x^{2}+b \right )}{2 c^{3}}\)	\(48\)
parallelrisch	\(\frac {2 c^{2} \ln \left (c \,x^{2}+b \right ) x^{4}+4 \ln \left (c \,x^{2}+b \right ) x^{2} b c +4 b c \,x^{2}+2 b^{2} \ln \left (c \,x^{2}+b \right )+3 b^{2}}{4 c^{3} \left (c \,x^{2}+b \right )^{2}}\)	\(72\)

[In]

int(x^11/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.41 \[ \int \frac {x^{11}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {4 \, b c x^{2} + 3 \, b^{2} + 2 \, {\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \log \left (c x^{2} + b\right )}{4 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} \]

[In]

integrate(x^11/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08 \[ \int \frac {x^{11}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 b^{2} + 4 b c x^{2}}{4 b^{2} c^{3} + 8 b c^{4} x^{2} + 4 c^{5} x^{4}} + \frac {\log {\left (b + c x^{2} \right )}}{2 c^{3}} \]

[In]

integrate(x**11/(c*x**4+b*x**2)**3,x)

[Out]

(3*b**2 + 4*b*c*x**2)/(4*b**2*c**3 + 8*b*c**4*x**2 + 4*c**5*x**4) + log(b + c*x**2)/(2*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.12 \[ \int \frac {x^{11}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {4 \, b c x^{2} + 3 \, b^{2}}{4 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} + \frac {\log \left (c x^{2} + b\right )}{2 \, c^{3}} \]

[In]

integrate(x^11/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86 \[ \int \frac {x^{11}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{3}} - \frac {3 \, c x^{4} + 2 \, b x^{2}}{4 \, {\left (c x^{2} + b\right )}^{2} c^{2}} \]

[In]

integrate(x^11/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.88 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {x^{11}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {3\,b^2}{4\,c^3}+\frac {b\,x^2}{c^2}}{b^2+2\,b\,c\,x^2+c^2\,x^4}+\frac {\ln \left (c\,x^2+b\right )}{2\,c^3} \]

[In]

int(x^11/(b*x^2 + c*x^4)^3,x)

[Out]

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