\(\int \frac {x^{13}}{(b x^2+c x^4)^3} \, dx\) [206]

Optimal result

Integrand size = 17, antiderivative size = 65 \[ \int \frac {x^{13}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x^2}{2 c^3}+\frac {b^3}{4 c^4 \left (b+c x^2\right )^2}-\frac {3 b^2}{2 c^4 \left (b+c x^2\right )}-\frac {3 b \log \left (b+c x^2\right )}{2 c^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, 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} \[ \int \frac {x^{13}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {b^3}{4 c^4 \left (b+c x^2\right )^2}-\frac {3 b^2}{2 c^4 \left (b+c x^2\right )}-\frac {3 b \log \left (b+c x^2\right )}{2 c^4}+\frac {x^2}{2 c^3} \]

[In]

Int[x^13/(b*x^2 + c*x^4)^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {x^{13}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {-2 c x^2+\frac {b^2 \left (5 b+6 c x^2\right )}{\left (b+c x^2\right )^2}+6 b \log \left (b+c x^2\right )}{4 c^4} \]

[In]

Integrate[x^13/(b*x^2 + c*x^4)^3,x]

[Out]

-1/4*(-2*c*x^2 + (b^2*(5*b + 6*c*x^2))/(b + c*x^2)^2 + 6*b*Log[b + c*x^2])/c^4

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {x^{2}}{2 c^{3}}+\frac {-\frac {3 b^{2} x^{2}}{2}-\frac {5 b^{3}}{4 c}}{c^{3} \left (c \,x^{2}+b \right )^{2}}-\frac {3 b \ln \left (c \,x^{2}+b \right )}{2 c^{4}}\)	\(54\)
norman	\(\frac {\frac {x^{11}}{2 c}-\frac {3 b^{2} x^{7}}{c^{3}}-\frac {9 b^{3} x^{5}}{4 c^{4}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}-\frac {3 b \ln \left (c \,x^{2}+b \right )}{2 c^{4}}\)	\(60\)
default	\(\frac {x^{2}}{2 c^{3}}-\frac {b \left (-\frac {b^{2}}{2 c \left (c \,x^{2}+b \right )^{2}}+\frac {3 \ln \left (c \,x^{2}+b \right )}{c}+\frac {3 b}{c \left (c \,x^{2}+b \right )}\right )}{2 c^{3}}\)	\(62\)
parallelrisch	\(-\frac {-2 c^{3} x^{6}+6 \ln \left (c \,x^{2}+b \right ) x^{4} b \,c^{2}+12 \ln \left (c \,x^{2}+b \right ) x^{2} b^{2} c +12 b^{2} c \,x^{2}+6 b^{3} \ln \left (c \,x^{2}+b \right )+9 b^{3}}{4 c^{4} \left (c \,x^{2}+b \right )^{2}}\)	\(85\)

[In]

int(x^13/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.40 \[ \int \frac {x^{13}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {2 \, c^{3} x^{6} + 4 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} - 5 \, b^{3} - 6 \, {\left (b c^{2} x^{4} + 2 \, b^{2} c x^{2} + b^{3}\right )} \log \left (c x^{2} + b\right )}{4 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} \]

[In]

integrate(x^13/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {x^{13}}{\left (b x^2+c x^4\right )^3} \, dx=- \frac {3 b \log {\left (b + c x^{2} \right )}}{2 c^{4}} + \frac {- 5 b^{3} - 6 b^{2} c x^{2}}{4 b^{2} c^{4} + 8 b c^{5} x^{2} + 4 c^{6} x^{4}} + \frac {x^{2}}{2 c^{3}} \]

[In]

integrate(x**13/(c*x**4+b*x**2)**3,x)

[Out]

-3*b*log(b + c*x**2)/(2*c**4) + (-5*b**3 - 6*b**2*c*x**2)/(4*b**2*c**4 + 8*b*c**5*x**2 + 4*c**6*x**4) + x**2/(
2*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \frac {x^{13}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {6 \, b^{2} c x^{2} + 5 \, b^{3}}{4 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} + \frac {x^{2}}{2 \, c^{3}} - \frac {3 \, b \log \left (c x^{2} + b\right )}{2 \, c^{4}} \]

[In]

integrate(x^13/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \frac {x^{13}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x^{2}}{2 \, c^{3}} - \frac {3 \, b \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{4}} + \frac {9 \, b c^{2} x^{4} + 12 \, b^{2} c x^{2} + 4 \, b^{3}}{4 \, {\left (c x^{2} + b\right )}^{2} c^{4}} \]

[In]

integrate(x^13/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

1/2*x^2/c^3 - 3/2*b*log(abs(c*x^2 + b))/c^4 + 1/4*(9*b*c^2*x^4 + 12*b^2*c*x^2 + 4*b^3)/((c*x^2 + b)^2*c^4)

Mupad [B] (verification not implemented)

Time = 12.89 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {x^{13}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x^2}{2\,c^3}-\frac {\frac {5\,b^3}{4\,c}+\frac {3\,b^2\,x^2}{2}}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}-\frac {3\,b\,\ln \left (c\,x^2+b\right )}{2\,c^4} \]

[In]

int(x^13/(b*x^2 + c*x^4)^3,x)

[Out]

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