\(\int \frac {x^{11}}{(a+c x^4)^2} \, dx\) [658]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int \frac {x^{11}}{\left (a+c x^4\right )^2} \, dx=\frac {x^4}{4 c^2}-\frac {a^2}{4 c^3 \left (a+c x^4\right )}-\frac {a \log \left (a+c x^4\right )}{2 c^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {x^{11}}{\left (a+c x^4\right )^2} \, dx=-\frac {a^2}{4 c^3 \left (a+c x^4\right )}-\frac {a \log \left (a+c x^4\right )}{2 c^3}+\frac {x^4}{4 c^2} \]

[In]

Int[x^11/(a + c*x^4)^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \frac {x^{11}}{\left (a+c x^4\right )^2} \, dx=\frac {c x^4-\frac {a^2}{a+c x^4}-2 a \log \left (a+c x^4\right )}{4 c^3} \]

[In]

Integrate[x^11/(a + c*x^4)^2,x]

[Out]

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

Maple [A] (verified)

Time = 4.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {x^{4}}{4 c^{2}}-\frac {a^{2}}{4 c^{3} \left (x^{4} c +a \right )}-\frac {a \ln \left (x^{4} c +a \right )}{2 c^{3}}\)	\(41\)
norman	\(\frac {\frac {x^{8}}{4 c}-\frac {a^{2}}{2 c^{3}}}{x^{4} c +a}-\frac {a \ln \left (x^{4} c +a \right )}{2 c^{3}}\)	\(43\)
default	\(\frac {x^{4}}{4 c^{2}}-\frac {a \left (\frac {a}{2 c \left (x^{4} c +a \right )}+\frac {\ln \left (x^{4} c +a \right )}{c}\right )}{2 c^{2}}\)	\(44\)
parallelrisch	\(-\frac {-c^{2} x^{8}+2 \ln \left (x^{4} c +a \right ) x^{4} a c +2 \ln \left (x^{4} c +a \right ) a^{2}+2 a^{2}}{4 c^{3} \left (x^{4} c +a \right )}\)	\(57\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.22 \[ \int \frac {x^{11}}{\left (a+c x^4\right )^2} \, dx=\frac {c^{2} x^{8} + a c x^{4} - a^{2} - 2 \, {\left (a c x^{4} + a^{2}\right )} \log \left (c x^{4} + a\right )}{4 \, {\left (c^{4} x^{4} + a c^{3}\right )}} \]

[In]

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

[Out]

1/4*(c^2*x^8 + a*c*x^4 - a^2 - 2*(a*c*x^4 + a^2)*log(c*x^4 + a))/(c^4*x^4 + a*c^3)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \frac {x^{11}}{\left (a+c x^4\right )^2} \, dx=- \frac {a^{2}}{4 a c^{3} + 4 c^{4} x^{4}} - \frac {a \log {\left (a + c x^{4} \right )}}{2 c^{3}} + \frac {x^{4}}{4 c^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \frac {x^{11}}{\left (a+c x^4\right )^2} \, dx=\frac {x^{4}}{4 \, c^{2}} - \frac {a^{2}}{4 \, {\left (c^{4} x^{4} + a c^{3}\right )}} - \frac {a \log \left (c x^{4} + a\right )}{2 \, c^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {x^{11}}{\left (a+c x^4\right )^2} \, dx=\frac {x^{4}}{4 \, c^{2}} - \frac {a \log \left ({\left | c x^{4} + a \right |}\right )}{2 \, c^{3}} + \frac {2 \, a c x^{4} + a^{2}}{4 \, {\left (c x^{4} + a\right )} c^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98 \[ \int \frac {x^{11}}{\left (a+c x^4\right )^2} \, dx=\frac {x^4}{4\,c^2}-\frac {a^2}{4\,\left (c^4\,x^4+a\,c^3\right )}-\frac {a\,\ln \left (c\,x^4+a\right )}{2\,c^3} \]

[In]

int(x^11/(a + c*x^4)^2,x)

[Out]

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