\(\int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx\) [477]

Optimal result

Integrand size = 24, antiderivative size = 57 \[ \int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {a x^2}{b^3}+\frac {x^4}{4 b^2}+\frac {a^3}{2 b^4 \left (a+b x^2\right )}+\frac {3 a^2 \log \left (a+b x^2\right )}{2 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 272, 45} \[ \int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a^3}{2 b^4 \left (a+b x^2\right )}+\frac {3 a^2 \log \left (a+b x^2\right )}{2 b^4}-\frac {a x^2}{b^3}+\frac {x^4}{4 b^2} \]

[In]

Int[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

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

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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}& = b^2 \int \frac {x^7}{\left (a b+b^2 x^2\right )^2} \, dx \\ & = \frac {1}{2} b^2 \text {Subst}\left (\int \frac {x^3}{\left (a b+b^2 x\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} b^2 \text {Subst}\left (\int \left (-\frac {2 a}{b^5}+\frac {x}{b^4}-\frac {a^3}{b^5 (a+b x)^2}+\frac {3 a^2}{b^5 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a x^2}{b^3}+\frac {x^4}{4 b^2}+\frac {a^3}{2 b^4 \left (a+b x^2\right )}+\frac {3 a^2 \log \left (a+b x^2\right )}{2 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {-4 a b x^2+b^2 x^4+\frac {2 a^3}{a+b x^2}+6 a^2 \log \left (a+b x^2\right )}{4 b^4} \]

[In]

Integrate[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {-\frac {1}{2} b \,x^{4}+2 a \,x^{2}}{2 b^{3}}+\frac {3 a^{2} \ln \left (b \,x^{2}+a \right )}{2 b^{4}}+\frac {a^{3}}{2 b^{4} \left (b \,x^{2}+a \right )}\)	\(53\)
norman	\(\frac {\frac {x^{6}}{4 b}-\frac {3 a \,x^{4}}{4 b^{2}}+\frac {3 a^{3}}{2 b^{4}}}{b \,x^{2}+a}+\frac {3 a^{2} \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(54\)
risch	\(\frac {x^{4}}{4 b^{2}}-\frac {a \,x^{2}}{b^{3}}+\frac {a^{2}}{b^{4}}+\frac {a^{3}}{2 b^{4} \left (b \,x^{2}+a \right )}+\frac {3 a^{2} \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(59\)
parallelrisch	\(\frac {b^{3} x^{6}-3 b^{2} x^{4} a +6 \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b +6 \ln \left (b \,x^{2}+a \right ) a^{3}+6 a^{3}}{4 b^{4} \left (b \,x^{2}+a \right )}\)	\(67\)

[In]

int(x^7/(b^2*x^4+2*a*b*x^2+a^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.23 \[ \int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {b^{3} x^{6} - 3 \, a b^{2} x^{4} - 4 \, a^{2} b x^{2} + 2 \, a^{3} + 6 \, {\left (a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{5} x^{2} + a b^{4}\right )}} \]

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a^{3}}{2 a b^{4} + 2 b^{5} x^{2}} + \frac {3 a^{2} \log {\left (a + b x^{2} \right )}}{2 b^{4}} - \frac {a x^{2}}{b^{3}} + \frac {x^{4}}{4 b^{2}} \]

[In]

integrate(x**7/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a^{3}}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} + \frac {3 \, a^{2} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac {b x^{4} - 4 \, a x^{2}}{4 \, b^{3}} \]

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18 \[ \int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {3 \, a^{2} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} + \frac {b^{2} x^{4} - 4 \, a b x^{2}}{4 \, b^{4}} - \frac {3 \, a^{2} b x^{2} + 2 \, a^{3}}{2 \, {\left (b x^{2} + a\right )} b^{4}} \]

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {x^7}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {x^4}{4\,b^2}+\frac {a^3}{2\,b\,\left (b^4\,x^2+a\,b^3\right )}-\frac {a\,x^2}{b^3}+\frac {3\,a^2\,\ln \left (b\,x^2+a\right )}{2\,b^4} \]

[In]

int(x^7/(a^2 + b^2*x^4 + 2*a*b*x^2),x)

[Out]

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