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

Optimal. Leaf size=71 \[ \frac {a^3}{6 b^4 \left (a+b x^2\right )^3}-\frac {3 a^2}{4 b^4 \left (a+b x^2\right )^2}+\frac {3 a}{2 b^4 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^4} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 266, 43} \[ \frac {a^3}{6 b^4 \left (a+b x^2\right )^3}-\frac {3 a^2}{4 b^4 \left (a+b x^2\right )^2}+\frac {3 a}{2 b^4 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {x^7}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac {1}{2} b^4 \operatorname {Subst}\left (\int \frac {x^3}{\left (a b+b^2 x\right )^4} \, dx,x,x^2\right )\\ &=\frac {1}{2} b^4 \operatorname {Subst}\left (\int \left (-\frac {a^3}{b^7 (a+b x)^4}+\frac {3 a^2}{b^7 (a+b x)^3}-\frac {3 a}{b^7 (a+b x)^2}+\frac {1}{b^7 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac {a^3}{6 b^4 \left (a+b x^2\right )^3}-\frac {3 a^2}{4 b^4 \left (a+b x^2\right )^2}+\frac {3 a}{2 b^4 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 50, normalized size = 0.70 \[ \frac {\frac {a \left (11 a^2+27 a b x^2+18 b^2 x^4\right )}{\left (a+b x^2\right )^3}+6 \log \left (a+b x^2\right )}{12 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(11*a^2 + 27*a*b*x^2 + 18*b^2*x^4))/(a + b*x^2)^3 + 6*Log[a + b*x^2])/(12*b^4)

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fricas [A]  time = 0.87, size = 102, normalized size = 1.44 \[ \frac {18 \, a b^{2} x^{4} + 27 \, a^{2} b x^{2} + 11 \, a^{3} + 6 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(18*a*b^2*x^4 + 27*a^2*b*x^2 + 11*a^3 + 6*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*log(b*x^2 + a))/(b^
7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)

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giac [A]  time = 0.17, size = 53, normalized size = 0.75 \[ \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} - \frac {11 \, b^{2} x^{6} + 15 \, a b x^{4} + 6 \, a^{2} x^{2}}{12 \, {\left (b x^{2} + a\right )}^{3} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(abs(b*x^2 + a))/b^4 - 1/12*(11*b^2*x^6 + 15*a*b*x^4 + 6*a^2*x^2)/((b*x^2 + a)^3*b^3)

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maple [A]  time = 0.01, size = 64, normalized size = 0.90 \[ \frac {a^{3}}{6 \left (b \,x^{2}+a \right )^{3} b^{4}}-\frac {3 a^{2}}{4 \left (b \,x^{2}+a \right )^{2} b^{4}}+\frac {3 a}{2 \left (b \,x^{2}+a \right ) b^{4}}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.31, size = 77, normalized size = 1.08 \[ \frac {18 \, a b^{2} x^{4} + 27 \, a^{2} b x^{2} + 11 \, a^{3}}{12 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(18*a*b^2*x^4 + 27*a^2*b*x^2 + 11*a^3)/(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4) + 1/2*log(b*x^2
+ a)/b^4

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mupad [B]  time = 4.33, size = 75, normalized size = 1.06 \[ \frac {\frac {11\,a^3}{12\,b^4}+\frac {3\,a\,x^4}{2\,b^2}+\frac {9\,a^2\,x^2}{4\,b^3}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6}+\frac {\ln \left (b\,x^2+a\right )}{2\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((11*a^3)/(12*b^4) + (3*a*x^4)/(2*b^2) + (9*a^2*x^2)/(4*b^3))/(a^3 + b^3*x^6 + 3*a^2*b*x^2 + 3*a*b^2*x^4) + lo
g(a + b*x^2)/(2*b^4)

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sympy [A]  time = 0.47, size = 76, normalized size = 1.07 \[ \frac {11 a^{3} + 27 a^{2} b x^{2} + 18 a b^{2} x^{4}}{12 a^{3} b^{4} + 36 a^{2} b^{5} x^{2} + 36 a b^{6} x^{4} + 12 b^{7} x^{6}} + \frac {\log {\left (a + b x^{2} \right )}}{2 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*a**3 + 27*a**2*b*x**2 + 18*a*b**2*x**4)/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6)
+ log(a + b*x**2)/(2*b**4)

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