3.464 \(\int \frac {x^7}{(-5+x^2)^3} \, dx\)

Optimal. Leaf size=46 \[ \frac {x^2}{2}+\frac {75}{2 \left (5-x^2\right )}-\frac {125}{4 \left (5-x^2\right )^2}+\frac {15}{2} \log \left (5-x^2\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {266, 43} \[ \frac {x^2}{2}+\frac {75}{2 \left (5-x^2\right )}-\frac {125}{4 \left (5-x^2\right )^2}+\frac {15}{2} \log \left (5-x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^7/(-5 + x^2)^3,x]

[Out]

x^2/2 - 125/(4*(5 - x^2)^2) + 75/(2*(5 - x^2)) + (15*Log[5 - x^2])/2

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 (-5+x^2\right )^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3}{(-5+x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (1+\frac {125}{(-5+x)^3}+\frac {75}{(-5+x)^2}+\frac {15}{-5+x}\right ) \, dx,x,x^2\right )\\ &=\frac {x^2}{2}-\frac {125}{4 \left (5-x^2\right )^2}+\frac {75}{2 \left (5-x^2\right )}+\frac {15}{2} \log \left (5-x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.78 \[ \frac {1}{4} \left (2 x^2-\frac {150}{x^2-5}-\frac {125}{\left (x^2-5\right )^2}+30 \log \left (x^2-5\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(-5 + x^2)^3,x]

[Out]

(2*x^2 - 125/(-5 + x^2)^2 - 150/(-5 + x^2) + 30*Log[-5 + x^2])/4

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IntegrateAlgebraic [A]  time = 0.02, size = 39, normalized size = 0.85 \[ \frac {15}{2} \log \left (x^2-5\right )+\frac {2 x^6-20 x^4-100 x^2+625}{4 \left (x^2-5\right )^2} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^7/(-5 + x^2)^3,x]

[Out]

(625 - 100*x^2 - 20*x^4 + 2*x^6)/(4*(-5 + x^2)^2) + (15*Log[-5 + x^2])/2

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fricas [A]  time = 0.76, size = 49, normalized size = 1.07 \[ \frac {2 \, x^{6} - 20 \, x^{4} - 100 \, x^{2} + 30 \, {\left (x^{4} - 10 \, x^{2} + 25\right )} \log \left (x^{2} - 5\right ) + 625}{4 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^2-5)^3,x, algorithm="fricas")

[Out]

1/4*(2*x^6 - 20*x^4 - 100*x^2 + 30*(x^4 - 10*x^2 + 25)*log(x^2 - 5) + 625)/(x^4 - 10*x^2 + 25)

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giac [A]  time = 0.60, size = 36, normalized size = 0.78 \[ \frac {1}{2} \, x^{2} - \frac {5 \, {\left (9 \, x^{4} - 60 \, x^{2} + 100\right )}}{4 \, {\left (x^{2} - 5\right )}^{2}} + \frac {15}{2} \, \log \left ({\left | x^{2} - 5 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^2-5)^3,x, algorithm="giac")

[Out]

1/2*x^2 - 5/4*(9*x^4 - 60*x^2 + 100)/(x^2 - 5)^2 + 15/2*log(abs(x^2 - 5))

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maple [A]  time = 0.33, size = 30, normalized size = 0.65

method	result	size
norman	\(\frac {-75 x^{2}+\frac {1}{2} x^{6}+\frac {1125}{4}}{\left (x^{2}-5\right )^{2}}+\frac {15 \ln \left (x^{2}-5\right )}{2}\)	\(30\)
risch	\(\frac {x^{2}}{2}+\frac {-\frac {75 x^{2}}{2}+\frac {625}{4}}{\left (x^{2}-5\right )^{2}}+\frac {15 \ln \left (x^{2}-5\right )}{2}\)	\(30\)
default	\(\frac {x^{2}}{2}-\frac {125}{4 \left (x^{2}-5\right )^{2}}-\frac {75}{2 \left (x^{2}-5\right )}+\frac {15 \ln \left (x^{2}-5\right )}{2}\)	\(33\)
meijerg	\(\frac {x^{2} \left (\frac {4}{25} x^{4}-\frac {18}{5} x^{2}+12\right )}{8 \left (-\frac {x^{2}}{5}+1\right )^{2}}+\frac {15 \ln \left (-\frac {x^{2}}{5}+1\right )}{2}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(x^2-5)^3,x,method=_RETURNVERBOSE)

[Out]

(-75*x^2+1/2*x^6+1125/4)/(x^2-5)^2+15/2*ln(x^2-5)

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maxima [A]  time = 0.47, size = 35, normalized size = 0.76 \[ \frac {1}{2} \, x^{2} - \frac {25 \, {\left (6 \, x^{2} - 25\right )}}{4 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} + \frac {15}{2} \, \log \left (x^{2} - 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^2-5)^3,x, algorithm="maxima")

[Out]

1/2*x^2 - 25/4*(6*x^2 - 25)/(x^4 - 10*x^2 + 25) + 15/2*log(x^2 - 5)

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mupad [B]  time = 0.06, size = 35, normalized size = 0.76 \[ \frac {15\,\ln \left (x^2-5\right )}{2}-\frac {\frac {75\,x^2}{2}-\frac {625}{4}}{x^4-10\,x^2+25}+\frac {x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(x^2 - 5)^3,x)

[Out]

(15*log(x^2 - 5))/2 - ((75*x^2)/2 - 625/4)/(x^4 - 10*x^2 + 25) + x^2/2

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sympy [A]  time = 0.12, size = 32, normalized size = 0.70 \[ \frac {x^{2}}{2} + \frac {625 - 150 x^{2}}{4 x^{4} - 40 x^{2} + 100} + \frac {15 \log {\left (x^{2} - 5 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(x**2-5)**3,x)

[Out]

x**2/2 + (625 - 150*x**2)/(4*x**4 - 40*x**2 + 100) + 15*log(x**2 - 5)/2

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