3.2.72 \(\int \frac {x^7}{(a+b x^2)^3} \, dx\) [172]

Optimal. Leaf size=65 \[ \frac {x^2}{2 b^3}+\frac {a^3}{4 b^4 \left (a+b x^2\right )^2}-\frac {3 a^2}{2 b^4 \left (a+b x^2\right )}-\frac {3 a \log \left (a+b x^2\right )}{2 b^4} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \begin {gather*} \frac {a^3}{4 b^4 \left (a+b x^2\right )^2}-\frac {3 a^2}{2 b^4 \left (a+b x^2\right )}-\frac {3 a \log \left (a+b x^2\right )}{2 b^4}+\frac {x^2}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/(a + b*x^2)^3,x]

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 48, normalized size = 0.74 \begin {gather*} -\frac {-2 b x^2+\frac {a^2 \left (5 a+6 b x^2\right )}{\left (a+b x^2\right )^2}+6 a \log \left (a+b x^2\right )}{4 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a + b*x^2)^3,x]

[Out]

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

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Maple [A]
time = 0.04, size = 62, normalized size = 0.95

method	result	size
norman	\(\frac {\frac {x^{6}}{2 b}-\frac {9 a^{3}}{4 b^{4}}-\frac {3 a^{2} x^{2}}{b^{3}}}{\left (b \,x^{2}+a \right )^{2}}-\frac {3 a \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(54\)
risch	\(\frac {x^{2}}{2 b^{3}}+\frac {-\frac {3 a^{2} x^{2}}{2}-\frac {5 a^{3}}{4 b}}{b^{3} \left (b \,x^{2}+a \right )^{2}}-\frac {3 a \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(54\)
default	\(\frac {x^{2}}{2 b^{3}}-\frac {a \left (\frac {3 a}{b \left (b \,x^{2}+a \right )}+\frac {3 \ln \left (b \,x^{2}+a \right )}{b}-\frac {a^{2}}{2 b \left (b \,x^{2}+a \right )^{2}}\right )}{2 b^{3}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 66, normalized size = 1.02 \begin {gather*} -\frac {6 \, a^{2} b x^{2} + 5 \, a^{3}}{4 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} + \frac {x^{2}}{2 \, b^{3}} - \frac {3 \, a \log \left (b x^{2} + a\right )}{2 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.87, size = 91, normalized size = 1.40 \begin {gather*} \frac {2 \, b^{3} x^{6} + 4 \, a b^{2} x^{4} - 4 \, a^{2} b x^{2} - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{4} + 2 \, a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.17, size = 68, normalized size = 1.05 \begin {gather*} - \frac {3 a \log {\left (a + b x^{2} \right )}}{2 b^{4}} + \frac {- 5 a^{3} - 6 a^{2} b x^{2}}{4 a^{2} b^{4} + 8 a b^{5} x^{2} + 4 b^{6} x^{4}} + \frac {x^{2}}{2 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(b*x**2+a)**3,x)

[Out]

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

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Giac [A]
time = 1.11, size = 62, normalized size = 0.95 \begin {gather*} \frac {x^{2}}{2 \, b^{3}} - \frac {3 \, a \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} + \frac {9 \, a b^{2} x^{4} + 12 \, a^{2} b x^{2} + 4 \, a^{3}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^2+a)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.75, size = 68, normalized size = 1.05 \begin {gather*} \frac {x^2}{2\,b^3}-\frac {\frac {5\,a^3}{4\,b}+\frac {3\,a^2\,x^2}{2}}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}-\frac {3\,a\,\ln \left (b\,x^2+a\right )}{2\,b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(a + b*x^2)^3,x)

[Out]

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

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