3.2.73 \(\int \frac {x^5}{(a+b x^2)^3} \, dx\) [173]

Optimal. Leaf size=49 \[ -\frac {a^2}{4 b^3 \left (a+b x^2\right )^2}+\frac {a}{b^3 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 49, 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^2}{4 b^3 \left (a+b x^2\right )^2}+\frac {a}{b^3 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^2)^3,x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 39, normalized size = 0.80 \begin {gather*} \frac {\frac {a \left (3 a+4 b x^2\right )}{\left (a+b x^2\right )^2}+2 \log \left (a+b x^2\right )}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^2)^3,x]

[Out]

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

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Maple [A]
time = 0.03, size = 46, normalized size = 0.94

method	result	size
norman	\(\frac {\frac {a \,x^{2}}{b^{2}}+\frac {3 a^{2}}{4 b^{3}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(42\)
risch	\(\frac {\frac {a \,x^{2}}{b^{2}}+\frac {3 a^{2}}{4 b^{3}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(42\)
default	\(-\frac {a^{2}}{4 b^{3} \left (b \,x^{2}+a \right )^{2}}+\frac {a}{b^{3} \left (b \,x^{2}+a \right )}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 55, normalized size = 1.12 \begin {gather*} \frac {4 \, a b x^{2} + 3 \, a^{2}}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.46, size = 69, normalized size = 1.41 \begin {gather*} \frac {4 \, a b x^{2} + 3 \, a^{2} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.13, size = 53, normalized size = 1.08 \begin {gather*} \frac {3 a^{2} + 4 a b x^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {\log {\left (a + b x^{2} \right )}}{2 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**2+a)**3,x)

[Out]

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

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Giac [A]
time = 1.48, size = 42, normalized size = 0.86 \begin {gather*} \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{3}} - \frac {3 \, b x^{4} + 2 \, a x^{2}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.06, size = 52, normalized size = 1.06 \begin {gather*} \frac {\frac {3\,a^2}{4\,b^3}+\frac {a\,x^2}{b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\ln \left (b\,x^2+a\right )}{2\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^2)^3,x)

[Out]

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

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