3.3.32 \(\int \frac {x^3}{(a-b x^2)^2} \, dx\) [232]

Optimal. Leaf size=35 \[ \frac {a}{2 b^2 \left (a-b x^2\right )}+\frac {\log \left (a-b x^2\right )}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {272, 45} \begin {gather*} \frac {a}{2 b^2 \left (a-b x^2\right )}+\frac {\log \left (a-b x^2\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(a - b*x^2)^2,x]

[Out]

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

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

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Mathematica [A]
time = 0.01, size = 29, normalized size = 0.83 \begin {gather*} \frac {\frac {a}{a-b x^2}+\log \left (a-b x^2\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a - b*x^2)^2,x]

[Out]

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

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Maple [A]
time = 0.03, size = 32, normalized size = 0.91

method	result	size
default	\(\frac {a}{2 b^{2} \left (-b \,x^{2}+a \right )}+\frac {\ln \left (-b \,x^{2}+a \right )}{2 b^{2}}\)	\(32\)
norman	\(\frac {a}{2 b^{2} \left (-b \,x^{2}+a \right )}+\frac {\ln \left (-b \,x^{2}+a \right )}{2 b^{2}}\)	\(32\)
risch	\(\frac {a}{2 b^{2} \left (-b \,x^{2}+a \right )}+\frac {\ln \left (-b \,x^{2}+a \right )}{2 b^{2}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 35, normalized size = 1.00 \begin {gather*} -\frac {a}{2 \, {\left (b^{3} x^{2} - a b^{2}\right )}} + \frac {\log \left (b x^{2} - a\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.33, size = 42, normalized size = 1.20 \begin {gather*} \frac {{\left (b x^{2} - a\right )} \log \left (b x^{2} - a\right ) - a}{2 \, {\left (b^{3} x^{2} - a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.08, size = 29, normalized size = 0.83 \begin {gather*} - \frac {a}{- 2 a b^{2} + 2 b^{3} x^{2}} + \frac {\log {\left (- a + b x^{2} \right )}}{2 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.24, size = 53, normalized size = 1.51 \begin {gather*} -\frac {\frac {\log \left (\frac {{\left | b x^{2} - a \right |}}{{\left (b x^{2} - a\right )}^{2} {\left | b \right |}}\right )}{b} + \frac {a}{{\left (b x^{2} - a\right )} b}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 32, normalized size = 0.91 \begin {gather*} \frac {\ln \left (b\,x^2-a\right )}{2\,b^2}+\frac {a}{2\,b^2\,\left (a-b\,x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a - b*x^2)^2,x)

[Out]

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

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