3.3.25 \(\int \frac {x^3}{a-b x^2} \, dx\) [225]

Optimal. Leaf size=28 \[ -\frac {x^2}{2 b}-\frac {a \log \left (a-b x^2\right )}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 28, 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 \log \left (a-b x^2\right )}{2 b^2}-\frac {x^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(a - b*x^2),x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 28, normalized size = 1.00 \begin {gather*} -\frac {x^2}{2 b}-\frac {a \log \left (a-b x^2\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a - b*x^2),x]

[Out]

-1/2*x^2/b - (a*Log[a - b*x^2])/(2*b^2)

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Maple [A]
time = 0.02, size = 25, normalized size = 0.89

method	result	size
default	\(-\frac {x^{2}}{2 b}-\frac {a \ln \left (-b \,x^{2}+a \right )}{2 b^{2}}\)	\(25\)
norman	\(-\frac {x^{2}}{2 b}-\frac {a \ln \left (-b \,x^{2}+a \right )}{2 b^{2}}\)	\(25\)
risch	\(-\frac {x^{2}}{2 b}-\frac {a \ln \left (-b \,x^{2}+a \right )}{2 b^{2}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 25, normalized size = 0.89 \begin {gather*} -\frac {x^{2}}{2 \, b} - \frac {a \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),x, algorithm="maxima")

[Out]

-1/2*x^2/b - 1/2*a*log(b*x^2 - a)/b^2

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Fricas [A]
time = 0.71, size = 23, normalized size = 0.82 \begin {gather*} -\frac {b x^{2} + a \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),x, algorithm="fricas")

[Out]

-1/2*(b*x^2 + a*log(b*x^2 - a))/b^2

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Sympy [A]
time = 0.05, size = 22, normalized size = 0.79 \begin {gather*} - \frac {a \log {\left (- a + b x^{2} \right )}}{2 b^{2}} - \frac {x^{2}}{2 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.24, size = 26, normalized size = 0.93 \begin {gather*} -\frac {x^{2}}{2 \, b} - \frac {a \log \left ({\left | b x^{2} - a \right |}\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 23, normalized size = 0.82 \begin {gather*} -\frac {b\,x^2+a\,\ln \left (b\,x^2-a\right )}{2\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a - b*x^2),x)

[Out]

-(b*x^2 + a*log(b*x^2 - a))/(2*b^2)

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