3.2.15 \(\int \frac {x^2 (b+2 c x^3)}{-a+b x^3+c x^6} \, dx\) [115]

Optimal. Leaf size=19 \[ \frac {1}{3} \log \left (a-b x^3-c x^6\right ) \]

[Out]

1/3*ln(-c*x^6-b*x^3+a)

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Rubi [A]
time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1482, 642} \begin {gather*} \frac {1}{3} \log \left (a-b x^3-c x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(b + 2*c*x^3))/(-a + b*x^3 + c*x^6),x]

[Out]

Log[a - b*x^3 - c*x^6]/3

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1482

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x]
&& EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps

\begin {align*} \int \frac {x^2 \left (b+2 c x^3\right )}{-a+b x^3+c x^6} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {b+2 c x}{-a+b x+c x^2} \, dx,x,x^3\right )\\ &=\frac {1}{3} \log \left (a-b x^3-c x^6\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} \frac {1}{3} \log \left (-a+b x^3+c x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(b + 2*c*x^3))/(-a + b*x^3 + c*x^6),x]

[Out]

Log[-a + b*x^3 + c*x^6]/3

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Maple [A]
time = 0.02, size = 18, normalized size = 0.95

method	result	size
default	\(\frac {\ln \left (-c \,x^{6}-b \,x^{3}+a \right )}{3}\)	\(18\)
norman	\(\frac {\ln \left (-c \,x^{6}-b \,x^{3}+a \right )}{3}\)	\(18\)
risch	\(\frac {\ln \left (-c \,x^{6}-b \,x^{3}+a \right )}{3}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(2*c*x^3+b)/(c*x^6+b*x^3-a),x,method=_RETURNVERBOSE)

[Out]

1/3*ln(-c*x^6-b*x^3+a)

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Maxima [A]
time = 0.27, size = 17, normalized size = 0.89 \begin {gather*} \frac {1}{3} \, \log \left (c x^{6} + b x^{3} - a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2*c*x^3+b)/(c*x^6+b*x^3-a),x, algorithm="maxima")

[Out]

1/3*log(c*x^6 + b*x^3 - a)

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Fricas [A]
time = 0.33, size = 17, normalized size = 0.89 \begin {gather*} \frac {1}{3} \, \log \left (c x^{6} + b x^{3} - a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2*c*x^3+b)/(c*x^6+b*x^3-a),x, algorithm="fricas")

[Out]

1/3*log(c*x^6 + b*x^3 - a)

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Sympy [A]
time = 0.21, size = 14, normalized size = 0.74 \begin {gather*} \frac {\log {\left (- a + b x^{3} + c x^{6} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(2*c*x**3+b)/(c*x**6+b*x**3-a),x)

[Out]

log(-a + b*x**3 + c*x**6)/3

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Giac [A]
time = 4.08, size = 18, normalized size = 0.95 \begin {gather*} \frac {1}{3} \, \log \left ({\left | c x^{6} + b x^{3} - a \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2*c*x^3+b)/(c*x^6+b*x^3-a),x, algorithm="giac")

[Out]

1/3*log(abs(c*x^6 + b*x^3 - a))

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Mupad [B]
time = 0.06, size = 17, normalized size = 0.89 \begin {gather*} \frac {\ln \left (c\,x^6+b\,x^3-a\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(b + 2*c*x^3))/(b*x^3 - a + c*x^6),x)

[Out]

log(b*x^3 - a + c*x^6)/3

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