3.1.78 \(\int \frac {x^2 (A+B x^3)}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=41 \[ \frac {a B-A b}{3 b^2 \left (a+b x^3\right )}+\frac {B \log \left (a+b x^3\right )}{3 b^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {444, 43} \begin {gather*} \frac {B \log \left (a+b x^3\right )}{3 b^2}-\frac {A b-a B}{3 b^2 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

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

Rule 43

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 444

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 41, normalized size = 1.00 \begin {gather*} \frac {a B-A b}{3 b^2 \left (a+b x^3\right )}+\frac {B \log \left (a+b x^3\right )}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^2*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

IntegrateAlgebraic[(x^2*(A + B*x^3))/(a + b*x^3)^2, x]

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fricas [A]  time = 0.56, size = 44, normalized size = 1.07 \begin {gather*} \frac {B a - A b + {\left (B b x^{3} + B a\right )} \log \left (b x^{3} + a\right )}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 65, normalized size = 1.59 \begin {gather*} -\frac {B {\left (\frac {\log \left (\frac {{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x^{3} + a\right )} b}\right )}}{3 \, b} - \frac {A}{3 \, {\left (b x^{3} + a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 47, normalized size = 1.15 \begin {gather*} -\frac {A}{3 \left (b \,x^{3}+a \right ) b}+\frac {B a}{3 \left (b \,x^{3}+a \right ) b^{2}}+\frac {B \ln \left (b \,x^{3}+a \right )}{3 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x^3+A)/(b*x^3+a)^2,x)

[Out]

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

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maxima [A]  time = 0.58, size = 40, normalized size = 0.98 \begin {gather*} \frac {B a - A b}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} + \frac {B \log \left (b x^{3} + a\right )}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.35, size = 37, normalized size = 0.90 \begin {gather*} \frac {B\,\ln \left (b\,x^3+a\right )}{3\,b^2}-\frac {A\,b-B\,a}{3\,b^2\,\left (b\,x^3+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(A + B*x^3))/(a + b*x^3)^2,x)

[Out]

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

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sympy [A]  time = 1.23, size = 36, normalized size = 0.88 \begin {gather*} \frac {B \log {\left (a + b x^{3} \right )}}{3 b^{2}} + \frac {- A b + B a}{3 a b^{2} + 3 b^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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