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

Optimal. Leaf size=35 \[ \frac {(A b-a B) \log \left (a+b x^3\right )}{3 b^2}+\frac {B x^3}{3 b} \]

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Rubi [A]  time = 0.03, antiderivative size = 35, 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 {(A b-a B) \log \left (a+b x^3\right )}{3 b^2}+\frac {B x^3}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x^3))/(a + b*x^3),x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.89 \begin {gather*} \frac {(A b-a B) \log \left (a+b x^3\right )+b B x^3}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*B*x^3 + (A*b - a*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 )}{a+b x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 30, normalized size = 0.86 \begin {gather*} \frac {B b x^{3} - {\left (B a - A b\right )} \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),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 32, normalized size = 0.91 \begin {gather*} \frac {B x^{3}}{3 \, b} - \frac {{\left (B a - A b\right )} \log \left ({\left | b x^{3} + a \right |}\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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 40, normalized size = 1.14 \begin {gather*} \frac {B \,x^{3}}{3 b}+\frac {A \ln \left (b \,x^{3}+a \right )}{3 b}-\frac {B a \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),x)

[Out]

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

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maxima [A]  time = 0.54, size = 31, normalized size = 0.89 \begin {gather*} \frac {B x^{3}}{3 \, b} - \frac {{\left (B a - A b\right )} \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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 31, normalized size = 0.89 \begin {gather*} \frac {B\,x^3}{3\,b}+\frac {\ln \left (b\,x^3+a\right )\,\left (A\,b-B\,a\right )}{3\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(A + B*x^3))/(a + b*x^3),x)

[Out]

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

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sympy [A]  time = 0.88, size = 27, normalized size = 0.77 \begin {gather*} \frac {B x^{3}}{3 b} - \frac {\left (- A b + B a\right ) \log {\left (a + b x^{3} \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),x)

[Out]

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

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