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

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

[Out]

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

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Rubi [A]  time = 0.0327442, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {444, 43} \[ \frac{(A b-a B) \log \left (a+b x^3\right )}{3 b^2}+\frac{B x^3}{3 b} \]

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 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]

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])

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*}

Mathematica [A]  time = 0.0122157, size = 31, normalized size = 0.89 \[ \frac{(A b-a B) \log \left (a+b x^3\right )+b B x^3}{3 b^2} \]

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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Maple [A]  time = 0.001, size = 40, normalized size = 1.1 \begin{align*}{\frac{B{x}^{3}}{3\,b}}+{\frac{\ln \left ( b{x}^{3}+a \right ) A}{3\,b}}-{\frac{\ln \left ( b{x}^{3}+a \right ) Ba}{3\,{b}^{2}}} \end{align*}

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 = 1.14153, size = 42, normalized size = 1.2 \begin{align*} \frac{B x^{3}}{3 \, b} - \frac{{\left (B a - A b\right )} \log \left (b x^{3} + a\right )}{3 \, b^{2}} \end{align*}

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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Fricas [A]  time = 1.38283, size = 65, normalized size = 1.86 \begin{align*} \frac{B b x^{3} -{\left (B a - A b\right )} \log \left (b x^{3} + a\right )}{3 \, b^{2}} \end{align*}

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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Sympy [A]  time = 0.691102, size = 27, normalized size = 0.77 \begin{align*} \frac{B x^{3}}{3 b} - \frac{\left (- A b + B a\right ) \log{\left (a + b x^{3} \right )}}{3 b^{2}} \end{align*}

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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Giac [A]  time = 1.11574, size = 43, normalized size = 1.23 \begin{align*} \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{align*}

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