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

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

[Out]

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

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Rubi [A]  time = 0.0299063, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {444, 43} \[ \frac{(A b-a B) \log \left (a+b x^2\right )}{2 b^2}+\frac{B x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x^2))/(a + b*x^2),x]

[Out]

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

Mathematica [A]  time = 0.0108656, size = 31, normalized size = 0.89 \[ \frac{(A b-a B) \log \left (a+b x^2\right )+b B x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 40, normalized size = 1.1 \begin{align*}{\frac{B{x}^{2}}{2\,b}}+{\frac{\ln \left ( b{x}^{2}+a \right ) A}{2\,b}}-{\frac{\ln \left ( b{x}^{2}+a \right ) Ba}{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.991694, size = 42, normalized size = 1.2 \begin{align*} \frac{B x^{2}}{2 \, b} - \frac{{\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.32872, size = 65, normalized size = 1.86 \begin{align*} \frac{B b x^{2} -{\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.375242, size = 27, normalized size = 0.77 \begin{align*} \frac{B x^{2}}{2 b} - \frac{\left (- A b + B a\right ) \log{\left (a + b x^{2} \right )}}{2 b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16417, size = 43, normalized size = 1.23 \begin{align*} \frac{B x^{2}}{2 \, b} - \frac{{\left (B a - A b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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