\(\int \frac {x (A+B x^2)}{a+b x^2} \, dx\) [61]

Optimal result

Integrand size = 18, antiderivative size = 35 \[ \int \frac {x \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {B x^2}{2 b}+\frac {(A b-a B) \log \left (a+b x^2\right )}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {455, 45} \[ \int \frac {x \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {(A b-a B) \log \left (a+b x^2\right )}{2 b^2}+\frac {B x^2}{2 b} \]

[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 45

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 455

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {A+B x}{a+b x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {x \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {b B x^2+(A b-a B) \log \left (a+b x^2\right )}{2 b^2} \]

[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)

Maple [A] (verified)

Time = 2.46 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {B \,x^{2}}{2 b}+\frac {\left (A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(32\)
norman	\(\frac {B \,x^{2}}{2 b}+\frac {\left (A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(32\)
parallelrisch	\(\frac {b B \,x^{2}+A \ln \left (b \,x^{2}+a \right ) b -B \ln \left (b \,x^{2}+a \right ) a}{2 b^{2}}\)	\(36\)
risch	\(\frac {B \,x^{2}}{2 b}+\frac {\ln \left (b \,x^{2}+a \right ) A}{2 b}-\frac {\ln \left (b \,x^{2}+a \right ) B a}{2 b^{2}}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {x \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {B b x^{2} - {\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {x \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {B x^{2}}{2 b} - \frac {\left (- A b + B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{2}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {x \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {B x^{2}}{2 \, b} - \frac {{\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int \frac {x \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {B x^{2}}{2 \, b} - \frac {{\left (B a - A b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {x \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {B\,x^2}{2\,b}+\frac {\ln \left (b\,x^2+a\right )\,\left (A\,b-B\,a\right )}{2\,b^2} \]

[In]

int((x*(A + B*x^2))/(a + b*x^2),x)

[Out]

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