3.183 \(\int \frac{x (a+b x^2)^2}{(c+d x^2)^2} \, dx\)

Optimal. Leaf size=62 \[ -\frac{(b c-a d)^2}{2 d^3 \left (c+d x^2\right )}-\frac{b (b c-a d) \log \left (c+d x^2\right )}{d^3}+\frac{b^2 x^2}{2 d^2} \]

[Out]

(b^2*x^2)/(2*d^2) - (b*c - a*d)^2/(2*d^3*(c + d*x^2)) - (b*(b*c - a*d)*Log[c + d*x^2])/d^3

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Rubi [A]  time = 0.058084, antiderivative size = 62, 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} \[ -\frac{(b c-a d)^2}{2 d^3 \left (c+d x^2\right )}-\frac{b (b c-a d) \log \left (c+d x^2\right )}{d^3}+\frac{b^2 x^2}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*x^2)^2)/(c + d*x^2)^2,x]

[Out]

(b^2*x^2)/(2*d^2) - (b*c - a*d)^2/(2*d^3*(c + d*x^2)) - (b*(b*c - a*d)*Log[c + d*x^2])/d^3

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

Mathematica [A]  time = 0.0447732, size = 56, normalized size = 0.9 \[ \frac{-\frac{(b c-a d)^2}{c+d x^2}+2 b (a d-b c) \log \left (c+d x^2\right )+b^2 d x^2}{2 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*x^2)^2)/(c + d*x^2)^2,x]

[Out]

(b^2*d*x^2 - (b*c - a*d)^2/(c + d*x^2) + 2*b*(-(b*c) + a*d)*Log[c + d*x^2])/(2*d^3)

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Maple [A]  time = 0.008, size = 97, normalized size = 1.6 \begin{align*}{\frac{{b}^{2}{x}^{2}}{2\,{d}^{2}}}+{\frac{b\ln \left ( d{x}^{2}+c \right ) a}{{d}^{2}}}-{\frac{{b}^{2}\ln \left ( d{x}^{2}+c \right ) c}{{d}^{3}}}-{\frac{{a}^{2}}{2\,d \left ( d{x}^{2}+c \right ) }}+{\frac{abc}{{d}^{2} \left ( d{x}^{2}+c \right ) }}-{\frac{{b}^{2}{c}^{2}}{2\,{d}^{3} \left ( d{x}^{2}+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2+a)^2/(d*x^2+c)^2,x)

[Out]

1/2*b^2*x^2/d^2+1/d^2*b*ln(d*x^2+c)*a-1/d^3*b^2*ln(d*x^2+c)*c-1/2/d/(d*x^2+c)*a^2+1/d^2/(d*x^2+c)*a*b*c-1/2/d^
3/(d*x^2+c)*b^2*c^2

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Maxima [A]  time = 0.987961, size = 100, normalized size = 1.61 \begin{align*} \frac{b^{2} x^{2}}{2 \, d^{2}} - \frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{2 \,{\left (d^{4} x^{2} + c d^{3}\right )}} - \frac{{\left (b^{2} c - a b d\right )} \log \left (d x^{2} + c\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^2/d^2 - 1/2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/(d^4*x^2 + c*d^3) - (b^2*c - a*b*d)*log(d*x^2 + c)/d^3

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Fricas [A]  time = 1.41026, size = 200, normalized size = 3.23 \begin{align*} \frac{b^{2} d^{2} x^{4} + b^{2} c d x^{2} - b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} - 2 \,{\left (b^{2} c^{2} - a b c d +{\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \log \left (d x^{2} + c\right )}{2 \,{\left (d^{4} x^{2} + c d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*d^2*x^4 + b^2*c*d*x^2 - b^2*c^2 + 2*a*b*c*d - a^2*d^2 - 2*(b^2*c^2 - a*b*c*d + (b^2*c*d - a*b*d^2)*x^
2)*log(d*x^2 + c))/(d^4*x^2 + c*d^3)

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Sympy [A]  time = 0.911734, size = 68, normalized size = 1.1 \begin{align*} \frac{b^{2} x^{2}}{2 d^{2}} + \frac{b \left (a d - b c\right ) \log{\left (c + d x^{2} \right )}}{d^{3}} - \frac{a^{2} d^{2} - 2 a b c d + b^{2} c^{2}}{2 c d^{3} + 2 d^{4} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**2+a)**2/(d*x**2+c)**2,x)

[Out]

b**2*x**2/(2*d**2) + b*(a*d - b*c)*log(c + d*x**2)/d**3 - (a**2*d**2 - 2*a*b*c*d + b**2*c**2)/(2*c*d**3 + 2*d*
*4*x**2)

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Giac [A]  time = 1.16857, size = 149, normalized size = 2.4 \begin{align*} \frac{{\left (d x^{2} + c\right )} b^{2}}{2 \, d^{3}} + \frac{{\left (b^{2} c - a b d\right )} \log \left (\frac{{\left | d x^{2} + c \right |}}{{\left (d x^{2} + c\right )}^{2}{\left | d \right |}}\right )}{d^{3}} - \frac{\frac{b^{2} c^{2} d}{d x^{2} + c} - \frac{2 \, a b c d^{2}}{d x^{2} + c} + \frac{a^{2} d^{3}}{d x^{2} + c}}{2 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d*x^2 + c)*b^2/d^3 + (b^2*c - a*b*d)*log(abs(d*x^2 + c)/((d*x^2 + c)^2*abs(d)))/d^3 - 1/2*(b^2*c^2*d/(d*x
^2 + c) - 2*a*b*c*d^2/(d*x^2 + c) + a^2*d^3/(d*x^2 + c))/d^4