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

Optimal. Leaf size=70 \[ \frac{1}{2 \left (c+d x^2\right ) (b c-a d)}+\frac{b \log \left (a+b x^2\right )}{2 (b c-a d)^2}-\frac{b \log \left (c+d x^2\right )}{2 (b c-a d)^2} \]

[Out]

1/(2*(b*c - a*d)*(c + d*x^2)) + (b*Log[a + b*x^2])/(2*(b*c - a*d)^2) - (b*Log[c + d*x^2])/(2*(b*c - a*d)^2)

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Rubi [A]  time = 0.0541384, antiderivative size = 70, 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, 44} \[ \frac{1}{2 \left (c+d x^2\right ) (b c-a d)}+\frac{b \log \left (a+b x^2\right )}{2 (b c-a d)^2}-\frac{b \log \left (c+d x^2\right )}{2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[x/((a + b*x^2)*(c + d*x^2)^2),x]

[Out]

1/(2*(b*c - a*d)*(c + d*x^2)) + (b*Log[a + b*x^2])/(2*(b*c - a*d)^2) - (b*Log[c + d*x^2])/(2*(b*c - a*d)^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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.02779, size = 66, normalized size = 0.94 \[ \frac{b \left (c+d x^2\right ) \log \left (a+b x^2\right )-a d-b \left (c+d x^2\right ) \log \left (c+d x^2\right )+b c}{2 \left (c+d x^2\right ) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 90, normalized size = 1.3 \begin{align*} -{\frac{b\ln \left ( d{x}^{2}+c \right ) }{2\, \left ( ad-bc \right ) ^{2}}}-{\frac{ad}{2\, \left ( ad-bc \right ) ^{2} \left ( d{x}^{2}+c \right ) }}+{\frac{bc}{2\, \left ( ad-bc \right ) ^{2} \left ( d{x}^{2}+c \right ) }}+{\frac{b\ln \left ( b{x}^{2}+a \right ) }{2\, \left ( ad-bc \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05275, size = 134, normalized size = 1.91 \begin{align*} \frac{b \log \left (b x^{2} + a\right )}{2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} - \frac{b \log \left (d x^{2} + c\right )}{2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} + \frac{1}{2 \,{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58463, size = 217, normalized size = 3.1 \begin{align*} \frac{b c - a d +{\left (b d x^{2} + b c\right )} \log \left (b x^{2} + a\right ) -{\left (b d x^{2} + b c\right )} \log \left (d x^{2} + c\right )}{2 \,{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} +{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.18226, size = 248, normalized size = 3.54 \begin{align*} - \frac{b \log{\left (x^{2} + \frac{- \frac{a^{3} b d^{3}}{\left (a d - b c\right )^{2}} + \frac{3 a^{2} b^{2} c d^{2}}{\left (a d - b c\right )^{2}} - \frac{3 a b^{3} c^{2} d}{\left (a d - b c\right )^{2}} + a b d + \frac{b^{4} c^{3}}{\left (a d - b c\right )^{2}} + b^{2} c}{2 b^{2} d} \right )}}{2 \left (a d - b c\right )^{2}} + \frac{b \log{\left (x^{2} + \frac{\frac{a^{3} b d^{3}}{\left (a d - b c\right )^{2}} - \frac{3 a^{2} b^{2} c d^{2}}{\left (a d - b c\right )^{2}} + \frac{3 a b^{3} c^{2} d}{\left (a d - b c\right )^{2}} + a b d - \frac{b^{4} c^{3}}{\left (a d - b c\right )^{2}} + b^{2} c}{2 b^{2} d} \right )}}{2 \left (a d - b c\right )^{2}} - \frac{1}{2 a c d - 2 b c^{2} + x^{2} \left (2 a d^{2} - 2 b c d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15498, size = 115, normalized size = 1.64 \begin{align*} \frac{b d \log \left ({\left | b - \frac{b c}{d x^{2} + c} + \frac{a d}{d x^{2} + c} \right |}\right )}{2 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}} + \frac{d}{2 \,{\left (b c d - a d^{2}\right )}{\left (d x^{2} + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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