3.255 \(\int \frac{x}{(a+b x^2) (c+d x^2)^3} \, dx\)

Optimal. Leaf size=98 \[ \frac{b^2 \log \left (a+b x^2\right )}{2 (b c-a d)^3}-\frac{b^2 \log \left (c+d x^2\right )}{2 (b c-a d)^3}+\frac{b}{2 \left (c+d x^2\right ) (b c-a d)^2}+\frac{1}{4 \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0749927, antiderivative size = 98, 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} \[ \frac{b^2 \log \left (a+b x^2\right )}{2 (b c-a d)^3}-\frac{b^2 \log \left (c+d x^2\right )}{2 (b c-a d)^3}+\frac{b}{2 \left (c+d x^2\right ) (b c-a d)^2}+\frac{1}{4 \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x/((a + b*x^2)*(c + d*x^2)^3),x]

[Out]

1/(4*(b*c - a*d)*(c + d*x^2)^2) + b/(2*(b*c - a*d)^2*(c + d*x^2)) + (b^2*Log[a + b*x^2])/(2*(b*c - a*d)^3) - (
b^2*Log[c + d*x^2])/(2*(b*c - a*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 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 )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(a+b x) (c+d x)^3} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{b^3}{(b c-a d)^3 (a+b x)}-\frac{d}{(b c-a d) (c+d x)^3}-\frac{b d}{(b c-a d)^2 (c+d x)^2}-\frac{b^2 d}{(b c-a d)^3 (c+d x)}\right ) \, dx,x,x^2\right )\\ &=\frac{1}{4 (b c-a d) \left (c+d x^2\right )^2}+\frac{b}{2 (b c-a d)^2 \left (c+d x^2\right )}+\frac{b^2 \log \left (a+b x^2\right )}{2 (b c-a d)^3}-\frac{b^2 \log \left (c+d x^2\right )}{2 (b c-a d)^3}\\ \end{align*}

Mathematica [A]  time = 0.0468274, size = 98, normalized size = 1. \[ \frac{b^2 \log \left (a+b x^2\right )}{2 (b c-a d)^3}-\frac{b^2 \log \left (c+d x^2\right )}{2 (b c-a d)^3}+\frac{b}{2 \left (c+d x^2\right ) (b c-a d)^2}-\frac{1}{4 \left (c+d x^2\right )^2 (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((a + b*x^2)*(c + d*x^2)^3),x]

[Out]

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

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Maple [A]  time = 0.01, size = 176, normalized size = 1.8 \begin{align*}{\frac{\ln \left ( d{x}^{2}+c \right ){b}^{2}}{2\, \left ( ad-bc \right ) ^{3}}}-{\frac{{a}^{2}{d}^{2}}{4\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{cabd}{2\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{{b}^{2}{c}^{2}}{4\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{bda}{2\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) }}-{\frac{{b}^{2}c}{2\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) }}-{\frac{{b}^{2}\ln \left ( b{x}^{2}+a \right ) }{2\, \left ( ad-bc \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x^2+a)/(d*x^2+c)^3,x)

[Out]

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

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Maxima [B]  time = 1.03754, size = 285, normalized size = 2.91 \begin{align*} \frac{b^{2} \log \left (b x^{2} + a\right )}{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}} - \frac{b^{2} \log \left (d x^{2} + c\right )}{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}} + \frac{2 \, b d x^{2} + 3 \, b c - a d}{4 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c 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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.56677, size = 506, normalized size = 5.16 \begin{align*} \frac{3 \, b^{2} c^{2} - 4 \, a b c d + a^{2} d^{2} + 2 \,{\left (b^{2} c d - a b d^{2}\right )} x^{2} + 2 \,{\left (b^{2} d^{2} x^{4} + 2 \, b^{2} c d x^{2} + b^{2} c^{2}\right )} \log \left (b x^{2} + a\right ) - 2 \,{\left (b^{2} d^{2} x^{4} + 2 \, b^{2} c d x^{2} + b^{2} c^{2}\right )} \log \left (d x^{2} + c\right )}{4 \,{\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3} +{\left (b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} x^{4} + 2 \,{\left (b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} + 3 \, a^{2} b c^{2} d^{3} - a^{3} c d^{4}\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)^3,x, algorithm="fricas")

[Out]

1/4*(3*b^2*c^2 - 4*a*b*c*d + a^2*d^2 + 2*(b^2*c*d - a*b*d^2)*x^2 + 2*(b^2*d^2*x^4 + 2*b^2*c*d*x^2 + b^2*c^2)*l
og(b*x^2 + a) - 2*(b^2*d^2*x^4 + 2*b^2*c*d*x^2 + b^2*c^2)*log(d*x^2 + c))/(b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c
^3*d^2 - a^3*c^2*d^3 + (b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 + 3*a^2*b*c*d^4 - a^3*d^5)*x^4 + 2*(b^3*c^4*d - 3*a*b^2*
c^3*d^2 + 3*a^2*b*c^2*d^3 - a^3*c*d^4)*x^2)

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Sympy [B]  time = 4.26234, size = 391, normalized size = 3.99 \begin{align*} \frac{b^{2} \log{\left (x^{2} + \frac{- \frac{a^{4} b^{2} d^{4}}{\left (a d - b c\right )^{3}} + \frac{4 a^{3} b^{3} c d^{3}}{\left (a d - b c\right )^{3}} - \frac{6 a^{2} b^{4} c^{2} d^{2}}{\left (a d - b c\right )^{3}} + \frac{4 a b^{5} c^{3} d}{\left (a d - b c\right )^{3}} + a b^{2} d - \frac{b^{6} c^{4}}{\left (a d - b c\right )^{3}} + b^{3} c}{2 b^{3} d} \right )}}{2 \left (a d - b c\right )^{3}} - \frac{b^{2} \log{\left (x^{2} + \frac{\frac{a^{4} b^{2} d^{4}}{\left (a d - b c\right )^{3}} - \frac{4 a^{3} b^{3} c d^{3}}{\left (a d - b c\right )^{3}} + \frac{6 a^{2} b^{4} c^{2} d^{2}}{\left (a d - b c\right )^{3}} - \frac{4 a b^{5} c^{3} d}{\left (a d - b c\right )^{3}} + a b^{2} d + \frac{b^{6} c^{4}}{\left (a d - b c\right )^{3}} + b^{3} c}{2 b^{3} d} \right )}}{2 \left (a d - b c\right )^{3}} + \frac{- a d + 3 b c + 2 b d x^{2}}{4 a^{2} c^{2} d^{2} - 8 a b c^{3} d + 4 b^{2} c^{4} + x^{4} \left (4 a^{2} d^{4} - 8 a b c d^{3} + 4 b^{2} c^{2} d^{2}\right ) + x^{2} \left (8 a^{2} c d^{3} - 16 a b c^{2} d^{2} + 8 b^{2} c^{3} 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)**3,x)

[Out]

b**2*log(x**2 + (-a**4*b**2*d**4/(a*d - b*c)**3 + 4*a**3*b**3*c*d**3/(a*d - b*c)**3 - 6*a**2*b**4*c**2*d**2/(a
*d - b*c)**3 + 4*a*b**5*c**3*d/(a*d - b*c)**3 + a*b**2*d - b**6*c**4/(a*d - b*c)**3 + b**3*c)/(2*b**3*d))/(2*(
a*d - b*c)**3) - b**2*log(x**2 + (a**4*b**2*d**4/(a*d - b*c)**3 - 4*a**3*b**3*c*d**3/(a*d - b*c)**3 + 6*a**2*b
**4*c**2*d**2/(a*d - b*c)**3 - 4*a*b**5*c**3*d/(a*d - b*c)**3 + a*b**2*d + b**6*c**4/(a*d - b*c)**3 + b**3*c)/
(2*b**3*d))/(2*(a*d - b*c)**3) + (-a*d + 3*b*c + 2*b*d*x**2)/(4*a**2*c**2*d**2 - 8*a*b*c**3*d + 4*b**2*c**4 +
x**4*(4*a**2*d**4 - 8*a*b*c*d**3 + 4*b**2*c**2*d**2) + x**2*(8*a**2*c*d**3 - 16*a*b*c**2*d**2 + 8*b**2*c**3*d)
)

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Giac [A]  time = 1.15733, size = 235, normalized size = 2.4 \begin{align*} \frac{b^{3} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \,{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )}} - \frac{b^{2} d \log \left ({\left | d x^{2} + c \right |}\right )}{2 \,{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )}} + \frac{3 \, b^{2} c^{2} - 4 \, a b c d + a^{2} d^{2} + 2 \,{\left (b^{2} c d - a b d^{2}\right )} x^{2}}{4 \,{\left (d x^{2} + c\right )}^{2}{\left (b c - a d\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="giac")

[Out]

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