\(\int \frac {x}{(a+b x^2)^2 (c+d x^2)} \, dx\) [292]

Optimal result

Integrand size = 20, antiderivative size = 70 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {1}{2 (b c-a d) \left (a+b x^2\right )}-\frac {d \log \left (a+b x^2\right )}{2 (b c-a d)^2}+\frac {d \log \left (c+d x^2\right )}{2 (b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {455, 46} \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {1}{2 \left (a+b x^2\right ) (b c-a d)}-\frac {d \log \left (a+b x^2\right )}{2 (b c-a d)^2}+\frac {d \log \left (c+d x^2\right )}{2 (b c-a d)^2} \]

[In]

Int[x/((a + b*x^2)^2*(c + d*x^2)),x]

[Out]

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

Rule 46

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {-b c+a d-d \left (a+b x^2\right ) \log \left (a+b x^2\right )+d \left (a+b x^2\right ) \log \left (c+d x^2\right )}{2 (b c-a d)^2 \left (a+b x^2\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03

method	result	size
default	\(-\frac {b \left (\frac {d \ln \left (b \,x^{2}+a \right )}{b}-\frac {a d -b c}{b \left (b \,x^{2}+a \right )}\right )}{2 \left (a d -b c \right )^{2}}+\frac {d \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right )^{2}}\)	\(72\)
risch	\(\frac {1}{2 \left (a d -b c \right ) \left (b \,x^{2}+a \right )}-\frac {d \ln \left (-b \,x^{2}-a \right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {d \ln \left (d \,x^{2}+c \right )}{2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}\)	\(94\)
norman	\(-\frac {b \,x^{2}}{2 \left (a d -b c \right ) a \left (b \,x^{2}+a \right )}-\frac {d \ln \left (b \,x^{2}+a \right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {d \ln \left (d \,x^{2}+c \right )}{2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}\)	\(98\)
parallelrisch	\(-\frac {\ln \left (b \,x^{2}+a \right ) x^{2} b^{2} d -\ln \left (d \,x^{2}+c \right ) x^{2} b^{2} d +\ln \left (b \,x^{2}+a \right ) a b d -\ln \left (d \,x^{2}+c \right ) a b d -a b d +b^{2} c}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (b \,x^{2}+a \right ) b}\)	\(107\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.47 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {b c - a d + {\left (b d x^{2} + a d\right )} \log \left (b x^{2} + a\right ) - {\left (b d x^{2} + a d\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (56) = 112\).

Time = 1.03 (sec) , antiderivative size = 248, normalized size of antiderivative = 3.54 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {d \log {\left (x^{2} + \frac {- \frac {a^{3} d^{4}}{\left (a d - b c\right )^{2}} + \frac {3 a^{2} b c d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 a b^{2} c^{2} d^{2}}{\left (a d - b c\right )^{2}} + a d^{2} + \frac {b^{3} c^{3} d}{\left (a d - b c\right )^{2}} + b c d}{2 b d^{2}} \right )}}{2 \left (a d - b c\right )^{2}} - \frac {d \log {\left (x^{2} + \frac {\frac {a^{3} d^{4}}{\left (a d - b c\right )^{2}} - \frac {3 a^{2} b c d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 a b^{2} c^{2} d^{2}}{\left (a d - b c\right )^{2}} + a d^{2} - \frac {b^{3} c^{3} d}{\left (a d - b c\right )^{2}} + b c d}{2 b d^{2}} \right )}}{2 \left (a d - b c\right )^{2}} + \frac {1}{2 a^{2} d - 2 a b c + x^{2} \cdot \left (2 a b d - 2 b^{2} c\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.41 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {d \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} + \frac {d \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} - \frac {1}{2 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.21 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {b d \log \left ({\left | \frac {b c}{b x^{2} + a} - \frac {a d}{b x^{2} + a} + d \right |}\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} - \frac {b}{2 \, {\left (b^{2} c - a b d\right )} {\left (b x^{2} + a\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.11 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.30 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {b\,c-a\,\left (d-d\,\mathrm {atan}\left (\frac {a\,d\,x^2\,1{}\mathrm {i}-b\,c\,x^2\,1{}\mathrm {i}}{2\,a\,c+a\,d\,x^2+b\,c\,x^2}\right )\,2{}\mathrm {i}\right )+b\,d\,x^2\,\mathrm {atan}\left (\frac {a\,d\,x^2\,1{}\mathrm {i}-b\,c\,x^2\,1{}\mathrm {i}}{2\,a\,c+a\,d\,x^2+b\,c\,x^2}\right )\,2{}\mathrm {i}}{2\,a^3\,d^2-4\,a^2\,b\,c\,d+2\,a^2\,b\,d^2\,x^2+2\,a\,b^2\,c^2-4\,a\,b^2\,c\,d\,x^2+2\,b^3\,c^2\,x^2} \]

[In]

int(x/((a + b*x^2)^2*(c + d*x^2)),x)

[Out]

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