\(\int \frac {x}{(a+b x^2) (c+d x^2)} \, dx\) [232]

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {\log \left (a+b x^2\right )}{2 (b c-a d)}-\frac {\log \left (c+d x^2\right )}{2 (b c-a d)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {455, 36, 31} \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {\log \left (a+b x^2\right )}{2 (b c-a d)}-\frac {\log \left (c+d x^2\right )}{2 (b c-a d)} \]

[In]

Int[x/((a + b*x^2)*(c + d*x^2)),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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) (c+d x)} \, dx,x,x^2\right ) \\ & = \frac {b \text {Subst}\left (\int \frac {1}{a+b x} \, dx,x,x^2\right )}{2 (b c-a d)}-\frac {d \text {Subst}\left (\int \frac {1}{c+d x} \, dx,x,x^2\right )}{2 (b c-a d)} \\ & = \frac {\log \left (a+b x^2\right )}{2 (b c-a d)}-\frac {\log \left (c+d x^2\right )}{2 (b c-a d)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.69 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {\log \left (a+b x^2\right )-\log \left (c+d x^2\right )}{2 b c-2 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(-\frac {\ln \left (b \,x^{2}+a \right )-\ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right )}\)	\(32\)
default	\(-\frac {\ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right )}+\frac {\ln \left (d \,x^{2}+c \right )}{2 a d -2 b c}\)	\(42\)
norman	\(-\frac {\ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right )}+\frac {\ln \left (d \,x^{2}+c \right )}{2 a d -2 b c}\)	\(42\)
risch	\(\frac {\ln \left (d \,x^{2}+c \right )}{2 a d -2 b c}-\frac {\ln \left (-b \,x^{2}-a \right )}{2 \left (a d -b c \right )}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.69 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {\log \left (b x^{2} + a\right ) - \log \left (d x^{2} + c\right )}{2 \, {\left (b c - a d\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (36) = 72\).

Time = 0.53 (sec) , antiderivative size = 138, normalized size of antiderivative = 3.07 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {\log {\left (x^{2} + \frac {- \frac {a^{2} d^{2}}{a d - b c} + \frac {2 a b c d}{a d - b c} + a d - \frac {b^{2} c^{2}}{a d - b c} + b c}{2 b d} \right )}}{2 \left (a d - b c\right )} - \frac {\log {\left (x^{2} + \frac {\frac {a^{2} d^{2}}{a d - b c} - \frac {2 a b c d}{a d - b c} + a d + \frac {b^{2} c^{2}}{a d - b c} + b c}{2 b d} \right )}}{2 \left (a d - b c\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {\log \left (b x^{2} + a\right )}{2 \, {\left (b c - a d\right )}} - \frac {\log \left (d x^{2} + c\right )}{2 \, {\left (b c - a d\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.13 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {b \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (b^{2} c - a b d\right )}} - \frac {d \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b c d - a d^{2}\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 3.29 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {8\,b^2\,d^2\,x^2}{\left (2\,a\,d-2\,b\,c\right )\,\left (\frac {32\,a\,b^2\,c\,d^2}{4\,a^2\,d^2-8\,a\,b\,c\,d+4\,b^2\,c^2}+\frac {16\,a\,b^2\,d^3\,x^2}{4\,a^2\,d^2-8\,a\,b\,c\,d+4\,b^2\,c^2}+\frac {16\,b^3\,c\,d^2\,x^2}{4\,a^2\,d^2-8\,a\,b\,c\,d+4\,b^2\,c^2}\right )}\right )}{2\,a\,d-2\,b\,c} \]

[In]

int(x/((a + b*x^2)*(c + d*x^2)),x)

[Out]

(2*atanh((8*b^2*d^2*x^2)/((2*a*d - 2*b*c)*((32*a*b^2*c*d^2)/(4*a^2*d^2 + 4*b^2*c^2 - 8*a*b*c*d) + (16*a*b^2*d^
3*x^2)/(4*a^2*d^2 + 4*b^2*c^2 - 8*a*b*c*d) + (16*b^3*c*d^2*x^2)/(4*a^2*d^2 + 4*b^2*c^2 - 8*a*b*c*d)))))/(2*a*d
- 2*b*c)