\(\int \frac {x^3}{(a+b x^2) (c+d x^2)} \, dx\) [230]

Optimal result

Integrand size = 22, antiderivative size = 53 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {a \log \left (a+b x^2\right )}{2 b (b c-a d)}+\frac {c \log \left (c+d x^2\right )}{2 d (b c-a d)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 78} \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {c \log \left (c+d x^2\right )}{2 d (b c-a d)}-\frac {a \log \left (a+b x^2\right )}{2 b (b c-a d)} \]

[In]

Int[x^3/((a + b*x^2)*(c + d*x^2)),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {a d \log \left (a+b x^2\right )-b c \log \left (c+d x^2\right )}{2 b^2 c d-2 a b d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(\frac {\ln \left (b \,x^{2}+a \right ) a d -c \ln \left (d \,x^{2}+c \right ) b}{2 \left (a d -b c \right ) b d}\)	\(43\)
default	\(\frac {a \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right ) b}-\frac {c \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right ) d}\)	\(50\)
norman	\(\frac {a \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right ) b}-\frac {c \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right ) d}\)	\(50\)
risch	\(-\frac {c \ln \left (-d \,x^{2}-c \right )}{2 \left (a d -b c \right ) d}+\frac {a \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right ) b}\)	\(53\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {a d \log \left (b x^{2} + a\right ) - b c \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c d - a b d^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (39) = 78\).

Time = 1.35 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.72 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {a \log {\left (x^{2} + \frac {\frac {a^{3} d^{2}}{b \left (a d - b c\right )} - \frac {2 a^{2} c d}{a d - b c} + \frac {a b c^{2}}{a d - b c} + 2 a c}{a d + b c} \right )}}{2 b \left (a d - b c\right )} - \frac {c \log {\left (x^{2} + \frac {- \frac {a^{2} c d}{a d - b c} + \frac {2 a b c^{2}}{a d - b c} + 2 a c - \frac {b^{2} c^{3}}{d \left (a d - b c\right )}}{a d + b c} \right )}}{2 d \left (a d - b c\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {a \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c - a b d\right )}} + \frac {c \log \left (d x^{2} + c\right )}{2 \, {\left (b c d - a d^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {a \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (b^{2} c - a b d\right )}} + \frac {c \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b c d - a d^{2}\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.48 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {a\,\ln \left (b\,x^2+a\right )}{2\,b^2\,c-2\,a\,b\,d}-\frac {c\,\ln \left (d\,x^2+c\right )}{2\,a\,d^2-2\,b\,c\,d} \]

[In]

int(x^3/((a + b*x^2)*(c + d*x^2)),x)

[Out]

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