\(\int \frac {x^3}{(a+b x^2)^2 (c+d x^2)} \, dx\) [290]

Optimal result

Integrand size = 22, antiderivative size = 74 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {a}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {c \log \left (a+b x^2\right )}{2 (b c-a d)^2}-\frac {c \log \left (c+d x^2\right )}{2 (b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, 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} \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {a}{2 b \left (a+b x^2\right ) (b c-a d)}+\frac {c \log \left (a+b x^2\right )}{2 (b c-a d)^2}-\frac {c \log \left (c+d x^2\right )}{2 (b c-a d)^2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {a}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {c \log \left (a+b x^2\right )}{2 (b c-a d)^2}-\frac {c \log \left (c+d x^2\right )}{2 (b c-a d)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {c \ln \left (b \,x^{2}+a \right )-\frac {\left (a d -b c \right ) a}{b \left (b \,x^{2}+a \right )}}{2 \left (a d -b c \right )^{2}}-\frac {c \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right )^{2}}\)	\(69\)
norman	\(\frac {x^{2}}{2 \left (a d -b c \right ) \left (b \,x^{2}+a \right )}+\frac {c \ln \left (b \,x^{2}+a \right )}{2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}-\frac {c \ln \left (d \,x^{2}+c \right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(94\)
risch	\(-\frac {a}{2 \left (a d -b c \right ) b \left (b \,x^{2}+a \right )}+\frac {c \ln \left (b \,x^{2}+a \right )}{2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}-\frac {c \ln \left (-d \,x^{2}-c \right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(98\)
parallelrisch	\(\frac {\ln \left (b \,x^{2}+a \right ) x^{2} b^{2} c -\ln \left (d \,x^{2}+c \right ) x^{2} b^{2} c +\ln \left (b \,x^{2}+a \right ) a b c -\ln \left (d \,x^{2}+c \right ) a b c -a^{2} d +a b c}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (b \,x^{2}+a \right ) b}\)	\(107\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.58 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {a b c - a^{2} d + {\left (b^{2} c x^{2} + a b c\right )} \log \left (b x^{2} + a\right ) - {\left (b^{2} c x^{2} + a b c\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (58) = 116\).

Time = 1.08 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.42 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=- \frac {a}{2 a^{2} b d - 2 a b^{2} c + x^{2} \cdot \left (2 a b^{2} d - 2 b^{3} c\right )} - \frac {c \log {\left (x^{2} + \frac {- \frac {a^{3} c d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 a^{2} b c^{2} d^{2}}{\left (a d - b c\right )^{2}} - \frac {3 a b^{2} c^{3} d}{\left (a d - b c\right )^{2}} + a c d + \frac {b^{3} c^{4}}{\left (a d - b c\right )^{2}} + b c^{2}}{2 b c d} \right )}}{2 \left (a d - b c\right )^{2}} + \frac {c \log {\left (x^{2} + \frac {\frac {a^{3} c d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 a^{2} b c^{2} d^{2}}{\left (a d - b c\right )^{2}} + \frac {3 a b^{2} c^{3} d}{\left (a d - b c\right )^{2}} + a c d - \frac {b^{3} c^{4}}{\left (a d - b c\right )^{2}} + b c^{2}}{2 b c d} \right )}}{2 \left (a d - b c\right )^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.42 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {c \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} - \frac {c \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} + \frac {a}{2 \, {\left (a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.24 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {\frac {b^{2} c \log \left ({\left | \frac {b c}{b x^{2} + a} - \frac {a d}{b x^{2} + a} + d \right |}\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} - \frac {a b}{{\left (b^{2} c - a b d\right )} {\left (b x^{2} + a\right )}}}{2 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.50 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.32 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {a\,\left (b\,c+b\,c\,\mathrm {atan}\left (\frac {a\,d\,x^2\,1{}\mathrm {i}-b\,c\,x^2\,1{}\mathrm {i}}{2\,a\,c+a\,d\,x^2+b\,c\,x^2}\right )\,2{}\mathrm {i}\right )-a^2\,d+b^2\,c\,x^2\,\mathrm {atan}\left (\frac {a\,d\,x^2\,1{}\mathrm {i}-b\,c\,x^2\,1{}\mathrm {i}}{2\,a\,c+a\,d\,x^2+b\,c\,x^2}\right )\,2{}\mathrm {i}}{2\,a^3\,b\,d^2-4\,a^2\,b^2\,c\,d+2\,a^2\,b^2\,d^2\,x^2+2\,a\,b^3\,c^2-4\,a\,b^3\,c\,d\,x^2+2\,b^4\,c^2\,x^2} \]

[In]

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

[Out]

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