\(\int \frac {x^5}{(a+b x^2)^2 (c+d x^2)} \, dx\) [241]

Optimal result

Integrand size = 22, antiderivative size = 93 \[ \int \frac {x^5}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {a^2}{2 b^2 (b c-a d) \left (a+b x^2\right )}-\frac {a (2 b c-a d) \log \left (a+b x^2\right )}{2 b^2 (b c-a d)^2}+\frac {c^2 \log \left (c+d x^2\right )}{2 d (b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 93, 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, 90} \[ \int \frac {x^5}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {a^2}{2 b^2 \left (a+b x^2\right ) (b c-a d)}-\frac {a (2 b c-a d) \log \left (a+b x^2\right )}{2 b^2 (b c-a d)^2}+\frac {c^2 \log \left (c+d x^2\right )}{2 d (b c-a d)^2} \]

[In]

Int[x^5/((a + b*x^2)^2*(c + d*x^2)),x]

[Out]

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

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.98 \[ \int \frac {x^5}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {a^2 d (-b c+a d)+a d (-2 b c+a d) \left (a+b x^2\right ) \log \left (a+b x^2\right )+b^2 c^2 \left (a+b x^2\right ) \log \left (c+d x^2\right )}{2 b^2 d (b c-a d)^2 \left (a+b x^2\right )} \]

[In]

Integrate[x^5/((a + b*x^2)^2*(c + d*x^2)),x]

[Out]

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

Maple [A] (verified)

Time = 2.72 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {a \left (\frac {\left (-a d +2 b c \right ) \ln \left (b \,x^{2}+a \right )}{b^{2}}-\frac {\left (a d -b c \right ) a}{b^{2} \left (b \,x^{2}+a \right )}\right )}{2 \left (a d -b c \right )^{2}}+\frac {c^{2} \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right )^{2} d}\)	\(86\)
norman	\(\frac {a^{2}}{2 b^{2} \left (a d -b c \right ) \left (b \,x^{2}+a \right )}+\frac {c^{2} \ln \left (d \,x^{2}+c \right )}{2 d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {a \left (a d -2 b c \right ) \ln \left (b \,x^{2}+a \right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2}}\)	\(113\)
risch	\(\frac {a^{2}}{2 b^{2} \left (a d -b c \right ) \left (b \,x^{2}+a \right )}+\frac {a^{2} \ln \left (b \,x^{2}+a \right ) d}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2}}-\frac {a \ln \left (b \,x^{2}+a \right ) c}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b}+\frac {c^{2} \ln \left (-d \,x^{2}-c \right )}{2 d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(149\)
parallelrisch	\(\frac {\ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b \,d^{2}-2 \ln \left (b \,x^{2}+a \right ) x^{2} a \,b^{2} c d +\ln \left (d \,x^{2}+c \right ) x^{2} b^{3} c^{2}+\ln \left (b \,x^{2}+a \right ) a^{3} d^{2}-2 \ln \left (b \,x^{2}+a \right ) a^{2} b c d +\ln \left (d \,x^{2}+c \right ) a \,b^{2} c^{2}+a^{3} d^{2}-a^{2} b c d}{2 d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (b \,x^{2}+a \right ) b^{2}}\)	\(160\)

[In]

int(x^5/(b*x^2+a)^2/(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.74 \[ \int \frac {x^5}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {a^{2} b c d - a^{3} d^{2} + {\left (2 \, a^{2} b c d - a^{3} d^{2} + {\left (2 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - {\left (b^{3} c^{2} x^{2} + a b^{2} c^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (a b^{4} c^{2} d - 2 \, a^{2} b^{3} c d^{2} + a^{3} b^{2} d^{3} + {\left (b^{5} c^{2} d - 2 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} x^{2}\right )}} \]

[In]

integrate(x^5/(b*x^2+a)^2/(d*x^2+c),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {x^5}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate(x**5/(b*x**2+a)**2/(d*x**2+c),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.40 \[ \int \frac {x^5}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {c^{2} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}} - \frac {a^{2}}{2 \, {\left (a b^{3} c - a^{2} b^{2} d + {\left (b^{4} c - a b^{3} d\right )} x^{2}\right )}} - \frac {{\left (2 \, a b c - a^{2} d\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )}} \]

[In]

integrate(x^5/(b*x^2+a)^2/(d*x^2+c),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.63 \[ \int \frac {x^5}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {c^{2} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}} - \frac {{\left (2 \, a b c - a^{2} d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )}} + \frac {2 \, a b c x^{2} - a^{2} d x^{2} + a^{2} c}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} {\left (b x^{2} + a\right )}} \]

[In]

integrate(x^5/(b*x^2+a)^2/(d*x^2+c),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.50 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.82 \[ \int \frac {x^5}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {a^2}{2\,\left (d\,a^2\,b^2+d\,a\,b^3\,x^2-c\,a\,b^3-c\,b^4\,x^2\right )}+\frac {c^2\,\ln \left (d\,x^2+c\right )}{2\,a^2\,d^3-4\,a\,b\,c\,d^2+2\,b^2\,c^2\,d}+\frac {a^2\,d\,\ln \left (b\,x^2+a\right )}{2\,a^2\,b^2\,d^2-4\,a\,b^3\,c\,d+2\,b^4\,c^2}-\frac {2\,a\,b\,c\,\ln \left (b\,x^2+a\right )}{2\,a^2\,b^2\,d^2-4\,a\,b^3\,c\,d+2\,b^4\,c^2} \]

[In]

int(x^5/((a + b*x^2)^2*(c + d*x^2)),x)

[Out]

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