\(\int \frac {(a+b x^2)^2}{x (c+d x^2)^2} \, dx\) [185]

Optimal result

Integrand size = 22, antiderivative size = 67 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^2} \, dx=\frac {(b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {a^2 \log (x)}{c^2}-\frac {1}{2} \left (\frac {a^2}{c^2}-\frac {b^2}{d^2}\right ) \log \left (c+d x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, 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} \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^2} \, dx=-\frac {1}{2} \left (\frac {a^2}{c^2}-\frac {b^2}{d^2}\right ) \log \left (c+d x^2\right )+\frac {a^2 \log (x)}{c^2}+\frac {(b c-a d)^2}{2 c d^2 \left (c+d x^2\right )} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^2} \, dx=\frac {2 a^2 \log (x)+\frac {(b c-a d) \left (c (b c-a d)+(b c+a d) \left (c+d x^2\right ) \log \left (c+d x^2\right )\right )}{d^2 \left (c+d x^2\right )}}{2 c^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.58 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {a^{2} \ln \left (x \right )}{c^{2}}-\frac {\left (a d -b c \right ) \left (\frac {\left (a d +b c \right ) \ln \left (d \,x^{2}+c \right )}{d^{2}}-\frac {\left (a d -b c \right ) c}{d^{2} \left (d \,x^{2}+c \right )}\right )}{2 c^{2}}\)	\(67\)
norman	\(\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{2 c \,d^{2} \left (d \,x^{2}+c \right )}+\frac {a^{2} \ln \left (x \right )}{c^{2}}-\frac {\left (a^{2} d^{2}-b^{2} c^{2}\right ) \ln \left (d \,x^{2}+c \right )}{2 c^{2} d^{2}}\)	\(81\)
risch	\(\frac {a^{2}}{2 c \left (d \,x^{2}+c \right )}-\frac {a b}{d \left (d \,x^{2}+c \right )}+\frac {c \,b^{2}}{2 d^{2} \left (d \,x^{2}+c \right )}+\frac {a^{2} \ln \left (x \right )}{c^{2}}-\frac {\ln \left (d \,x^{2}+c \right ) a^{2}}{2 c^{2}}+\frac {\ln \left (d \,x^{2}+c \right ) b^{2}}{2 d^{2}}\)	\(94\)
parallelrisch	\(\frac {2 \ln \left (x \right ) x^{2} a^{2} d^{3}-\ln \left (d \,x^{2}+c \right ) x^{2} a^{2} d^{3}+\ln \left (d \,x^{2}+c \right ) x^{2} b^{2} c^{2} d +2 \ln \left (x \right ) a^{2} c \,d^{2}-\ln \left (d \,x^{2}+c \right ) a^{2} c \,d^{2}+\ln \left (d \,x^{2}+c \right ) b^{2} c^{3}+c \,a^{2} d^{2}-2 a b \,c^{2} d +b^{2} c^{3}}{2 c^{2} d^{2} \left (d \,x^{2}+c \right )}\)	\(136\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.73 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^2} \, dx=\frac {b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{3} - a^{2} c d^{2} + {\left (b^{2} c^{2} d - a^{2} d^{3}\right )} x^{2}\right )} \log \left (d x^{2} + c\right ) + 2 \, {\left (a^{2} d^{3} x^{2} + a^{2} c d^{2}\right )} \log \left (x\right )}{2 \, {\left (c^{2} d^{3} x^{2} + c^{3} d^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^2} \, dx=\frac {a^{2} \log {\left (x \right )}}{c^{2}} + \frac {a^{2} d^{2} - 2 a b c d + b^{2} c^{2}}{2 c^{2} d^{2} + 2 c d^{3} x^{2}} - \frac {\left (a d - b c\right ) \left (a d + b c\right ) \log {\left (\frac {c}{d} + x^{2} \right )}}{2 c^{2} d^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^2} \, dx=\frac {a^{2} \log \left (x^{2}\right )}{2 \, c^{2}} + \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{2 \, {\left (c d^{3} x^{2} + c^{2} d^{2}\right )}} + \frac {{\left (b^{2} c^{2} - a^{2} d^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, c^{2} d^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^2} \, dx=\frac {a^{2} \log \left (x^{2}\right )}{2 \, c^{2}} + \frac {{\left (b^{2} c^{2} - a^{2} d^{2}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, c^{2} d^{2}} - \frac {b^{2} c^{2} x^{2} - a^{2} d^{2} x^{2} + 2 \, a b c^{2} - 2 \, a^{2} c d}{2 \, {\left (d x^{2} + c\right )} c^{2} d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^2} \, dx=\frac {a^2\,\ln \left (x\right )}{c^2}+\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{2\,c\,d^2\,\left (d\,x^2+c\right )}-\frac {\ln \left (d\,x^2+c\right )\,\left (a^2\,d^2-b^2\,c^2\right )}{2\,c^2\,d^2} \]

[In]

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

[Out]

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