\(\int \frac {(a+b x^2)^2}{x (c+d x^2)} \, dx\) [173]

Optimal result

Integrand size = 22, antiderivative size = 51 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )} \, dx=\frac {b^2 x^2}{2 d}+\frac {a^2 \log (x)}{c}-\frac {(b c-a d)^2 \log \left (c+d x^2\right )}{2 c d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, 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, 84} \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )} \, dx=\frac {a^2 \log (x)}{c}-\frac {(b c-a d)^2 \log \left (c+d x^2\right )}{2 c d^2}+\frac {b^2 x^2}{2 d} \]

[In]

Int[(a + b*x^2)^2/(x*(c + d*x^2)),x]

[Out]

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

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )} \, dx=\frac {b^2 c d x^2+2 a^2 d^2 \log (x)-(b c-a d)^2 \log \left (c+d x^2\right )}{2 c d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.62 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {b^{2} x^{2}}{2 d}+\frac {a^{2} \ln \left (x \right )}{c}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (d \,x^{2}+c \right )}{2 c \,d^{2}}\)	\(59\)
norman	\(\frac {b^{2} x^{2}}{2 d}+\frac {a^{2} \ln \left (x \right )}{c}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (d \,x^{2}+c \right )}{2 c \,d^{2}}\)	\(59\)
risch	\(\frac {b^{2} x^{2}}{2 d}+\frac {a^{2} \ln \left (x \right )}{c}-\frac {\ln \left (d \,x^{2}+c \right ) a^{2}}{2 c}+\frac {\ln \left (d \,x^{2}+c \right ) a b}{d}-\frac {c \ln \left (d \,x^{2}+c \right ) b^{2}}{2 d^{2}}\)	\(69\)
parallelrisch	\(\frac {x^{2} b^{2} c d +2 a^{2} \ln \left (x \right ) d^{2}-\ln \left (d \,x^{2}+c \right ) a^{2} d^{2}+2 \ln \left (d \,x^{2}+c \right ) a b c d -\ln \left (d \,x^{2}+c \right ) b^{2} c^{2}}{2 c \,d^{2}}\)	\(75\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )} \, dx=\frac {b^{2} c d x^{2} + 2 \, a^{2} d^{2} \log \left (x\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, c d^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )} \, dx=\frac {a^{2} \log {\left (x \right )}}{c} + \frac {b^{2} x^{2}}{2 d} - \frac {\left (a d - b c\right )^{2} \log {\left (\frac {c}{d} + x^{2} \right )}}{2 c d^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )} \, dx=\frac {b^{2} x^{2}}{2 \, d} + \frac {a^{2} \log \left (x^{2}\right )}{2 \, c} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, c d^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )} \, dx=\frac {b^{2} x^{2}}{2 \, d} + \frac {a^{2} \log \left (x^{2}\right )}{2 \, c} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, c d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )} \, dx=\frac {b^2\,x^2}{2\,d}+\frac {a^2\,\ln \left (x\right )}{c}-\frac {\ln \left (d\,x^2+c\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,c\,d^2} \]

[In]

int((a + b*x^2)^2/(x*(c + d*x^2)),x)

[Out]

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