\(\int \frac {(c+d x^2)^2}{x (a+b x^2)} \, dx\) [214]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.04 (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} \[ \int \frac {\left (c+d x^2\right )^2}{x \left (a+b x^2\right )} \, dx=-\frac {(b c-a d)^2 \log \left (a+b x^2\right )}{2 a b^2}+\frac {c^2 \log (x)}{a}+\frac {d^2 x^2}{2 b} \]

[In]

Int[(c + d*x^2)^2/(x*(a + b*x^2)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98 \[ \int \frac {\left (c+d x^2\right )^2}{x \left (a+b x^2\right )} \, dx=\frac {a b d^2 x^2+2 b^2 c^2 \log (x)-(b c-a d)^2 \log \left (a+b x^2\right )}{2 a b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.65 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int \frac {\left (c+d x^2\right )^2}{x \left (a+b x^2\right )} \, dx=\frac {a b d^{2} x^{2} + 2 \, b^{2} c^{2} \log \left (x\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, a b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int \frac {\left (c+d x^2\right )^2}{x \left (a+b x^2\right )} \, dx=\frac {d^{2} x^{2}}{2 b} + \frac {c^{2} \log {\left (x \right )}}{a} - \frac {\left (a d - b c\right )^{2} \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a b^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \frac {\left (c+d x^2\right )^2}{x \left (a+b x^2\right )} \, dx=\frac {d^2\,x^2}{2\,b}+\frac {c^2\,\ln \left (x\right )}{a}-\frac {\ln \left (b\,x^2+a\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,a\,b^2} \]

[In]

int((c + d*x^2)^2/(x*(a + b*x^2)),x)

[Out]

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