\(\int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [78]

Optimal result

Integrand size = 33, antiderivative size = 92 \[ \int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {d \left (a+b x^2\right ) \log (x)}{a \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a b \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1264, 457, 78} \[ \int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {d \log (x) \left (a+b x^2\right )}{a \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 a b \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[(d + e*x^2)/(x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

(d*(a + b*x^2)*Log[x])/(a*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((b*d - a*e)*(a + b*x^2)*Log[a + b*x^2])/(2*a*b*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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]]

Rule 1264

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dis
t[(a + b*x^2 + c*x^4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(f*x)^m*(d + e*x^2)^q*(b/2
+ c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a b+b^2 x^2\right ) \int \frac {d+e x^2}{x \left (a b+b^2 x^2\right )} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (a b+b^2 x^2\right ) \text {Subst}\left (\int \frac {d+e x}{x \left (a b+b^2 x\right )} \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (a b+b^2 x^2\right ) \text {Subst}\left (\int \left (\frac {d}{a b x}+\frac {-b d+a e}{a b (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {d \left (a+b x^2\right ) \log (x)}{a \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a b \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.36 \[ \int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {1}{4} \left (\left (\frac {2 d}{a}-\frac {4 e}{b}\right ) \text {arctanh}\left (\frac {\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}{b x^2}\right )+\frac {d \left (-2 \log \left (x^2\right )+\log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+\log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{\sqrt {a^2}}\right ) \]

[In]

Integrate[(d + e*x^2)/(x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

(((2*d)/a - (4*e)/b)*ArcTanh[(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])/(b*x^2)] + (d*(-2*Log[x^2] + Log[Sqrt[a^2] - b*
x^2 - Sqrt[(a + b*x^2)^2]] + Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]]))/Sqrt[a^2])/4

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.34 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.52

method	result	size
pseudoelliptic	\(\frac {\left (d \ln \left (x^{2}\right ) b +\ln \left (b \,x^{2}+a \right ) a e -\ln \left (b \,x^{2}+a \right ) b d \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 a b}\)	\(48\)
default	\(\frac {\left (b \,x^{2}+a \right ) \left (2 d \ln \left (x \right ) b +\ln \left (b \,x^{2}+a \right ) a e -\ln \left (b \,x^{2}+a \right ) b d \right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a b}\)	\(57\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, d \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (a e -b d \right ) \ln \left (-b \,x^{2}-a \right )}{2 \left (b \,x^{2}+a \right ) a b}\)	\(76\)

[In]

int((e*x^2+d)/x/((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.36 \[ \int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {2 \, b d \log \left (x\right ) - {\left (b d - a e\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \]

[In]

integrate((e*x^2+d)/x/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {d + e x^{2}}{x \sqrt {\left (a + b x^{2}\right )^{2}}}\, dx \]

[In]

integrate((e*x**2+d)/x/((b*x**2+a)**2)**(1/2),x)

[Out]

Integral((d + e*x**2)/(x*sqrt((a + b*x**2)**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.38 \[ \int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {d \log \left (x^{2}\right )}{2 \, a} - \frac {{\left (b d - a e\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \]

[In]

integrate((e*x^2+d)/x/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65 \[ \int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {d \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, a} - \frac {{\left (b d \mathrm {sgn}\left (b x^{2} + a\right ) - a e \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a b} \]

[In]

integrate((e*x^2+d)/x/((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \frac {d+e x^2}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {e\,\ln \left (b^2\,x^2+a\,b\right )\,\mathrm {sign}\left (2\,b^2\,x^2+2\,a\,b\right )}{2\,\sqrt {b^2}}-\frac {d\,\ln \left (\frac {1}{x^2}\right )}{2\,\sqrt {a^2}}-\frac {d\,\ln \left (\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {a^2}+a^2+a\,b\,x^2\right )}{2\,\sqrt {a^2}} \]

[In]

int((d + e*x^2)/(x*((a + b*x^2)^2)^(1/2)),x)

[Out]

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