\(\int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [80]

Optimal result

Integrand size = 33, antiderivative size = 137 \[ \int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {d \left (a+b x^2\right )}{2 a x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) \left (a+b x^2\right ) \log (x)}{a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(b d-a e) \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 137, 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} \[ \int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\log (x) \left (a+b x^2\right ) (b d-a e)}{a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{2 a x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[(d + e*x^2)/(x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

-1/2*(d*(a + b*x^2))/(a*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((b*d - a*e)*(a + b*x^2)*Log[x])/(a^2*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^2*Sqrt[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^3 \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^2 \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^2}+\frac {-b d+a e}{a^2 b x}+\frac {b d-a e}{a^2 (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 )}{2 a x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) \left (a+b x^2\right ) \log (x)}{a^2 \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^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.20 \[ \int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {a^2 d-\sqrt {a^2} d \sqrt {\left (a+b x^2\right )^2}+2 a (b d-a e) x^2 \log \left (x^2\right )-\left (-a+\sqrt {a^2}\right ) (-b d+a e) x^2 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+\left (a+\sqrt {a^2}\right ) (-b d+a e) x^2 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )}{4 \left (a^2\right )^{3/2} x^2} \]

[In]

Integrate[(d + e*x^2)/(x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

(a^2*d - Sqrt[a^2]*d*Sqrt[(a + b*x^2)^2] + 2*a*(b*d - a*e)*x^2*Log[x^2] - (-a + Sqrt[a^2])*(-(b*d) + a*e)*x^2*
Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + (a + Sqrt[a^2])*(-(b*d) + a*e)*x^2*Log[Sqrt[a^2] + b*x^2 - Sqrt
[(a + b*x^2)^2]])/(4*(a^2)^(3/2)*x^2)

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 = 58, normalized size of antiderivative = 0.42

method	result	size
pseudoelliptic	\(-\frac {\left (x^{2} \left (a e -b d \right ) \ln \left (b \,x^{2}+a \right )-x^{2} \left (a e -b d \right ) \ln \left (x^{2}\right )+d a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 a^{2} x^{2}}\)	\(58\)
default	\(\frac {\left (b \,x^{2}+a \right ) \left (2 \ln \left (x \right ) a e \,x^{2}-2 \ln \left (x \right ) b d \,x^{2}-\ln \left (b \,x^{2}+a \right ) a e \,x^{2}+\ln \left (b \,x^{2}+a \right ) b d \,x^{2}-d a \right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} x^{2}}\)	\(79\)
risch	\(-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, d}{2 \left (b \,x^{2}+a \right ) a \,x^{2}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (a e -b d \right ) \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{2}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (a e -b d \right ) \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{2}}\)	\(106\)

[In]

int((e*x^2+d)/x^3/((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.35 \[ \int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {{\left (b d - a e\right )} x^{2} \log \left (b x^{2} + a\right ) - 2 \, {\left (b d - a e\right )} x^{2} \log \left (x\right ) - a d}{2 \, a^{2} x^{2}} \]

[In]

integrate((e*x^2+d)/x^3/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {d + e x^{2}}{x^{3} \sqrt {\left (a + b x^{2}\right )^{2}}}\, dx \]

[In]

integrate((e*x**2+d)/x**3/((b*x**2+a)**2)**(1/2),x)

[Out]

Integral((d + e*x**2)/(x**3*sqrt((a + b*x**2)**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.35 \[ \int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {{\left (b d - a e\right )} \log \left (b x^{2} + a\right )}{2 \, a^{2}} - \frac {{\left (b d - a e\right )} \log \left (x^{2}\right )}{2 \, a^{2}} - \frac {d}{2 \, a x^{2}} \]

[In]

integrate((e*x^2+d)/x^3/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.93 \[ \int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {{\left (b d \mathrm {sgn}\left (b x^{2} + a\right ) - a e \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {{\left (b^{2} d \mathrm {sgn}\left (b x^{2} + a\right ) - a b e \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2} b} + \frac {b d x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) - a e x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) - a d \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, a^{2} x^{2}} \]

[In]

integrate((e*x^2+d)/x^3/((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.32 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {d+e x^2}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {a\,b\,d\,\mathrm {atanh}\left (\frac {a^2+b\,a\,x^2}{\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}\right )}{2\,{\left (a^2\right )}^{3/2}}-\frac {e\,\ln \left (\frac {1}{x^2}\right )}{2\,\sqrt {a^2}}-\frac {d\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,a^2\,x^2}-\frac {e\,\ln \left (\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {a^2}+a^2+a\,b\,x^2\right )}{2\,\sqrt {a^2}} \]

[In]

int((d + e*x^2)/(x^3*((a + b*x^2)^2)^(1/2)),x)

[Out]

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