\(\int \frac {d+e x^2}{x (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [84]

Optimal result

Integrand size = 33, antiderivative size = 161 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {d}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b d-a e}{4 a b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {d \left (a+b x^2\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 161, 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} \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {b d-a e}{4 a b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {d}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {d \log (x) \left (a+b x^2\right )}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

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

[Out]

d/(2*a^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (b*d - a*e)/(4*a*b*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) +
(d*(a + b*x^2)*Log[x])/(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (d*(a + b*x^2)*Log[a + b*x^2])/(2*a^3*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 (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {d+e x^2}{x \left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \frac {d+e x}{x \left (a b+b^2 x\right )^3} \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \left (\frac {d}{a^3 b^3 x}+\frac {-b d+a e}{a b^3 (a+b x)^3}-\frac {d}{a^2 b^2 (a+b x)^2}-\frac {d}{a^3 b^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}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b d-a e}{4 a b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {d \left (a+b x^2\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.57 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {a \left (3 a b d-a^2 e+2 b^2 d x^2\right )+4 b d \left (a+b x^2\right )^2 \log (x)-2 b d \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^3 b \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.54

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (2 d b \left (b \,x^{2}+a \right )^{2} \ln \left (b \,x^{2}+a \right )-2 d b \left (b \,x^{2}+a \right )^{2} \ln \left (x^{2}\right )+a \left (-2 b^{2} d \,x^{2}+e \,a^{2}-3 d a b \right )\right )}{4 a^{3} b \left (b \,x^{2}+a \right )^{2}}\)	\(87\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b d \,x^{2}}{2 a^{2}}-\frac {a e -3 b d}{4 a b}\right )}{\left (b \,x^{2}+a \right )^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, d \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{3}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, d \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{3}}\)	\(111\)
default	\(\frac {\left (4 \ln \left (x \right ) b^{3} d \,x^{4}-2 \ln \left (b \,x^{2}+a \right ) b^{3} d \,x^{4}+8 \ln \left (x \right ) a \,b^{2} d \,x^{2}-4 \ln \left (b \,x^{2}+a \right ) a \,b^{2} d \,x^{2}+2 b^{2} d \,x^{2} a +4 \ln \left (x \right ) a^{2} b d -2 \ln \left (b \,x^{2}+a \right ) a^{2} b d -a^{3} e +3 a^{2} b d \right ) \left (b \,x^{2}+a \right )}{4 b \,a^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(133\)

[In]

int((e*x^2+d)/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.74 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {2 \, a b^{2} d x^{2} + 3 \, a^{2} b d - a^{3} e - 2 \, {\left (b^{3} d x^{4} + 2 \, a b^{2} d x^{2} + a^{2} b d\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left (b^{3} d x^{4} + 2 \, a b^{2} d x^{2} + a^{2} b d\right )} \log \left (x\right )}{4 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}} \]

[In]

integrate((e*x^2+d)/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/4*(2*a*b^2*d*x^2 + 3*a^2*b*d - a^3*e - 2*(b^3*d*x^4 + 2*a*b^2*d*x^2 + a^2*b*d)*log(b*x^2 + a) + 4*(b^3*d*x^4
+ 2*a*b^2*d*x^2 + a^2*b*d)*log(x))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b)

Sympy [F]

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

[In]

integrate((e*x**2+d)/x/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.55 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {1}{4} \, d {\left (\frac {2 \, b x^{2} + 3 \, a}{a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}} - \frac {2 \, \log \left (b x^{2} + a\right )}{a^{3}} + \frac {4 \, \log \left (x\right )}{a^{3}}\right )} - \frac {e}{4 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{2} + a^{2} b\right )}} \]

[In]

integrate((e*x^2+d)/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.66 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {d \log \left (x^{2}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {d \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b^{3} d x^{4} + 8 \, a b^{2} d x^{2} + 6 \, a^{2} b d - a^{3} e}{4 \, {\left (b x^{2} + a\right )}^{2} a^{3} b \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate((e*x^2+d)/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/2*d*log(x^2)/(a^3*sgn(b*x^2 + a)) - 1/2*d*log(abs(b*x^2 + a))/(a^3*sgn(b*x^2 + a)) + 1/4*(3*b^3*d*x^4 + 8*a*
b^2*d*x^2 + 6*a^2*b*d - a^3*e)/((b*x^2 + a)^2*a^3*b*sgn(b*x^2 + a))

Mupad [F(-1)]

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

[In]

int((d + e*x^2)/(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)),x)

[Out]

int((d + e*x^2)/(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)), x)