\(\int \frac {d+e x^2}{x^3 (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [86]

Optimal result

Integrand size = 33, antiderivative size = 223 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 b d-a e}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b d-a e}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{2 a^3 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(3 b d-a e) \left (a+b x^2\right ) \log (x)}{a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(3 b d-a e) \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 223, 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} \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {b d-a e}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\log (x) \left (a+b x^2\right ) (3 b d-a e)}{a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) (3 b d-a e) \log \left (a+b x^2\right )}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {2 b d-a e}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{2 a^3 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.58 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {a \left (-6 b^2 d x^4+a^2 \left (-2 d+3 e x^2\right )+a b \left (-9 d x^2+2 e x^4\right )\right )+4 (-3 b d+a e) x^2 \left (a+b x^2\right )^2 \log (x)+2 (3 b d-a e) x^2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^4 x^2 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]

[In]

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

[Out]

(a*(-6*b^2*d*x^4 + a^2*(-2*d + 3*e*x^2) + a*b*(-9*d*x^2 + 2*e*x^4)) + 4*(-3*b*d + a*e)*x^2*(a + b*x^2)^2*Log[x
] + 2*(3*b*d - a*e)*x^2*(a + b*x^2)^2*Log[a + b*x^2])/(4*a^4*x^2*(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 = 0.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.54

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (x^{2} \left (b \,x^{2}+a \right )^{2} \left (a e -3 b d \right ) \ln \left (b \,x^{2}+a \right )-x^{2} \left (b \,x^{2}+a \right )^{2} \left (a e -3 b d \right ) \ln \left (x^{2}\right )+a \left (\left (-a b e +3 b^{2} d \right ) x^{4}-\frac {3 a \left (a e -3 b d \right ) x^{2}}{2}+d \,a^{2}\right )\right )}{2 \left (b \,x^{2}+a \right )^{2} x^{2} a^{4}}\)	\(120\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b \left (a e -3 b d \right ) x^{4}}{2 a^{3}}+\frac {3 \left (a e -3 b d \right ) x^{2}}{4 a^{2}}-\frac {d}{2 a}\right )}{\left (b \,x^{2}+a \right )^{3} x^{2}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (a e -3 b d \right ) \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{4}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (a e -3 b d \right ) \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{4}}\)	\(141\)
default	\(\frac {\left (4 \ln \left (x \right ) a \,b^{2} e \,x^{6}-12 \ln \left (x \right ) b^{3} d \,x^{6}-2 \ln \left (b \,x^{2}+a \right ) a \,b^{2} e \,x^{6}+6 \ln \left (b \,x^{2}+a \right ) b^{3} d \,x^{6}+8 \ln \left (x \right ) a^{2} b e \,x^{4}-24 \ln \left (x \right ) a \,b^{2} d \,x^{4}-4 \ln \left (b \,x^{2}+a \right ) a^{2} b e \,x^{4}+12 \ln \left (b \,x^{2}+a \right ) a \,b^{2} d \,x^{4}+2 a^{2} b e \,x^{4}-6 a \,b^{2} d \,x^{4}+4 \ln \left (x \right ) a^{3} e \,x^{2}-12 \ln \left (x \right ) a^{2} b d \,x^{2}-2 \ln \left (b \,x^{2}+a \right ) a^{3} e \,x^{2}+6 \ln \left (b \,x^{2}+a \right ) a^{2} b d \,x^{2}+3 a^{3} e \,x^{2}-9 a^{2} b d \,x^{2}-2 a^{3} d \right ) \left (b \,x^{2}+a \right )}{4 x^{2} a^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(249\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.92 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 \, {\left (3 \, a b^{2} d - a^{2} b e\right )} x^{4} + 2 \, a^{3} d + 3 \, {\left (3 \, a^{2} b d - a^{3} e\right )} x^{2} - 2 \, {\left ({\left (3 \, b^{3} d - a b^{2} e\right )} x^{6} + 2 \, {\left (3 \, a b^{2} d - a^{2} b e\right )} x^{4} + {\left (3 \, a^{2} b d - a^{3} e\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left ({\left (3 \, b^{3} d - a b^{2} e\right )} x^{6} + 2 \, {\left (3 \, a b^{2} d - a^{2} b e\right )} x^{4} + {\left (3 \, a^{2} b d - a^{3} e\right )} x^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}} \]

[In]

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

[Out]

-1/4*(2*(3*a*b^2*d - a^2*b*e)*x^4 + 2*a^3*d + 3*(3*a^2*b*d - a^3*e)*x^2 - 2*((3*b^3*d - a*b^2*e)*x^6 + 2*(3*a*
b^2*d - a^2*b*e)*x^4 + (3*a^2*b*d - a^3*e)*x^2)*log(b*x^2 + a) + 4*((3*b^3*d - a*b^2*e)*x^6 + 2*(3*a*b^2*d - a
^2*b*e)*x^4 + (3*a^2*b*d - a^3*e)*x^2)*log(x))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.62 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {1}{4} \, d {\left (\frac {6 \, b^{2} x^{4} + 9 \, a b x^{2} + 2 \, a^{2}}{a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}} - \frac {6 \, b \log \left (b x^{2} + a\right )}{a^{4}} + \frac {12 \, b \log \left (x\right )}{a^{4}}\right )} + \frac {1}{4} \, e {\left (\frac {2 \, b x^{2} + 3 \, a}{a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}} - \frac {2 \, \log \left (b x^{2} + a\right )}{a^{3}} + \frac {4 \, \log \left (x\right )}{a^{3}}\right )} \]

[In]

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

[Out]

-1/4*d*((6*b^2*x^4 + 9*a*b*x^2 + 2*a^2)/(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2) - 6*b*log(b*x^2 + a)/a^4 + 12*b*
log(x)/a^4) + 1/4*e*((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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.82 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {{\left (3 \, b d - a e\right )} \log \left (x^{2}\right )}{2 \, a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {{\left (3 \, b^{2} d - a b e\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4} b \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {9 \, b^{3} d x^{4} - 3 \, a b^{2} e x^{4} + 22 \, a b^{2} d x^{2} - 8 \, a^{2} b e x^{2} + 14 \, a^{2} b d - 6 \, a^{3} e}{4 \, {\left (b x^{2} + a\right )}^{2} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b d x^{2} - a e x^{2} - a d}{2 \, a^{4} x^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

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

[Out]

-1/2*(3*b*d - a*e)*log(x^2)/(a^4*sgn(b*x^2 + a)) + 1/2*(3*b^2*d - a*b*e)*log(abs(b*x^2 + a))/(a^4*b*sgn(b*x^2
+ a)) - 1/4*(9*b^3*d*x^4 - 3*a*b^2*e*x^4 + 22*a*b^2*d*x^2 - 8*a^2*b*e*x^2 + 14*a^2*b*d - 6*a^3*e)/((b*x^2 + a)
^2*a^4*sgn(b*x^2 + a)) + 1/2*(3*b*d*x^2 - a*e*x^2 - a*d)/(a^4*x^2*sgn(b*x^2 + a))

Mupad [F(-1)]

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

[In]

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

[Out]

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