\(\int \frac {d+e x^2}{x^2 (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [85]

Optimal result

Integrand size = 33, antiderivative size = 190 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {(7 b d-3 a e) x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 (5 b d-a e) \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

-1/8*(-3*a*e+7*b*d)*x/a^3/((b*x^2+a)^2)^(1/2)-1/4*(-a*e+b*d)*x/a^2/(b*x^2+a)/((b*x^2+a)^2)^(1/2)-d*(b*x^2+a)/a
^3/x/((b*x^2+a)^2)^(1/2)-3/8*(-a*e+5*b*d)*(b*x^2+a)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)/b^(1/2)/((b*x^2+a)^2)^(1
/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1264, 467, 464, 211} \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {x (b d-a e)}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (5 b d-a e)}{8 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x (7 b d-3 a e)}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

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

[Out]

-1/8*((7*b*d - 3*a*e)*x)/(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((b*d - a*e)*x)/(4*a^2*(a + b*x^2)*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4]) - (d*(a + b*x^2))/(a^3*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (3*(5*b*d - a*e)*(a + b*x^2
)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 467

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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^2 \left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \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 ) \int \frac {-\frac {4 d}{a b}+\frac {3 (b d-a e) x^2}{a^2 b}}{x^2 \left (a b+b^2 x^2\right )^2} \, dx}{4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {(7 b d-3 a e) x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \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 ) \int \frac {\frac {8 d}{a^2 b^2}-\frac {(7 b d-3 a e) x^2}{a^3 b^2}}{x^2 \left (a b+b^2 x^2\right )} \, dx}{8 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {(7 b d-3 a e) x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{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 )}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (3 (5 b d-a e) \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {(7 b d-3 a e) x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{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 )}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 (5 b d-a e) \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.65 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {a} \sqrt {b} \left (-15 b^2 d x^4+a^2 \left (-8 d+5 e x^2\right )+a b \left (-25 d x^2+3 e x^4\right )\right )+3 (-5 b d+a e) x \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b} x \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]

[In]

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

[Out]

(Sqrt[a]*Sqrt[b]*(-15*b^2*d*x^4 + a^2*(-8*d + 5*e*x^2) + a*b*(-25*d*x^2 + 3*e*x^4)) + 3*(-5*b*d + a*e)*x*(a +
b*x^2)^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b]*x*(a + b*x^2)*Sqrt[(a + b*x^2)^2])

Maple [C] (verified)

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

Time = 1.22 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {3 b \left (a e -5 b d \right ) x^{4}}{8 a^{3}}+\frac {5 \left (a e -5 b d \right ) x^{2}}{8 a^{2}}-\frac {d}{a}\right )}{\left (b \,x^{2}+a \right )^{3} x}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{2} b +e^{2} a^{2}-10 a b d e +25 b^{2} d^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{7} b +2 e^{2} a^{2}-20 a b d e +50 b^{2} d^{2}\right ) x +\left (-e \,a^{5}+5 a^{4} b d \right ) \textit {\_R} \right )\right )}{16 \left (b \,x^{2}+a \right )}\)	\(176\)
default	\(-\frac {\left (-3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{2} e \,x^{5}+15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{3} d \,x^{5}-3 \sqrt {a b}\, a b e \,x^{4}+15 \sqrt {a b}\, b^{2} d \,x^{4}-6 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b e \,x^{3}+30 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{2} d \,x^{3}-5 \sqrt {a b}\, a^{2} e \,x^{2}+25 \sqrt {a b}\, a b d \,x^{2}-3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{3} e x +15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b d x +8 \sqrt {a b}\, a^{2} d \right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, x \,a^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(206\)

[In]

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

[Out]

((b*x^2+a)^2)^(1/2)/(b*x^2+a)^3*(3/8*b*(a*e-5*b*d)/a^3*x^4+5/8/a^2*(a*e-5*b*d)*x^2-d/a)/x+3/16*((b*x^2+a)^2)^(
1/2)/(b*x^2+a)*sum(_R*ln((3*_R^2*a^7*b+2*a^2*e^2-20*a*b*d*e+50*b^2*d^2)*x+(-a^5*e+5*a^4*b*d)*_R),_R=RootOf(_Z^
2*a^7*b+a^2*e^2-10*a*b*d*e+25*b^2*d^2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.76 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [-\frac {16 \, a^{3} b d + 6 \, {\left (5 \, a b^{3} d - a^{2} b^{2} e\right )} x^{4} + 10 \, {\left (5 \, a^{2} b^{2} d - a^{3} b e\right )} x^{2} - 3 \, {\left ({\left (5 \, b^{3} d - a b^{2} e\right )} x^{5} + 2 \, {\left (5 \, a b^{2} d - a^{2} b e\right )} x^{3} + {\left (5 \, a^{2} b d - a^{3} e\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, -\frac {8 \, a^{3} b d + 3 \, {\left (5 \, a b^{3} d - a^{2} b^{2} e\right )} x^{4} + 5 \, {\left (5 \, a^{2} b^{2} d - a^{3} b e\right )} x^{2} + 3 \, {\left ({\left (5 \, b^{3} d - a b^{2} e\right )} x^{5} + 2 \, {\left (5 \, a b^{2} d - a^{2} b e\right )} x^{3} + {\left (5 \, a^{2} b d - a^{3} e\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \]

[In]

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

[Out]

[-1/16*(16*a^3*b*d + 6*(5*a*b^3*d - a^2*b^2*e)*x^4 + 10*(5*a^2*b^2*d - a^3*b*e)*x^2 - 3*((5*b^3*d - a*b^2*e)*x
^5 + 2*(5*a*b^2*d - a^2*b*e)*x^3 + (5*a^2*b*d - a^3*e)*x)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 +
a)))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x), -1/8*(8*a^3*b*d + 3*(5*a*b^3*d - a^2*b^2*e)*x^4 + 5*(5*a^2*b^2*
d - a^3*b*e)*x^2 + 3*((5*b^3*d - a*b^2*e)*x^5 + 2*(5*a*b^2*d - a^2*b*e)*x^3 + (5*a^2*b*d - a^3*e)*x)*sqrt(a*b)
*arctan(sqrt(a*b)*x/a))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.71 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {1}{8} \, d {\left (\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 8 \, a^{2}}{a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x} + \frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}}\right )} + \frac {1}{8} \, e {\left (\frac {3 \, b x^{3} + 5 \, a x}{a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}} + \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}}\right )} \]

[In]

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

[Out]

-1/8*d*((15*b^2*x^4 + 25*a*b*x^2 + 8*a^2)/(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x) + 15*b*arctan(b*x/sqrt(a*b))/(sq
rt(a*b)*a^3)) + 1/8*e*((3*b*x^3 + 5*a*x)/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4) + 3*arctan(b*x/sqrt(a*b))/(sqrt(a*b
)*a^2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.59 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (5 \, b d - a e\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {d}{a^{3} x \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {7 \, b^{2} d x^{3} - 3 \, a b e x^{3} + 9 \, a b d x - 5 \, a^{2} e x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

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

[Out]

-3/8*(5*b*d - a*e)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*sgn(b*x^2 + a)) - d/(a^3*x*sgn(b*x^2 + a)) - 1/8*(7*b^
2*d*x^3 - 3*a*b*e*x^3 + 9*a*b*d*x - 5*a^2*e*x)/((b*x^2 + a)^2*a^3*sgn(b*x^2 + a))

Mupad [F(-1)]

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

[In]

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

[Out]

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