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

Optimal. Leaf size=101 \[ -\frac {d \left (a+b x^2\right )}{a x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) (b d-a e) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1250, 453, 205} \[ -\frac {\left (a+b x^2\right ) (b d-a e) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{a x \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 453

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 1250

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

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Mathematica [A]  time = 0.03, size = 72, normalized size = 0.71 \[ \frac {\left (a+b x^2\right ) \left (\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (a e x-b d x)-\sqrt {a} \sqrt {b} d\right )}{a^{3/2} \sqrt {b} x \sqrt {\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 105, normalized size = 1.04 \[ \left [\frac {\sqrt {-a b} {\left (b d - a e\right )} x \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 2 \, a b d}{2 \, a^{2} b x}, -\frac {\sqrt {a b} {\left (b d - a e\right )} x \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + a b d}{a^{2} b x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-a*b)*(b*d - a*e)*x*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 2*a*b*d)/(a^2*b*x), -(sqrt(a*b)
*(b*d - a*e)*x*arctan(sqrt(a*b)*x/a) + a*b*d)/(a^2*b*x)]

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giac [A]  time = 0.34, size = 62, normalized size = 0.61 \[ -\frac {{\left (b d \mathrm {sgn}\left (b x^{2} + a\right ) - a e \mathrm {sgn}\left (b x^{2} + a\right )\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {d \mathrm {sgn}\left (b x^{2} + a\right )}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*d*sgn(b*x^2 + a) - a*e*sgn(b*x^2 + a))*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a) - d*sgn(b*x^2 + a)/(a*x)

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maple [A]  time = 0.01, size = 67, normalized size = 0.66 \[ -\frac {\left (b \,x^{2}+a \right ) \left (-a e x \arctan \left (\frac {b x}{\sqrt {a b}}\right )+b d x \arctan \left (\frac {b x}{\sqrt {a b}}\right )+\sqrt {a b}\, d \right )}{\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {a b}\, a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.22, size = 37, normalized size = 0.37 \[ -\frac {{\left (b d - a e\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {d}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*d - a*e)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a) - d/(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {e\,x^2+d}{x^2\,\sqrt {{\left (b\,x^2+a\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.37, size = 82, normalized size = 0.81 \[ - \frac {\sqrt {- \frac {1}{a^{3} b}} \left (a e - b d\right ) \log {\left (- a^{2} \sqrt {- \frac {1}{a^{3} b}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{a^{3} b}} \left (a e - b d\right ) \log {\left (a^{2} \sqrt {- \frac {1}{a^{3} b}} + x \right )}}{2} - \frac {d}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**3*b))*(a*e - b*d)*log(-a**2*sqrt(-1/(a**3*b)) + x)/2 + sqrt(-1/(a**3*b))*(a*e - b*d)*log(a**2*sqr
t(-1/(a**3*b)) + x)/2 - d/(a*x)

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