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

[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])

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Rubi [A]  time = 0.0629733, antiderivative size = 101, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0313869, 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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Maple [A]  time = 0.01, size = 67, normalized size = 0.7 \begin{align*} -{\frac{b{x}^{2}+a}{ax} \left ( -\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ) xae+\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ) xbd+d\sqrt{ab} \right ){\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}{\frac{1}{\sqrt{ab}}}} \end{align*}

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(b*x/(a*b)^(1/2))*x*a*e+arctan(b*x/(a*b)^(1/2))*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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.57388, size = 230, normalized size = 2.28 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 0.514622, size = 82, normalized size = 0.81 \begin{align*} - \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} \end{align*}

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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Giac [A]  time = 1.11881, size = 84, normalized size = 0.83 \begin{align*} -\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} \end{align*}

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)