3.80 \(\int \frac{d+e x^2}{x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=137 \[ -\frac{\log (x) \left (a+b x^2\right ) (b d-a e)}{a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d \left (a+b x^2\right )}{2 a x^2 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0981535, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1250, 446, 77} \[ -\frac{\log (x) \left (a+b x^2\right ) (b d-a e)}{a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d \left (a+b x^2\right )}{2 a x^2 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(a + b*x^2))/(2*a*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((b*d - a*e)*(a + b*x^2)*Log[x])/(a^2*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4]) + ((b*d - a*e)*(a + b*x^2)*Log[a + b*x^2])/(2*a^2*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 446

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 77

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]))))

Rubi steps

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

Mathematica [A]  time = 0.0327839, size = 70, normalized size = 0.51 \[ \frac{\left (a+b x^2\right ) \left (2 x^2 \log (x) (a e-b d)+x^2 (b d-a e) \log \left (a+b x^2\right )-a d\right )}{2 a^2 x^2 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 79, normalized size = 0.6 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \left ( 2\,\ln \left ( x \right ){x}^{2}ae-2\,\ln \left ( x \right ){x}^{2}bd-\ln \left ( b{x}^{2}+a \right ){x}^{2}ae+\ln \left ( b{x}^{2}+a \right ){x}^{2}bd-ad \right ) }{2\,{a}^{2}{x}^{2}}{\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)/x^3/((b*x^2+a)^2)^(1/2),x)

[Out]

1/2*(b*x^2+a)*(2*ln(x)*x^2*a*e-2*ln(x)*x^2*b*d-ln(b*x^2+a)*x^2*a*e+ln(b*x^2+a)*x^2*b*d-a*d)/((b*x^2+a)^2)^(1/2
)/a^2/x^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^3/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53572, size = 109, normalized size = 0.8 \begin{align*} \frac{{\left (b d - a e\right )} x^{2} \log \left (b x^{2} + a\right ) - 2 \,{\left (b d - a e\right )} x^{2} \log \left (x\right ) - a d}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/x^3/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.878829, size = 41, normalized size = 0.3 \begin{align*} - \frac{d}{2 a x^{2}} + \frac{\left (a e - b d\right ) \log{\left (x \right )}}{a^{2}} - \frac{\left (a e - b d\right ) \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)/x**3/((b*x**2+a)**2)**(1/2),x)

[Out]

-d/(2*a*x**2) + (a*e - b*d)*log(x)/a**2 - (a*e - b*d)*log(a/b + x**2)/(2*a**2)

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Giac [A]  time = 1.1387, size = 177, normalized size = 1.29 \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 )} \log \left (x^{2}\right )}{2 \, a^{2}} + \frac{{\left (b^{2} d \mathrm{sgn}\left (b x^{2} + a\right ) - a b e \mathrm{sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2} b} + \frac{b d x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) - a x^{2} e \mathrm{sgn}\left (b x^{2} + a\right ) - a d \mathrm{sgn}\left (b x^{2} + a\right )}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/x^3/((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

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