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

Optimal. Leaf size=92 \[ \frac{d \log (x) \left (a+b x^2\right )}{a \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 a b \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

(d*(a + b*x^2)*Log[x])/(a*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*b*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.0717011, antiderivative size = 92, 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, 72} \[ \frac{d \log (x) \left (a+b x^2\right )}{a \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 a b \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(a + b*x^2)*Log[x])/(a*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*b*S
qrt[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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{d+e x^2}{x \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 \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 \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}+\frac{-b d+a e}{a b (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 ) \log (x)}{a \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 b \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0209118, size = 54, normalized size = 0.59 \[ \frac{\left (a+b x^2\right ) \left ((a e-b d) \log \left (a+b x^2\right )+2 b d \log (x)\right )}{2 a b \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 57, normalized size = 0.6 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \left ( 2\,d\ln \left ( x \right ) b+\ln \left ( b{x}^{2}+a \right ) ae-\ln \left ( b{x}^{2}+a \right ) bd \right ) }{2\,ab}{\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/((b*x^2+a)^2)^(1/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.49825, size = 74, normalized size = 0.8 \begin{align*} \frac{2 \, b d \log \left (x\right ) -{\left (b d - a e\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.731905, size = 26, normalized size = 0.28 \begin{align*} \frac{d \log{\left (x \right )}}{a} + \frac{\left (a e - b d\right ) \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.11569, size = 82, normalized size = 0.89 \begin{align*} \frac{d \log \left (x^{2}\right ) \mathrm{sgn}\left (b x^{2} + a\right )}{2 \, a} - \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 ({\left | b x^{2} + a \right |}\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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