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

Optimal. Leaf size=83 \[ \frac{\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{e \sqrt{a^2+2 a b x^2+b^2 x^4}}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.072228, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1247, 640, 608, 31} \[ \frac{\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{e \sqrt{a^2+2 a b x^2+b^2 x^4}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0212931, size = 51, normalized size = 0.61 \[ \frac{\left (a+b x^2\right ) \left ((b d-a e) \log \left (a+b x^2\right )+b e x^2\right )}{2 b^2 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 55, normalized size = 0.7 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( -{x}^{2}eb+\ln \left ( b{x}^{2}+a \right ) ae-\ln \left ( b{x}^{2}+a \right ) bd \right ) }{2\,{b}^{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(x*(e*x^2+d)/((b*x^2+a)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 0.99303, size = 92, normalized size = 1.11 \begin{align*} \frac{1}{2} \, \sqrt{\frac{1}{b^{2}}} d \log \left (x^{2} + \frac{a}{b}\right ) - \frac{1}{2} \,{\left (\frac{a \sqrt{\frac{1}{b^{2}}} \log \left (x^{2} + \frac{a}{b}\right )}{b} - \frac{\sqrt{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}{b^{2}}\right )} e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b^(-2))*d*log(x^2 + a/b) - 1/2*(a*sqrt(b^(-2))*log(x^2 + a/b)/b - sqrt(b^2*x^4 + 2*a*b*x^2 + a^2)/b^2
)*e

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Fricas [A]  time = 1.52182, size = 65, normalized size = 0.78 \begin{align*} \frac{b e x^{2} +{\left (b d - a e\right )} \log \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.432277, size = 27, normalized size = 0.33 \begin{align*} \frac{e x^{2}}{2 b} - \frac{\left (a e - b d\right ) \log{\left (a + b x^{2} \right )}}{2 b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13645, size = 57, normalized size = 0.69 \begin{align*} \frac{1}{2} \,{\left (\frac{x^{2} e}{b} + \frac{{\left (b d - a e\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{b^{2}}\right )} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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