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

Optimal. Leaf size=161 \[ \frac{b d-a e}{4 a b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{d}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{d \log (x) \left (a+b x^2\right )}{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.118266, antiderivative size = 161, 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} \[ \frac{b d-a e}{4 a b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{d}{2 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{d \log (x) \left (a+b x^2\right )}{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0433481, size = 92, normalized size = 0.57 \[ \frac{a \left (a^2 (-e)+3 a b d+2 b^2 d x^2\right )+4 b d \log (x) \left (a+b x^2\right )^2-2 b d \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^3 b \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.021, size = 133, normalized size = 0.8 \begin{align*}{\frac{ \left ( 4\,\ln \left ( x \right ){x}^{4}{b}^{3}d-2\,\ln \left ( b{x}^{2}+a \right ){x}^{4}{b}^{3}d+8\,\ln \left ( x \right ){x}^{2}a{b}^{2}d-4\,\ln \left ( b{x}^{2}+a \right ){x}^{2}a{b}^{2}d+2\,{b}^{2}d{x}^{2}a+4\,\ln \left ( x \right ){a}^{2}bd-2\,\ln \left ( b{x}^{2}+a \right ){a}^{2}bd-{a}^{3}e+3\,{a}^{2}bd \right ) \left ( b{x}^{2}+a \right ) }{4\,b{a}^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/4*(4*ln(x)*x^4*b^3*d-2*ln(b*x^2+a)*x^4*b^3*d+8*ln(x)*x^2*a*b^2*d-4*ln(b*x^2+a)*x^2*a*b^2*d+2*b^2*d*x^2*a+4*l
n(x)*a^2*b*d-2*ln(b*x^2+a)*a^2*b*d-a^3*e+3*a^2*b*d)*(b*x^2+a)/b/a^3/((b*x^2+a)^2)^(3/2)

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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/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56315, size = 250, normalized size = 1.55 \begin{align*} \frac{2 \, a b^{2} d x^{2} + 3 \, a^{2} b d - a^{3} e - 2 \,{\left (b^{3} d x^{4} + 2 \, a b^{2} d x^{2} + a^{2} b d\right )} \log \left (b x^{2} + a\right ) + 4 \,{\left (b^{3} d x^{4} + 2 \, a b^{2} d x^{2} + a^{2} b d\right )} \log \left (x\right )}{4 \,{\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/4*(2*a*b^2*d*x^2 + 3*a^2*b*d - a^3*e - 2*(b^3*d*x^4 + 2*a*b^2*d*x^2 + a^2*b*d)*log(b*x^2 + a) + 4*(b^3*d*x^4
+ 2*a*b^2*d*x^2 + a^2*b*d)*log(x))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x^{2}}{x \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)/x/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x