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

Optimal. Leaf size=74 \[ \frac{1}{2} x^2 \left (e (a e+2 b d)+c d^2\right )+d \log (x) (2 a e+b d)-\frac{a d^2}{2 x^2}+\frac{1}{4} e x^4 (b e+2 c d)+\frac{1}{6} c e^2 x^6 \]

[Out]

-(a*d^2)/(2*x^2) + ((c*d^2 + e*(2*b*d + a*e))*x^2)/2 + (e*(2*c*d + b*e)*x^4)/4 + (c*e^2*x^6)/6 + d*(b*d + 2*a*
e)*Log[x]

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Rubi [A]  time = 0.0957302, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {1251, 893} \[ \frac{1}{2} x^2 \left (e (a e+2 b d)+c d^2\right )+d \log (x) (2 a e+b d)-\frac{a d^2}{2 x^2}+\frac{1}{4} e x^4 (b e+2 c d)+\frac{1}{6} c e^2 x^6 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*d^2)/(2*x^2) + ((c*d^2 + e*(2*b*d + a*e))*x^2)/2 + (e*(2*c*d + b*e)*x^4)/4 + (c*e^2*x^6)/6 + d*(b*d + 2*a*
e)*Log[x]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(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] &&
IntegerQ[(m - 1)/2]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^2 \left (a+b x^2+c x^4\right )}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(d+e x)^2 \left (a+b x+c x^2\right )}{x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (c d^2 \left (1+\frac{e (2 b d+a e)}{c d^2}\right )+\frac{a d^2}{x^2}+\frac{d (b d+2 a e)}{x}+e (2 c d+b e) x+c e^2 x^2\right ) \, dx,x,x^2\right )\\ &=-\frac{a d^2}{2 x^2}+\frac{1}{2} \left (c d^2+e (2 b d+a e)\right ) x^2+\frac{1}{4} e (2 c d+b e) x^4+\frac{1}{6} c e^2 x^6+d (b d+2 a e) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0422363, size = 71, normalized size = 0.96 \[ \frac{1}{12} \left (6 x^2 \left (e (a e+2 b d)+c d^2\right )+12 d \log (x) (2 a e+b d)-\frac{6 a d^2}{x^2}+3 e x^4 (b e+2 c d)+2 c e^2 x^6\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*a*d^2)/x^2 + 6*(c*d^2 + e*(2*b*d + a*e))*x^2 + 3*e*(2*c*d + b*e)*x^4 + 2*c*e^2*x^6 + 12*d*(b*d + 2*a*e)*L
og[x])/12

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Maple [A]  time = 0.007, size = 76, normalized size = 1. \begin{align*}{\frac{c{e}^{2}{x}^{6}}{6}}+{\frac{{x}^{4}b{e}^{2}}{4}}+{\frac{{x}^{4}cde}{2}}+{\frac{{x}^{2}a{e}^{2}}{2}}+{x}^{2}bde+{\frac{{x}^{2}c{d}^{2}}{2}}+2\,\ln \left ( x \right ) ade+\ln \left ( x \right ) b{d}^{2}-{\frac{a{d}^{2}}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c*e^2*x^6+1/4*x^4*b*e^2+1/2*x^4*c*d*e+1/2*x^2*a*e^2+x^2*b*d*e+1/2*x^2*c*d^2+2*ln(x)*a*d*e+ln(x)*b*d^2-1/2*
a*d^2/x^2

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Maxima [A]  time = 0.935074, size = 99, normalized size = 1.34 \begin{align*} \frac{1}{6} \, c e^{2} x^{6} + \frac{1}{4} \,{\left (2 \, c d e + b e^{2}\right )} x^{4} + \frac{1}{2} \,{\left (c d^{2} + 2 \, b d e + a e^{2}\right )} x^{2} + \frac{1}{2} \,{\left (b d^{2} + 2 \, a d e\right )} \log \left (x^{2}\right ) - \frac{a d^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c*e^2*x^6 + 1/4*(2*c*d*e + b*e^2)*x^4 + 1/2*(c*d^2 + 2*b*d*e + a*e^2)*x^2 + 1/2*(b*d^2 + 2*a*d*e)*log(x^2)
- 1/2*a*d^2/x^2

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Fricas [A]  time = 1.70462, size = 173, normalized size = 2.34 \begin{align*} \frac{2 \, c e^{2} x^{8} + 3 \,{\left (2 \, c d e + b e^{2}\right )} x^{6} + 6 \,{\left (c d^{2} + 2 \, b d e + a e^{2}\right )} x^{4} + 12 \,{\left (b d^{2} + 2 \, a d e\right )} x^{2} \log \left (x\right ) - 6 \, a d^{2}}{12 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*c*e^2*x^8 + 3*(2*c*d*e + b*e^2)*x^6 + 6*(c*d^2 + 2*b*d*e + a*e^2)*x^4 + 12*(b*d^2 + 2*a*d*e)*x^2*log(x
) - 6*a*d^2)/x^2

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Sympy [A]  time = 0.410208, size = 71, normalized size = 0.96 \begin{align*} - \frac{a d^{2}}{2 x^{2}} + \frac{c e^{2} x^{6}}{6} + d \left (2 a e + b d\right ) \log{\left (x \right )} + x^{4} \left (\frac{b e^{2}}{4} + \frac{c d e}{2}\right ) + x^{2} \left (\frac{a e^{2}}{2} + b d e + \frac{c d^{2}}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d**2/(2*x**2) + c*e**2*x**6/6 + d*(2*a*e + b*d)*log(x) + x**4*(b*e**2/4 + c*d*e/2) + x**2*(a*e**2/2 + b*d*e
+ c*d**2/2)

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Giac [A]  time = 1.08428, size = 131, normalized size = 1.77 \begin{align*} \frac{1}{6} \, c x^{6} e^{2} + \frac{1}{2} \, c d x^{4} e + \frac{1}{4} \, b x^{4} e^{2} + \frac{1}{2} \, c d^{2} x^{2} + b d x^{2} e + \frac{1}{2} \, a x^{2} e^{2} + \frac{1}{2} \,{\left (b d^{2} + 2 \, a d e\right )} \log \left (x^{2}\right ) - \frac{b d^{2} x^{2} + 2 \, a d x^{2} e + a d^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c*x^6*e^2 + 1/2*c*d*x^4*e + 1/4*b*x^4*e^2 + 1/2*c*d^2*x^2 + b*d*x^2*e + 1/2*a*x^2*e^2 + 1/2*(b*d^2 + 2*a*d
*e)*log(x^2) - 1/2*(b*d^2*x^2 + 2*a*d*x^2*e + a*d^2)/x^2