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

Optimal. Leaf size=132 \[ \frac{(b d-2 a e) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 \sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac{d \log \left (d+e x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}+\frac{d \log \left (a+b x^2+c x^4\right )}{4 \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.157051, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1251, 800, 634, 618, 206, 628} \[ \frac{(b d-2 a e) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 \sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac{d \log \left (d+e x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}+\frac{d \log \left (a+b x^2+c x^4\right )}{4 \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

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

Rubi steps

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

Mathematica [A]  time = 0.07163, size = 114, normalized size = 0.86 \[ \frac{d \sqrt{4 a c-b^2} \left (2 \log \left (d+e x^2\right )-\log \left (a+b x^2+c x^4\right )\right )+2 (b d-2 a e) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{4 \sqrt{4 a c-b^2} \left (e (b d-a e)-c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 176, normalized size = 1.3 \begin{align*}{\frac{d\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) }{4\,a{e}^{2}-4\,deb+4\,c{d}^{2}}}+{\frac{ae}{a{e}^{2}-deb+c{d}^{2}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{bd}{2\,a{e}^{2}-2\,deb+2\,c{d}^{2}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{d\ln \left ( e{x}^{2}+d \right ) }{2\,a{e}^{2}-2\,deb+2\,c{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d*ln(c*x^4+b*x^2+a)/(a*e^2-b*d*e+c*d^2)+1/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-
b^2)^(1/2))*a*e-1/2/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b*d-1/2*d*ln(e
*x^2+d)/(a*e^2-b*d*e+c*d^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(x^3/(e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 112.082, size = 716, normalized size = 5.42 \begin{align*} \left [\frac{{\left (b^{2} - 4 \, a c\right )} d \log \left (c x^{4} + b x^{2} + a\right ) - 2 \,{\left (b^{2} - 4 \, a c\right )} d \log \left (e x^{2} + d\right ) - \sqrt{b^{2} - 4 \, a c}{\left (b d - 2 \, a e\right )} \log \left (\frac{2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c -{\left (2 \, c x^{2} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right )}{4 \,{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} -{\left (b^{3} - 4 \, a b c\right )} d e +{\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}, \frac{{\left (b^{2} - 4 \, a c\right )} d \log \left (c x^{4} + b x^{2} + a\right ) - 2 \,{\left (b^{2} - 4 \, a c\right )} d \log \left (e x^{2} + d\right ) + 2 \, \sqrt{-b^{2} + 4 \, a c}{\left (b d - 2 \, a e\right )} \arctan \left (-\frac{{\left (2 \, c x^{2} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{4 \,{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} -{\left (b^{3} - 4 \, a b c\right )} d e +{\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((b^2 - 4*a*c)*d*log(c*x^4 + b*x^2 + a) - 2*(b^2 - 4*a*c)*d*log(e*x^2 + d) - sqrt(b^2 - 4*a*c)*(b*d - 2*a
*e)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c - (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)))/((b^2*c
- 4*a*c^2)*d^2 - (b^3 - 4*a*b*c)*d*e + (a*b^2 - 4*a^2*c)*e^2), 1/4*((b^2 - 4*a*c)*d*log(c*x^4 + b*x^2 + a) - 2
*(b^2 - 4*a*c)*d*log(e*x^2 + d) + 2*sqrt(-b^2 + 4*a*c)*(b*d - 2*a*e)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/
(b^2 - 4*a*c)))/((b^2*c - 4*a*c^2)*d^2 - (b^3 - 4*a*b*c)*d*e + (a*b^2 - 4*a^2*c)*e^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.13567, size = 180, normalized size = 1.36 \begin{align*} -\frac{d e \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{2} e - b d e^{2} + a e^{3}\right )}} + \frac{d \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\left (c d^{2} - b d e + a e^{2}\right )}} - \frac{{\left (b d - 2 \, a e\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*d*e*log(abs(x^2*e + d))/(c*d^2*e - b*d*e^2 + a*e^3) + 1/4*d*log(c*x^4 + b*x^2 + a)/(c*d^2 - b*d*e + a*e^2
) - 1/2*(b*d - 2*a*e)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((c*d^2 - b*d*e + a*e^2)*sqrt(-b^2 + 4*a*c))