3.297 \(\int \frac{x^5}{(d+e x^2) (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=158 \[ -\frac{\left (-a b e-2 a c d+b^2 d\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c \sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac{d^2 \log \left (d+e x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac{(b d-a e) \log \left (a+b x^2+c x^4\right )}{4 c \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.2606, antiderivative size = 158, 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, 1628, 634, 618, 206, 628} \[ -\frac{\left (-a b e-2 a c d+b^2 d\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c \sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac{d^2 \log \left (d+e x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac{(b d-a e) \log \left (a+b x^2+c x^4\right )}{4 c \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((d + e*x^2)*(a + b*x^2 + c*x^4)),x]

[Out]

-((b^2*d - 2*a*c*d - a*b*e)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c*Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e +
a*e^2)) + (d^2*Log[d + e*x^2])/(2*e*(c*d^2 - b*d*e + a*e^2)) - ((b*d - a*e)*Log[a + b*x^2 + c*x^4])/(4*c*(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 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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^5}{\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^2}{(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^2}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac{-a d-(b d-a e) 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^2 \log \left (d+e x^2\right )}{2 e \left (c d^2-b d e+a e^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-a d-(b d-a e) 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^2 \log \left (d+e x^2\right )}{2 e \left (c d^2-b d e+a e^2\right )}-\frac{(b d-a e) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c \left (c d^2-b d e+a e^2\right )}+\frac{\left (b^2 d-2 a c d-a b e\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c \left (c d^2-b d e+a e^2\right )}\\ &=\frac{d^2 \log \left (d+e x^2\right )}{2 e \left (c d^2-b d e+a e^2\right )}-\frac{(b d-a e) \log \left (a+b x^2+c x^4\right )}{4 c \left (c d^2-b d e+a e^2\right )}-\frac{\left (b^2 d-2 a c d-a b e\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{\left (b^2 d-2 a c d-a b e\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c \sqrt{b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac{d^2 \log \left (d+e x^2\right )}{2 e \left (c d^2-b d e+a e^2\right )}-\frac{(b d-a e) \log \left (a+b x^2+c x^4\right )}{4 c \left (c d^2-b d e+a e^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.107997, size = 139, normalized size = 0.88 \[ -\frac{\sqrt{4 a c-b^2} \left (e (b d-a e) \log \left (a+b x^2+c x^4\right )-2 c d^2 \log \left (d+e x^2\right )\right )+2 e \left (a b e+2 a c d+b^2 (-d)\right ) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{4 c e \sqrt{4 a c-b^2} \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((d + e*x^2)*(a + b*x^2 + c*x^4)),x]

[Out]

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

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Maple [A]  time = 0.007, size = 289, normalized size = 1.8 \begin{align*}{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) ae}{ \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ) c}}-{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) bd}{ \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ) c}}-{\frac{ad}{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{abe}{ \left ( 2\,a{e}^{2}-2\,deb+2\,c{d}^{2} \right ) c}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{b}^{2}d}{ \left ( 2\,a{e}^{2}-2\,deb+2\,c{d}^{2} \right ) c}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{d}^{2}\ln \left ( e{x}^{2}+d \right ) }{2\,e \left ( a{e}^{2}-deb+c{d}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(e*x^2+d)/(c*x^4+b*x^2+a),x)

[Out]

1/4/(a*e^2-b*d*e+c*d^2)/c*ln(c*x^4+b*x^2+a)*a*e-1/4/(a*e^2-b*d*e+c*d^2)/c*ln(c*x^4+b*x^2+a)*b*d-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*d-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/c*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^2/c*d+1/2*d^2*ln(e*x^2+d)/e/(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^5/(e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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^5/(e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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Giac [A]  time = 1.17576, size = 212, normalized size = 1.34 \begin{align*} \frac{d^{2} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{2} e - b d e^{2} + a e^{3}\right )}} - \frac{{\left (b d - a e\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\left (c^{2} d^{2} - b c d e + a c e^{2}\right )}} + \frac{{\left (b^{2} d - 2 \, a c d - a b e\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

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