3.295 \(\int \frac{x^9}{(d+e x^2) (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=230 \[ -\frac{\left (a^2 c e-a b^2 e-2 a b c d+b^3 d\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3 \left (a e^2-b d e+c d^2\right )}-\frac{\left (3 a^2 b c e+2 a^2 c^2 d-4 a b^2 c d-a b^3 e+b^4 d\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac{d^4 \log \left (d+e x^2\right )}{2 e^3 \left (a e^2-b d e+c d^2\right )}-\frac{x^2 (b e+c d)}{2 c^2 e^2}+\frac{x^4}{4 c e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.487482, antiderivative size = 230, 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} \[ -\frac{\left (a^2 c e-a b^2 e-2 a b c d+b^3 d\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3 \left (a e^2-b d e+c d^2\right )}-\frac{\left (3 a^2 b c e+2 a^2 c^2 d-4 a b^2 c d-a b^3 e+b^4 d\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac{d^4 \log \left (d+e x^2\right )}{2 e^3 \left (a e^2-b d e+c d^2\right )}-\frac{x^2 (b e+c d)}{2 c^2 e^2}+\frac{x^4}{4 c e} \]

Antiderivative was successfully verified.

[In]

Int[x^9/((d + e*x^2)*(a + b*x^2 + c*x^4)),x]

[Out]

-((c*d + b*e)*x^2)/(2*c^2*e^2) + x^4/(4*c*e) - ((b^4*d - 4*a*b^2*c*d + 2*a^2*c^2*d - a*b^3*e + 3*a^2*b*c*e)*Ar
cTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^3*Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2)) + (d^4*Log[d + e*x^2
])/(2*e^3*(c*d^2 - b*d*e + a*e^2)) - ((b^3*d - 2*a*b*c*d - a*b^2*e + a^2*c*e)*Log[a + b*x^2 + c*x^4])/(4*c^3*(
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^9}{\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^4}{(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{-c d-b e}{c^2 e^2}+\frac{x}{c e}+\frac{d^4}{e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac{-a \left (b^2 d-a c d-a b e\right )-\left (b^3 d-2 a b c d-a b^2 e+a^2 c e\right ) x}{c^2 \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{(c d+b e) x^2}{2 c^2 e^2}+\frac{x^4}{4 c e}+\frac{d^4 \log \left (d+e x^2\right )}{2 e^3 \left (c d^2-b d e+a e^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-a \left (b^2 d-a c d-a b e\right )-\left (b^3 d-2 a b c d-a b^2 e+a^2 c e\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c^2 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{(c d+b e) x^2}{2 c^2 e^2}+\frac{x^4}{4 c e}+\frac{d^4 \log \left (d+e x^2\right )}{2 e^3 \left (c d^2-b d e+a e^2\right )}-\frac{\left (b^3 d-2 a b c d-a b^2 e+a^2 c e\right ) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3 \left (c d^2-b d e+a e^2\right )}+\frac{\left (b^4 d-4 a b^2 c d+2 a^2 c^2 d-a b^3 e+3 a^2 b c e\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{(c d+b e) x^2}{2 c^2 e^2}+\frac{x^4}{4 c e}+\frac{d^4 \log \left (d+e x^2\right )}{2 e^3 \left (c d^2-b d e+a e^2\right )}-\frac{\left (b^3 d-2 a b c d-a b^2 e+a^2 c e\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3 \left (c d^2-b d e+a e^2\right )}-\frac{\left (b^4 d-4 a b^2 c d+2 a^2 c^2 d-a b^3 e+3 a^2 b c 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^3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{(c d+b e) x^2}{2 c^2 e^2}+\frac{x^4}{4 c e}-\frac{\left (b^4 d-4 a b^2 c d+2 a^2 c^2 d-a b^3 e+3 a^2 b c e\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \sqrt{b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac{d^4 \log \left (d+e x^2\right )}{2 e^3 \left (c d^2-b d e+a e^2\right )}-\frac{\left (b^3 d-2 a b c d-a b^2 e+a^2 c e\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3 \left (c d^2-b d e+a e^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.25112, size = 228, normalized size = 0.99 \[ \frac{1}{4} \left (\frac{\left (-a^2 c e+a b^2 e+2 a b c d+b^3 (-d)\right ) \log \left (a+b x^2+c x^4\right )}{c^3 \left (e (a e-b d)+c d^2\right )}-\frac{2 \left (3 a^2 b c e+2 a^2 c^2 d-4 a b^2 c d-a b^3 e+b^4 d\right ) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{c^3 \sqrt{4 a c-b^2} \left (e (b d-a e)-c d^2\right )}+\frac{2 d^4 \log \left (d+e x^2\right )}{e^3 \left (e (a e-b d)+c d^2\right )}-\frac{2 x^2 (b e+c d)}{c^2 e^2}+\frac{x^4}{c e}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/((d + e*x^2)*(a + b*x^2 + c*x^4)),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.014, size = 538, normalized size = 2.3 \begin{align*}{\frac{{x}^{4}}{4\,ce}}-{\frac{b{x}^{2}}{2\,{c}^{2}e}}-{\frac{d{x}^{2}}{2\,{e}^{2}c}}-{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){a}^{2}e}{ \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ){c}^{2}}}+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) a{b}^{2}e}{ \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ){c}^{3}}}+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) abd}{ \left ( 2\,a{e}^{2}-2\,deb+2\,c{d}^{2} \right ){c}^{2}}}-{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{3}d}{ \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ){c}^{3}}}+{\frac{3\,{a}^{2}be}{ \left ( 2\,a{e}^{2}-2\,deb+2\,c{d}^{2} \right ){c}^{2}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{a}^{2}d}{ \left ( a{e}^{2}-deb+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}}}}}-2\,{\frac{a{b}^{2}d}{ \left ( a{e}^{2}-deb+c{d}^{2} \right ){c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{a{b}^{3}e}{ \left ( 2\,a{e}^{2}-2\,deb+2\,c{d}^{2} \right ){c}^{3}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{b}^{4}d}{ \left ( 2\,a{e}^{2}-2\,deb+2\,c{d}^{2} \right ){c}^{3}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{d}^{4}\ln \left ( e{x}^{2}+d \right ) }{2\,{e}^{3} \left ( a{e}^{2}-deb+c{d}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(e*x^2+d)/(c*x^4+b*x^2+a),x)

[Out]

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

________________________________________________________________________________________

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^9/(e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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^9/(e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.16437, size = 319, normalized size = 1.39 \begin{align*} \frac{d^{4} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{2} e^{3} - b d e^{4} + a e^{5}\right )}} - \frac{{\left (b^{3} d - 2 \, a b c d - a b^{2} e + a^{2} c e\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\left (c^{4} d^{2} - b c^{3} d e + a c^{3} e^{2}\right )}} + \frac{{\left (b^{4} d - 4 \, a b^{2} c d + 2 \, a^{2} c^{2} d - a b^{3} e + 3 \, a^{2} b c e\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (c^{4} d^{2} - b c^{3} d e + a c^{3} e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (c x^{4} e - 2 \, c d x^{2} - 2 \, b x^{2} e\right )} e^{\left (-2\right )}}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

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