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

Optimal. Leaf size=165 \[ \frac {x^2 \left (-\left (x^2 \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )-b (a f+c d)+2 a c e\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right ) \left (-2 b c (3 a f+c d)+4 a c^2 e+b^3 f\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac {f \log \left (a+b x^2+c x^4\right )}{4 c^2} \]

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Rubi [A]  time = 0.29, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1663, 1644, 634, 618, 206, 628} \begin {gather*} \frac {x^2 \left (x^2 \left (-\left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )-b (a f+c d)+2 a c e\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right ) \left (-2 b c (3 a f+c d)+4 a c^2 e+b^3 f\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac {f \log \left (a+b x^2+c x^4\right )}{4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 1644

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[Po
lynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g +
(2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x
+ c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(d + e*x)*Q + g*(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c
*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && N
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[p] ||  !IntegerQ[m
] ||  !RationalQ[a, b, c, d, e]) &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2,
0]))

Rule 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 175, normalized size = 1.06 \begin {gather*} \frac {\frac {2 \left (-2 a^2 c f+a \left (b^2 f-b c \left (e+3 f x^2\right )+2 c^2 \left (d+e x^2\right )\right )+b x^2 \left (b^2 f-b c e+c^2 d\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {2 \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right ) \left (-2 b c (3 a f+c d)+4 a c^2 e+b^3 f\right )}{\left (4 a c-b^2\right )^{3/2}}+f \log \left (a+b x^2+c x^4\right )}{4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (d+e x^2+f x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 1.00, size = 970, normalized size = 5.88 \begin {gather*} \left [\frac {2 \, {\left ({\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} d - {\left (b^{4} c - 6 \, a b^{2} c^{2} + 8 \, a^{2} c^{3}\right )} e + {\left (b^{5} - 7 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} f\right )} x^{2} - {\left (2 \, a b c^{2} d - 4 \, a^{2} c^{2} e + {\left (2 \, b c^{3} d - 4 \, a c^{3} e - {\left (b^{3} c - 6 \, a b c^{2}\right )} f\right )} x^{4} + {\left (2 \, b^{2} c^{2} d - 4 \, a b c^{2} e - {\left (b^{4} - 6 \, a b^{2} c\right )} f\right )} x^{2} - {\left (a b^{3} - 6 \, a^{2} b c\right )} f\right )} \sqrt {b^{2} - 4 \, a c} \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 (a b^{2} c^{2} - 4 \, a^{2} c^{3}\right )} d - 2 \, {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} e + 2 \, {\left (a b^{4} - 6 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} f + {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} f x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} f x^{2} + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} f\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} + {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{4} + {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x^{2}\right )}}, \frac {2 \, {\left ({\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} d - {\left (b^{4} c - 6 \, a b^{2} c^{2} + 8 \, a^{2} c^{3}\right )} e + {\left (b^{5} - 7 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} f\right )} x^{2} - 2 \, {\left (2 \, a b c^{2} d - 4 \, a^{2} c^{2} e + {\left (2 \, b c^{3} d - 4 \, a c^{3} e - {\left (b^{3} c - 6 \, a b c^{2}\right )} f\right )} x^{4} + {\left (2 \, b^{2} c^{2} d - 4 \, a b c^{2} e - {\left (b^{4} - 6 \, a b^{2} c\right )} f\right )} x^{2} - {\left (a b^{3} - 6 \, a^{2} b c\right )} f\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + 4 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3}\right )} d - 2 \, {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} e + 2 \, {\left (a b^{4} - 6 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} f + {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} f x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} f x^{2} + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} f\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} + {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{4} + {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*((b^3*c^2 - 4*a*b*c^3)*d - (b^4*c - 6*a*b^2*c^2 + 8*a^2*c^3)*e + (b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*f)*x
^2 - (2*a*b*c^2*d - 4*a^2*c^2*e + (2*b*c^3*d - 4*a*c^3*e - (b^3*c - 6*a*b*c^2)*f)*x^4 + (2*b^2*c^2*d - 4*a*b*c
^2*e - (b^4 - 6*a*b^2*c)*f)*x^2 - (a*b^3 - 6*a^2*b*c)*f)*sqrt(b^2 - 4*a*c)*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)) + 4*(a*b^2*c^2 - 4*a^2*c^3)*d - 2*(a*b^3*c - 4*a
^2*b*c^2)*e + 2*(a*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*f + ((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*f*x^4 + (b^5 - 8*a*b
^3*c + 16*a^2*b*c^2)*f*x^2 + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*f)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^2 - 8*a^2*
b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^2),
1/4*(2*((b^3*c^2 - 4*a*b*c^3)*d - (b^4*c - 6*a*b^2*c^2 + 8*a^2*c^3)*e + (b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*f)*x
^2 - 2*(2*a*b*c^2*d - 4*a^2*c^2*e + (2*b*c^3*d - 4*a*c^3*e - (b^3*c - 6*a*b*c^2)*f)*x^4 + (2*b^2*c^2*d - 4*a*b
*c^2*e - (b^4 - 6*a*b^2*c)*f)*x^2 - (a*b^3 - 6*a^2*b*c)*f)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2
+ 4*a*c)/(b^2 - 4*a*c)) + 4*(a*b^2*c^2 - 4*a^2*c^3)*d - 2*(a*b^3*c - 4*a^2*b*c^2)*e + 2*(a*b^4 - 6*a^2*b^2*c +
8*a^3*c^2)*f + ((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*f*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*f*x^2 + (a*b^4 -
8*a^2*b^2*c + 16*a^3*c^2)*f)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8*a*
b^2*c^4 + 16*a^2*c^5)*x^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^2)]

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giac [A]  time = 1.84, size = 195, normalized size = 1.18 \begin {gather*} \frac {{\left (2 \, b c^{2} d - b^{3} f + 6 \, a b c f - 4 \, a c^{2} e\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {f \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{2}} + \frac {2 \, a c^{2} d + a b^{2} f - 2 \, a^{2} c f - a b c e + {\left (b c^{2} d + b^{3} f - 3 \, a b c f - b^{2} c e + 2 \, a c^{2} e\right )} x^{2}}{2 \, {\left (c x^{4} + b x^{2} + a\right )} {\left (b^{2} - 4 \, a c\right )} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*c^2*d - b^3*f + 6*a*b*c*f - 4*a*c^2*e)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^2*c^2 - 4*a*c^3)*
sqrt(-b^2 + 4*a*c)) + 1/4*f*log(c*x^4 + b*x^2 + a)/c^2 + 1/2*(2*a*c^2*d + a*b^2*f - 2*a^2*c*f - a*b*c*e + (b*c
^2*d + b^3*f - 3*a*b*c*f - b^2*c*e + 2*a*c^2*e)*x^2)/((c*x^4 + b*x^2 + a)*(b^2 - 4*a*c)*c^2)

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maple [B]  time = 0.02, size = 336, normalized size = 2.04 \begin {gather*} -\frac {3 a b f \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}} c}+\frac {2 a e \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {b^{3} f \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \left (4 a c -b^{2}\right )^{\frac {3}{2}} c^{2}}-\frac {b d \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {a f \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{\left (4 a c -b^{2}\right ) c}-\frac {b^{2} f \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 \left (4 a c -b^{2}\right ) c^{2}}+\frac {\frac {\left (3 a b c f -2 a \,c^{2} e -b^{3} f +b^{2} c e -b \,c^{2} d \right ) x^{2}}{\left (4 a c -b^{2}\right ) c^{2}}+\frac {\left (2 a c f -b^{2} f +b c e -2 c^{2} d \right ) a}{\left (4 a c -b^{2}\right ) c^{2}}}{2 c \,x^{4}+2 b \,x^{2}+2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 2.72, size = 1651, normalized size = 10.01

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((a*(2*c^2*d + b^2*f - 2*a*c*f - b*c*e))/(2*c^2*(4*a*c - b^2)) + (x^2*(b^3*f + 2*a*c^2*e + b*c^2*d - b^2*c*e
- 3*a*b*c*f))/(2*c^2*(4*a*c - b^2)))/(a + b*x^2 + c*x^4) - (log(a + b*x^2 + c*x^4)*(2*b^6*f - 128*a^3*c^3*f +
96*a^2*b^2*c^2*f - 24*a*b^4*c*f))/(2*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)) - (atan(((8*
a*c^3*(4*a*c - b^2)^3 - 2*b^2*c^2*(4*a*c - b^2)^3)*((((8*a*f + (8*a*c^2*(2*b^6*f - 128*a^3*c^3*f + 96*a^2*b^2*
c^2*f - 24*a*b^4*c*f))/(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4))*(b^3*f + 4*a*c^2*e - 2*b*c^
2*d - 6*a*b*c*f))/(8*c^2*(4*a*c - b^2)^(3/2)) + (a*(2*b^6*f - 128*a^3*c^3*f + 96*a^2*b^2*c^2*f - 24*a*b^4*c*f)
*(b^3*f + 4*a*c^2*e - 2*b*c^2*d - 6*a*b*c*f))/((4*a*c - b^2)^(3/2)*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 1
92*a^2*b^2*c^4)))/(a*(4*a*c - b^2)) - x^2*(((((6*b^3*c^2*f + 8*a*c^4*e - 4*b*c^4*d - 28*a*b*c^3*f)/(4*a*c^3 -
b^2*c^2) + ((8*b^3*c^4 - 32*a*b*c^5)*(2*b^6*f - 128*a^3*c^3*f + 96*a^2*b^2*c^2*f - 24*a*b^4*c*f))/(2*(4*a*c^3
- b^2*c^2)*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)))*(b^3*f + 4*a*c^2*e - 2*b*c^2*d - 6*a*b
*c*f))/(8*c^2*(4*a*c - b^2)^(3/2)) + ((8*b^3*c^4 - 32*a*b*c^5)*(2*b^6*f - 128*a^3*c^3*f + 96*a^2*b^2*c^2*f - 2
4*a*b^4*c*f)*(b^3*f + 4*a*c^2*e - 2*b*c^2*d - 6*a*b*c*f))/(16*c^2*(4*a*c - b^2)^(3/2)*(4*a*c^3 - b^2*c^2)*(256
*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)))/(a*(4*a*c - b^2)) + (b*((b^3*f^2 - 5*a*b*c*f^2 + 2*a*
c^2*e*f - b*c^2*d*f)/(4*a*c^3 - b^2*c^2) + (((6*b^3*c^2*f + 8*a*c^4*e - 4*b*c^4*d - 28*a*b*c^3*f)/(4*a*c^3 - b
^2*c^2) + ((8*b^3*c^4 - 32*a*b*c^5)*(2*b^6*f - 128*a^3*c^3*f + 96*a^2*b^2*c^2*f - 24*a*b^4*c*f))/(2*(4*a*c^3 -
b^2*c^2)*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)))*(2*b^6*f - 128*a^3*c^3*f + 96*a^2*b^2*c
^2*f - 24*a*b^4*c*f))/(2*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)) - (((b^3*c^4)/2 - 2*a*b*c
^5)*(b^3*f + 4*a*c^2*e - 2*b*c^2*d - 6*a*b*c*f)^2)/(c^4*(4*a*c - b^2)^3*(4*a*c^3 - b^2*c^2))))/(2*a*(4*a*c - b
^2)^(3/2))) + (b*(((8*a*f + (8*a*c^2*(2*b^6*f - 128*a^3*c^3*f + 96*a^2*b^2*c^2*f - 24*a*b^4*c*f))/(256*a^3*c^5
- 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4))*(2*b^6*f - 128*a^3*c^3*f + 96*a^2*b^2*c^2*f - 24*a*b^4*c*f))/(
2*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)) + (a*f^2)/c^2 - (a*(b^3*f + 4*a*c^2*e - 2*b*c^2*
d - 6*a*b*c*f)^2)/(c^2*(4*a*c - b^2)^3)))/(2*a*(4*a*c - b^2)^(3/2))))/(b^6*f^2 + 16*a^2*c^4*e^2 + 4*b^2*c^4*d^
2 + 36*a^2*b^2*c^2*f^2 - 12*a*b^4*c*f^2 - 4*b^4*c^2*d*f + 24*a*b^2*c^3*d*f + 8*a*b^3*c^2*e*f - 48*a^2*b*c^3*e*
f - 16*a*b*c^4*d*e))*(b^3*f + 4*a*c^2*e - 2*b*c^2*d - 6*a*b*c*f))/(2*c^2*(4*a*c - b^2)^(3/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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