3.2.4 \(\int \frac {x (A+B x^2)}{a+b x^2+c x^4} \, dx\) [104]

Optimal. Leaf size=71 \[ \frac {(b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {B \log \left (a+b x^2+c x^4\right )}{4 c} \]

[Out]

1/4*B*ln(c*x^4+b*x^2+a)/c+1/2*(-2*A*c+B*b)*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/c/(-4*a*c+b^2)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1261, 648, 632, 212, 642} \begin {gather*} \frac {(b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {B \log \left (a+b x^2+c x^4\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x^2))/(a + b*x^2 + c*x^4),x]

[Out]

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

Rule 212

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

Rule 632

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 642

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 648

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 1261

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 71, normalized size = 1.00 \begin {gather*} \frac {-\frac {2 (b B-2 A c) \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+B \log \left (a+b x^2+c x^4\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(A + B*x^2))/(a + b*x^2 + c*x^4),x]

[Out]

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

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Maple [A]
time = 0.06, size = 65, normalized size = 0.92

method	result	size
default	\(\frac {B \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c}+\frac {\left (A -\frac {B b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}\)	\(65\)
risch	\(\frac {\ln \left (\left (-8 c^{2} a A +2 A \,b^{2} c +4 a b B c -b^{3} B -\sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, b \right ) x^{2}-2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, a \right ) a B}{4 a c -b^{2}}-\frac {\ln \left (\left (-8 c^{2} a A +2 A \,b^{2} c +4 a b B c -b^{3} B -\sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, b \right ) x^{2}-2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, a \right ) b^{2} B}{4 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-8 c^{2} a A +2 A \,b^{2} c +4 a b B c -b^{3} B -\sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, b \right ) x^{2}-2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, a \right ) \sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}}{4 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-8 c^{2} a A +2 A \,b^{2} c +4 a b B c -b^{3} B +\sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, b \right ) x^{2}+2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, a \right ) a B}{4 a c -b^{2}}-\frac {\ln \left (\left (-8 c^{2} a A +2 A \,b^{2} c +4 a b B c -b^{3} B +\sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, b \right ) x^{2}+2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, a \right ) b^{2} B}{4 c \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (-8 c^{2} a A +2 A \,b^{2} c +4 a b B c -b^{3} B +\sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, b \right ) x^{2}+2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}\, a \right ) \sqrt {-\left (4 a c -b^{2}\right ) \left (2 A c -b B \right )^{2}}}{4 c \left (4 a c -b^{2}\right )}\)	\(689\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x^2+A)/(c*x^4+b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

1/4*B*ln(c*x^4+b*x^2+a)/c+(A-1/2*B*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))

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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*(B*x^2+A)/(c*x^4+b*x^2+a),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 deta

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Fricas [A]
time = 0.37, size = 219, normalized size = 3.08 \begin {gather*} \left [-\frac {{\left (B b - 2 \, A c\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 ) - {\left (B b^{2} - 4 \, B a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac {2 \, {\left (B b - 2 \, A c\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 ) + {\left (B b^{2} - 4 \, B a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (61) = 122\).
time = 5.04, size = 287, normalized size = 4.04 \begin {gather*} \left (\frac {B}{4 c} - \frac {\left (- 2 A c + B b\right ) \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{2} + \frac {- A b + 2 B a - 8 a c \left (\frac {B}{4 c} - \frac {\left (- 2 A c + B b\right ) \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )}\right ) + 2 b^{2} \left (\frac {B}{4 c} - \frac {\left (- 2 A c + B b\right ) \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )}\right )}{- 2 A c + B b} \right )} + \left (\frac {B}{4 c} + \frac {\left (- 2 A c + B b\right ) \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{2} + \frac {- A b + 2 B a - 8 a c \left (\frac {B}{4 c} + \frac {\left (- 2 A c + B b\right ) \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )}\right ) + 2 b^{2} \left (\frac {B}{4 c} + \frac {\left (- 2 A c + B b\right ) \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )}\right )}{- 2 A c + B b} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x**2+A)/(c*x**4+b*x**2+a),x)

[Out]

(B/(4*c) - (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)))*log(x**2 + (-A*b + 2*B*a - 8*a*c*(B/(4*c)
- (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2))) + 2*b**2*(B/(4*c) - (-2*A*c + B*b)*sqrt(-4*a*c + b*
*2)/(4*c*(4*a*c - b**2))))/(-2*A*c + B*b)) + (B/(4*c) + (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)
))*log(x**2 + (-A*b + 2*B*a - 8*a*c*(B/(4*c) + (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2))) + 2*b*
*2*(B/(4*c) + (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2))))/(-2*A*c + B*b))

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Giac [A]
time = 5.40, size = 67, normalized size = 0.94 \begin {gather*} \frac {B \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c} - \frac {{\left (B b - 2 \, A c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

1/4*B*log(c*x^4 + b*x^2 + a)/c - 1/2*(B*b - 2*A*c)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c
)*c)

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Mupad [B]
time = 0.50, size = 606, normalized size = 8.54 \begin {gather*} -\frac {\ln \left (c\,x^4+b\,x^2+a\right )\,\left (2\,B\,b^2-8\,B\,a\,c\right )}{2\,\left (16\,a\,c^2-4\,b^2\,c\right )}-\frac {\mathrm {atan}\left (\frac {2\,\left (4\,a\,c-b^2\right )\,\left (x^2\,\left (\frac {\frac {\left (2\,A\,c-B\,b\right )\,\left (6\,B\,b\,c-4\,A\,c^2+\frac {4\,b\,c^2\,\left (2\,B\,b^2-8\,B\,a\,c\right )}{16\,a\,c^2-4\,b^2\,c}\right )}{8\,c\,\sqrt {4\,a\,c-b^2}}+\frac {b\,c\,\left (2\,B\,b^2-8\,B\,a\,c\right )\,\left (2\,A\,c-B\,b\right )}{2\,\left (16\,a\,c^2-4\,b^2\,c\right )\,\sqrt {4\,a\,c-b^2}}}{a}+\frac {b\,\left (B^2\,b-A\,B\,c-\frac {b\,{\left (2\,A\,c-B\,b\right )}^2}{2\,\left (4\,a\,c-b^2\right )}+\frac {\left (2\,B\,b^2-8\,B\,a\,c\right )\,\left (6\,B\,b\,c-4\,A\,c^2+\frac {4\,b\,c^2\,\left (2\,B\,b^2-8\,B\,a\,c\right )}{16\,a\,c^2-4\,b^2\,c}\right )}{2\,\left (16\,a\,c^2-4\,b^2\,c\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )+\frac {\frac {\left (8\,B\,a\,c+\frac {8\,a\,c^2\,\left (2\,B\,b^2-8\,B\,a\,c\right )}{16\,a\,c^2-4\,b^2\,c}\right )\,\left (2\,A\,c-B\,b\right )}{8\,c\,\sqrt {4\,a\,c-b^2}}+\frac {a\,c\,\left (2\,B\,b^2-8\,B\,a\,c\right )\,\left (2\,A\,c-B\,b\right )}{\left (16\,a\,c^2-4\,b^2\,c\right )\,\sqrt {4\,a\,c-b^2}}}{a}+\frac {b\,\left (B^2\,a+\frac {\left (2\,B\,b^2-8\,B\,a\,c\right )\,\left (8\,B\,a\,c+\frac {8\,a\,c^2\,\left (2\,B\,b^2-8\,B\,a\,c\right )}{16\,a\,c^2-4\,b^2\,c}\right )}{2\,\left (16\,a\,c^2-4\,b^2\,c\right )}-\frac {a\,{\left (2\,A\,c-B\,b\right )}^2}{4\,a\,c-b^2}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )}{4\,A^2\,c^2-4\,A\,B\,b\,c+B^2\,b^2}\right )\,\left (2\,A\,c-B\,b\right )}{2\,c\,\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(A + B*x^2))/(a + b*x^2 + c*x^4),x)

[Out]

- (log(a + b*x^2 + c*x^4)*(2*B*b^2 - 8*B*a*c))/(2*(16*a*c^2 - 4*b^2*c)) - (atan((2*(4*a*c - b^2)*(x^2*((((2*A*
c - B*b)*(6*B*b*c - 4*A*c^2 + (4*b*c^2*(2*B*b^2 - 8*B*a*c))/(16*a*c^2 - 4*b^2*c)))/(8*c*(4*a*c - b^2)^(1/2)) +
(b*c*(2*B*b^2 - 8*B*a*c)*(2*A*c - B*b))/(2*(16*a*c^2 - 4*b^2*c)*(4*a*c - b^2)^(1/2)))/a + (b*(B^2*b - A*B*c -
(b*(2*A*c - B*b)^2)/(2*(4*a*c - b^2)) + ((2*B*b^2 - 8*B*a*c)*(6*B*b*c - 4*A*c^2 + (4*b*c^2*(2*B*b^2 - 8*B*a*c
))/(16*a*c^2 - 4*b^2*c)))/(2*(16*a*c^2 - 4*b^2*c))))/(2*a*(4*a*c - b^2)^(1/2))) + (((8*B*a*c + (8*a*c^2*(2*B*b
^2 - 8*B*a*c))/(16*a*c^2 - 4*b^2*c))*(2*A*c - B*b))/(8*c*(4*a*c - b^2)^(1/2)) + (a*c*(2*B*b^2 - 8*B*a*c)*(2*A*
c - B*b))/((16*a*c^2 - 4*b^2*c)*(4*a*c - b^2)^(1/2)))/a + (b*(B^2*a + ((2*B*b^2 - 8*B*a*c)*(8*B*a*c + (8*a*c^2
*(2*B*b^2 - 8*B*a*c))/(16*a*c^2 - 4*b^2*c)))/(2*(16*a*c^2 - 4*b^2*c)) - (a*(2*A*c - B*b)^2)/(4*a*c - b^2)))/(2
*a*(4*a*c - b^2)^(1/2))))/(4*A^2*c^2 + B^2*b^2 - 4*A*B*b*c))*(2*A*c - B*b))/(2*c*(4*a*c - b^2)^(1/2))

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