3.2.71 \(\int \frac {x (A+B x^2)}{\sqrt {a+b x^2+c x^4}} \, dx\) [171]

Optimal. Leaf size=76 \[ \frac {B \sqrt {a+b x^2+c x^4}}{2 c}-\frac {(b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 c^{3/2}} \]

[Out]

-1/4*(-2*A*c+B*b)*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(3/2)+1/2*B*(c*x^4+b*x^2+a)^(1/2)/c

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1261, 654, 635, 212} \begin {gather*} \frac {B \sqrt {a+b x^2+c x^4}}{2 c}-\frac {(b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sqrt[a + b*x^2 + c*x^4])/(2*c) - ((b*B - 2*A*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])
/(4*c^(3/2))

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 635

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

Rule 654

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

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 )}{\sqrt {a+b x^2+c x^4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {A+B x}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac {B \sqrt {a+b x^2+c x^4}}{2 c}+\frac {(-b B+2 A c) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{4 c}\\ &=\frac {B \sqrt {a+b x^2+c x^4}}{2 c}+\frac {(-b B+2 A c) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^2}{\sqrt {a+b x^2+c x^4}}\right )}{2 c}\\ &=\frac {B \sqrt {a+b x^2+c x^4}}{2 c}-\frac {(b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 c^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.27, size = 78, normalized size = 1.03 \begin {gather*} \frac {B \sqrt {a+b x^2+c x^4}}{2 c}+\frac {(b B-2 A c) \log \left (b c+2 c^2 x^2-2 c^{3/2} \sqrt {a+b x^2+c x^4}\right )}{4 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.05, size = 94, normalized size = 1.24

method	result	size
risch	\(\frac {B \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 c}+\frac {A \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) b B}{4 c^{\frac {3}{2}}}\)	\(93\)
elliptic	\(\frac {B \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 c}+\frac {A \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) b B}{4 c^{\frac {3}{2}}}\)	\(93\)
default	\(B \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 c}-\frac {b \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\right )+\frac {A \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {c}}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x^2+A)/(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

B*(1/2*(c*x^4+b*x^2+a)^(1/2)/c-1/4*b/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2)))+1/2*A*ln((1/2*b+
c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))/c^(1/2)

________________________________________________________________________________________

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)^(1/2),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 deta

________________________________________________________________________________________

Fricas [A]
time = 0.38, size = 178, normalized size = 2.34 \begin {gather*} \left [\frac {4 \, \sqrt {c x^{4} + b x^{2} + a} B c - {\left (B b - 2 \, A c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right )}{8 \, c^{2}}, \frac {2 \, \sqrt {c x^{4} + b x^{2} + a} B c + {\left (B b - 2 \, A c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right )}{4 \, c^{2}}\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)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(4*sqrt(c*x^4 + b*x^2 + a)*B*c - (B*b - 2*A*c)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 +
b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c))/c^2, 1/4*(2*sqrt(c*x^4 + b*x^2 + a)*B*c + (B*b - 2*A*c)*sqrt(-c)*ar
ctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)))/c^2]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (A + B x^{2}\right )}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \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)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 4.06, size = 69, normalized size = 0.91 \begin {gather*} \frac {\sqrt {c x^{4} + b x^{2} + a} B}{2 \, c} + \frac {{\left (B b - 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{4 \, c^{\frac {3}{2}}} \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)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(c*x^4 + b*x^2 + a)*B/c + 1/4*(B*b - 2*A*c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c)
- b))/c^(3/2)

________________________________________________________________________________________

Mupad [B]
time = 1.05, size = 92, normalized size = 1.21 \begin {gather*} \frac {A\,\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )}{2\,\sqrt {c}}+\frac {B\,\sqrt {c\,x^4+b\,x^2+a}}{2\,c}-\frac {B\,b\,\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )}{4\,c^{3/2}} \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)^(1/2),x)

[Out]

(A*log((a + b*x^2 + c*x^4)^(1/2) + (b/2 + c*x^2)/c^(1/2)))/(2*c^(1/2)) + (B*(a + b*x^2 + c*x^4)^(1/2))/(2*c) -
(B*b*log((a + b*x^2 + c*x^4)^(1/2) + (b/2 + c*x^2)/c^(1/2)))/(4*c^(3/2))

________________________________________________________________________________________