3.2.73 \(\int \frac {A+B x^2}{x^3 \sqrt {a+b x^2+c x^4}} \, dx\) [173]

Optimal. Leaf size=80 \[ -\frac {A \sqrt {a+b x^2+c x^4}}{2 a x^2}+\frac {(A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 a^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1265, 820, 738, 212} \begin {gather*} \frac {(A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 a^{3/2}}-\frac {A \sqrt {a+b x^2+c x^4}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(A*Sqrt[a + b*x^2 + c*x^4])/(a*x^2) + ((A*b - 2*a*B)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*
x^4])])/(4*a^(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 738

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

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Dist[
(b*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[S
implify[m + 2*p + 3], 0]

Rule 1265

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]

Rubi steps

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

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Mathematica [A]
time = 0.32, size = 81, normalized size = 1.01 \begin {gather*} -\frac {A \sqrt {a+b x^2+c x^4}}{2 a x^2}+\frac {(-A b+2 a B) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{2 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(A*Sqrt[a + b*x^2 + c*x^4])/(a*x^2) + ((-(A*b) + 2*a*B)*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/S
qrt[a]])/(2*a^(3/2))

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Maple [A]
time = 0.06, size = 105, normalized size = 1.31

method	result	size
risch	\(-\frac {A \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a \,x^{2}}+\frac {\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) A b}{4 a^{\frac {3}{2}}}-\frac {B \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 \sqrt {a}}\)	\(104\)
elliptic	\(-\frac {A \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a \,x^{2}}+\frac {\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) A b}{4 a^{\frac {3}{2}}}-\frac {B \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 \sqrt {a}}\)	\(104\)
default	\(A \left (-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\right )-\frac {B \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 \sqrt {a}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^3/(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Fricas [A]
time = 0.40, size = 197, normalized size = 2.46 \begin {gather*} \left [-\frac {{\left (2 \, B a - A b\right )} \sqrt {a} x^{2} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, \sqrt {c x^{4} + b x^{2} + a} A a}{8 \, a^{2} x^{2}}, \frac {{\left (2 \, B a - A b\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2} + a} A a}{4 \, a^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^3/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x^{2}}{x^{3} \sqrt {a + b x^{2} + c x^{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**3/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral((A + B*x**2)/(x**3*sqrt(a + b*x**2 + c*x**4)), x)

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Giac [A]
time = 3.70, size = 124, normalized size = 1.55 \begin {gather*} \frac {{\left (2 \, B a - A b\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a} + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} A b + 2 \, A a \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^3/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.78, size = 103, normalized size = 1.29 \begin {gather*} \frac {A\,b\,\mathrm {atanh}\left (\frac {\frac {b\,x^2}{2}+a}{\sqrt {a}\,\sqrt {c\,x^4+b\,x^2+a}}\right )}{4\,a^{3/2}}-\frac {B\,\ln \left (2\,a+2\,\sqrt {a}\,\sqrt {c\,x^4+b\,x^2+a}+b\,x^2\right )}{2\,\sqrt {a}}-\frac {A\,\sqrt {c\,x^4+b\,x^2+a}}{2\,a\,x^2}-\frac {B\,\ln \left (\frac {1}{x^2}\right )}{2\,\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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