3.10.86 \(\int \frac {1}{x (a+b x^2+c x^4)^{3/2}} \, dx\) [986]

Optimal. Leaf size=89 \[ \frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {\tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 a^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1128, 754, 12, 738, 212} \begin {gather*} \frac {-2 a c+b^2+b c x^2}{a \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {\tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 754

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(b
*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e +
a*e^2))), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 1128

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

Rubi steps

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

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Mathematica [A]
time = 0.38, size = 124, normalized size = 1.39 \begin {gather*} -\frac {\left (b^2-2 a c+b c x^2\right ) \sqrt {a+b x^2+c x^4}}{a \left (4 a^2 c-b^2 x^2 \left (b+c x^2\right )+a \left (-b^2+4 b c x^2+4 c^2 x^4\right )\right )}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.04, size = 99, normalized size = 1.11

method	result	size
default	\(\frac {1}{2 a \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (2 c \,x^{2}+b \right )}{2 a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 a^{\frac {3}{2}}}\)	\(99\)
elliptic	\(\frac {1}{2 a \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (2 c \,x^{2}+b \right )}{2 a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 a^{\frac {3}{2}}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a/(c*x^4+b*x^2+a)^(1/2)-1/2*b/a*(2*c*x^2+b)/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)-1/2/a^(3/2)*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(1/x/(c*x^4+b*x^2+a)^(3/2),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 [B] Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (77) = 154\).
time = 0.40, size = 389, normalized size = 4.37 \begin {gather*} \left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} x^{2}\right )} \sqrt {a} \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 \, {\left (a b c x^{2} + a b^{2} - 2 \, a^{2} c\right )} \sqrt {c x^{4} + b x^{2} + a}}{4 \, {\left (a^{3} b^{2} - 4 \, a^{4} c + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{4} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )}}, \frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} x^{2}\right )} \sqrt {-a} \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 \, {\left (a b c x^{2} + a b^{2} - 2 \, a^{2} c\right )} \sqrt {c x^{4} + b x^{2} + a}}{2 \, {\left (a^{3} b^{2} - 4 \, a^{4} c + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{4} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((b^2*c - 4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*x^2)*sqrt(a)*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*(a*b*c*x^2 + a*b^2 - 2*a^2*c)*sqrt(c*x
^4 + b*x^2 + a))/(a^3*b^2 - 4*a^4*c + (a^2*b^2*c - 4*a^3*c^2)*x^4 + (a^2*b^3 - 4*a^3*b*c)*x^2), 1/2*(((b^2*c -
4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*x^2)*sqrt(-a)*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*(a*b*c*x^2 + a*b^2 - 2*a^2*c)*sqrt(c*x^4 + b*x^2 + a))/(a^3*b^2 - 4
*a^4*c + (a^2*b^2*c - 4*a^3*c^2)*x^4 + (a^2*b^3 - 4*a^3*b*c)*x^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x*(a + b*x**2 + c*x**4)**(3/2)), x)

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Giac [A]
time = 3.24, size = 110, normalized size = 1.24 \begin {gather*} \frac {\frac {a b c x^{2}}{a^{2} b^{2} - 4 \, a^{3} c} + \frac {a b^{2} - 2 \, a^{2} c}{a^{2} b^{2} - 4 \, a^{3} c}}{\sqrt {c x^{4} + b x^{2} + a}} + \frac {\arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*b*c*x^2/(a^2*b^2 - 4*a^3*c) + (a*b^2 - 2*a^2*c)/(a^2*b^2 - 4*a^3*c))/sqrt(c*x^4 + b*x^2 + a) + arctan(-(sqr
t(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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