3.8.30 \(\int \frac {\sqrt {a+b x^2+c x^4}}{x^7} \, dx\)

Optimal. Leaf size=116 \[ -\frac {b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{5/2}}+\frac {b \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a^2 x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6} \]

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Rubi [A]  time = 0.10, antiderivative size = 116, 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 = {1114, 730, 720, 724, 206} \begin {gather*} -\frac {b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{5/2}}+\frac {b \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a^2 x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^7,x]

[Out]

(b*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*a^2*x^4) - (a + b*x^2 + c*x^4)^(3/2)/(6*a*x^6) - (b*(b^2 - 4*a*c
)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(32*a^(5/2))

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 720

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

Rule 724

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 730

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

Rule 1114

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 {\sqrt {a+b x^2+c x^4}}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6}-\frac {b \operatorname {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^2\right )}{4 a}\\ &=\frac {b \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a^2 x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6}+\frac {\left (b \left (b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{32 a^2}\\ &=\frac {b \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a^2 x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6}-\frac {\left (b \left (b^2-4 a c\right )\right ) \operatorname {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 )}{16 a^2}\\ &=\frac {b \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a^2 x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6}-\frac {b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 108, normalized size = 0.93 \begin {gather*} -\frac {b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{5/2}}-\frac {\sqrt {a+b x^2+c x^4} \left (8 a^2+2 a x^2 \left (b+4 c x^2\right )-3 b^2 x^4\right )}{48 a^2 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2 + c*x^4]/x^7,x]

[Out]

-1/48*(Sqrt[a + b*x^2 + c*x^4]*(8*a^2 - 3*b^2*x^4 + 2*a*x^2*(b + 4*c*x^2)))/(a^2*x^6) - (b*(b^2 - 4*a*c)*ArcTa
nh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(32*a^(5/2))

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IntegrateAlgebraic [A]  time = 0.57, size = 108, normalized size = 0.93 \begin {gather*} \frac {\left (b^3-4 a b c\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{16 a^{5/2}}+\frac {\sqrt {a+b x^2+c x^4} \left (-8 a^2-2 a b x^2-8 a c x^4+3 b^2 x^4\right )}{48 a^2 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + b*x^2 + c*x^4]/x^7,x]

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(-8*a^2 - 2*a*b*x^2 + 3*b^2*x^4 - 8*a*c*x^4))/(48*a^2*x^6) + ((b^3 - 4*a*b*c)*ArcTanh
[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/(16*a^(5/2))

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fricas [A]  time = 0.99, size = 261, normalized size = 2.25 \begin {gather*} \left [-\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {a} x^{6} \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 (2 \, a^{2} b x^{2} - {\left (3 \, a b^{2} - 8 \, a^{2} c\right )} x^{4} + 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, a^{3} x^{6}}, \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {-a} x^{6} \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 (2 \, a^{2} b x^{2} - {\left (3 \, a b^{2} - 8 \, a^{2} c\right )} x^{4} + 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, a^{3} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^7,x, algorithm="fricas")

[Out]

[-1/192*(3*(b^3 - 4*a*b*c)*sqrt(a)*x^6*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*(2*a^2*b*x^2 - (3*a*b^2 - 8*a^2*c)*x^4 + 8*a^3)*sqrt(c*x^4 + b*x^2 + a))/(a^3
*x^6), 1/96*(3*(b^3 - 4*a*b*c)*sqrt(-a)*x^6*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*(2*a^2*b*x^2 - (3*a*b^2 - 8*a^2*c)*x^4 + 8*a^3)*sqrt(c*x^4 + b*x^2 + a))/(a^3*x^6)]

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giac [B]  time = 0.30, size = 359, normalized size = 3.09 \begin {gather*} \frac {{\left (b^{3} - 4 \, a b c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{16 \, \sqrt {-a} a^{2}} - \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} b^{3} - 12 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a b c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{2} c^{\frac {3}{2}} - 8 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a b^{3} - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{2} b c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{2} b^{2} \sqrt {c} - 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{2} b^{3} - 36 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{3} b c - 16 \, a^{4} c^{\frac {3}{2}}}{48 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{3} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^7,x, algorithm="giac")

[Out]

1/16*(b^3 - 4*a*b*c)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/48*(3*(sqrt(
c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*b^3 - 12*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a*b*c - 48*(sqrt(c)*x^2
- sqrt(c*x^4 + b*x^2 + a))^4*a^2*c^(3/2) - 8*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a*b^3 - 48*(sqrt(c)*x^
2 - sqrt(c*x^4 + b*x^2 + a))^3*a^2*b*c - 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^2*b^2*sqrt(c) - 3*(sqr
t(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^2*b^3 - 36*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^3*b*c - 16*a^4*c^(3
/2))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^3*a^2)

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maple [B]  time = 0.01, size = 222, normalized size = 1.91 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} c \,x^{2}}{16 a^{3}}+\frac {b c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{8 a^{\frac {3}{2}}}-\frac {b^{3} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{32 a^{\frac {5}{2}}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b c}{8 a^{2}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3}}{16 a^{3}}-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}{16 a^{3} x^{2}}+\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b}{8 a^{2} x^{4}}-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{6 a \,x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(1/2)/x^7,x)

[Out]

-1/6*(c*x^4+b*x^2+a)^(3/2)/a/x^6+1/8*b/a^2/x^4*(c*x^4+b*x^2+a)^(3/2)-1/16*b^2/a^3/x^2*(c*x^4+b*x^2+a)^(3/2)+1/
16*b^3/a^3*(c*x^4+b*x^2+a)^(1/2)-1/32*b^3/a^(5/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)+1/16*b^2
/a^3*c*(c*x^4+b*x^2+a)^(1/2)*x^2-1/8*b/a^2*c*(c*x^4+b*x^2+a)^(1/2)+1/8*b/a^(3/2)*c*ln((b*x^2+2*a+2*(c*x^4+b*x^
2+a)^(1/2)*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((c*x^4+b*x^2+a)^(1/2)/x^7,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, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^4+b\,x^2+a}}{x^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(1/2)/x^7,x)

[Out]

int((a + b*x^2 + c*x^4)^(1/2)/x^7, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(1/2)/x**7,x)

[Out]

Integral(sqrt(a + b*x**2 + c*x**4)/x**7, x)

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