3.962 \(\int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=145 \[ \frac {b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{7/2}}-\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}+\frac {5 b \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\sqrt {a+b x^2+c x^4}}{6 a x^6} \]

[Out]

1/32*b*(-12*a*c+5*b^2)*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/a^(7/2)-1/6*(c*x^4+b*x^2+a)^(1/2
)/a/x^6+5/24*b*(c*x^4+b*x^2+a)^(1/2)/a^2/x^4-1/48*(-16*a*c+15*b^2)*(c*x^4+b*x^2+a)^(1/2)/a^3/x^2

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Rubi [A]  time = 0.17, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1114, 744, 834, 806, 724, 206} \[ -\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}+\frac {b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{7/2}}+\frac {5 b \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\sqrt {a+b x^2+c x^4}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

-Sqrt[a + b*x^2 + c*x^4]/(6*a*x^6) + (5*b*Sqrt[a + b*x^2 + c*x^4])/(24*a^2*x^4) - ((15*b^2 - 16*a*c)*Sqrt[a +
b*x^2 + c*x^4])/(48*a^3*x^2) + (b*(5*b^2 - 12*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])
/(32*a^(7/2))

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 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 744

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[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((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[Sim
plify[m + 2*p + 3], 0]

Rule 834

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))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

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 {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{6 a x^6}-\frac {\operatorname {Subst}\left (\int \frac {\frac {5 b}{2}+2 c x}{x^3 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{6 a}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {5 b \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{4} \left (15 b^2-16 a c\right )+\frac {5 b c x}{2}}{x^2 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{12 a^2}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {5 b \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}-\frac {\left (b \left (5 b^2-12 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^3}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {5 b \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}+\frac {\left (b \left (5 b^2-12 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^3}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {5 b \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}+\frac {b \left (5 b^2-12 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^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 112, normalized size = 0.77 \[ \frac {b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{7/2}}+\frac {\sqrt {a+b x^2+c x^4} \left (-8 a^2+2 a \left (5 b x^2+8 c x^4\right )-15 b^2 x^4\right )}{48 a^3 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(-8*a^2 - 15*b^2*x^4 + 2*a*(5*b*x^2 + 8*c*x^4)))/(48*a^3*x^6) + (b*(5*b^2 - 12*a*c)*A
rcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(32*a^(7/2))

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fricas [A]  time = 0.92, size = 265, normalized size = 1.83 \[ \left [-\frac {3 \, {\left (5 \, b^{3} - 12 \, a b c\right )} \sqrt {a} x^{6} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, {\left (10 \, a^{2} b x^{2} - {\left (15 \, a b^{2} - 16 \, a^{2} c\right )} x^{4} - 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, a^{4} x^{6}}, -\frac {3 \, {\left (5 \, b^{3} - 12 \, a b c\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) - 2 \, {\left (10 \, a^{2} b x^{2} - {\left (15 \, a b^{2} - 16 \, a^{2} c\right )} x^{4} - 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, a^{4} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/192*(3*(5*b^3 - 12*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*(10*a^2*b*x^2 - (15*a*b^2 - 16*a^2*c)*x^4 - 8*a^3)*sqrt(c*x^4 + b*x^2 + a)
)/(a^4*x^6), -1/96*(3*(5*b^3 - 12*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*(10*a^2*b*x^2 - (15*a*b^2 - 16*a^2*c)*x^4 - 8*a^3)*sqrt(c*x^4 + b*x^2 + a))/(
a^4*x^6)]

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giac [B]  time = 0.26, size = 335, normalized size = 2.31 \[ -\frac {{\left (5 \, b^{3} - 12 \, a b c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{16 \, \sqrt {-a} a^{3}} + \frac {15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} b^{3} - 36 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a b c - 40 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a b^{3} + 96 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{2} b c + 96 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{3} c^{\frac {3}{2}} + 33 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{2} b^{3} + 36 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{3} b c + 48 \, a^{3} b^{2} \sqrt {c} - 32 \, a^{4} c^{\frac {3}{2}}}{48 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{3} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(5*b^3 - 12*a*b*c)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^3) + 1/48*(15*(
sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*b^3 - 36*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a*b*c - 40*(sqrt(c
)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a*b^3 + 96*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^2*b*c + 96*(sqrt(c)*
x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^3*c^(3/2) + 33*(sqrt(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 + 48*a^3*b^2*sqrt(c) - 32*a^4*c^(3/2))/(((sqrt(c)*x^2 - sqrt(c*x^4 +
b*x^2 + a))^2 - a)^3*a^3)

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maple [A]  time = 0.02, size = 176, normalized size = 1.21 \[ -\frac {3 b c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{8 a^{\frac {5}{2}}}+\frac {5 b^{3} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{32 a^{\frac {7}{2}}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, c}{3 a^{2} x^{2}}-\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2}}{16 a^{3} x^{2}}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{24 a^{2} x^{4}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{6 a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(c*x^4+b*x^2+a)^(1/2)/a/x^6+5/24*b*(c*x^4+b*x^2+a)^(1/2)/a^2/x^4-5/16*b^2/a^3/x^2*(c*x^4+b*x^2+a)^(1/2)+5
/32*b^3/a^(7/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-3/8*b/a^(5/2)*c*ln((b*x^2+2*a+2*(c*x^4+b*x
^2+a)^(1/2)*a^(1/2))/x^2)+1/3*c/a^2/x^2*(c*x^4+b*x^2+a)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(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 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 \[ \int \frac {1}{x^7\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{7} \sqrt {a + b x^{2} + c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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