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

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

[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))

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Rubi [A]  time = 0.165968, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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 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]

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 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 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 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 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])

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*}

Mathematica [A]  time = 0.0800423, size = 112, normalized size = 0.77 \[ \frac{\sqrt{a+b x^2+c x^4} \left (-8 a^2+2 a \left (5 b x^2+8 c x^4\right )-15 b^2 x^4\right )}{48 a^3 x^6}+\frac{b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{32 a^{7/2}} \]

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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Maple [A]  time = 0.172, size = 176, normalized size = 1.2 \begin{align*} -{\frac{1}{6\,a{x}^{6}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,b}{24\,{a}^{2}{x}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{5\,{b}^{2}}{16\,{x}^{2}{a}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,{b}^{3}}{32}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{7}{2}}}}-{\frac{3\,bc}{8}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{c}{3\,{a}^{2}{x}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}} \end{align*}

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((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)-3/8*b/a^(5/2)*c*ln((2*a+b*x^2+2*a^(1/2)*(c
*x^4+b*x^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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 1.83151, size = 612, normalized size = 4.22 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{7} \sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{4} + b x^{2} + a} x^{7}}\,{d x} \end{align*}

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]

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