3.10.75 \(\int \frac {1}{x^7 \sqrt {-a+b x^2+c x^4}} \, dx\) [975]

Optimal. Leaf size=154 \[ \frac {\sqrt {-a+b x^2+c x^4}}{6 a x^6}+\frac {5 b \sqrt {-a+b x^2+c x^4}}{24 a^2 x^4}+\frac {\left (15 b^2+16 a c\right ) \sqrt {-a+b x^2+c x^4}}{48 a^3 x^2}-\frac {b \left (5 b^2+12 a c\right ) \tan ^{-1}\left (\frac {2 a-b x^2}{2 \sqrt {a} \sqrt {-a+b x^2+c x^4}}\right )}{32 a^{7/2}} \]

[Out]

-1/32*b*(12*a*c+5*b^2)*arctan(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 = 0.11, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1128, 758, 848, 820, 738, 210} \begin {gather*} -\frac {b \left (12 a c+5 b^2\right ) \text {ArcTan}\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 (16 a c+15 b^2\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} \end {gather*}

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)*ArcTan[(2*a - b*x^2)/(2*Sqrt[a]*Sqrt[-a + b*x^2 + c*x^4])
])/(32*a^(7/2))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 758

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 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 848

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 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^7 \sqrt {-a+b x^2+c x^4}} \, dx &=\frac {1}{2} \text {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 {\text {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 {\text {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 ) \text {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 ) \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 )}{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 ) \tan ^{-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 = 0.32, size = 114, normalized size = 0.74 \begin {gather*} \frac {\sqrt {-a+b x^2+c x^4} \left (8 a^2+10 a b x^2+15 b^2 x^4+16 a c x^4\right )}{48 a^3 x^6}+\frac {\left (-5 b^3-12 a b c\right ) \tan ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {-a+b x^2+c x^4}}{\sqrt {a}}\right )}{16 a^{7/2}} \end {gather*}

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 + 10*a*b*x^2 + 15*b^2*x^4 + 16*a*c*x^4))/(48*a^3*x^6) + ((-5*b^3 - 12*a*b*c)*
ArcTan[(Sqrt[c]*x^2 - Sqrt[-a + b*x^2 + c*x^4])/Sqrt[a]])/(16*a^(7/2))

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Maple [A]
time = 0.05, size = 202, normalized size = 1.31

method	result	size
risch	\(-\frac {\left (-c \,x^{4}-b \,x^{2}+a \right ) \left (16 c \,x^{4} a +15 b^{2} x^{4}+10 a b \,x^{2}+8 a^{2}\right )}{48 a^{3} x^{6} \sqrt {c \,x^{4}+b \,x^{2}-a}}-\frac {3 b c \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{8 a^{2} \sqrt {-a}}-\frac {5 b^{3} \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{32 a^{3} \sqrt {-a}}\)	\(167\)
default	\(\frac {\sqrt {c \,x^{4}+b \,x^{2}-a}}{6 a \,x^{6}}+\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}-a}}{24 a^{2} x^{4}}+\frac {5 b^{2} \sqrt {c \,x^{4}+b \,x^{2}-a}}{16 a^{3} x^{2}}-\frac {5 b^{3} \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{32 a^{3} \sqrt {-a}}-\frac {3 b c \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{8 a^{2} \sqrt {-a}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}-a}}{3 a^{2} x^{2}}\)	\(202\)
elliptic	\(\frac {\sqrt {c \,x^{4}+b \,x^{2}-a}}{6 a \,x^{6}}+\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}-a}}{24 a^{2} x^{4}}+\frac {5 b^{2} \sqrt {c \,x^{4}+b \,x^{2}-a}}{16 a^{3} x^{2}}-\frac {5 b^{3} \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{32 a^{3} \sqrt {-a}}-\frac {3 b c \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{8 a^{2} \sqrt {-a}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}-a}}{3 a^{2} x^{2}}\)	\(202\)

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,method=_RETURNVERBOSE)

[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^3/(-a)^(1/2)*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^2*c/(-a)^(1/2)*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 [A]
time = 0.48, size = 179, normalized size = 1.16 \begin {gather*} -\frac {5 \, b^{3} \arcsin \left (-\frac {b}{\sqrt {b^{2} + 4 \, a c}} + \frac {2 \, a}{\sqrt {b^{2} + 4 \, a c} x^{2}}\right )}{32 \, a^{\frac {7}{2}}} - \frac {3 \, b c \arcsin \left (-\frac {b}{\sqrt {b^{2} + 4 \, a c}} + \frac {2 \, a}{\sqrt {b^{2} + 4 \, a c} x^{2}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {5 \, \sqrt {c x^{4} + b x^{2} - a} b^{2}}{16 \, a^{3} x^{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}{24 \, a^{2} x^{4}} + \frac {\sqrt {c x^{4} + b x^{2} - a}}{6 \, a x^{6}} \end {gather*}

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]

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

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Fricas [A]
time = 0.41, size = 272, normalized size = 1.77 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{7} \sqrt {- a + b x^{2} + c x^{4}}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (131) = 262\).
time = 7.70, size = 344, normalized size = 2.23 \begin {gather*} \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 \, a^{\frac {7}{2}}} - \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}} \end {gather*}

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))/a^(7/2) - 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 - s
qrt(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 - sqr
t(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 - s
qrt(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^7\,\sqrt {c\,x^4+b\,x^2-a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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