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

Optimal. Leaf size=108 \[ \frac {16 c^3 \sqrt {b x^2+c x^4}}{35 b^4 x^2}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{35 b^3 x^4}+\frac {6 c \sqrt {b x^2+c x^4}}{35 b^2 x^6}-\frac {\sqrt {b x^2+c x^4}}{7 b x^8} \]

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Rubi [A]  time = 0.17, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2016, 2014} \begin {gather*} \frac {16 c^3 \sqrt {b x^2+c x^4}}{35 b^4 x^2}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{35 b^3 x^4}+\frac {6 c \sqrt {b x^2+c x^4}}{35 b^2 x^6}-\frac {\sqrt {b x^2+c x^4}}{7 b x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*Sqrt[b*x^2 + c*x^4]),x]

[Out]

-Sqrt[b*x^2 + c*x^4]/(7*b*x^8) + (6*c*Sqrt[b*x^2 + c*x^4])/(35*b^2*x^6) - (8*c^2*Sqrt[b*x^2 + c*x^4])/(35*b^3*
x^4) + (16*c^3*Sqrt[b*x^2 + c*x^4])/(35*b^4*x^2)

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx &=-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}-\frac {(6 c) \int \frac {1}{x^5 \sqrt {b x^2+c x^4}} \, dx}{7 b}\\ &=-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}+\frac {6 c \sqrt {b x^2+c x^4}}{35 b^2 x^6}+\frac {\left (24 c^2\right ) \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx}{35 b^2}\\ &=-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}+\frac {6 c \sqrt {b x^2+c x^4}}{35 b^2 x^6}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{35 b^3 x^4}-\frac {\left (16 c^3\right ) \int \frac {1}{x \sqrt {b x^2+c x^4}} \, dx}{35 b^3}\\ &=-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}+\frac {6 c \sqrt {b x^2+c x^4}}{35 b^2 x^6}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{35 b^3 x^4}+\frac {16 c^3 \sqrt {b x^2+c x^4}}{35 b^4 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 57, normalized size = 0.53 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-5 b^3+6 b^2 c x^2-8 b c^2 x^4+16 c^3 x^6\right )}{35 b^4 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-5*b^3 + 6*b^2*c*x^2 - 8*b*c^2*x^4 + 16*c^3*x^6))/(35*b^4*x^8)

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IntegrateAlgebraic [A]  time = 0.16, size = 57, normalized size = 0.53 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-5 b^3+6 b^2 c x^2-8 b c^2 x^4+16 c^3 x^6\right )}{35 b^4 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-5*b^3 + 6*b^2*c*x^2 - 8*b*c^2*x^4 + 16*c^3*x^6))/(35*b^4*x^8)

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fricas [A]  time = 1.26, size = 53, normalized size = 0.49 \begin {gather*} \frac {{\left (16 \, c^{3} x^{6} - 8 \, b c^{2} x^{4} + 6 \, b^{2} c x^{2} - 5 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, b^{4} x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(16*c^3*x^6 - 8*b*c^2*x^4 + 6*b^2*c*x^2 - 5*b^3)*sqrt(c*x^4 + b*x^2)/(b^4*x^8)

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giac [A]  time = 0.20, size = 123, normalized size = 1.14 \begin {gather*} \frac {70 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{3} c^{\frac {3}{2}} + 84 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{2} b c + 35 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} b^{2} \sqrt {c} + 5 \, b^{3}}{35 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(70*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^3*c^(3/2) + 84*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^2*b*c + 35*(sq
rt(c)*x^2 - sqrt(c*x^4 + b*x^2))*b^2*sqrt(c) + 5*b^3)/(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^7

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maple [A]  time = 0.01, size = 61, normalized size = 0.56 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+8 b \,c^{2} x^{4}-6 b^{2} c \,x^{2}+5 b^{3}\right )}{35 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{4} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(c*x^2+b)*(-16*c^3*x^6+8*b*c^2*x^4-6*b^2*c*x^2+5*b^3)/x^6/b^4/(c*x^4+b*x^2)^(1/2)

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maxima [A]  time = 1.41, size = 92, normalized size = 0.85 \begin {gather*} \frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{35 \, b^{4} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{35 \, b^{3} x^{4}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} c}{35 \, b^{2} x^{6}} - \frac {\sqrt {c x^{4} + b x^{2}}}{7 \, b x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/35*sqrt(c*x^4 + b*x^2)*c^3/(b^4*x^2) - 8/35*sqrt(c*x^4 + b*x^2)*c^2/(b^3*x^4) + 6/35*sqrt(c*x^4 + b*x^2)*c/
(b^2*x^6) - 1/7*sqrt(c*x^4 + b*x^2)/(b*x^8)

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mupad [B]  time = 4.27, size = 92, normalized size = 0.85 \begin {gather*} \frac {6\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,b^2\,x^6}-\frac {\sqrt {c\,x^4+b\,x^2}}{7\,b\,x^8}-\frac {8\,c^2\,\sqrt {c\,x^4+b\,x^2}}{35\,b^3\,x^4}+\frac {16\,c^3\,\sqrt {c\,x^4+b\,x^2}}{35\,b^4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

(6*c*(b*x^2 + c*x^4)^(1/2))/(35*b^2*x^6) - (b*x^2 + c*x^4)^(1/2)/(7*b*x^8) - (8*c^2*(b*x^2 + c*x^4)^(1/2))/(35
*b^3*x^4) + (16*c^3*(b*x^2 + c*x^4)^(1/2))/(35*b^4*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{7} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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