3.2.62 \(\int \frac {1}{x^5 (b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=130 \[ \frac {128 c^3 \sqrt {b x^2+c x^4}}{35 b^5 x^2}-\frac {64 c^2 \sqrt {b x^2+c x^4}}{35 b^4 x^4}+\frac {48 c \sqrt {b x^2+c x^4}}{35 b^3 x^6}-\frac {8 \sqrt {b x^2+c x^4}}{7 b^2 x^8}+\frac {1}{b x^6 \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.23, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2015, 2016, 2014} \begin {gather*} \frac {128 c^3 \sqrt {b x^2+c x^4}}{35 b^5 x^2}-\frac {64 c^2 \sqrt {b x^2+c x^4}}{35 b^4 x^4}+\frac {48 c \sqrt {b x^2+c x^4}}{35 b^3 x^6}-\frac {8 \sqrt {b x^2+c x^4}}{7 b^2 x^8}+\frac {1}{b x^6 \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

1/(b*x^6*Sqrt[b*x^2 + c*x^4]) - (8*Sqrt[b*x^2 + c*x^4])/(7*b^2*x^8) + (48*c*Sqrt[b*x^2 + c*x^4])/(35*b^3*x^6)
- (64*c^2*Sqrt[b*x^2 + c*x^4])/(35*b^4*x^4) + (128*c^3*Sqrt[b*x^2 + c*x^4])/(35*b^5*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 2015

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] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n, j
] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (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^5 \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{b x^6 \sqrt {b x^2+c x^4}}+\frac {8 \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx}{b}\\ &=\frac {1}{b x^6 \sqrt {b x^2+c x^4}}-\frac {8 \sqrt {b x^2+c x^4}}{7 b^2 x^8}-\frac {(48 c) \int \frac {1}{x^5 \sqrt {b x^2+c x^4}} \, dx}{7 b^2}\\ &=\frac {1}{b x^6 \sqrt {b x^2+c x^4}}-\frac {8 \sqrt {b x^2+c x^4}}{7 b^2 x^8}+\frac {48 c \sqrt {b x^2+c x^4}}{35 b^3 x^6}+\frac {\left (192 c^2\right ) \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx}{35 b^3}\\ &=\frac {1}{b x^6 \sqrt {b x^2+c x^4}}-\frac {8 \sqrt {b x^2+c x^4}}{7 b^2 x^8}+\frac {48 c \sqrt {b x^2+c x^4}}{35 b^3 x^6}-\frac {64 c^2 \sqrt {b x^2+c x^4}}{35 b^4 x^4}-\frac {\left (128 c^3\right ) \int \frac {1}{x \sqrt {b x^2+c x^4}} \, dx}{35 b^4}\\ &=\frac {1}{b x^6 \sqrt {b x^2+c x^4}}-\frac {8 \sqrt {b x^2+c x^4}}{7 b^2 x^8}+\frac {48 c \sqrt {b x^2+c x^4}}{35 b^3 x^6}-\frac {64 c^2 \sqrt {b x^2+c x^4}}{35 b^4 x^4}+\frac {128 c^3 \sqrt {b x^2+c x^4}}{35 b^5 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 68, normalized size = 0.52 \begin {gather*} \frac {-5 b^4+8 b^3 c x^2-16 b^2 c^2 x^4+64 b c^3 x^6+128 c^4 x^8}{35 b^5 x^6 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(-5*b^4 + 8*b^3*c*x^2 - 16*b^2*c^2*x^4 + 64*b*c^3*x^6 + 128*c^4*x^8)/(35*b^5*x^6*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.30, size = 77, normalized size = 0.59 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-5 b^4+8 b^3 c x^2-16 b^2 c^2 x^4+64 b c^3 x^6+128 c^4 x^8\right )}{35 b^5 x^8 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^5*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-5*b^4 + 8*b^3*c*x^2 - 16*b^2*c^2*x^4 + 64*b*c^3*x^6 + 128*c^4*x^8))/(35*b^5*x^8*(b + c*
x^2))

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fricas [A]  time = 2.05, size = 76, normalized size = 0.58 \begin {gather*} \frac {{\left (128 \, c^{4} x^{8} + 64 \, b c^{3} x^{6} - 16 \, b^{2} c^{2} x^{4} + 8 \, b^{3} c x^{2} - 5 \, b^{4}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, {\left (b^{5} c x^{10} + b^{6} x^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

1/35*(128*c^4*x^8 + 64*b*c^3*x^6 - 16*b^2*c^2*x^4 + 8*b^3*c*x^2 - 5*b^4)*sqrt(c*x^4 + b*x^2)/(b^5*c*x^10 + b^6
*x^8)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^5), x)

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maple [A]  time = 0.01, size = 72, normalized size = 0.55 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-128 c^{4} x^{8}-64 c^{3} x^{6} b +16 c^{2} x^{4} b^{2}-8 c \,x^{2} b^{3}+5 b^{4}\right )}{35 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{5} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/35*(c*x^2+b)*(-128*c^4*x^8-64*b*c^3*x^6+16*b^2*c^2*x^4-8*b^3*c*x^2+5*b^4)/x^4/b^5/(c*x^4+b*x^2)^(3/2)

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maxima [A]  time = 1.47, size = 113, normalized size = 0.87 \begin {gather*} \frac {128 \, c^{4} x^{2}}{35 \, \sqrt {c x^{4} + b x^{2}} b^{5}} + \frac {64 \, c^{3}}{35 \, \sqrt {c x^{4} + b x^{2}} b^{4}} - \frac {16 \, c^{2}}{35 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}} + \frac {8 \, c}{35 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{4}} - \frac {1}{7 \, \sqrt {c x^{4} + b x^{2}} b x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

128/35*c^4*x^2/(sqrt(c*x^4 + b*x^2)*b^5) + 64/35*c^3/(sqrt(c*x^4 + b*x^2)*b^4) - 16/35*c^2/(sqrt(c*x^4 + b*x^2
)*b^3*x^2) + 8/35*c/(sqrt(c*x^4 + b*x^2)*b^2*x^4) - 1/7/(sqrt(c*x^4 + b*x^2)*b*x^6)

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mupad [B]  time = 4.41, size = 114, normalized size = 0.88 \begin {gather*} \frac {13\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,b^3\,x^6}-\frac {\sqrt {c\,x^4+b\,x^2}}{7\,b^2\,x^8}-\frac {29\,c^2\,\sqrt {c\,x^4+b\,x^2}}{35\,b^4\,x^4}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {93\,c^3}{35\,b^4}+\frac {128\,c^4\,x^2}{35\,b^5}\right )}{x^2\,\left (c\,x^2+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(b*x^2 + c*x^4)^(3/2)),x)

[Out]

(13*c*(b*x^2 + c*x^4)^(1/2))/(35*b^3*x^6) - (b*x^2 + c*x^4)^(1/2)/(7*b^2*x^8) - (29*c^2*(b*x^2 + c*x^4)^(1/2))
/(35*b^4*x^4) + ((b*x^2 + c*x^4)^(1/2)*((93*c^3)/(35*b^4) + (128*c^4*x^2)/(35*b^5)))/(x^2*(b + c*x^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{5} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(1/(x**5*(x**2*(b + c*x**2))**(3/2)), x)

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