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

Optimal. Leaf size=52 \[ \frac {2 c \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6}-\frac {\left (b x^2+c x^4\right )^{3/2}}{5 b x^8} \]

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Rubi [A]  time = 0.08, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, 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 {2 c \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6}-\frac {\left (b x^2+c x^4\right )^{3/2}}{5 b x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^7,x]

[Out]

-(b*x^2 + c*x^4)^(3/2)/(5*b*x^8) + (2*c*(b*x^2 + c*x^4)^(3/2))/(15*b^2*x^6)

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 {\sqrt {b x^2+c x^4}}{x^7} \, dx &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{5 b x^8}-\frac {(2 c) \int \frac {\sqrt {b x^2+c x^4}}{x^5} \, dx}{5 b}\\ &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{5 b x^8}+\frac {2 c \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.67 \begin {gather*} \frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (2 c x^2-3 b\right )}{15 b^2 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^7,x]

[Out]

((x^2*(b + c*x^2))^(3/2)*(-3*b + 2*c*x^2))/(15*b^2*x^8)

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IntegrateAlgebraic [A]  time = 0.13, size = 46, normalized size = 0.88 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-3 b^2-b c x^2+2 c^2 x^4\right )}{15 b^2 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-3*b^2 - b*c*x^2 + 2*c^2*x^4))/(15*b^2*x^6)

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fricas [A]  time = 0.58, size = 42, normalized size = 0.81 \begin {gather*} \frac {{\left (2 \, c^{2} x^{4} - b c x^{2} - 3 \, b^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*c^2*x^4 - b*c*x^2 - 3*b^2)*sqrt(c*x^4 + b*x^2)/(b^2*x^6)

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giac [B]  time = 0.22, size = 120, normalized size = 2.31 \begin {gather*} \frac {4 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{2} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) - b^{3} c^{\frac {5}{2}} \mathrm {sgn}\relax (x)\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b))^6*c^(5/2)*sgn(x) + 5*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b*c^(5/2)*sgn(x) +
5*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^2*c^(5/2)*sgn(x) - b^3*c^(5/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 -
b)^5

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maple [A]  time = 0.00, size = 39, normalized size = 0.75 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-2 c \,x^{2}+3 b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(c*x^2+b)*(-2*c*x^2+3*b)*(c*x^4+b*x^2)^(1/2)/b^2/x^6

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maxima [A]  time = 1.42, size = 65, normalized size = 1.25 \begin {gather*} \frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{15 \, b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{15 \, b x^{4}} - \frac {\sqrt {c x^{4} + b x^{2}}}{5 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*sqrt(c*x^4 + b*x^2)*c^2/(b^2*x^2) - 1/15*sqrt(c*x^4 + b*x^2)*c/(b*x^4) - 1/5*sqrt(c*x^4 + b*x^2)/x^6

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mupad [B]  time = 4.26, size = 41, normalized size = 0.79 \begin {gather*} -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (3\,b^2+b\,c\,x^2-2\,c^2\,x^4\right )}{15\,b^2\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(1/2)/x^7,x)

[Out]

-((b*x^2 + c*x^4)^(1/2)*(3*b^2 - 2*c^2*x^4 + b*c*x^2))/(15*b^2*x^6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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