3.245 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^{11}} \, dx\)

Optimal. Leaf size=52 \[ \frac {2 c \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}}-\frac {\left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}} \]

[Out]

-1/7*(c*x^4+b*x^2)^(5/2)/b/x^12+2/35*c*(c*x^4+b*x^2)^(5/2)/b^2/x^10

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Rubi [A]  time = 0.09, 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} \[ \frac {2 c \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}}-\frac {\left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^11,x]

[Out]

-(b*x^2 + c*x^4)^(5/2)/(7*b*x^12) + (2*c*(b*x^2 + c*x^4)^(5/2))/(35*b^2*x^10)

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 {\left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx &=-\frac {\left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}-\frac {(2 c) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx}{7 b}\\ &=-\frac {\left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}+\frac {2 c \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.67 \[ \frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (2 c x^2-5 b\right )}{35 b^2 x^{12}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^11,x]

[Out]

((x^2*(b + c*x^2))^(5/2)*(-5*b + 2*c*x^2))/(35*b^2*x^12)

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fricas [A]  time = 0.62, size = 53, normalized size = 1.02 \[ \frac {{\left (2 \, c^{3} x^{6} - b c^{2} x^{4} - 8 \, b^{2} c x^{2} - 5 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, b^{2} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^11,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.25, size = 178, normalized size = 3.42 \[ \frac {4 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{2} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 14 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{3} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 7 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{4} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) - b^{5} c^{\frac {7}{2}} \mathrm {sgn}\relax (x)\right )}}{35 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^11,x, algorithm="giac")

[Out]

4/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b))^10*c^(7/2)*sgn(x) + 35*(sqrt(c)*x - sqrt(c*x^2 + b))^8*b*c^(7/2)*sgn(x)
+ 70*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b^2*c^(7/2)*sgn(x) + 14*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^3*c^(7/2)*sgn(
x) + 7*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^4*c^(7/2)*sgn(x) - b^5*c^(7/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))
^2 - b)^7

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maple [A]  time = 0.00, size = 39, normalized size = 0.75 \[ -\frac {\left (c \,x^{2}+b \right ) \left (-2 c \,x^{2}+5 b \right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^11,x)

[Out]

-1/35*(c*x^2+b)*(-2*c*x^2+5*b)*(c*x^4+b*x^2)^(3/2)/x^10/b^2

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maxima [B]  time = 1.46, size = 105, normalized size = 2.02 \[ \frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{35 \, b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c^{2}}{35 \, b x^{4}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} c}{140 \, x^{6}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{28 \, x^{8}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{4 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^11,x, algorithm="maxima")

[Out]

2/35*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^2) - 1/35*sqrt(c*x^4 + b*x^2)*c^2/(b*x^4) + 3/140*sqrt(c*x^4 + b*x^2)*c/x^
6 + 3/28*sqrt(c*x^4 + b*x^2)*b/x^8 - 1/4*(c*x^4 + b*x^2)^(3/2)/x^10

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mupad [B]  time = 4.58, size = 87, normalized size = 1.67 \[ \frac {2\,c^3\,\sqrt {c\,x^4+b\,x^2}}{35\,b^2\,x^2}-\frac {8\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,x^6}-\frac {c^2\,\sqrt {c\,x^4+b\,x^2}}{35\,b\,x^4}-\frac {b\,\sqrt {c\,x^4+b\,x^2}}{7\,x^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^11,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{11}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**11,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**11, x)

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