3.549 \(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^{11}} \, dx\)

Optimal. Leaf size=79 \[ -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )} \]

[Out]

-1/10*a*((b*x^2+a)^2)^(1/2)/x^10/(b*x^2+a)-1/8*b*((b*x^2+a)^2)^(1/2)/x^8/(b*x^2+a)

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Rubi [A]  time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \[ -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^11,x]

[Out]

-(a*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10*x^10*(a + b*x^2)) - (b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*x^8*(a + b
*x^2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1111

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] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps

\begin {align*} \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^{11}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx,x,x^2\right )\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \frac {a b+b^2 x}{x^6} \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \left (\frac {a b}{x^6}+\frac {b^2}{x^5}\right ) \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )}\\ &=-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.49 \[ -\frac {\sqrt {\left (a+b x^2\right )^2} \left (4 a+5 b x^2\right )}{40 x^{10} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^11,x]

[Out]

-1/40*(Sqrt[(a + b*x^2)^2]*(4*a + 5*b*x^2))/(x^10*(a + b*x^2))

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fricas [A]  time = 0.74, size = 15, normalized size = 0.19 \[ -\frac {5 \, b x^{2} + 4 \, a}{40 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^11,x, algorithm="fricas")

[Out]

-1/40*(5*b*x^2 + 4*a)/x^10

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giac [A]  time = 0.16, size = 31, normalized size = 0.39 \[ -\frac {5 \, b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, a \mathrm {sgn}\left (b x^{2} + a\right )}{40 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^11,x, algorithm="giac")

[Out]

-1/40*(5*b*x^2*sgn(b*x^2 + a) + 4*a*sgn(b*x^2 + a))/x^10

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maple [A]  time = 0.00, size = 36, normalized size = 0.46 \[ -\frac {\left (5 b \,x^{2}+4 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{40 \left (b \,x^{2}+a \right ) x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^2+a)^2)^(1/2)/x^11,x)

[Out]

-1/40*(5*b*x^2+4*a)*((b*x^2+a)^2)^(1/2)/x^10/(b*x^2+a)

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maxima [A]  time = 1.31, size = 15, normalized size = 0.19 \[ -\frac {5 \, b x^{2} + 4 \, a}{40 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^11,x, algorithm="maxima")

[Out]

-1/40*(5*b*x^2 + 4*a)/x^10

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mupad [B]  time = 4.21, size = 35, normalized size = 0.44 \[ -\frac {\left (5\,b\,x^2+4\,a\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{40\,x^{10}\,\left (b\,x^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2)^(1/2)/x^11,x)

[Out]

-((4*a + 5*b*x^2)*((a + b*x^2)^2)^(1/2))/(40*x^10*(a + b*x^2))

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sympy [A]  time = 0.22, size = 15, normalized size = 0.19 \[ \frac {- 4 a - 5 b x^{2}}{40 x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x**2+a)**2)**(1/2)/x**11,x)

[Out]

(-4*a - 5*b*x**2)/(40*x**10)

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