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]

-(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))

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Rubi [A]  time = 0.0581569, antiderivative size = 79, normalized size of antiderivative = 1., 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 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])

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 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])

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*}

Mathematica [A]  time = 0.0080135, 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]

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

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Maple [A]  time = 0.044, size = 36, normalized size = 0.5 \begin{align*} -{\frac{5\,b{x}^{2}+4\,a}{40\,{x}^{10} \left ( b{x}^{2}+a \right ) }\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.43148, size = 38, normalized size = 0.48 \begin{align*} -\frac{5 \, b x^{2} + 4 \, a}{40 \, x^{10}} \end{align*}

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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Sympy [A]  time = 0.345309, size = 15, normalized size = 0.19 \begin{align*} - \frac{4 a + 5 b x^{2}}{40 x^{10}} \end{align*}

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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Giac [A]  time = 1.15097, size = 42, normalized size = 0.53 \begin{align*} -\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}} \end{align*}

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