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

Optimal. Leaf size=72 \[ \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{12 a^2 x^6}-\frac{\left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 a x^6} \]

[Out]

-((a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*a*x^6) + (a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(12*a^2*x^6)

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Rubi [A]  time = 0.0155318, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {1110} \[ \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{12 a^2 x^6}-\frac{\left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^7,x]

[Out]

-((a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*a*x^6) + (a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(12*a^2*x^6)

Rule 1110

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*x^2
+ c*x^4)^(p + 1))/(4*a*d*(p + 1)*(2*p + 1)), x] - Simp[((d*x)^(m + 1)*(2*a + b*x^2)*(a + b*x^2 + c*x^4)^p)/(4*
a*d*(2*p + 1)), x] /; FreeQ[{a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && EqQ[m + 4*p + 5,
0] && NeQ[p, -2^(-1)]

Rubi steps

\begin{align*} \int \frac{\sqrt{a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx &=-\frac{\left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 a x^6}+\frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{12 a^2 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0085409, size = 39, normalized size = 0.54 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (2 a+3 b x^2\right )}{12 x^6 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^7,x]

[Out]

-(Sqrt[(a + b*x^2)^2]*(2*a + 3*b*x^2))/(12*x^6*(a + b*x^2))

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Maple [A]  time = 0.043, size = 36, normalized size = 0.5 \begin{align*} -{\frac{3\,b{x}^{2}+2\,a}{12\,{x}^{6} \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^7,x)

[Out]

-1/12*(3*b*x^2+2*a)*((b*x^2+a)^2)^(1/2)/x^6/(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^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.42886, size = 36, normalized size = 0.5 \begin{align*} -\frac{3 \, b x^{2} + 2 \, a}{12 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b*x^2 + 2*a)/x^6

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Sympy [A]  time = 0.306986, size = 15, normalized size = 0.21 \begin{align*} - \frac{2 a + 3 b x^{2}}{12 x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a + 3*b*x**2)/(12*x**6)

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Giac [A]  time = 1.14877, size = 42, normalized size = 0.58 \begin{align*} -\frac{3 \, b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 2 \, a \mathrm{sgn}\left (b x^{2} + a\right )}{12 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b*x^2*sgn(b*x^2 + a) + 2*a*sgn(b*x^2 + a))/x^6