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

Optimal. Leaf size=71 \[ -\frac{a \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac{b \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0186558, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {646, 43} \[ -\frac{a \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac{b \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0069367, size = 33, normalized size = 0.46 \[ -\frac{\sqrt{(a+b x)^2} (2 a+3 b x)}{6 x^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 30, normalized size = 0.4 \begin{align*} -{\frac{3\,bx+2\,a}{6\,{x}^{3} \left ( bx+a \right ) }\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*b*x+2*a)*((b*x+a)^2)^(1/2)/x^3/(b*x+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+a)^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89047, size = 32, normalized size = 0.45 \begin{align*} -\frac{3 \, b x + 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*b*x + 2*a)/x^3

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Sympy [A]  time = 0.888519, size = 14, normalized size = 0.2 \begin{align*} - \frac{2 a + 3 b x}{6 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a + 3*b*x)/(6*x**3)

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Giac [A]  time = 1.24129, size = 54, normalized size = 0.76 \begin{align*} \frac{b^{3} \mathrm{sgn}\left (b x + a\right )}{6 \, a^{2}} - \frac{3 \, b x \mathrm{sgn}\left (b x + a\right ) + 2 \, a \mathrm{sgn}\left (b x + a\right )}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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