### 3.146 $$\int \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{x^5} \, dx$$

Optimal. Leaf size=71 $-\frac{a \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac{b \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}$

[Out]

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

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Rubi [A]  time = 0.0185848, 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} $-\frac{a \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac{b \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/x^5,x]

[Out]

-(a*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^4*(a + b*x)) - (b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(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^5} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{a b+b^2 x}{x^5} \, dx}{a b+b^2 x}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (\frac{a b}{x^5}+\frac{b^2}{x^4}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{a \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac{b \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0079465, size = 33, normalized size = 0.46 $-\frac{\sqrt{(a+b x)^2} (3 a+4 b x)}{12 x^4 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/x^5,x]

[Out]

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

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Maple [A]  time = 0.044, size = 30, normalized size = 0.4 \begin{align*} -{\frac{4\,bx+3\,a}{12\,{x}^{4} \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^5,x)

[Out]

-1/12*(4*b*x+3*a)*((b*x+a)^2)^(1/2)/x^4/(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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88232, size = 34, normalized size = 0.48 \begin{align*} -\frac{4 \, b x + 3 \, a}{12 \, x^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

-1/12*(4*b*x + 3*a)/x^4

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Sympy [A]  time = 0.665221, size = 14, normalized size = 0.2 \begin{align*} - \frac{3 a + 4 b x}{12 x^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)**2)**(1/2)/x**5,x)

[Out]

-(3*a + 4*b*x)/(12*x**4)

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Giac [A]  time = 1.24318, size = 54, normalized size = 0.76 \begin{align*} -\frac{b^{4} \mathrm{sgn}\left (b x + a\right )}{12 \, a^{3}} - \frac{4 \, b x \mathrm{sgn}\left (b x + a\right ) + 3 \, a \mathrm{sgn}\left (b x + a\right )}{12 \, x^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

-1/12*b^4*sgn(b*x + a)/a^3 - 1/12*(4*b*x*sgn(b*x + a) + 3*a*sgn(b*x + a))/x^4