3.158 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=37 \[ -\frac{(a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.0136598, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {646, 37} \[ -\frac{(a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^5,x]

[Out]

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

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 37

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

Rubi steps

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

Mathematica [A]  time = 0.0119043, size = 53, normalized size = 1.43 \[ -\frac{\sqrt{(a+b x)^2} \left (4 a^2 b x+a^3+6 a b^2 x^2+4 b^3 x^3\right )}{4 x^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^5,x]

[Out]

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

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Maple [B]  time = 0.166, size = 50, normalized size = 1.4 \begin{align*} -{\frac{4\,{b}^{3}{x}^{3}+6\,a{b}^{2}{x}^{2}+4\,b{a}^{2}x+{a}^{3}}{4\,{x}^{4} \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^5,x)

[Out]

-1/4*(4*b^3*x^3+6*a*b^2*x^2+4*a^2*b*x+a^3)*((b*x+a)^2)^(3/2)/x^4/(b*x+a)^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95822, size = 73, normalized size = 1.97 \begin{align*} -\frac{4 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 4 \, a^{2} b x + a^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^5,x, algorithm="fricas")

[Out]

-1/4*(4*b^3*x^3 + 6*a*b^2*x^2 + 4*a^2*b*x + a^3)/x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/x**5,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/x**5, x)

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Giac [B]  time = 1.37638, size = 99, normalized size = 2.68 \begin{align*} -\frac{b^{4} \mathrm{sgn}\left (b x + a\right )}{4 \, a} - \frac{4 \, b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \, a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 4 \, a^{2} b x \mathrm{sgn}\left (b x + a\right ) + a^{3} \mathrm{sgn}\left (b x + a\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^5,x, algorithm="giac")

[Out]

-1/4*b^4*sgn(b*x + a)/a - 1/4*(4*b^3*x^3*sgn(b*x + a) + 6*a*b^2*x^2*sgn(b*x + a) + 4*a^2*b*x*sgn(b*x + a) + a^
3*sgn(b*x + a))/x^4