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

Optimal. Leaf size=142 \[ -\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{3 a b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{3 a^2 b \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.0344887, antiderivative size = 142, 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} \[ -\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{3 a b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{3 a^2 b \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))) + (3*a*b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (
b^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a + b*x)) + (3*a^2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[x])/(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{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^3}{x^2} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (3 a b^5+\frac{a^3 b^3}{x^2}+\frac{3 a^2 b^4}{x}+b^6 x\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{3 a b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2} \log (x)}{a+b x}\\ \end{align*}

Mathematica [A]  time = 0.0171186, size = 56, normalized size = 0.39 \[ \frac{\sqrt{(a+b x)^2} \left (6 a^2 b x \log (x)-2 a^3+6 a b^2 x^2+b^3 x^3\right )}{2 x (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.222, size = 53, normalized size = 0.4 \begin{align*}{\frac{{b}^{3}{x}^{3}+6\,b{a}^{2}\ln \left ( x \right ) x+6\,a{b}^{2}{x}^{2}-2\,{a}^{3}}{2\, \left ( bx+a \right ) ^{3}x} \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^2,x)

[Out]

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

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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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88336, size = 78, normalized size = 0.55 \begin{align*} \frac{b^{3} x^{3} + 6 \, a b^{2} x^{2} + 6 \, a^{2} b x \log \left (x\right ) - 2 \, a^{3}}{2 \, x} \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^2,x, algorithm="fricas")

[Out]

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

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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^{2}}\, 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**2,x)

[Out]

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

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Giac [A]  time = 1.28843, size = 77, normalized size = 0.54 \begin{align*} \frac{1}{2} \, b^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, a b^{2} x \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{a^{3} \mathrm{sgn}\left (b x + a\right )}{x} \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^2,x, algorithm="giac")

[Out]

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