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

Optimal. Leaf size=141 \[ -\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^3 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{3 a b^2 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.0355548, antiderivative size = 141, 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} \[ -\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^3 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{3 a b^2 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0155769, size = 55, normalized size = 0.39 \[ -\frac{\sqrt{(a+b x)^2} \left (6 a^2 b x+a^3-6 a b^2 x^2 \log (x)-2 b^3 x^3\right )}{2 x^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.209, size = 54, normalized size = 0.4 \begin{align*}{\frac{6\,{b}^{2}a\ln \left ( x \right ){x}^{2}+2\,{b}^{3}{x}^{3}-6\,b{a}^{2}x-{a}^{3}}{2\, \left ( bx+a \right ) ^{3}{x}^{2}} \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^3,x)

[Out]

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

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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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99319, size = 81, normalized size = 0.57 \begin{align*} \frac{2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} \log \left (x\right ) - 6 \, a^{2} b x - a^{3}}{2 \, x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

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

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Giac [A]  time = 1.24739, size = 76, normalized size = 0.54 \begin{align*} b^{3} x \mathrm{sgn}\left (b x + a\right ) + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{6 \, a^{2} b x \mathrm{sgn}\left (b x + a\right ) + a^{3} \mathrm{sgn}\left (b x + a\right )}{2 \, x^{2}} \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^3,x, algorithm="giac")

[Out]

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