### 3.154 $$\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{x} \, dx$$

Optimal. Leaf size=143 $\frac{3 a^2 b x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{3 a b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{a^3 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}$

[Out]

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

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Rubi [A]  time = 0.0341437, antiderivative size = 143, 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} $\frac{3 a^2 b x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{3 a b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{a^3 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0159683, size = 52, normalized size = 0.36 $\frac{\sqrt{(a+b x)^2} \left (b x \left (18 a^2+9 a b x+2 b^2 x^2\right )+6 a^3 \log (x)\right )}{6 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(b*x*(18*a^2 + 9*a*b*x + 2*b^2*x^2) + 6*a^3*Log[x]))/(6*(a + b*x))

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Maple [A]  time = 0.219, size = 51, normalized size = 0.4 \begin{align*}{\frac{2\,{b}^{3}{x}^{3}+9\,a{b}^{2}{x}^{2}+6\,{a}^{3}\ln \left ( x \right ) +18\,b{a}^{2}x}{6\, \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,x)

[Out]

1/6*((b*x+a)^2)^(3/2)*(2*b^3*x^3+9*a*b^2*x^2+6*a^3*ln(x)+18*b*a^2*x)/(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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96813, size = 73, normalized size = 0.51 \begin{align*} \frac{1}{3} \, b^{3} x^{3} + \frac{3}{2} \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3} \log \left (x\right ) \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,x, algorithm="fricas")

[Out]

1/3*b^3*x^3 + 3/2*a*b^2*x^2 + 3*a^2*b*x + a^3*log(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}\, 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,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/x, x)

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Giac [A]  time = 1.13416, size = 76, normalized size = 0.53 \begin{align*} \frac{1}{3} \, b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b x \mathrm{sgn}\left (b x + a\right ) + a^{3} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) \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,x, algorithm="giac")

[Out]

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