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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 142, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.02, size = 56, normalized size = 0.39 \[ \frac {\sqrt {(a+b x)^2} \left (-2 a^3+6 a^2 b x \log (x)+6 a b^2 x^2+b^3 x^3\right )}{2 x (a+b x)} \]

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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fricas [A]  time = 0.92, size = 36, normalized size = 0.25 \[ \frac {b^{3} x^{3} + 6 \, a b^{2} x^{2} + 6 \, a^{2} b x \log \relax (x) - 2 \, a^{3}}{2 \, x} \]

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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giac [A]  time = 0.21, size = 57, normalized size = 0.40 \[ \frac {1}{2} \, b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a b^{2} x \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {a^{3} \mathrm {sgn}\left (b x + a\right )}{x} \]

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

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maple [A]  time = 0.05, size = 53, normalized size = 0.37 \[ \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (b^{3} x^{3}+6 a^{2} b x \ln \relax (x )+6 a \,b^{2} x^{2}-2 a^{3}\right )}{2 \left (b x +a \right )^{3} x} \]

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

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maxima [A]  time = 1.34, size = 140, normalized size = 0.99 \[ 3 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} a^{2} b \log \left (2 \, b^{2} x + 2 \, a b\right ) - 3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a^{2} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2} x + \frac {9}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a b - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}{x} \]

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]

3*(-1)^(2*b^2*x + 2*a*b)*a^2*b*log(2*b^2*x + 2*a*b) - 3*(-1)^(2*a*b*x + 2*a^2)*a^2*b*log(2*a*b*x/abs(x) + 2*a^
2/abs(x)) + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2*x + 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*b - (b^2*x^2 + 2*a*b
*x + a^2)^(3/2)/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{2}}\, dx \]

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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