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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 141, 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} \[ -\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^3 x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {3 a b^2 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

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 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^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*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.39 \[ -\frac {\sqrt {(a+b x)^2} \left (a^3+6 a^2 b x-6 a b^2 x^2 \log (x)-2 b^3 x^3\right )}{2 x^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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]))/(x^2*(a + b*x))

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fricas [A]  time = 0.98, size = 37, normalized size = 0.26 \[ \frac {2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} \log \relax (x) - 6 \, a^{2} b x - a^{3}}{2 \, x^{2}} \]

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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giac [A]  time = 0.16, size = 56, normalized size = 0.40 \[ b^{3} x \mathrm {sgn}\left (b x + a\right ) + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {6 \, a^{2} b x \mathrm {sgn}\left (b x + a\right ) + a^{3} \mathrm {sgn}\left (b x + a\right )}{2 \, x^{2}} \]

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

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maple [A]  time = 0.06, size = 54, normalized size = 0.38 \[ \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (6 a \,b^{2} x^{2} \ln \relax (x )+2 b^{3} x^{3}-6 a^{2} b x -a^{3}\right )}{2 \left (b x +a \right )^{3} x^{2}} \]

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

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maxima [B]  time = 1.45, size = 200, normalized size = 1.42 \[ 3 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} a b^{2} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3} x}{2 \, a} + \frac {9}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}}{2 \, a^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b}{2 \, a x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{2 \, a^{2} x^{2}} \]

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]

3*(-1)^(2*b^2*x + 2*a*b)*a*b^2*log(2*b^2*x + 2*a*b) - 3*(-1)^(2*a*b*x + 2*a^2)*a*b^2*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^3*x/a + 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2 + 1/2*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*b^2/a^2 - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b/(a*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(5
/2)/(a^2*x^2)

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

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^3,x)

[Out]

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

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

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**3,x)

[Out]

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

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