3.2.35 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=145 \[ -\frac {3 a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^3 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}-\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 145, 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} \begin {gather*} -\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {3 a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^3 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^4,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(a + b*x)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^2*(a + b
*x)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x)) + (b^3*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^4} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^3}{x^4} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^3 b^3}{x^4}+\frac {3 a^2 b^4}{x^3}+\frac {3 a b^5}{x^2}+\frac {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}}{3 x^3 (a+b x)}-\frac {3 a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^3 \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 = 57, normalized size = 0.39 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (a \left (2 a^2+9 a b x+18 b^2 x^2\right )-6 b^3 x^3 \log (x)\right )}{6 x^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^4,x]

[Out]

-1/6*(Sqrt[(a + b*x)^2]*(a*(2*a^2 + 9*a*b*x + 18*b^2*x^2) - 6*b^3*x^3*Log[x]))/(x^3*(a + b*x))

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IntegrateAlgebraic [A]  time = 1.03, size = 268, normalized size = 1.85 \begin {gather*} -\frac {1}{2} \left (b^2\right )^{3/2} \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-\frac {1}{2} \left (b^2\right )^{3/2} \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )+b^3 \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )+\frac {a \sqrt {a^2+2 a b x+b^2 x^2} \left (-2 a^2 b-9 a b^2 x-18 b^3 x^2\right )+a \sqrt {b^2} \left (2 a^3+11 a^2 b x+27 a b^2 x^2+18 b^3 x^3\right )}{6 x^3 \left (a b+b^2 x\right )-6 \sqrt {b^2} x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^4,x]

[Out]

(a*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-2*a^2*b - 9*a*b^2*x - 18*b^3*x^2) + a*Sqrt[b^2]*(2*a^3 + 11*a^2*b*x + 27*a*
b^2*x^2 + 18*b^3*x^3))/(6*x^3*(a*b + b^2*x) - 6*Sqrt[b^2]*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + b^3*ArcTanh[(Sq
rt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] - ((b^2)^(3/2)*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2
*x^2]])/2 - ((b^2)^(3/2)*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2

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fricas [A]  time = 0.39, size = 37, normalized size = 0.26 \begin {gather*} \frac {6 \, b^{3} x^{3} \log \relax (x) - 18 \, a b^{2} x^{2} - 9 \, a^{2} b x - 2 \, a^{3}}{6 \, x^{3}} \end {gather*}

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^4,x, algorithm="fricas")

[Out]

1/6*(6*b^3*x^3*log(x) - 18*a*b^2*x^2 - 9*a^2*b*x - 2*a^3)/x^3

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giac [A]  time = 0.16, size = 59, normalized size = 0.41 \begin {gather*} b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {18 \, a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 9 \, a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} \mathrm {sgn}\left (b x + a\right )}{6 \, x^{3}} \end {gather*}

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^4,x, algorithm="giac")

[Out]

b^3*log(abs(x))*sgn(b*x + a) - 1/6*(18*a*b^2*x^2*sgn(b*x + a) + 9*a^2*b*x*sgn(b*x + a) + 2*a^3*sgn(b*x + a))/x
^3

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maple [A]  time = 0.06, size = 54, normalized size = 0.37 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (6 b^{3} x^{3} \ln \relax (x )-18 a \,b^{2} x^{2}-9 a^{2} b x -2 a^{3}\right )}{6 \left (b x +a \right )^{3} x^{3}} \end {gather*}

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^4,x)

[Out]

1/6*((b*x+a)^2)^(3/2)*(6*b^3*ln(x)*x^3-18*a*b^2*x^2-9*a^2*b*x-2*a^3)/(b*x+a)^3/x^3

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maxima [B]  time = 1.42, size = 229, normalized size = 1.58 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4} x}{2 \, a^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}}{2 \, a} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{3}}{6 \, a^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}}{2 \, a^{2} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{3 \, a^{2} x^{3}} \end {gather*}

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^4,x, algorithm="maxima")

[Out]

(-1)^(2*b^2*x + 2*a*b)*b^3*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x)
) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4*x/a^2 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3/a - 1/6*(b^2*x^2 + 2*a
*b*x + a^2)^(3/2)*b^3/a^3 - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2/(a^2*x) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5
/2)*b/(a^3*x^2) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)/(a^2*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{x^4} \,d x \end {gather*}

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^4,x)

[Out]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{4}}\, dx \end {gather*}

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**4,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/x**4, x)

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