3.1.33 \(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x} \, dx\)

Optimal. Leaf size=160 \[ \frac {a b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {a^2 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {a^3 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1355, 266, 43} \begin {gather*} \frac {b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {a b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {a^2 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {a^3 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x,x]

[Out]

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (a b+b^2 x^3\right )^3}{x} \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \operatorname {Subst}\left (\int \frac {\left (a b+b^2 x\right )^3}{x} \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \operatorname {Subst}\left (\int \left (3 a^2 b^4+\frac {a^3 b^3}{x}+3 a b^5 x+b^6 x^2\right ) \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {a^2 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {a b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 60, normalized size = 0.38 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (18 a^3 \log (x)+b x^3 \left (18 a^2+9 a b x^3+2 b^2 x^6\right )\right )}{18 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.44, size = 256, normalized size = 1.60 \begin {gather*} \frac {1}{36} \sqrt {a^2+2 a b x^3+b^2 x^6} \left (11 a^2+7 a b x^3+2 b^2 x^6\right )+\frac {1}{36} \left (-18 a^2 \sqrt {b^2} x^3-9 a b \sqrt {b^2} x^6-2 \left (b^2\right )^{3/2} x^9\right )+\frac {1}{6} a^3 \log \left (\sqrt {a^2+2 a b x^3+b^2 x^6}-a-\sqrt {b^2} x^3\right )-\frac {a^3 \left (\sqrt {b^2}+b\right ) \log \left (\sqrt {a^2+2 a b x^3+b^2 x^6}+a-\sqrt {b^2} x^3\right )}{6 b}-\frac {a^3 \sqrt {b^2} \log \left (b \sqrt {a^2+2 a b x^3+b^2 x^6}-a b-b \sqrt {b^2} x^3\right )}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x,x]

[Out]

(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(11*a^2 + 7*a*b*x^3 + 2*b^2*x^6))/36 + (-18*a^2*Sqrt[b^2]*x^3 - 9*a*b*Sqrt[b^
2]*x^6 - 2*(b^2)^(3/2)*x^9)/36 + (a^3*Log[-a - Sqrt[b^2]*x^3 + Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]])/6 - (a^3*(b +
Sqrt[b^2])*Log[a - Sqrt[b^2]*x^3 + Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]])/(6*b) - (a^3*Sqrt[b^2]*Log[-(a*b) - b*Sq
rt[b^2]*x^3 + b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]])/(6*b)

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fricas [A]  time = 1.09, size = 32, normalized size = 0.20 \begin {gather*} \frac {1}{9} \, b^{3} x^{9} + \frac {1}{2} \, a b^{2} x^{6} + a^{2} b x^{3} + a^{3} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.43, size = 65, normalized size = 0.41 \begin {gather*} \frac {1}{9} \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{2} \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 57, normalized size = 0.36 \begin {gather*} \frac {\left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}} \left (2 b^{3} x^{9}+9 a \,b^{2} x^{6}+18 a^{2} b \,x^{3}+18 a^{3} \ln \relax (x )\right )}{18 \left (b \,x^{3}+a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x,x)

[Out]

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

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maxima [A]  time = 0.52, size = 152, normalized size = 0.95 \begin {gather*} \frac {1}{6} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a b x^{3} + \frac {1}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a^{3} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {1}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2} + \frac {1}{9} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x,x, algorithm="maxima")

[Out]

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

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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^3+b^2\,x^6\right )}^{3/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x, 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^{3}\right )^{2}\right )^{\frac {3}{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x,x)

[Out]

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

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