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

Optimal. Leaf size=163 \[ \frac {3 a^2 b x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {3 a b^2 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {b^3 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1355, 270} \begin {gather*} \frac {b^3 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}+\frac {3 a b^2 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {3 a^2 b x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*x^2*(a + b*x^3)) + (3*a^2*b*x*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a +
b*x^3) + (3*a*b^2*x^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*(a + b*x^3)) + (b^3*x^7*Sqrt[a^2 + 2*a*b*x^3 + b^2*x
^6])/(7*(a + b*x^3))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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^3} \, 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^3} \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (3 a^2 b^4+\frac {a^3 b^3}{x^3}+3 a b^5 x^3+b^6 x^6\right ) \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}+\frac {3 a^2 b x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {3 a b^2 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {b^3 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.37 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (-14 a^3+84 a^2 b x^3+21 a b^2 x^6+4 b^3 x^9\right )}{28 x^2 \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^3,x]

[Out]

(Sqrt[(a + b*x^3)^2]*(-14*a^3 + 84*a^2*b*x^3 + 21*a*b^2*x^6 + 4*b^3*x^9))/(28*x^2*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 19.88, size = 61, normalized size = 0.37 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (-14 a^3+84 a^2 b x^3+21 a b^2 x^6+4 b^3 x^9\right )}{28 x^2 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^3)^2]*(-14*a^3 + 84*a^2*b*x^3 + 21*a*b^2*x^6 + 4*b^3*x^9))/(28*x^2*(a + b*x^3))

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fricas [A]  time = 1.47, size = 37, normalized size = 0.23 \begin {gather*} \frac {4 \, b^{3} x^{9} + 21 \, a b^{2} x^{6} + 84 \, a^{2} b x^{3} - 14 \, a^{3}}{28 \, x^{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^3,x, algorithm="fricas")

[Out]

1/28*(4*b^3*x^9 + 21*a*b^2*x^6 + 84*a^2*b*x^3 - 14*a^3)/x^2

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giac [A]  time = 0.38, size = 65, normalized size = 0.40 \begin {gather*} \frac {1}{7} \, b^{3} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{4} \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, a^{2} b x \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{2 \, x^{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^3,x, algorithm="giac")

[Out]

1/7*b^3*x^7*sgn(b*x^3 + a) + 3/4*a*b^2*x^4*sgn(b*x^3 + a) + 3*a^2*b*x*sgn(b*x^3 + a) - 1/2*a^3*sgn(b*x^3 + a)/
x^2

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maple [A]  time = 0.01, size = 58, normalized size = 0.36 \begin {gather*} -\frac {\left (-4 b^{3} x^{9}-21 a \,b^{2} x^{6}-84 a^{2} b \,x^{3}+14 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}}}{28 \left (b \,x^{3}+a \right )^{3} x^{2}} \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^3,x)

[Out]

-1/28*(-4*b^3*x^9-21*a*b^2*x^6-84*a^2*b*x^3+14*a^3)*((b*x^3+a)^2)^(3/2)/x^2/(b*x^3+a)^3

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maxima [A]  time = 0.50, size = 37, normalized size = 0.23 \begin {gather*} \frac {4 \, b^{3} x^{9} + 21 \, a b^{2} x^{6} + 84 \, a^{2} b x^{3} - 14 \, a^{3}}{28 \, x^{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^3,x, algorithm="maxima")

[Out]

1/28*(4*b^3*x^9 + 21*a*b^2*x^6 + 84*a^2*b*x^3 - 14*a^3)/x^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^3} \,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^3,x)

[Out]

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

[Out]

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

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