3.1.40 \(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=165 \[ -\frac {3 a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}+\frac {b^3 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 165, 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 {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {b^3 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{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^8,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(7*x^7*(a + b*x^3)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*x^4*
(a + b*x^3)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x*(a + b*x^3)) + (b^3*x^2*Sqrt[a^2 + 2*a*b*x^3 + b^2
*x^6])/(2*(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^8} \, 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^8} \, 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 (\frac {a^3 b^3}{x^8}+\frac {3 a^2 b^4}{x^5}+\frac {3 a b^5}{x^2}+b^6 x\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}}{7 x^7 \left (a+b x^3\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {b^3 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 61, normalized size = 0.37 \begin {gather*} -\frac {\sqrt {\left (a+b x^3\right )^2} \left (4 a^3+21 a^2 b x^3+84 a b^2 x^6-14 b^3 x^9\right )}{28 x^7 \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^8,x]

[Out]

-1/28*(Sqrt[(a + b*x^3)^2]*(4*a^3 + 21*a^2*b*x^3 + 84*a*b^2*x^6 - 14*b^3*x^9))/(x^7*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 20.10, size = 61, normalized size = 0.37 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (-4 a^3-21 a^2 b x^3-84 a b^2 x^6+14 b^3 x^9\right )}{28 x^7 \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^8,x]

[Out]

(Sqrt[(a + b*x^3)^2]*(-4*a^3 - 21*a^2*b*x^3 - 84*a*b^2*x^6 + 14*b^3*x^9))/(28*x^7*(a + b*x^3))

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fricas [A]  time = 1.92, size = 37, normalized size = 0.22 \begin {gather*} \frac {14 \, b^{3} x^{9} - 84 \, a b^{2} x^{6} - 21 \, a^{2} b x^{3} - 4 \, a^{3}}{28 \, x^{7}} \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^8,x, algorithm="fricas")

[Out]

1/28*(14*b^3*x^9 - 84*a*b^2*x^6 - 21*a^2*b*x^3 - 4*a^3)/x^7

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giac [A]  time = 0.47, size = 70, normalized size = 0.42 \begin {gather*} \frac {1}{2} \, b^{3} x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {84 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 21 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 4 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{28 \, x^{7}} \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^8,x, algorithm="giac")

[Out]

1/2*b^3*x^2*sgn(b*x^3 + a) - 1/28*(84*a*b^2*x^6*sgn(b*x^3 + a) + 21*a^2*b*x^3*sgn(b*x^3 + a) + 4*a^3*sgn(b*x^3
+ a))/x^7

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maple [A]  time = 0.01, size = 58, normalized size = 0.35 \begin {gather*} -\frac {\left (-14 b^{3} x^{9}+84 a \,b^{2} x^{6}+21 a^{2} b \,x^{3}+4 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}}}{28 \left (b \,x^{3}+a \right )^{3} x^{7}} \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^8,x)

[Out]

-1/28*(-14*b^3*x^9+84*a*b^2*x^6+21*a^2*b*x^3+4*a^3)*((b*x^3+a)^2)^(3/2)/x^7/(b*x^3+a)^3

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maxima [A]  time = 0.74, size = 37, normalized size = 0.22 \begin {gather*} \frac {14 \, b^{3} x^{9} - 84 \, a b^{2} x^{6} - 21 \, a^{2} b x^{3} - 4 \, a^{3}}{28 \, x^{7}} \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^8,x, algorithm="maxima")

[Out]

1/28*(14*b^3*x^9 - 84*a*b^2*x^6 - 21*a^2*b*x^3 - 4*a^3)/x^7

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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^8} \,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^8,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^8, 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^{8}}\, 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**8,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**8, x)

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