3.37 \(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=165 \[ -\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {3 a b^2 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {b^3 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )} \]

[Out]

-1/4*a^3*((b*x^3+a)^2)^(1/2)/x^4/(b*x^3+a)-3*a^2*b*((b*x^3+a)^2)^(1/2)/x/(b*x^3+a)+3/2*a*b^2*x^2*((b*x^3+a)^2)
^(1/2)/(b*x^3+a)+1/5*b^3*x^5*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

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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} \[ \frac {b^3 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {3 a b^2 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^5,x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.37 \[ \frac {\sqrt {\left (a+b x^3\right )^2} \left (-5 a^3-60 a^2 b x^3+30 a b^2 x^6+4 b^3 x^9\right )}{20 x^4 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^5,x]

[Out]

(Sqrt[(a + b*x^3)^2]*(-5*a^3 - 60*a^2*b*x^3 + 30*a*b^2*x^6 + 4*b^3*x^9))/(20*x^4*(a + b*x^3))

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fricas [A]  time = 0.88, size = 37, normalized size = 0.22 \[ \frac {4 \, b^{3} x^{9} + 30 \, a b^{2} x^{6} - 60 \, a^{2} b x^{3} - 5 \, a^{3}}{20 \, x^{4}} \]

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^5,x, algorithm="fricas")

[Out]

1/20*(4*b^3*x^9 + 30*a*b^2*x^6 - 60*a^2*b*x^3 - 5*a^3)/x^4

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giac [A]  time = 0.36, size = 69, normalized size = 0.42 \[ \frac {1}{5} \, b^{3} x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{2} \, a b^{2} x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {12 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{4 \, x^{4}} \]

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^5,x, algorithm="giac")

[Out]

1/5*b^3*x^5*sgn(b*x^3 + a) + 3/2*a*b^2*x^2*sgn(b*x^3 + a) - 1/4*(12*a^2*b*x^3*sgn(b*x^3 + a) + a^3*sgn(b*x^3 +
a))/x^4

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maple [A]  time = 0.01, size = 58, normalized size = 0.35 \[ -\frac {\left (-4 b^{3} x^{9}-30 a \,b^{2} x^{6}+60 a^{2} b \,x^{3}+5 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}}}{20 \left (b \,x^{3}+a \right )^{3} x^{4}} \]

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^5,x)

[Out]

-1/20*(-4*b^3*x^9-30*a*b^2*x^6+60*a^2*b*x^3+5*a^3)*((b*x^3+a)^2)^(3/2)/x^4/(b*x^3+a)^3

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maxima [A]  time = 0.53, size = 37, normalized size = 0.22 \[ \frac {4 \, b^{3} x^{9} + 30 \, a b^{2} x^{6} - 60 \, a^{2} b x^{3} - 5 \, a^{3}}{20 \, x^{4}} \]

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^5,x, algorithm="maxima")

[Out]

1/20*(4*b^3*x^9 + 30*a*b^2*x^6 - 60*a^2*b*x^3 - 5*a^3)/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{x^5} \,d x \]

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^5,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{5}}\, dx \]

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**5,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**5, x)

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