3.164 \(\int \frac {(b x^2+c x^4)^3}{x^8} \, dx\)

Optimal. Leaf size=34 \[ -\frac {b^3}{x}+3 b^2 c x+b c^2 x^3+\frac {c^3 x^5}{5} \]

[Out]

-b^3/x+3*b^2*c*x+b*c^2*x^3+1/5*c^3*x^5

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1584, 270} \[ 3 b^2 c x-\frac {b^3}{x}+b c^2 x^3+\frac {c^3 x^5}{5} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^8,x]

[Out]

-(b^3/x) + 3*b^2*c*x + b*c^2*x^3 + (c^3*x^5)/5

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 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {\left (b x^2+c x^4\right )^3}{x^8} \, dx &=\int \frac {\left (b+c x^2\right )^3}{x^2} \, dx\\ &=\int \left (3 b^2 c+\frac {b^3}{x^2}+3 b c^2 x^2+c^3 x^4\right ) \, dx\\ &=-\frac {b^3}{x}+3 b^2 c x+b c^2 x^3+\frac {c^3 x^5}{5}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 34, normalized size = 1.00 \[ -\frac {b^3}{x}+3 b^2 c x+b c^2 x^3+\frac {c^3 x^5}{5} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^8,x]

[Out]

-(b^3/x) + 3*b^2*c*x + b*c^2*x^3 + (c^3*x^5)/5

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fricas [A]  time = 0.83, size = 36, normalized size = 1.06 \[ \frac {c^{3} x^{6} + 5 \, b c^{2} x^{4} + 15 \, b^{2} c x^{2} - 5 \, b^{3}}{5 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^8,x, algorithm="fricas")

[Out]

1/5*(c^3*x^6 + 5*b*c^2*x^4 + 15*b^2*c*x^2 - 5*b^3)/x

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giac [A]  time = 0.15, size = 32, normalized size = 0.94 \[ \frac {1}{5} \, c^{3} x^{5} + b c^{2} x^{3} + 3 \, b^{2} c x - \frac {b^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^8,x, algorithm="giac")

[Out]

1/5*c^3*x^5 + b*c^2*x^3 + 3*b^2*c*x - b^3/x

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maple [A]  time = 0.00, size = 33, normalized size = 0.97 \[ \frac {c^{3} x^{5}}{5}+b \,c^{2} x^{3}+3 b^{2} c x -\frac {b^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^8,x)

[Out]

-b^3/x+3*b^2*c*x+b*c^2*x^3+1/5*c^3*x^5

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maxima [A]  time = 1.38, size = 32, normalized size = 0.94 \[ \frac {1}{5} \, c^{3} x^{5} + b c^{2} x^{3} + 3 \, b^{2} c x - \frac {b^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^8,x, algorithm="maxima")

[Out]

1/5*c^3*x^5 + b*c^2*x^3 + 3*b^2*c*x - b^3/x

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mupad [B]  time = 0.04, size = 32, normalized size = 0.94 \[ \frac {c^3\,x^5}{5}-\frac {b^3}{x}+b\,c^2\,x^3+3\,b^2\,c\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^3/x^8,x)

[Out]

(c^3*x^5)/5 - b^3/x + b*c^2*x^3 + 3*b^2*c*x

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sympy [A]  time = 0.12, size = 29, normalized size = 0.85 \[ - \frac {b^{3}}{x} + 3 b^{2} c x + b c^{2} x^{3} + \frac {c^{3} x^{5}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**3/x**8,x)

[Out]

-b**3/x + 3*b**2*c*x + b*c**2*x**3 + c**3*x**5/5

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