3.1.53 \(\int \frac {(b x^2+c x^4)^3}{x^{15}} \, dx\)

Optimal. Leaf size=19 \[ -\frac {\left (b+c x^2\right )^4}{8 b x^8} \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1584, 264} \begin {gather*} -\frac {\left (b+c x^2\right )^4}{8 b x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^15,x]

[Out]

-(b + c*x^2)^4/(8*b*x^8)

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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^{15}} \, dx &=\int \frac {\left (b+c x^2\right )^3}{x^9} \, dx\\ &=-\frac {\left (b+c x^2\right )^4}{8 b x^8}\\ \end {align*}

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Mathematica [B]  time = 0.01, size = 43, normalized size = 2.26 \begin {gather*} -\frac {b^3}{8 x^8}-\frac {b^2 c}{2 x^6}-\frac {3 b c^2}{4 x^4}-\frac {c^3}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^15,x]

[Out]

-1/8*b^3/x^8 - (b^2*c)/(2*x^6) - (3*b*c^2)/(4*x^4) - c^3/(2*x^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x^2+c x^4\right )^3}{x^{15}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(b*x^2 + c*x^4)^3/x^15,x]

[Out]

IntegrateAlgebraic[(b*x^2 + c*x^4)^3/x^15, x]

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fricas [B]  time = 0.77, size = 35, normalized size = 1.84 \begin {gather*} -\frac {4 \, c^{3} x^{6} + 6 \, b c^{2} x^{4} + 4 \, b^{2} c x^{2} + b^{3}}{8 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^15,x, algorithm="fricas")

[Out]

-1/8*(4*c^3*x^6 + 6*b*c^2*x^4 + 4*b^2*c*x^2 + b^3)/x^8

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giac [B]  time = 0.15, size = 35, normalized size = 1.84 \begin {gather*} -\frac {4 \, c^{3} x^{6} + 6 \, b c^{2} x^{4} + 4 \, b^{2} c x^{2} + b^{3}}{8 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^15,x, algorithm="giac")

[Out]

-1/8*(4*c^3*x^6 + 6*b*c^2*x^4 + 4*b^2*c*x^2 + b^3)/x^8

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maple [B]  time = 0.00, size = 36, normalized size = 1.89 \begin {gather*} -\frac {c^{3}}{2 x^{2}}-\frac {3 b \,c^{2}}{4 x^{4}}-\frac {b^{2} c}{2 x^{6}}-\frac {b^{3}}{8 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^15,x)

[Out]

-1/2*c^3/x^2-1/2*b^2*c/x^6-3/4*b*c^2/x^4-1/8*b^3/x^8

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maxima [B]  time = 1.34, size = 35, normalized size = 1.84 \begin {gather*} -\frac {4 \, c^{3} x^{6} + 6 \, b c^{2} x^{4} + 4 \, b^{2} c x^{2} + b^{3}}{8 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^15,x, algorithm="maxima")

[Out]

-1/8*(4*c^3*x^6 + 6*b*c^2*x^4 + 4*b^2*c*x^2 + b^3)/x^8

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mupad [B]  time = 0.03, size = 37, normalized size = 1.95 \begin {gather*} -\frac {\frac {b^3}{8}+\frac {b^2\,c\,x^2}{2}+\frac {3\,b\,c^2\,x^4}{4}+\frac {c^3\,x^6}{2}}{x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^3/x^15,x)

[Out]

-(b^3/8 + (c^3*x^6)/2 + (b^2*c*x^2)/2 + (3*b*c^2*x^4)/4)/x^8

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sympy [B]  time = 0.31, size = 37, normalized size = 1.95 \begin {gather*} \frac {- b^{3} - 4 b^{2} c x^{2} - 6 b c^{2} x^{4} - 4 c^{3} x^{6}}{8 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**3/x**15,x)

[Out]

(-b**3 - 4*b**2*c*x**2 - 6*b*c**2*x**4 - 4*c**3*x**6)/(8*x**8)

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