3.173 \(\int \frac {(b x^2+c x^4)^3}{x^{17}} \, dx\)

Optimal. Leaf size=40 \[ \frac {c \left (b+c x^2\right )^4}{40 b^2 x^8}-\frac {\left (b+c x^2\right )^4}{10 b x^{10}} \]

[Out]

-1/10*(c*x^2+b)^4/b/x^10+1/40*c*(c*x^2+b)^4/b^2/x^8

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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1584, 266, 45, 37} \[ \frac {c \left (b+c x^2\right )^4}{40 b^2 x^8}-\frac {\left (b+c x^2\right )^4}{10 b x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^17,x]

[Out]

-(b + c*x^2)^4/(10*b*x^10) + (c*(b + c*x^2)^4)/(40*b^2*x^8)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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^{17}} \, dx &=\int \frac {\left (b+c x^2\right )^3}{x^{11}} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(b+c x)^3}{x^6} \, dx,x,x^2\right )\\ &=-\frac {\left (b+c x^2\right )^4}{10 b x^{10}}-\frac {c \operatorname {Subst}\left (\int \frac {(b+c x)^3}{x^5} \, dx,x,x^2\right )}{10 b}\\ &=-\frac {\left (b+c x^2\right )^4}{10 b x^{10}}+\frac {c \left (b+c x^2\right )^4}{40 b^2 x^8}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 43, normalized size = 1.08 \[ -\frac {b^3}{10 x^{10}}-\frac {3 b^2 c}{8 x^8}-\frac {b c^2}{2 x^6}-\frac {c^3}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^17,x]

[Out]

-1/10*b^3/x^10 - (3*b^2*c)/(8*x^8) - (b*c^2)/(2*x^6) - c^3/(4*x^4)

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fricas [A]  time = 0.70, size = 37, normalized size = 0.92 \[ -\frac {10 \, c^{3} x^{6} + 20 \, b c^{2} x^{4} + 15 \, b^{2} c x^{2} + 4 \, b^{3}}{40 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^17,x, algorithm="fricas")

[Out]

-1/40*(10*c^3*x^6 + 20*b*c^2*x^4 + 15*b^2*c*x^2 + 4*b^3)/x^10

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giac [A]  time = 0.15, size = 37, normalized size = 0.92 \[ -\frac {10 \, c^{3} x^{6} + 20 \, b c^{2} x^{4} + 15 \, b^{2} c x^{2} + 4 \, b^{3}}{40 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^17,x, algorithm="giac")

[Out]

-1/40*(10*c^3*x^6 + 20*b*c^2*x^4 + 15*b^2*c*x^2 + 4*b^3)/x^10

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maple [A]  time = 0.00, size = 36, normalized size = 0.90 \[ -\frac {c^{3}}{4 x^{4}}-\frac {b \,c^{2}}{2 x^{6}}-\frac {3 b^{2} c}{8 x^{8}}-\frac {b^{3}}{10 x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^17,x)

[Out]

-3/8*b^2*c/x^8-1/2*b*c^2/x^6-1/4*c^3/x^4-1/10*b^3/x^10

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maxima [A]  time = 1.31, size = 37, normalized size = 0.92 \[ -\frac {10 \, c^{3} x^{6} + 20 \, b c^{2} x^{4} + 15 \, b^{2} c x^{2} + 4 \, b^{3}}{40 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^17,x, algorithm="maxima")

[Out]

-1/40*(10*c^3*x^6 + 20*b*c^2*x^4 + 15*b^2*c*x^2 + 4*b^3)/x^10

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mupad [B]  time = 0.03, size = 37, normalized size = 0.92 \[ -\frac {\frac {b^3}{10}+\frac {3\,b^2\,c\,x^2}{8}+\frac {b\,c^2\,x^4}{2}+\frac {c^3\,x^6}{4}}{x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^3/x^17,x)

[Out]

-(b^3/10 + (c^3*x^6)/4 + (3*b^2*c*x^2)/8 + (b*c^2*x^4)/2)/x^10

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sympy [A]  time = 0.32, size = 39, normalized size = 0.98 \[ \frac {- 4 b^{3} - 15 b^{2} c x^{2} - 20 b c^{2} x^{4} - 10 c^{3} x^{6}}{40 x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**3/x**17,x)

[Out]

(-4*b**3 - 15*b**2*c*x**2 - 20*b*c**2*x**4 - 10*c**3*x**6)/(40*x**10)

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