3.2.67 \(\int \frac {(b x^2+c x^4)^3}{x^{11}} \, dx\) [167]

Optimal. Leaf size=40 \[ -\frac {b^3}{4 x^4}-\frac {3 b^2 c}{2 x^2}+\frac {c^3 x^2}{2}+3 b c^2 \log (x) \]

[Out]

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

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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 = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1598, 272, 45} \begin {gather*} -\frac {b^3}{4 x^4}-\frac {3 b^2 c}{2 x^2}+3 b c^2 \log (x)+\frac {c^3 x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^11,x]

[Out]

-1/4*b^3/x^4 - (3*b^2*c)/(2*x^2) + (c^3*x^2)/2 + 3*b*c^2*Log[x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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 1598

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^{11}} \, dx &=\int \frac {\left (b+c x^2\right )^3}{x^5} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {(b+c x)^3}{x^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (c^3+\frac {b^3}{x^3}+\frac {3 b^2 c}{x^2}+\frac {3 b c^2}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac {b^3}{4 x^4}-\frac {3 b^2 c}{2 x^2}+\frac {c^3 x^2}{2}+3 b c^2 \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 40, normalized size = 1.00 \begin {gather*} -\frac {b^3}{4 x^4}-\frac {3 b^2 c}{2 x^2}+\frac {c^3 x^2}{2}+3 b c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^11,x]

[Out]

-1/4*b^3/x^4 - (3*b^2*c)/(2*x^2) + (c^3*x^2)/2 + 3*b*c^2*Log[x]

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Maple [A]
time = 0.08, size = 35, normalized size = 0.88

method	result	size
default	\(-\frac {b^{3}}{4 x^{4}}-\frac {3 b^{2} c}{2 x^{2}}+\frac {c^{3} x^{2}}{2}+3 b \,c^{2} \ln \left (x \right )\)	\(35\)
risch	\(\frac {c^{3} x^{2}}{2}+\frac {-\frac {3}{2} b^{2} c \,x^{2}-\frac {1}{4} b^{3}}{x^{4}}+3 b \,c^{2} \ln \left (x \right )\)	\(37\)
norman	\(\frac {-\frac {1}{4} b^{3} x^{6}+\frac {1}{2} c^{3} x^{12}-\frac {3}{2} b^{2} c \,x^{8}}{x^{10}}+3 b \,c^{2} \ln \left (x \right )\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^11,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 37, normalized size = 0.92 \begin {gather*} \frac {1}{2} \, c^{3} x^{2} + \frac {3}{2} \, b c^{2} \log \left (x^{2}\right ) - \frac {6 \, b^{2} c x^{2} + b^{3}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^11,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 39, normalized size = 0.98 \begin {gather*} \frac {2 \, c^{3} x^{6} + 12 \, b c^{2} x^{4} \log \left (x\right ) - 6 \, b^{2} c x^{2} - b^{3}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^11,x, algorithm="fricas")

[Out]

1/4*(2*c^3*x^6 + 12*b*c^2*x^4*log(x) - 6*b^2*c*x^2 - b^3)/x^4

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Sympy [A]
time = 0.09, size = 37, normalized size = 0.92 \begin {gather*} 3 b c^{2} \log {\left (x \right )} + \frac {c^{3} x^{2}}{2} + \frac {- b^{3} - 6 b^{2} c x^{2}}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**3/x**11,x)

[Out]

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

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Giac [A]
time = 3.89, size = 46, normalized size = 1.15 \begin {gather*} \frac {1}{2} \, c^{3} x^{2} + \frac {3}{2} \, b c^{2} \log \left (x^{2}\right ) - \frac {9 \, b c^{2} x^{4} + 6 \, b^{2} c x^{2} + b^{3}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^11,x, algorithm="giac")

[Out]

1/2*c^3*x^2 + 3/2*b*c^2*log(x^2) - 1/4*(9*b*c^2*x^4 + 6*b^2*c*x^2 + b^3)/x^4

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Mupad [B]
time = 0.03, size = 37, normalized size = 0.92 \begin {gather*} \frac {c^3\,x^2}{2}-\frac {\frac {b^3}{4}+\frac {3\,c\,b^2\,x^2}{2}}{x^4}+3\,b\,c^2\,\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^3/x^11,x)

[Out]

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

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