3.165 \(\int \frac{(b x^2+c x^4)^3}{x^9} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0291111, antiderivative size = 40, normalized size of antiderivative = 1., 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 = {1584, 266, 43} \[ 3 b^2 c \log (x)-\frac{b^3}{2 x^2}+\frac{3}{2} b c^2 x^2+\frac{c^3 x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^9,x]

[Out]

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

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]

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 43

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])

Rubi steps

\begin{align*} \int \frac{\left (b x^2+c x^4\right )^3}{x^9} \, dx &=\int \frac{\left (b+c x^2\right )^3}{x^3} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(b+c x)^3}{x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (3 b c^2+\frac{b^3}{x^2}+\frac{3 b^2 c}{x}+c^3 x\right ) \, dx,x,x^2\right )\\ &=-\frac{b^3}{2 x^2}+\frac{3}{2} b c^2 x^2+\frac{c^3 x^4}{4}+3 b^2 c \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0063887, size = 40, normalized size = 1. \[ 3 b^2 c \log (x)-\frac{b^3}{2 x^2}+\frac{3}{2} b c^2 x^2+\frac{c^3 x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^9,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 35, normalized size = 0.9 \begin{align*} -{\frac{{b}^{3}}{2\,{x}^{2}}}+{\frac{3\,b{c}^{2}{x}^{2}}{2}}+{\frac{{c}^{3}{x}^{4}}{4}}+3\,{b}^{2}c\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^9,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.01006, size = 49, normalized size = 1.22 \begin{align*} \frac{1}{4} \, c^{3} x^{4} + \frac{3}{2} \, b c^{2} x^{2} + \frac{3}{2} \, b^{2} c \log \left (x^{2}\right ) - \frac{b^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^9,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.50259, size = 85, normalized size = 2.12 \begin{align*} \frac{c^{3} x^{6} + 6 \, b c^{2} x^{4} + 12 \, b^{2} c x^{2} \log \left (x\right ) - 2 \, b^{3}}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^9,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.326276, size = 37, normalized size = 0.92 \begin{align*} - \frac{b^{3}}{2 x^{2}} + 3 b^{2} c \log{\left (x \right )} + \frac{3 b c^{2} x^{2}}{2} + \frac{c^{3} x^{4}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**3/x**9,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.28012, size = 62, normalized size = 1.55 \begin{align*} \frac{1}{4} \, c^{3} x^{4} + \frac{3}{2} \, b c^{2} x^{2} + \frac{3}{2} \, b^{2} c \log \left (x^{2}\right ) - \frac{3 \, b^{2} c x^{2} + b^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^9,x, algorithm="giac")

[Out]

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