3.230 \(\int \frac{c+d x^3+e x^6+f x^9}{x^{10} (a+b x^3)} \, dx\)

Optimal. Leaf size=128 \[ \frac{\log \left (a+b x^3\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{3 a^4}-\frac{\log (x) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{a^4}-\frac{a^2 e-a b d+b^2 c}{3 a^3 x^3}+\frac{b c-a d}{6 a^2 x^6}-\frac{c}{9 a x^9} \]

[Out]

-c/(9*a*x^9) + (b*c - a*d)/(6*a^2*x^6) - (b^2*c - a*b*d + a^2*e)/(3*a^3*x^3) - ((b^3*c - a*b^2*d + a^2*b*e - a
^3*f)*Log[x])/a^4 + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a + b*x^3])/(3*a^4)

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Rubi [A]  time = 0.161649, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {1821, 1620} \[ \frac{\log \left (a+b x^3\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{3 a^4}-\frac{\log (x) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{a^4}-\frac{a^2 e-a b d+b^2 c}{3 a^3 x^3}+\frac{b c-a d}{6 a^2 x^6}-\frac{c}{9 a x^9} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^10*(a + b*x^3)),x]

[Out]

-c/(9*a*x^9) + (b*c - a*d)/(6*a^2*x^6) - (b^2*c - a*b*d + a^2*e)/(3*a^3*x^3) - ((b^3*c - a*b^2*d + a^2*b*e - a
^3*f)*Log[x])/a^4 + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a + b*x^3])/(3*a^4)

Rule 1821

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

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin{align*} \int \frac{c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{c+d x+e x^2+f x^3}{x^4 (a+b x)} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{c}{a x^4}+\frac{-b c+a d}{a^2 x^3}+\frac{b^2 c-a b d+a^2 e}{a^3 x^2}+\frac{-b^3 c+a b^2 d-a^2 b e+a^3 f}{a^4 x}-\frac{b \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{a^4 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=-\frac{c}{9 a x^9}+\frac{b c-a d}{6 a^2 x^6}-\frac{b^2 c-a b d+a^2 e}{3 a^3 x^3}-\frac{\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log (x)}{a^4}+\frac{\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{3 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0643549, size = 128, normalized size = 1. \[ \frac{\log \left (a+b x^3\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{3 a^4}+\frac{\log (x) \left (-a^2 b e+a^3 f+a b^2 d-b^3 c\right )}{a^4}+\frac{a^2 (-e)+a b d-b^2 c}{3 a^3 x^3}+\frac{b c-a d}{6 a^2 x^6}-\frac{c}{9 a x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^10*(a + b*x^3)),x]

[Out]

-c/(9*a*x^9) + (b*c - a*d)/(6*a^2*x^6) + (-(b^2*c) + a*b*d - a^2*e)/(3*a^3*x^3) + ((-(b^3*c) + a*b^2*d - a^2*b
*e + a^3*f)*Log[x])/a^4 + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a + b*x^3])/(3*a^4)

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Maple [A]  time = 0.007, size = 161, normalized size = 1.3 \begin{align*} -{\frac{\ln \left ( b{x}^{3}+a \right ) f}{3\,a}}+{\frac{\ln \left ( b{x}^{3}+a \right ) be}{3\,{a}^{2}}}-{\frac{\ln \left ( b{x}^{3}+a \right ){b}^{2}d}{3\,{a}^{3}}}+{\frac{\ln \left ( b{x}^{3}+a \right ){b}^{3}c}{3\,{a}^{4}}}-{\frac{c}{9\,a{x}^{9}}}-{\frac{d}{6\,a{x}^{6}}}+{\frac{bc}{6\,{a}^{2}{x}^{6}}}-{\frac{e}{3\,a{x}^{3}}}+{\frac{bd}{3\,{x}^{3}{a}^{2}}}-{\frac{{b}^{2}c}{3\,{a}^{3}{x}^{3}}}+{\frac{\ln \left ( x \right ) f}{a}}-{\frac{\ln \left ( x \right ) be}{{a}^{2}}}+{\frac{\ln \left ( x \right ){b}^{2}d}{{a}^{3}}}-{\frac{\ln \left ( x \right ){b}^{3}c}{{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^9+e*x^6+d*x^3+c)/x^10/(b*x^3+a),x)

[Out]

-1/3/a*ln(b*x^3+a)*f+1/3/a^2*ln(b*x^3+a)*b*e-1/3/a^3*ln(b*x^3+a)*b^2*d+1/3/a^4*ln(b*x^3+a)*b^3*c-1/9*c/a/x^9-1
/6/a/x^6*d+1/6/a^2/x^6*b*c-1/3/a/x^3*e+1/3/a^2/x^3*b*d-1/3/a^3/x^3*b^2*c+1/a*ln(x)*f-1/a^2*ln(x)*b*e+1/a^3*ln(
x)*b^2*d-1/a^4*ln(x)*b^3*c

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Maxima [A]  time = 0.95327, size = 169, normalized size = 1.32 \begin{align*} \frac{{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{4}} - \frac{{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x^{3}\right )}{3 \, a^{4}} - \frac{6 \,{\left (b^{2} c - a b d + a^{2} e\right )} x^{6} - 3 \,{\left (a b c - a^{2} d\right )} x^{3} + 2 \, a^{2} c}{18 \, a^{3} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^10/(b*x^3+a),x, algorithm="maxima")

[Out]

1/3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(b*x^3 + a)/a^4 - 1/3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(x^3)/
a^4 - 1/18*(6*(b^2*c - a*b*d + a^2*e)*x^6 - 3*(a*b*c - a^2*d)*x^3 + 2*a^2*c)/(a^3*x^9)

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Fricas [A]  time = 1.47109, size = 269, normalized size = 2.1 \begin{align*} \frac{6 \,{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{9} \log \left (b x^{3} + a\right ) - 18 \,{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{9} \log \left (x\right ) - 6 \,{\left (a b^{2} c - a^{2} b d + a^{3} e\right )} x^{6} - 2 \, a^{3} c + 3 \,{\left (a^{2} b c - a^{3} d\right )} x^{3}}{18 \, a^{4} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^10/(b*x^3+a),x, algorithm="fricas")

[Out]

1/18*(6*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^9*log(b*x^3 + a) - 18*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^9*lo
g(x) - 6*(a*b^2*c - a^2*b*d + a^3*e)*x^6 - 2*a^3*c + 3*(a^2*b*c - a^3*d)*x^3)/(a^4*x^9)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**9+e*x**6+d*x**3+c)/x**10/(b*x**3+a),x)

[Out]

Timed out

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Giac [A]  time = 1.06691, size = 248, normalized size = 1.94 \begin{align*} -\frac{{\left (b^{3} c - a b^{2} d - a^{3} f + a^{2} b e\right )} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac{{\left (b^{4} c - a b^{3} d - a^{3} b f + a^{2} b^{2} e\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{4} b} + \frac{11 \, b^{3} c x^{9} - 11 \, a b^{2} d x^{9} - 11 \, a^{3} f x^{9} + 11 \, a^{2} b x^{9} e - 6 \, a b^{2} c x^{6} + 6 \, a^{2} b d x^{6} - 6 \, a^{3} x^{6} e + 3 \, a^{2} b c x^{3} - 3 \, a^{3} d x^{3} - 2 \, a^{3} c}{18 \, a^{4} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^10/(b*x^3+a),x, algorithm="giac")

[Out]

-(b^3*c - a*b^2*d - a^3*f + a^2*b*e)*log(abs(x))/a^4 + 1/3*(b^4*c - a*b^3*d - a^3*b*f + a^2*b^2*e)*log(abs(b*x
^3 + a))/(a^4*b) + 1/18*(11*b^3*c*x^9 - 11*a*b^2*d*x^9 - 11*a^3*f*x^9 + 11*a^2*b*x^9*e - 6*a*b^2*c*x^6 + 6*a^2
*b*d*x^6 - 6*a^3*x^6*e + 3*a^2*b*c*x^3 - 3*a^3*d*x^3 - 2*a^3*c)/(a^4*x^9)