3.258 \(\int \frac {c+d x^3+e x^6+f x^9}{x^{10} (a+b x^3)^2} \, dx\)

Optimal. Leaf size=175 \[ \frac {2 b c-a d}{6 a^3 x^6}-\frac {c}{9 a^2 x^9}-\frac {a^2 e-2 a b d+3 b^2 c}{3 a^4 x^3}+\frac {\log \left (a+b x^3\right ) \left (a^3 (-f)+2 a^2 b e-3 a b^2 d+4 b^3 c\right )}{3 a^5}-\frac {\log (x) \left (a^3 (-f)+2 a^2 b e-3 a b^2 d+4 b^3 c\right )}{a^5}-\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{3 a^4 \left (a+b x^3\right )} \]

[Out]

-1/9*c/a^2/x^9+1/6*(-a*d+2*b*c)/a^3/x^6+1/3*(-a^2*e+2*a*b*d-3*b^2*c)/a^4/x^3+1/3*(a^3*f-a^2*b*e+a*b^2*d-b^3*c)
/a^4/(b*x^3+a)-(-a^3*f+2*a^2*b*e-3*a*b^2*d+4*b^3*c)*ln(x)/a^5+1/3*(-a^3*f+2*a^2*b*e-3*a*b^2*d+4*b^3*c)*ln(b*x^
3+a)/a^5

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Rubi [A]  time = 0.20, antiderivative size = 175, normalized size of antiderivative = 1.00, 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} \[ -\frac {a^2 b e+a^3 (-f)-a b^2 d+b^3 c}{3 a^4 \left (a+b x^3\right )}+\frac {\log \left (a+b x^3\right ) \left (2 a^2 b e+a^3 (-f)-3 a b^2 d+4 b^3 c\right )}{3 a^5}-\frac {\log (x) \left (2 a^2 b e+a^3 (-f)-3 a b^2 d+4 b^3 c\right )}{a^5}-\frac {a^2 e-2 a b d+3 b^2 c}{3 a^4 x^3}+\frac {2 b c-a d}{6 a^3 x^6}-\frac {c}{9 a^2 x^9} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^10*(a + b*x^3)^2),x]

[Out]

-c/(9*a^2*x^9) + (2*b*c - a*d)/(6*a^3*x^6) - (3*b^2*c - 2*a*b*d + a^2*e)/(3*a^4*x^3) - (b^3*c - a*b^2*d + a^2*
b*e - a^3*f)/(3*a^4*(a + b*x^3)) - ((4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*Log[x])/a^5 + ((4*b^3*c - 3*a*b^
2*d + 2*a^2*b*e - a^3*f)*Log[a + b*x^3])/(3*a^5)

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]

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]]

Rubi steps

\begin {align*} \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{x^4 (a+b x)^2} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {c}{a^2 x^4}+\frac {-2 b c+a d}{a^3 x^3}+\frac {3 b^2 c-2 a b d+a^2 e}{a^4 x^2}+\frac {-4 b^3 c+3 a b^2 d-2 a^2 b e+a^3 f}{a^5 x}-\frac {b \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{a^4 (a+b x)^2}-\frac {b \left (-4 b^3 c+3 a b^2 d-2 a^2 b e+a^3 f\right )}{a^5 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {c}{9 a^2 x^9}+\frac {2 b c-a d}{6 a^3 x^6}-\frac {3 b^2 c-2 a b d+a^2 e}{3 a^4 x^3}-\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{3 a^4 \left (a+b x^3\right )}-\frac {\left (4 b^3 c-3 a b^2 d+2 a^2 b e-a^3 f\right ) \log (x)}{a^5}+\frac {\left (4 b^3 c-3 a b^2 d+2 a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{3 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 160, normalized size = 0.91 \[ \frac {-\frac {2 a^3 c}{x^9}-\frac {6 a \left (a^2 e-2 a b d+3 b^2 c\right )}{x^3}-\frac {3 a^2 (a d-2 b c)}{x^6}+\frac {6 a \left (a^3 f-a^2 b e+a b^2 d-b^3 c\right )}{a+b x^3}+6 \log \left (a+b x^3\right ) \left (a^3 (-f)+2 a^2 b e-3 a b^2 d+4 b^3 c\right )+18 \log (x) \left (a^3 f-2 a^2 b e+3 a b^2 d-4 b^3 c\right )}{18 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^10*(a + b*x^3)^2),x]

[Out]

((-2*a^3*c)/x^9 - (3*a^2*(-2*b*c + a*d))/x^6 - (6*a*(3*b^2*c - 2*a*b*d + a^2*e))/x^3 + (6*a*(-(b^3*c) + a*b^2*
d - a^2*b*e + a^3*f))/(a + b*x^3) + 18*(-4*b^3*c + 3*a*b^2*d - 2*a^2*b*e + a^3*f)*Log[x] + 6*(4*b^3*c - 3*a*b^
2*d + 2*a^2*b*e - a^3*f)*Log[a + b*x^3])/(18*a^5)

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fricas [A]  time = 0.62, size = 261, normalized size = 1.49 \[ -\frac {6 \, {\left (4 \, a b^{3} c - 3 \, a^{2} b^{2} d + 2 \, a^{3} b e - a^{4} f\right )} x^{9} + 3 \, {\left (4 \, a^{2} b^{2} c - 3 \, a^{3} b d + 2 \, a^{4} e\right )} x^{6} + 2 \, a^{4} c - {\left (4 \, a^{3} b c - 3 \, a^{4} d\right )} x^{3} - 6 \, {\left ({\left (4 \, b^{4} c - 3 \, a b^{3} d + 2 \, a^{2} b^{2} e - a^{3} b f\right )} x^{12} + {\left (4 \, a b^{3} c - 3 \, a^{2} b^{2} d + 2 \, a^{3} b e - a^{4} f\right )} x^{9}\right )} \log \left (b x^{3} + a\right ) + 18 \, {\left ({\left (4 \, b^{4} c - 3 \, a b^{3} d + 2 \, a^{2} b^{2} e - a^{3} b f\right )} x^{12} + {\left (4 \, a b^{3} c - 3 \, a^{2} b^{2} d + 2 \, a^{3} b e - a^{4} f\right )} x^{9}\right )} \log \relax (x)}{18 \, {\left (a^{5} b x^{12} + a^{6} x^{9}\right )}} \]

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)^2,x, algorithm="fricas")

[Out]

-1/18*(6*(4*a*b^3*c - 3*a^2*b^2*d + 2*a^3*b*e - a^4*f)*x^9 + 3*(4*a^2*b^2*c - 3*a^3*b*d + 2*a^4*e)*x^6 + 2*a^4
*c - (4*a^3*b*c - 3*a^4*d)*x^3 - 6*((4*b^4*c - 3*a*b^3*d + 2*a^2*b^2*e - a^3*b*f)*x^12 + (4*a*b^3*c - 3*a^2*b^
2*d + 2*a^3*b*e - a^4*f)*x^9)*log(b*x^3 + a) + 18*((4*b^4*c - 3*a*b^3*d + 2*a^2*b^2*e - a^3*b*f)*x^12 + (4*a*b
^3*c - 3*a^2*b^2*d + 2*a^3*b*e - a^4*f)*x^9)*log(x))/(a^5*b*x^12 + a^6*x^9)

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giac [A]  time = 0.20, size = 275, normalized size = 1.57 \[ -\frac {{\left (4 \, b^{3} c - 3 \, a b^{2} d - a^{3} f + 2 \, a^{2} b e\right )} \log \left ({\left | x \right |}\right )}{a^{5}} + \frac {{\left (4 \, b^{4} c - 3 \, a b^{3} d - a^{3} b f + 2 \, a^{2} b^{2} e\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{5} b} - \frac {4 \, b^{4} c x^{3} - 3 \, a b^{3} d x^{3} - a^{3} b f x^{3} + 2 \, a^{2} b^{2} x^{3} e + 5 \, a b^{3} c - 4 \, a^{2} b^{2} d - 2 \, a^{4} f + 3 \, a^{3} b e}{3 \, {\left (b x^{3} + a\right )} a^{5}} + \frac {44 \, b^{3} c x^{9} - 33 \, a b^{2} d x^{9} - 11 \, a^{3} f x^{9} + 22 \, a^{2} b x^{9} e - 18 \, a b^{2} c x^{6} + 12 \, a^{2} b d x^{6} - 6 \, a^{3} x^{6} e + 6 \, a^{2} b c x^{3} - 3 \, a^{3} d x^{3} - 2 \, a^{3} c}{18 \, a^{5} x^{9}} \]

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)^2,x, algorithm="giac")

[Out]

-(4*b^3*c - 3*a*b^2*d - a^3*f + 2*a^2*b*e)*log(abs(x))/a^5 + 1/3*(4*b^4*c - 3*a*b^3*d - a^3*b*f + 2*a^2*b^2*e)
*log(abs(b*x^3 + a))/(a^5*b) - 1/3*(4*b^4*c*x^3 - 3*a*b^3*d*x^3 - a^3*b*f*x^3 + 2*a^2*b^2*x^3*e + 5*a*b^3*c -
4*a^2*b^2*d - 2*a^4*f + 3*a^3*b*e)/((b*x^3 + a)*a^5) + 1/18*(44*b^3*c*x^9 - 33*a*b^2*d*x^9 - 11*a^3*f*x^9 + 22
*a^2*b*x^9*e - 18*a*b^2*c*x^6 + 12*a^2*b*d*x^6 - 6*a^3*x^6*e + 6*a^2*b*c*x^3 - 3*a^3*d*x^3 - 2*a^3*c)/(a^5*x^9
)

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maple [A]  time = 0.06, size = 229, normalized size = 1.31 \[ \frac {f}{3 \left (b \,x^{3}+a \right ) a}-\frac {b e}{3 \left (b \,x^{3}+a \right ) a^{2}}+\frac {f \ln \relax (x )}{a^{2}}-\frac {f \ln \left (b \,x^{3}+a \right )}{3 a^{2}}+\frac {b^{2} d}{3 \left (b \,x^{3}+a \right ) a^{3}}-\frac {2 b e \ln \relax (x )}{a^{3}}+\frac {2 b e \ln \left (b \,x^{3}+a \right )}{3 a^{3}}-\frac {b^{3} c}{3 \left (b \,x^{3}+a \right ) a^{4}}+\frac {3 b^{2} d \ln \relax (x )}{a^{4}}-\frac {b^{2} d \ln \left (b \,x^{3}+a \right )}{a^{4}}-\frac {4 b^{3} c \ln \relax (x )}{a^{5}}+\frac {4 b^{3} c \ln \left (b \,x^{3}+a \right )}{3 a^{5}}-\frac {e}{3 a^{2} x^{3}}+\frac {2 b d}{3 a^{3} x^{3}}-\frac {b^{2} c}{a^{4} x^{3}}-\frac {d}{6 a^{2} x^{6}}+\frac {b c}{3 a^{3} x^{6}}-\frac {c}{9 a^{2} x^{9}} \]

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)^2,x)

[Out]

-1/3/a^2*ln(b*x^3+a)*f+2/3*b/a^3*ln(b*x^3+a)*e-b^2/a^4*ln(b*x^3+a)*d+4/3*b^3/a^5*ln(b*x^3+a)*c+1/3/a/(b*x^3+a)
*f-1/3*b/a^2/(b*x^3+a)*e+1/3*b^2/a^3/(b*x^3+a)*d-1/3*b^3/a^4/(b*x^3+a)*c-1/9/a^2*c/x^9-1/6/a^2/x^6*d+1/3/a^3/x
^6*b*c-1/3/a^2/x^3*e+2/3/a^3/x^3*b*d-1/a^4/x^3*b^2*c+1/a^2*ln(x)*f-2/a^3*ln(x)*b*e+3/a^4*ln(x)*b^2*d-4/a^5*ln(
x)*b^3*c

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maxima [A]  time = 1.43, size = 181, normalized size = 1.03 \[ -\frac {6 \, {\left (4 \, b^{3} c - 3 \, a b^{2} d + 2 \, a^{2} b e - a^{3} f\right )} x^{9} + 3 \, {\left (4 \, a b^{2} c - 3 \, a^{2} b d + 2 \, a^{3} e\right )} x^{6} + 2 \, a^{3} c - {\left (4 \, a^{2} b c - 3 \, a^{3} d\right )} x^{3}}{18 \, {\left (a^{4} b x^{12} + a^{5} x^{9}\right )}} + \frac {{\left (4 \, b^{3} c - 3 \, a b^{2} d + 2 \, a^{2} b e - a^{3} f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{5}} - \frac {{\left (4 \, b^{3} c - 3 \, a b^{2} d + 2 \, a^{2} b e - a^{3} f\right )} \log \left (x^{3}\right )}{3 \, a^{5}} \]

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)^2,x, algorithm="maxima")

[Out]

-1/18*(6*(4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*x^9 + 3*(4*a*b^2*c - 3*a^2*b*d + 2*a^3*e)*x^6 + 2*a^3*c - (
4*a^2*b*c - 3*a^3*d)*x^3)/(a^4*b*x^12 + a^5*x^9) + 1/3*(4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*log(b*x^3 + a
)/a^5 - 1/3*(4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*log(x^3)/a^5

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mupad [B]  time = 5.08, size = 175, normalized size = 1.00 \[ \frac {\ln \left (b\,x^3+a\right )\,\left (-f\,a^3+2\,e\,a^2\,b-3\,d\,a\,b^2+4\,c\,b^3\right )}{3\,a^5}-\frac {\frac {c}{9\,a}+\frac {x^9\,\left (-f\,a^3+2\,e\,a^2\,b-3\,d\,a\,b^2+4\,c\,b^3\right )}{3\,a^4}+\frac {x^3\,\left (3\,a\,d-4\,b\,c\right )}{18\,a^2}+\frac {x^6\,\left (2\,e\,a^2-3\,d\,a\,b+4\,c\,b^2\right )}{6\,a^3}}{b\,x^{12}+a\,x^9}-\frac {\ln \relax (x)\,\left (-f\,a^3+2\,e\,a^2\,b-3\,d\,a\,b^2+4\,c\,b^3\right )}{a^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3 + e*x^6 + f*x^9)/(x^10*(a + b*x^3)^2),x)

[Out]

(log(a + b*x^3)*(4*b^3*c - a^3*f - 3*a*b^2*d + 2*a^2*b*e))/(3*a^5) - (c/(9*a) + (x^9*(4*b^3*c - a^3*f - 3*a*b^
2*d + 2*a^2*b*e))/(3*a^4) + (x^3*(3*a*d - 4*b*c))/(18*a^2) + (x^6*(4*b^2*c + 2*a^2*e - 3*a*b*d))/(6*a^3))/(a*x
^9 + b*x^12) - (log(x)*(4*b^3*c - a^3*f - 3*a*b^2*d + 2*a^2*b*e))/a^5

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**2,x)

[Out]

Timed out

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