\(\int \frac {c+d x^3+e x^6+f x^9}{x (a+b x^3)} \, dx\) [227]

Optimal result

Integrand size = 30, antiderivative size = 80 \[ \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )} \, dx=\frac {(b e-a f) x^3}{3 b^2}+\frac {f x^6}{6 b}+\frac {c \log (x)}{a}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{3 a b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1835, 1634} \[ \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )} \, dx=-\frac {\log \left (a+b x^3\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a b^3}+\frac {x^3 (b e-a f)}{3 b^2}+\frac {c \log (x)}{a}+\frac {f x^6}{6 b} \]

[In]

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

[Out]

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

Rule 1634

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 1835

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )} \, dx=\frac {a b x^3 \left (2 b e-2 a f+b f x^3\right )+6 b^3 c \log (x)-2 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{6 a b^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (-f \,x^{3} b +a f -b e \right )^{2}}{6 b^{3} f}+\frac {c \ln \left (x \right )}{a}+\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a \,b^{3}}\)	\(75\)
norman	\(-\frac {\left (a f -b e \right ) x^{3}}{3 b^{2}}+\frac {f \,x^{6}}{6 b}+\frac {c \ln \left (x \right )}{a}+\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a \,b^{3}}\)	\(75\)
parallelrisch	\(\frac {x^{6} a \,b^{2} f -2 a^{2} b f \,x^{3}+2 a \,b^{2} e \,x^{3}+6 c \ln \left (x \right ) b^{3}+2 \ln \left (b \,x^{3}+a \right ) a^{3} f -2 \ln \left (b \,x^{3}+a \right ) a^{2} b e +2 \ln \left (b \,x^{3}+a \right ) a \,b^{2} d -2 \ln \left (b \,x^{3}+a \right ) b^{3} c}{6 a \,b^{3}}\)	\(105\)
risch	\(\frac {f \,x^{6}}{6 b}-\frac {f a \,x^{3}}{3 b^{2}}+\frac {e \,x^{3}}{3 b}+\frac {f \,a^{2}}{6 b^{3}}-\frac {a e}{3 b^{2}}+\frac {e^{2}}{6 b f}+\frac {c \ln \left (x \right )}{a}+\frac {a^{2} \ln \left (-b \,x^{3}-a \right ) f}{3 b^{3}}-\frac {a \ln \left (-b \,x^{3}-a \right ) e}{3 b^{2}}+\frac {\ln \left (-b \,x^{3}-a \right ) d}{3 b}-\frac {\ln \left (-b \,x^{3}-a \right ) c}{3 a}\)	\(136\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )} \, dx=\frac {a b^{2} f x^{6} + 6 \, b^{3} c \log \left (x\right ) + 2 \, {\left (a b^{2} e - a^{2} b f\right )} x^{3} - 2 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (b x^{3} + a\right )}{6 \, a b^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )} \, dx=x^{3} \left (- \frac {a f}{3 b^{2}} + \frac {e}{3 b}\right ) + \frac {f x^{6}}{6 b} + \frac {c \log {\left (x \right )}}{a} + \frac {\left (a^{3} f - a^{2} b e + a b^{2} d - b^{3} c\right ) \log {\left (\frac {a}{b} + x^{3} \right )}}{3 a b^{3}} \]

[In]

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

[Out]

x**3*(-a*f/(3*b**2) + e/(3*b)) + f*x**6/(6*b) + c*log(x)/a + (a**3*f - a**2*b*e + a*b**2*d - b**3*c)*log(a/b +
x**3)/(3*a*b**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )} \, dx=\frac {c \log \left (x^{3}\right )}{3 \, a} + \frac {b f x^{6} + 2 \, {\left (b e - a f\right )} x^{3}}{6 \, b^{2}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (b x^{3} + a\right )}{3 \, a b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )} \, dx=\frac {c \log \left ({\left | x \right |}\right )}{a} + \frac {b f x^{6} + 2 \, b e x^{3} - 2 \, a f x^{3}}{6 \, b^{2}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )} \, dx=x^3\,\left (\frac {e}{3\,b}-\frac {a\,f}{3\,b^2}\right )+\frac {f\,x^6}{6\,b}+\frac {c\,\ln \left (x\right )}{a}-\frac {\ln \left (b\,x^3+a\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a\,b^3} \]

[In]

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

[Out]

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