\(\int \frac {(c+d x+e x^2) (a+b x^3)}{x^3} \, dx\) [318]

Optimal result

Integrand size = 21, antiderivative size = 44 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^3} \, dx=-\frac {a c}{2 x^2}-\frac {a d}{x}+b c x+\frac {1}{2} b d x^2+\frac {1}{3} b e x^3+a e \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {1642} \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^3} \, dx=-\frac {a c}{2 x^2}-\frac {a d}{x}+a e \log (x)+b c x+\frac {1}{2} b d x^2+\frac {1}{3} b e x^3 \]

[In]

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

[Out]

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

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^3} \, dx=-\frac {a c}{2 x^2}-\frac {a d}{x}+b c x+\frac {1}{2} b d x^2+\frac {1}{3} b e x^3+a e \log (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {a c}{2 x^{2}}-\frac {a d}{x}+b c x +\frac {b d \,x^{2}}{2}+\frac {b e \,x^{3}}{3}+a e \ln \left (x \right )\)	\(39\)
risch	\(\frac {b e \,x^{3}}{3}+\frac {b d \,x^{2}}{2}+b c x +\frac {-a d x -\frac {1}{2} a c}{x^{2}}+a e \ln \left (x \right )\)	\(39\)
norman	\(\frac {b c \,x^{3}-\frac {1}{2} a c -a d x +\frac {1}{2} b d \,x^{4}+\frac {1}{3} b e \,x^{5}}{x^{2}}+a e \ln \left (x \right )\)	\(41\)
parallelrisch	\(\frac {2 b e \,x^{5}+3 b d \,x^{4}+6 a e \ln \left (x \right ) x^{2}+6 b c \,x^{3}-6 a d x -3 a c}{6 x^{2}}\)	\(46\)

[In]

int((e*x^2+d*x+c)*(b*x^3+a)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.02 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^3} \, dx=\frac {2 \, b e x^{5} + 3 \, b d x^{4} + 6 \, b c x^{3} + 6 \, a e x^{2} \log \left (x\right ) - 6 \, a d x - 3 \, a c}{6 \, x^{2}} \]

[In]

integrate((e*x^2+d*x+c)*(b*x^3+a)/x^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^3} \, dx=a e \log {\left (x \right )} + b c x + \frac {b d x^{2}}{2} + \frac {b e x^{3}}{3} + \frac {- a c - 2 a d x}{2 x^{2}} \]

[In]

integrate((e*x**2+d*x+c)*(b*x**3+a)/x**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^3} \, dx=\frac {1}{3} \, b e x^{3} + \frac {1}{2} \, b d x^{2} + b c x + a e \log \left (x\right ) - \frac {2 \, a d x + a c}{2 \, x^{2}} \]

[In]

integrate((e*x^2+d*x+c)*(b*x^3+a)/x^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^3} \, dx=\frac {1}{3} \, b e x^{3} + \frac {1}{2} \, b d x^{2} + b c x + a e \log \left ({\left | x \right |}\right ) - \frac {2 \, a d x + a c}{2 \, x^{2}} \]

[In]

integrate((e*x^2+d*x+c)*(b*x^3+a)/x^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^3} \, dx=a\,e\,\ln \left (x\right )-\frac {\frac {a\,c}{2}+a\,d\,x}{x^2}+b\,c\,x+\frac {b\,d\,x^2}{2}+\frac {b\,e\,x^3}{3} \]

[In]

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

[Out]

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