\(\int \frac {(c+d x+e x^2) (a+b x^3)}{x^2} \, dx\) [317]

Optimal result

Integrand size = 21, antiderivative size = 44 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^2} \, dx=-\frac {a c}{x}+a e x+\frac {1}{2} b c x^2+\frac {1}{3} b d x^3+\frac {1}{4} b e x^4+a d \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (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} \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^2} \, dx=-\frac {a c}{x}+a d \log (x)+a e x+\frac {1}{2} b c x^2+\frac {1}{3} b d x^3+\frac {1}{4} b e x^4 \]

[In]

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

[Out]

-((a*c)/x) + a*e*x + (b*c*x^2)/2 + (b*d*x^3)/3 + (b*e*x^4)/4 + a*d*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 (a e+\frac {a c}{x^2}+\frac {a d}{x}+b c x+b d x^2+b e x^3\right ) \, dx \\ & = -\frac {a c}{x}+a e x+\frac {1}{2} b c x^2+\frac {1}{3} b d x^3+\frac {1}{4} b e x^4+a d \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^2} \, dx=-\frac {a c}{x}+a e x+\frac {1}{2} b c x^2+\frac {1}{3} b d x^3+\frac {1}{4} b e x^4+a d \log (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

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

method	result	size
default	\(-\frac {a c}{x}+a e x +\frac {c b \,x^{2}}{2}+\frac {b d \,x^{3}}{3}+\frac {b e \,x^{4}}{4}+a d \ln \left (x \right )\)	\(39\)
risch	\(-\frac {a c}{x}+a e x +\frac {c b \,x^{2}}{2}+\frac {b d \,x^{3}}{3}+\frac {b e \,x^{4}}{4}+a d \ln \left (x \right )\)	\(39\)
norman	\(\frac {a e \,x^{2}-a c +\frac {1}{2} b c \,x^{3}+\frac {1}{3} b d \,x^{4}+\frac {1}{4} b e \,x^{5}}{x}+a d \ln \left (x \right )\)	\(43\)
parallelrisch	\(\frac {3 b e \,x^{5}+4 b d \,x^{4}+6 b c \,x^{3}+12 a d \ln \left (x \right ) x +12 a e \,x^{2}-12 a c}{12 x}\)	\(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.02 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^2} \, dx=\frac {3 \, b e x^{5} + 4 \, b d x^{4} + 6 \, b c x^{3} + 12 \, a e x^{2} + 12 \, a d x \log \left (x\right ) - 12 \, a c}{12 \, x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^2} \, dx=- \frac {a c}{x} + a d \log {\left (x \right )} + a e x + \frac {b c x^{2}}{2} + \frac {b d x^{3}}{3} + \frac {b e x^{4}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^2} \, dx=\frac {1}{4} \, b e x^{4} + \frac {1}{3} \, b d x^{3} + \frac {1}{2} \, b c x^{2} + a e x + a d \log \left (x\right ) - \frac {a c}{x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^2} \, dx=\frac {1}{4} \, b e x^{4} + \frac {1}{3} \, b d x^{3} + \frac {1}{2} \, b c x^{2} + a e x + a d \log \left ({\left | x \right |}\right ) - \frac {a c}{x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )}{x^2} \, dx=a\,d\,\ln \left (x\right )+a\,e\,x-\frac {a\,c}{x}+\frac {b\,c\,x^2}{2}+\frac {b\,d\,x^3}{3}+\frac {b\,e\,x^4}{4} \]

[In]

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

[Out]

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