\(\int \frac {a x+b x^3+c x^5}{x^2} \, dx\) [70]

Optimal result

Integrand size = 18, antiderivative size = 21 \[ \int \frac {a x+b x^3+c x^5}{x^2} \, dx=\frac {b x^2}{2}+\frac {c x^4}{4}+a \log (x) \]

[Out]

1/2*b*x^2+1/4*c*x^4+a*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14} \[ \int \frac {a x+b x^3+c x^5}{x^2} \, dx=a \log (x)+\frac {b x^2}{2}+\frac {c x^4}{4} \]

[In]

Int[(a*x + b*x^3 + c*x^5)/x^2,x]

[Out]

(b*x^2)/2 + (c*x^4)/4 + a*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a}{x}+b x+c x^3\right ) \, dx \\ & = \frac {b x^2}{2}+\frac {c x^4}{4}+a \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {a x+b x^3+c x^5}{x^2} \, dx=\frac {b x^2}{2}+\frac {c x^4}{4}+a \log (x) \]

[In]

Integrate[(a*x + b*x^3 + c*x^5)/x^2,x]

[Out]

(b*x^2)/2 + (c*x^4)/4 + a*Log[x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {b \,x^{2}}{2}+\frac {c \,x^{4}}{4}+a \ln \left (x \right )\)	\(18\)
parallelrisch	\(\frac {b \,x^{2}}{2}+\frac {c \,x^{4}}{4}+a \ln \left (x \right )\)	\(18\)
norman	\(\frac {\frac {1}{2} b \,x^{3}+\frac {1}{4} c \,x^{5}}{x}+a \ln \left (x \right )\)	\(23\)
risch	\(\frac {c \,x^{4}}{4}+\frac {b \,x^{2}}{2}+\frac {b^{2}}{4 c}+a \ln \left (x \right )\)	\(26\)

[In]

int((c*x^5+b*x^3+a*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/2*b*x^2+1/4*c*x^4+a*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {a x+b x^3+c x^5}{x^2} \, dx=\frac {1}{4} \, c x^{4} + \frac {1}{2} \, b x^{2} + a \log \left (x\right ) \]

[In]

integrate((c*x^5+b*x^3+a*x)/x^2,x, algorithm="fricas")

[Out]

1/4*c*x^4 + 1/2*b*x^2 + a*log(x)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {a x+b x^3+c x^5}{x^2} \, dx=a \log {\left (x \right )} + \frac {b x^{2}}{2} + \frac {c x^{4}}{4} \]

[In]

integrate((c*x**5+b*x**3+a*x)/x**2,x)

[Out]

a*log(x) + b*x**2/2 + c*x**4/4

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {a x+b x^3+c x^5}{x^2} \, dx=\frac {1}{4} \, c x^{4} + \frac {1}{2} \, b x^{2} + a \log \left (x\right ) \]

[In]

integrate((c*x^5+b*x^3+a*x)/x^2,x, algorithm="maxima")

[Out]

1/4*c*x^4 + 1/2*b*x^2 + a*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {a x+b x^3+c x^5}{x^2} \, dx=\frac {1}{4} \, c x^{4} + \frac {1}{2} \, b x^{2} + \frac {1}{2} \, a \log \left (x^{2}\right ) \]

[In]

integrate((c*x^5+b*x^3+a*x)/x^2,x, algorithm="giac")

[Out]

1/4*c*x^4 + 1/2*b*x^2 + 1/2*a*log(x^2)

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {a x+b x^3+c x^5}{x^2} \, dx=\frac {b\,x^2}{2}+\frac {c\,x^4}{4}+a\,\ln \left (x\right ) \]

[In]

int((a*x + b*x^3 + c*x^5)/x^2,x)

[Out]

(b*x^2)/2 + (c*x^4)/4 + a*log(x)