\(\int \frac {b x^2+c x^4}{x^7} \, dx\) [143]

Optimal result

Integrand size = 15, antiderivative size = 17 \[ \int \frac {b x^2+c x^4}{x^7} \, dx=-\frac {b}{4 x^4}-\frac {c}{2 x^2} \]

[Out]

-1/4*b/x^4-1/2*c/x^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14} \[ \int \frac {b x^2+c x^4}{x^7} \, dx=-\frac {b}{4 x^4}-\frac {c}{2 x^2} \]

[In]

Int[(b*x^2 + c*x^4)/x^7,x]

[Out]

-1/4*b/x^4 - c/(2*x^2)

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 {b}{x^5}+\frac {c}{x^3}\right ) \, dx \\ & = -\frac {b}{4 x^4}-\frac {c}{2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {b x^2+c x^4}{x^7} \, dx=-\frac {b}{4 x^4}-\frac {c}{2 x^2} \]

[In]

Integrate[(b*x^2 + c*x^4)/x^7,x]

[Out]

-1/4*b/x^4 - c/(2*x^2)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
gosper	\(-\frac {2 c \,x^{2}+b}{4 x^{4}}\)	\(14\)
default	\(-\frac {b}{4 x^{4}}-\frac {c}{2 x^{2}}\)	\(14\)
risch	\(\frac {-\frac {c \,x^{2}}{2}-\frac {b}{4}}{x^{4}}\)	\(15\)
parallelrisch	\(\frac {-2 c \,x^{2}-b}{4 x^{4}}\)	\(16\)
norman	\(\frac {-\frac {1}{4} b \,x^{2}-\frac {1}{2} c \,x^{4}}{x^{6}}\)	\(18\)

[In]

int((c*x^4+b*x^2)/x^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {b x^2+c x^4}{x^7} \, dx=-\frac {2 \, c x^{2} + b}{4 \, x^{4}} \]

[In]

integrate((c*x^4+b*x^2)/x^7,x, algorithm="fricas")

[Out]

-1/4*(2*c*x^2 + b)/x^4

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {b x^2+c x^4}{x^7} \, dx=\frac {- b - 2 c x^{2}}{4 x^{4}} \]

[In]

integrate((c*x**4+b*x**2)/x**7,x)

[Out]

(-b - 2*c*x**2)/(4*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {b x^2+c x^4}{x^7} \, dx=-\frac {2 \, c x^{2} + b}{4 \, x^{4}} \]

[In]

integrate((c*x^4+b*x^2)/x^7,x, algorithm="maxima")

[Out]

-1/4*(2*c*x^2 + b)/x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {b x^2+c x^4}{x^7} \, dx=-\frac {2 \, c x^{2} + b}{4 \, x^{4}} \]

[In]

integrate((c*x^4+b*x^2)/x^7,x, algorithm="giac")

[Out]

-1/4*(2*c*x^2 + b)/x^4

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {b x^2+c x^4}{x^7} \, dx=-\frac {2\,c\,x^2+b}{4\,x^4} \]

[In]

int((b*x^2 + c*x^4)/x^7,x)

[Out]

-(b + 2*c*x^2)/(4*x^4)