\(\int \frac {b x^2+c x^4}{x^{5/2}} \, dx\) [298]

Optimal result

Integrand size = 17, antiderivative size = 19 \[ \int \frac {b x^2+c x^4}{x^{5/2}} \, dx=2 b \sqrt {x}+\frac {2}{5} c x^{5/2} \]

[Out]

2/5*c*x^(5/2)+2*b*x^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \[ \int \frac {b x^2+c x^4}{x^{5/2}} \, dx=2 b \sqrt {x}+\frac {2}{5} c x^{5/2} \]

[In]

Int[(b*x^2 + c*x^4)/x^(5/2),x]

[Out]

2*b*Sqrt[x] + (2*c*x^(5/2))/5

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {b x^2+c x^4}{x^{5/2}} \, dx=\frac {2}{5} \left (5 b \sqrt {x}+c x^{5/2}\right ) \]

[In]

Integrate[(b*x^2 + c*x^4)/x^(5/2),x]

[Out]

(2*(5*b*Sqrt[x] + c*x^(5/2)))/5

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {2 c \,x^{\frac {5}{2}}}{5}+2 b \sqrt {x}\)	\(14\)
default	\(\frac {2 c \,x^{\frac {5}{2}}}{5}+2 b \sqrt {x}\)	\(14\)
gosper	\(\frac {2 \sqrt {x}\, \left (c \,x^{2}+5 b \right )}{5}\)	\(15\)
trager	\(\left (\frac {2 c \,x^{2}}{5}+2 b \right ) \sqrt {x}\)	\(15\)
risch	\(\frac {2 \sqrt {x}\, \left (c \,x^{2}+5 b \right )}{5}\)	\(15\)

[In]

int((c*x^4+b*x^2)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/5*c*x^(5/2)+2*b*x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {b x^2+c x^4}{x^{5/2}} \, dx=\frac {2}{5} \, {\left (c x^{2} + 5 \, b\right )} \sqrt {x} \]

[In]

integrate((c*x^4+b*x^2)/x^(5/2),x, algorithm="fricas")

[Out]

2/5*(c*x^2 + 5*b)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {b x^2+c x^4}{x^{5/2}} \, dx=2 b \sqrt {x} + \frac {2 c x^{\frac {5}{2}}}{5} \]

[In]

integrate((c*x**4+b*x**2)/x**(5/2),x)

[Out]

2*b*sqrt(x) + 2*c*x**(5/2)/5

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {b x^2+c x^4}{x^{5/2}} \, dx=\frac {2}{5} \, c x^{\frac {5}{2}} + 2 \, b \sqrt {x} \]

[In]

integrate((c*x^4+b*x^2)/x^(5/2),x, algorithm="maxima")

[Out]

2/5*c*x^(5/2) + 2*b*sqrt(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {b x^2+c x^4}{x^{5/2}} \, dx=\frac {2}{5} \, c x^{\frac {5}{2}} + 2 \, b \sqrt {x} \]

[In]

integrate((c*x^4+b*x^2)/x^(5/2),x, algorithm="giac")

[Out]

2/5*c*x^(5/2) + 2*b*sqrt(x)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {b x^2+c x^4}{x^{5/2}} \, dx=\frac {2\,\sqrt {x}\,\left (c\,x^2+5\,b\right )}{5} \]

[In]

int((b*x^2 + c*x^4)/x^(5/2),x)

[Out]

(2*x^(1/2)*(5*b + c*x^2))/5