\(\int \frac {b x^2+c x^4}{\sqrt {x}} \, dx\) [296]

Optimal result

Integrand size = 17, antiderivative size = 21 \[ \int \frac {b x^2+c x^4}{\sqrt {x}} \, dx=\frac {2}{5} b x^{5/2}+\frac {2}{9} c x^{9/2} \]

[Out]

2/5*b*x^(5/2)+2/9*c*x^(9/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \[ \int \frac {b x^2+c x^4}{\sqrt {x}} \, dx=\frac {2}{5} b x^{5/2}+\frac {2}{9} c x^{9/2} \]

[In]

Int[(b*x^2 + c*x^4)/Sqrt[x],x]

[Out]

(2*b*x^(5/2))/5 + (2*c*x^(9/2))/9

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 (b x^{3/2}+c x^{7/2}\right ) \, dx \\ & = \frac {2}{5} b x^{5/2}+\frac {2}{9} c x^{9/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {b x^2+c x^4}{\sqrt {x}} \, dx=\frac {2}{45} \left (9 b x^{5/2}+5 c x^{9/2}\right ) \]

[In]

Integrate[(b*x^2 + c*x^4)/Sqrt[x],x]

[Out]

(2*(9*b*x^(5/2) + 5*c*x^(9/2)))/45

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {2 b \,x^{\frac {5}{2}}}{5}+\frac {2 c \,x^{\frac {9}{2}}}{9}\)	\(14\)
default	\(\frac {2 b \,x^{\frac {5}{2}}}{5}+\frac {2 c \,x^{\frac {9}{2}}}{9}\)	\(14\)
gosper	\(\frac {2 x^{\frac {5}{2}} \left (5 c \,x^{2}+9 b \right )}{45}\)	\(16\)
trager	\(\frac {2 x^{\frac {5}{2}} \left (5 c \,x^{2}+9 b \right )}{45}\)	\(16\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (5 c \,x^{2}+9 b \right )}{45}\)	\(16\)

[In]

int((c*x^4+b*x^2)/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/5*b*x^(5/2)+2/9*c*x^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {b x^2+c x^4}{\sqrt {x}} \, dx=\frac {2}{45} \, {\left (5 \, c x^{4} + 9 \, b x^{2}\right )} \sqrt {x} \]

[In]

integrate((c*x^4+b*x^2)/x^(1/2),x, algorithm="fricas")

[Out]

2/45*(5*c*x^4 + 9*b*x^2)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {b x^2+c x^4}{\sqrt {x}} \, dx=\frac {2 b x^{\frac {5}{2}}}{5} + \frac {2 c x^{\frac {9}{2}}}{9} \]

[In]

integrate((c*x**4+b*x**2)/x**(1/2),x)

[Out]

2*b*x**(5/2)/5 + 2*c*x**(9/2)/9

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {b x^2+c x^4}{\sqrt {x}} \, dx=\frac {2}{9} \, c x^{\frac {9}{2}} + \frac {2}{5} \, b x^{\frac {5}{2}} \]

[In]

integrate((c*x^4+b*x^2)/x^(1/2),x, algorithm="maxima")

[Out]

2/9*c*x^(9/2) + 2/5*b*x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {b x^2+c x^4}{\sqrt {x}} \, dx=\frac {2}{9} \, c x^{\frac {9}{2}} + \frac {2}{5} \, b x^{\frac {5}{2}} \]

[In]

integrate((c*x^4+b*x^2)/x^(1/2),x, algorithm="giac")

[Out]

2/9*c*x^(9/2) + 2/5*b*x^(5/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {b x^2+c x^4}{\sqrt {x}} \, dx=\frac {2\,x^{5/2}\,\left (5\,c\,x^2+9\,b\right )}{45} \]

[In]

int((b*x^2 + c*x^4)/x^(1/2),x)

[Out]

(2*x^(5/2)*(9*b + 5*c*x^2))/45