\(\int \frac {(b x^2+c x^4)^2}{\sqrt {x}} \, dx\) [304]

Optimal result

Integrand size = 19, antiderivative size = 36 \[ \int \frac {\left (b x^2+c x^4\right )^2}{\sqrt {x}} \, dx=\frac {2}{9} b^2 x^{9/2}+\frac {4}{13} b c x^{13/2}+\frac {2}{17} c^2 x^{17/2} \]

[Out]

2/9*b^2*x^(9/2)+4/13*b*c*x^(13/2)+2/17*c^2*x^(17/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1598, 276} \[ \int \frac {\left (b x^2+c x^4\right )^2}{\sqrt {x}} \, dx=\frac {2}{9} b^2 x^{9/2}+\frac {4}{13} b c x^{13/2}+\frac {2}{17} c^2 x^{17/2} \]

[In]

Int[(b*x^2 + c*x^4)^2/Sqrt[x],x]

[Out]

(2*b^2*x^(9/2))/9 + (4*b*c*x^(13/2))/13 + (2*c^2*x^(17/2))/17

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int x^{7/2} \left (b+c x^2\right )^2 \, dx \\ & = \int \left (b^2 x^{7/2}+2 b c x^{11/2}+c^2 x^{15/2}\right ) \, dx \\ & = \frac {2}{9} b^2 x^{9/2}+\frac {4}{13} b c x^{13/2}+\frac {2}{17} c^2 x^{17/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.83 \[ \int \frac {\left (b x^2+c x^4\right )^2}{\sqrt {x}} \, dx=\frac {2 x^{9/2} \left (221 b^2+306 b c x^2+117 c^2 x^4\right )}{1989} \]

[In]

Integrate[(b*x^2 + c*x^4)^2/Sqrt[x],x]

[Out]

(2*x^(9/2)*(221*b^2 + 306*b*c*x^2 + 117*c^2*x^4))/1989

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {2 b^{2} x^{\frac {9}{2}}}{9}+\frac {4 b c \,x^{\frac {13}{2}}}{13}+\frac {2 c^{2} x^{\frac {17}{2}}}{17}\)	\(25\)
default	\(\frac {2 b^{2} x^{\frac {9}{2}}}{9}+\frac {4 b c \,x^{\frac {13}{2}}}{13}+\frac {2 c^{2} x^{\frac {17}{2}}}{17}\)	\(25\)
gosper	\(\frac {2 x^{\frac {9}{2}} \left (117 c^{2} x^{4}+306 b c \,x^{2}+221 b^{2}\right )}{1989}\)	\(27\)
trager	\(\frac {2 x^{\frac {9}{2}} \left (117 c^{2} x^{4}+306 b c \,x^{2}+221 b^{2}\right )}{1989}\)	\(27\)
risch	\(\frac {2 x^{\frac {9}{2}} \left (117 c^{2} x^{4}+306 b c \,x^{2}+221 b^{2}\right )}{1989}\)	\(27\)

[In]

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

[Out]

2/9*b^2*x^(9/2)+4/13*b*c*x^(13/2)+2/17*c^2*x^(17/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^2}{\sqrt {x}} \, dx=\frac {2}{1989} \, {\left (117 \, c^{2} x^{8} + 306 \, b c x^{6} + 221 \, b^{2} x^{4}\right )} \sqrt {x} \]

[In]

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

[Out]

2/1989*(117*c^2*x^8 + 306*b*c*x^6 + 221*b^2*x^4)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {\left (b x^2+c x^4\right )^2}{\sqrt {x}} \, dx=\frac {2 b^{2} x^{\frac {9}{2}}}{9} + \frac {4 b c x^{\frac {13}{2}}}{13} + \frac {2 c^{2} x^{\frac {17}{2}}}{17} \]

[In]

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

[Out]

2*b**2*x**(9/2)/9 + 4*b*c*x**(13/2)/13 + 2*c**2*x**(17/2)/17

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (b x^2+c x^4\right )^2}{\sqrt {x}} \, dx=\frac {2}{17} \, c^{2} x^{\frac {17}{2}} + \frac {4}{13} \, b c x^{\frac {13}{2}} + \frac {2}{9} \, b^{2} x^{\frac {9}{2}} \]

[In]

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

[Out]

2/17*c^2*x^(17/2) + 4/13*b*c*x^(13/2) + 2/9*b^2*x^(9/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (b x^2+c x^4\right )^2}{\sqrt {x}} \, dx=\frac {2}{17} \, c^{2} x^{\frac {17}{2}} + \frac {4}{13} \, b c x^{\frac {13}{2}} + \frac {2}{9} \, b^{2} x^{\frac {9}{2}} \]

[In]

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

[Out]

2/17*c^2*x^(17/2) + 4/13*b*c*x^(13/2) + 2/9*b^2*x^(9/2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \frac {\left (b x^2+c x^4\right )^2}{\sqrt {x}} \, dx=x^{9/2}\,\left (\frac {2\,b^2}{9}+\frac {4\,b\,c\,x^2}{13}+\frac {2\,c^2\,x^4}{17}\right ) \]

[In]

int((b*x^2 + c*x^4)^2/x^(1/2),x)

[Out]

x^(9/2)*((2*b^2)/9 + (2*c^2*x^4)/17 + (4*b*c*x^2)/13)