\(\int \frac {(b x^2+c x^4)^3}{\sqrt {x}} \, dx\) [312]

Optimal result

Integrand size = 19, antiderivative size = 51 \[ \int \frac {\left (b x^2+c x^4\right )^3}{\sqrt {x}} \, dx=\frac {2}{13} b^3 x^{13/2}+\frac {6}{17} b^2 c x^{17/2}+\frac {2}{7} b c^2 x^{21/2}+\frac {2}{25} c^3 x^{25/2} \]

[Out]

2/13*b^3*x^(13/2)+6/17*b^2*c*x^(17/2)+2/7*b*c^2*x^(21/2)+2/25*c^3*x^(25/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, 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} \[ \int \frac {\left (b x^2+c x^4\right )^3}{\sqrt {x}} \, dx=\frac {2}{13} b^3 x^{13/2}+\frac {6}{17} b^2 c x^{17/2}+\frac {2}{7} b c^2 x^{21/2}+\frac {2}{25} c^3 x^{25/2} \]

[In]

Int[(b*x^2 + c*x^4)^3/Sqrt[x],x]

[Out]

(2*b^3*x^(13/2))/13 + (6*b^2*c*x^(17/2))/17 + (2*b*c^2*x^(21/2))/7 + (2*c^3*x^(25/2))/25

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^{11/2} \left (b+c x^2\right )^3 \, dx \\ & = \int \left (b^3 x^{11/2}+3 b^2 c x^{15/2}+3 b c^2 x^{19/2}+c^3 x^{23/2}\right ) \, dx \\ & = \frac {2}{13} b^3 x^{13/2}+\frac {6}{17} b^2 c x^{17/2}+\frac {2}{7} b c^2 x^{21/2}+\frac {2}{25} c^3 x^{25/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int \frac {\left (b x^2+c x^4\right )^3}{\sqrt {x}} \, dx=\frac {2 x^{13/2} \left (2975 b^3+6825 b^2 c x^2+5525 b c^2 x^4+1547 c^3 x^6\right )}{38675} \]

[In]

Integrate[(b*x^2 + c*x^4)^3/Sqrt[x],x]

[Out]

(2*x^(13/2)*(2975*b^3 + 6825*b^2*c*x^2 + 5525*b*c^2*x^4 + 1547*c^3*x^6))/38675

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {2 b^{3} x^{\frac {13}{2}}}{13}+\frac {6 b^{2} c \,x^{\frac {17}{2}}}{17}+\frac {2 b \,c^{2} x^{\frac {21}{2}}}{7}+\frac {2 c^{3} x^{\frac {25}{2}}}{25}\)	\(36\)
default	\(\frac {2 b^{3} x^{\frac {13}{2}}}{13}+\frac {6 b^{2} c \,x^{\frac {17}{2}}}{17}+\frac {2 b \,c^{2} x^{\frac {21}{2}}}{7}+\frac {2 c^{3} x^{\frac {25}{2}}}{25}\)	\(36\)
gosper	\(\frac {2 x^{\frac {13}{2}} \left (1547 c^{3} x^{6}+5525 b \,c^{2} x^{4}+6825 b^{2} c \,x^{2}+2975 b^{3}\right )}{38675}\)	\(38\)
trager	\(\frac {2 x^{\frac {13}{2}} \left (1547 c^{3} x^{6}+5525 b \,c^{2} x^{4}+6825 b^{2} c \,x^{2}+2975 b^{3}\right )}{38675}\)	\(38\)
risch	\(\frac {2 x^{\frac {13}{2}} \left (1547 c^{3} x^{6}+5525 b \,c^{2} x^{4}+6825 b^{2} c \,x^{2}+2975 b^{3}\right )}{38675}\)	\(38\)

[In]

int((c*x^4+b*x^2)^3/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/13*b^3*x^(13/2)+6/17*b^2*c*x^(17/2)+2/7*b*c^2*x^(21/2)+2/25*c^3*x^(25/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int \frac {\left (b x^2+c x^4\right )^3}{\sqrt {x}} \, dx=\frac {2}{38675} \, {\left (1547 \, c^{3} x^{12} + 5525 \, b c^{2} x^{10} + 6825 \, b^{2} c x^{8} + 2975 \, b^{3} x^{6}\right )} \sqrt {x} \]

[In]

integrate((c*x^4+b*x^2)^3/x^(1/2),x, algorithm="fricas")

[Out]

2/38675*(1547*c^3*x^12 + 5525*b*c^2*x^10 + 6825*b^2*c*x^8 + 2975*b^3*x^6)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \frac {\left (b x^2+c x^4\right )^3}{\sqrt {x}} \, dx=\frac {2 b^{3} x^{\frac {13}{2}}}{13} + \frac {6 b^{2} c x^{\frac {17}{2}}}{17} + \frac {2 b c^{2} x^{\frac {21}{2}}}{7} + \frac {2 c^{3} x^{\frac {25}{2}}}{25} \]

[In]

integrate((c*x**4+b*x**2)**3/x**(1/2),x)

[Out]

2*b**3*x**(13/2)/13 + 6*b**2*c*x**(17/2)/17 + 2*b*c**2*x**(21/2)/7 + 2*c**3*x**(25/2)/25

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {\left (b x^2+c x^4\right )^3}{\sqrt {x}} \, dx=\frac {2}{25} \, c^{3} x^{\frac {25}{2}} + \frac {2}{7} \, b c^{2} x^{\frac {21}{2}} + \frac {6}{17} \, b^{2} c x^{\frac {17}{2}} + \frac {2}{13} \, b^{3} x^{\frac {13}{2}} \]

[In]

integrate((c*x^4+b*x^2)^3/x^(1/2),x, algorithm="maxima")

[Out]

2/25*c^3*x^(25/2) + 2/7*b*c^2*x^(21/2) + 6/17*b^2*c*x^(17/2) + 2/13*b^3*x^(13/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {\left (b x^2+c x^4\right )^3}{\sqrt {x}} \, dx=\frac {2}{25} \, c^{3} x^{\frac {25}{2}} + \frac {2}{7} \, b c^{2} x^{\frac {21}{2}} + \frac {6}{17} \, b^{2} c x^{\frac {17}{2}} + \frac {2}{13} \, b^{3} x^{\frac {13}{2}} \]

[In]

integrate((c*x^4+b*x^2)^3/x^(1/2),x, algorithm="giac")

[Out]

2/25*c^3*x^(25/2) + 2/7*b*c^2*x^(21/2) + 6/17*b^2*c*x^(17/2) + 2/13*b^3*x^(13/2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {\left (b x^2+c x^4\right )^3}{\sqrt {x}} \, dx=\frac {2\,b^3\,x^{13/2}}{13}+\frac {2\,c^3\,x^{25/2}}{25}+\frac {6\,b^2\,c\,x^{17/2}}{17}+\frac {2\,b\,c^2\,x^{21/2}}{7} \]

[In]

int((b*x^2 + c*x^4)^3/x^(1/2),x)

[Out]

(2*b^3*x^(13/2))/13 + (2*c^3*x^(25/2))/25 + (6*b^2*c*x^(17/2))/17 + (2*b*c^2*x^(21/2))/7