\(\int x^{5/2} (b x^2+c x^4)^3 \, dx\) [309]

Optimal result

Integrand size = 19, antiderivative size = 51 \[ \int x^{5/2} \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{19} b^3 x^{19/2}+\frac {6}{23} b^2 c x^{23/2}+\frac {2}{9} b c^2 x^{27/2}+\frac {2}{31} c^3 x^{31/2} \]

[Out]

2/19*b^3*x^(19/2)+6/23*b^2*c*x^(23/2)+2/9*b*c^2*x^(27/2)+2/31*c^3*x^(31/2)

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} \[ \int x^{5/2} \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{19} b^3 x^{19/2}+\frac {6}{23} b^2 c x^{23/2}+\frac {2}{9} b c^2 x^{27/2}+\frac {2}{31} c^3 x^{31/2} \]

[In]

Int[x^(5/2)*(b*x^2 + c*x^4)^3,x]

[Out]

(2*b^3*x^(19/2))/19 + (6*b^2*c*x^(23/2))/23 + (2*b*c^2*x^(27/2))/9 + (2*c^3*x^(31/2))/31

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^{17/2} \left (b+c x^2\right )^3 \, dx \\ & = \int \left (b^3 x^{17/2}+3 b^2 c x^{21/2}+3 b c^2 x^{25/2}+c^3 x^{29/2}\right ) \, dx \\ & = \frac {2}{19} b^3 x^{19/2}+\frac {6}{23} b^2 c x^{23/2}+\frac {2}{9} b c^2 x^{27/2}+\frac {2}{31} c^3 x^{31/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int x^{5/2} \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{19} b^3 x^{19/2}+\frac {6}{23} b^2 c x^{23/2}+\frac {2}{9} b c^2 x^{27/2}+\frac {2}{31} c^3 x^{31/2} \]

[In]

Integrate[x^(5/2)*(b*x^2 + c*x^4)^3,x]

[Out]

(2*b^3*x^(19/2))/19 + (6*b^2*c*x^(23/2))/23 + (2*b*c^2*x^(27/2))/9 + (2*c^3*x^(31/2))/31

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {2 b^{3} x^{\frac {19}{2}}}{19}+\frac {6 b^{2} c \,x^{\frac {23}{2}}}{23}+\frac {2 b \,c^{2} x^{\frac {27}{2}}}{9}+\frac {2 c^{3} x^{\frac {31}{2}}}{31}\)	\(36\)
default	\(\frac {2 b^{3} x^{\frac {19}{2}}}{19}+\frac {6 b^{2} c \,x^{\frac {23}{2}}}{23}+\frac {2 b \,c^{2} x^{\frac {27}{2}}}{9}+\frac {2 c^{3} x^{\frac {31}{2}}}{31}\)	\(36\)
gosper	\(\frac {2 x^{\frac {19}{2}} \left (3933 c^{3} x^{6}+13547 b \,c^{2} x^{4}+15903 b^{2} c \,x^{2}+6417 b^{3}\right )}{121923}\)	\(38\)
trager	\(\frac {2 x^{\frac {19}{2}} \left (3933 c^{3} x^{6}+13547 b \,c^{2} x^{4}+15903 b^{2} c \,x^{2}+6417 b^{3}\right )}{121923}\)	\(38\)
risch	\(\frac {2 x^{\frac {19}{2}} \left (3933 c^{3} x^{6}+13547 b \,c^{2} x^{4}+15903 b^{2} c \,x^{2}+6417 b^{3}\right )}{121923}\)	\(38\)

[In]

int(x^(5/2)*(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

2/19*b^3*x^(19/2)+6/23*b^2*c*x^(23/2)+2/9*b*c^2*x^(27/2)+2/31*c^3*x^(31/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int x^{5/2} \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{121923} \, {\left (3933 \, c^{3} x^{15} + 13547 \, b c^{2} x^{13} + 15903 \, b^{2} c x^{11} + 6417 \, b^{3} x^{9}\right )} \sqrt {x} \]

[In]

integrate(x^(5/2)*(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

2/121923*(3933*c^3*x^15 + 13547*b*c^2*x^13 + 15903*b^2*c*x^11 + 6417*b^3*x^9)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 1.56 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int x^{5/2} \left (b x^2+c x^4\right )^3 \, dx=\frac {2 b^{3} x^{\frac {19}{2}}}{19} + \frac {6 b^{2} c x^{\frac {23}{2}}}{23} + \frac {2 b c^{2} x^{\frac {27}{2}}}{9} + \frac {2 c^{3} x^{\frac {31}{2}}}{31} \]

[In]

integrate(x**(5/2)*(c*x**4+b*x**2)**3,x)

[Out]

2*b**3*x**(19/2)/19 + 6*b**2*c*x**(23/2)/23 + 2*b*c**2*x**(27/2)/9 + 2*c**3*x**(31/2)/31

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{5/2} \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{31} \, c^{3} x^{\frac {31}{2}} + \frac {2}{9} \, b c^{2} x^{\frac {27}{2}} + \frac {6}{23} \, b^{2} c x^{\frac {23}{2}} + \frac {2}{19} \, b^{3} x^{\frac {19}{2}} \]

[In]

integrate(x^(5/2)*(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

2/31*c^3*x^(31/2) + 2/9*b*c^2*x^(27/2) + 6/23*b^2*c*x^(23/2) + 2/19*b^3*x^(19/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{5/2} \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{31} \, c^{3} x^{\frac {31}{2}} + \frac {2}{9} \, b c^{2} x^{\frac {27}{2}} + \frac {6}{23} \, b^{2} c x^{\frac {23}{2}} + \frac {2}{19} \, b^{3} x^{\frac {19}{2}} \]

[In]

integrate(x^(5/2)*(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

2/31*c^3*x^(31/2) + 2/9*b*c^2*x^(27/2) + 6/23*b^2*c*x^(23/2) + 2/19*b^3*x^(19/2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{5/2} \left (b x^2+c x^4\right )^3 \, dx=\frac {2\,b^3\,x^{19/2}}{19}+\frac {2\,c^3\,x^{31/2}}{31}+\frac {6\,b^2\,c\,x^{23/2}}{23}+\frac {2\,b\,c^2\,x^{27/2}}{9} \]

[In]

int(x^(5/2)*(b*x^2 + c*x^4)^3,x)

[Out]

(2*b^3*x^(19/2))/19 + (2*c^3*x^(31/2))/31 + (6*b^2*c*x^(23/2))/23 + (2*b*c^2*x^(27/2))/9