\(\int x^{3/2} (a+c x^4)^3 \, dx\) [732]

Optimal result

Integrand size = 15, antiderivative size = 51 \[ \int x^{3/2} \left (a+c x^4\right )^3 \, dx=\frac {2}{5} a^3 x^{5/2}+\frac {6}{13} a^2 c x^{13/2}+\frac {2}{7} a c^2 x^{21/2}+\frac {2}{29} c^3 x^{29/2} \]

[Out]

2/5*a^3*x^(5/2)+6/13*a^2*c*x^(13/2)+2/7*a*c^2*x^(21/2)+2/29*c^3*x^(29/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {276} \[ \int x^{3/2} \left (a+c x^4\right )^3 \, dx=\frac {2}{5} a^3 x^{5/2}+\frac {6}{13} a^2 c x^{13/2}+\frac {2}{7} a c^2 x^{21/2}+\frac {2}{29} c^3 x^{29/2} \]

[In]

Int[x^(3/2)*(a + c*x^4)^3,x]

[Out]

(2*a^3*x^(5/2))/5 + (6*a^2*c*x^(13/2))/13 + (2*a*c^2*x^(21/2))/7 + (2*c^3*x^(29/2))/29

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 x^{3/2}+3 a^2 c x^{11/2}+3 a c^2 x^{19/2}+c^3 x^{27/2}\right ) \, dx \\ & = \frac {2}{5} a^3 x^{5/2}+\frac {6}{13} a^2 c x^{13/2}+\frac {2}{7} a c^2 x^{21/2}+\frac {2}{29} c^3 x^{29/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92 \[ \int x^{3/2} \left (a+c x^4\right )^3 \, dx=\frac {2 \left (2639 a^3 x^{5/2}+3045 a^2 c x^{13/2}+1885 a c^2 x^{21/2}+455 c^3 x^{29/2}\right )}{13195} \]

[In]

Integrate[x^(3/2)*(a + c*x^4)^3,x]

[Out]

(2*(2639*a^3*x^(5/2) + 3045*a^2*c*x^(13/2) + 1885*a*c^2*x^(21/2) + 455*c^3*x^(29/2)))/13195

Maple [A] (verified)

Time = 3.95 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {2 a^{3} x^{\frac {5}{2}}}{5}+\frac {6 a^{2} c \,x^{\frac {13}{2}}}{13}+\frac {2 a \,c^{2} x^{\frac {21}{2}}}{7}+\frac {2 c^{3} x^{\frac {29}{2}}}{29}\)	\(36\)
default	\(\frac {2 a^{3} x^{\frac {5}{2}}}{5}+\frac {6 a^{2} c \,x^{\frac {13}{2}}}{13}+\frac {2 a \,c^{2} x^{\frac {21}{2}}}{7}+\frac {2 c^{3} x^{\frac {29}{2}}}{29}\)	\(36\)
gosper	\(\frac {2 x^{\frac {5}{2}} \left (455 c^{3} x^{12}+1885 a \,c^{2} x^{8}+3045 a^{2} c \,x^{4}+2639 a^{3}\right )}{13195}\)	\(38\)
trager	\(\frac {2 x^{\frac {5}{2}} \left (455 c^{3} x^{12}+1885 a \,c^{2} x^{8}+3045 a^{2} c \,x^{4}+2639 a^{3}\right )}{13195}\)	\(38\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (455 c^{3} x^{12}+1885 a \,c^{2} x^{8}+3045 a^{2} c \,x^{4}+2639 a^{3}\right )}{13195}\)	\(38\)

[In]

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

[Out]

2/5*a^3*x^(5/2)+6/13*a^2*c*x^(13/2)+2/7*a*c^2*x^(21/2)+2/29*c^3*x^(29/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int x^{3/2} \left (a+c x^4\right )^3 \, dx=\frac {2}{13195} \, {\left (455 \, c^{3} x^{14} + 1885 \, a c^{2} x^{10} + 3045 \, a^{2} c x^{6} + 2639 \, a^{3} x^{2}\right )} \sqrt {x} \]

[In]

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

[Out]

2/13195*(455*c^3*x^14 + 1885*a*c^2*x^10 + 3045*a^2*c*x^6 + 2639*a^3*x^2)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int x^{3/2} \left (a+c x^4\right )^3 \, dx=\frac {2 a^{3} x^{\frac {5}{2}}}{5} + \frac {6 a^{2} c x^{\frac {13}{2}}}{13} + \frac {2 a c^{2} x^{\frac {21}{2}}}{7} + \frac {2 c^{3} x^{\frac {29}{2}}}{29} \]

[In]

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

[Out]

2*a**3*x**(5/2)/5 + 6*a**2*c*x**(13/2)/13 + 2*a*c**2*x**(21/2)/7 + 2*c**3*x**(29/2)/29

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{3/2} \left (a+c x^4\right )^3 \, dx=\frac {2}{29} \, c^{3} x^{\frac {29}{2}} + \frac {2}{7} \, a c^{2} x^{\frac {21}{2}} + \frac {6}{13} \, a^{2} c x^{\frac {13}{2}} + \frac {2}{5} \, a^{3} x^{\frac {5}{2}} \]

[In]

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

[Out]

2/29*c^3*x^(29/2) + 2/7*a*c^2*x^(21/2) + 6/13*a^2*c*x^(13/2) + 2/5*a^3*x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{3/2} \left (a+c x^4\right )^3 \, dx=\frac {2}{29} \, c^{3} x^{\frac {29}{2}} + \frac {2}{7} \, a c^{2} x^{\frac {21}{2}} + \frac {6}{13} \, a^{2} c x^{\frac {13}{2}} + \frac {2}{5} \, a^{3} x^{\frac {5}{2}} \]

[In]

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

[Out]

2/29*c^3*x^(29/2) + 2/7*a*c^2*x^(21/2) + 6/13*a^2*c*x^(13/2) + 2/5*a^3*x^(5/2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{3/2} \left (a+c x^4\right )^3 \, dx=\frac {2\,a^3\,x^{5/2}}{5}+\frac {2\,c^3\,x^{29/2}}{29}+\frac {6\,a^2\,c\,x^{13/2}}{13}+\frac {2\,a\,c^2\,x^{21/2}}{7} \]

[In]

int(x^(3/2)*(a + c*x^4)^3,x)

[Out]

(2*a^3*x^(5/2))/5 + (2*c^3*x^(29/2))/29 + (6*a^2*c*x^(13/2))/13 + (2*a*c^2*x^(21/2))/7