\(\int x^3 (a+c x^4)^{3/2} \, dx\) [785]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int x^3 \left (a+c x^4\right )^{3/2} \, dx=\frac {\left (a+c x^4\right )^{5/2}}{10 c} \]

[Out]

1/10*(c*x^4+a)^(5/2)/c

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {267} \[ \int x^3 \left (a+c x^4\right )^{3/2} \, dx=\frac {\left (a+c x^4\right )^{5/2}}{10 c} \]

[In]

Int[x^3*(a + c*x^4)^(3/2),x]

[Out]

(a + c*x^4)^(5/2)/(10*c)

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+c x^4\right )^{5/2}}{10 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+c x^4\right )^{3/2} \, dx=\frac {\left (a+c x^4\right )^{5/2}}{10 c} \]

[In]

Integrate[x^3*(a + c*x^4)^(3/2),x]

[Out]

(a + c*x^4)^(5/2)/(10*c)

Maple [A] (verified)

Time = 4.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {\left (x^{4} c +a \right )^{\frac {5}{2}}}{10 c}\)	\(15\)
derivativedivides	\(\frac {\left (x^{4} c +a \right )^{\frac {5}{2}}}{10 c}\)	\(15\)
default	\(\frac {\left (x^{4} c +a \right )^{\frac {5}{2}}}{10 c}\)	\(15\)
pseudoelliptic	\(\frac {\left (x^{4} c +a \right )^{\frac {5}{2}}}{10 c}\)	\(15\)
trager	\(\frac {\left (c^{2} x^{8}+2 a \,x^{4} c +a^{2}\right ) \sqrt {x^{4} c +a}}{10 c}\)	\(33\)
risch	\(\frac {\left (c^{2} x^{8}+2 a \,x^{4} c +a^{2}\right ) \sqrt {x^{4} c +a}}{10 c}\)	\(33\)

[In]

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

[Out]

1/10*(c*x^4+a)^(5/2)/c

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (14) = 28\).

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int x^3 \left (a+c x^4\right )^{3/2} \, dx=\frac {{\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \sqrt {c x^{4} + a}}{10 \, c} \]

[In]

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

[Out]

1/10*(c^2*x^8 + 2*a*c*x^4 + a^2)*sqrt(c*x^4 + a)/c

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (12) = 24\).

Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 3.33 \[ \int x^3 \left (a+c x^4\right )^{3/2} \, dx=\begin {cases} \frac {a^{2} \sqrt {a + c x^{4}}}{10 c} + \frac {a x^{4} \sqrt {a + c x^{4}}}{5} + \frac {c x^{8} \sqrt {a + c x^{4}}}{10} & \text {for}\: c \neq 0 \\\frac {a^{\frac {3}{2}} x^{4}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*sqrt(a + c*x**4)/(10*c) + a*x**4*sqrt(a + c*x**4)/5 + c*x**8*sqrt(a + c*x**4)/10, Ne(c, 0)), (
a**(3/2)*x**4/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int x^3 \left (a+c x^4\right )^{3/2} \, dx=\frac {{\left (c x^{4} + a\right )}^{\frac {5}{2}}}{10 \, c} \]

[In]

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

[Out]

1/10*(c*x^4 + a)^(5/2)/c

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int x^3 \left (a+c x^4\right )^{3/2} \, dx=\frac {{\left (c x^{4} + a\right )}^{\frac {5}{2}}}{10 \, c} \]

[In]

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

[Out]

1/10*(c*x^4 + a)^(5/2)/c

Mupad [B] (verification not implemented)

Time = 5.98 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int x^3 \left (a+c x^4\right )^{3/2} \, dx=\frac {{\left (c\,x^4+a\right )}^{5/2}}{10\,c} \]

[In]

int(x^3*(a + c*x^4)^(3/2),x)

[Out]

(a + c*x^4)^(5/2)/(10*c)