\(\int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx\) [1022]

Optimal result

Integrand size = 23, antiderivative size = 16 \[ \int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=\frac {x^4}{2 \sqrt {c x^4}} \]

[Out]

1/2*x^4/(c*x^4)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1, 15, 30} \[ \int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=\frac {x^4}{2 \sqrt {c x^4}} \]

[In]

Int[x^3/Sqrt[2 + 2*a - 2*(1 + a) + c*x^4],x]

[Out]

x^4/(2*Sqrt[c*x^4])

Rule 1

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n)^p, x] /; FreeQ[{a, b, n, p}, x] && EqQ[a
, 0]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3}{\sqrt {c x^4}} \, dx \\ & = \frac {x^2 \int x \, dx}{\sqrt {c x^4}} \\ & = \frac {x^4}{2 \sqrt {c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=\frac {x^4}{2 \sqrt {c x^4}} \]

[In]

Integrate[x^3/Sqrt[2 + 2*a - 2*(1 + a) + c*x^4],x]

[Out]

x^4/(2*Sqrt[c*x^4])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
gosper	\(\frac {x^{4}}{2 \sqrt {c \,x^{4}}}\)	\(13\)
derivativedivides	\(\frac {\sqrt {c \,x^{4}}}{2 c}\)	\(13\)
default	\(\frac {x^{4}}{2 \sqrt {c \,x^{4}}}\)	\(13\)
risch	\(\frac {x^{4}}{2 \sqrt {c \,x^{4}}}\)	\(13\)
pseudoelliptic	\(\frac {\sqrt {c \,x^{4}}}{2 c}\)	\(13\)
trager	\(\frac {\left (x -1\right ) \left (x +1\right ) \sqrt {c \,x^{4}}}{2 c \,x^{2}}\)	\(22\)

[In]

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

[Out]

1/2*x^4/(c*x^4)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=\frac {\sqrt {c x^{4}}}{2 \, c} \]

[In]

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

[Out]

1/2*sqrt(c*x^4)/c

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=\frac {x^{4}}{2 \sqrt {c x^{4}}} \]

[In]

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

[Out]

x**4/(2*sqrt(c*x**4))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=\frac {\sqrt {c x^{4}}}{2 \, c} \]

[In]

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

[Out]

1/2*sqrt(c*x^4)/c

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.50 \[ \int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=\frac {x^{2}}{2 \, \sqrt {c}} \]

[In]

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

[Out]

1/2*x^2/sqrt(c)

Mupad [B] (verification not implemented)

Time = 13.39 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {x^3}{\sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=\frac {\sqrt {x^4}}{2\,\sqrt {c}} \]

[In]

int(x^3/(c*x^4)^(1/2),x)

[Out]

(x^4)^(1/2)/(2*c^(1/2))