\(\int x \sqrt {a+c x^4} \, dx\) [771]

Optimal result

Integrand size = 13, antiderivative size = 50 \[ \int x \sqrt {a+c x^4} \, dx=\frac {1}{4} x^2 \sqrt {a+c x^4}+\frac {a \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 201, 223, 212} \[ \int x \sqrt {a+c x^4} \, dx=\frac {a \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 \sqrt {c}}+\frac {1}{4} x^2 \sqrt {a+c x^4} \]

[In]

Int[x*Sqrt[a + c*x^4],x]

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \sqrt {a+c x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{4} x^2 \sqrt {a+c x^4}+\frac {1}{4} a \text {Subst}\left (\int \frac {1}{\sqrt {a+c x^2}} \, dx,x,x^2\right ) \\ & = \frac {1}{4} x^2 \sqrt {a+c x^4}+\frac {1}{4} a \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {a+c x^4}}\right ) \\ & = \frac {1}{4} x^2 \sqrt {a+c x^4}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.02 \[ \int x \sqrt {a+c x^4} \, dx=\frac {1}{4} x^2 \sqrt {a+c x^4}+\frac {a \log \left (\sqrt {c} x^2+\sqrt {a+c x^4}\right )}{4 \sqrt {c}} \]

[In]

Integrate[x*Sqrt[a + c*x^4],x]

[Out]

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

Maple [A] (verified)

Time = 4.61 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {x^{2} \sqrt {x^{4} c +a}}{4}+\frac {a \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{4 \sqrt {c}}\)	\(40\)
risch	\(\frac {x^{2} \sqrt {x^{4} c +a}}{4}+\frac {a \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{4 \sqrt {c}}\)	\(40\)
elliptic	\(\frac {x^{2} \sqrt {x^{4} c +a}}{4}+\frac {a \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{4 \sqrt {c}}\)	\(40\)
pseudoelliptic	\(\frac {\sqrt {x^{4} c +a}\, x^{2} \sqrt {c}+\operatorname {arctanh}\left (\frac {\sqrt {x^{4} c +a}}{x^{2} \sqrt {c}}\right ) a}{4 \sqrt {c}}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.04 \[ \int x \sqrt {a+c x^4} \, dx=\left [\frac {2 \, \sqrt {c x^{4} + a} c x^{2} + a \sqrt {c} \log \left (-2 \, c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right )}{8 \, c}, \frac {\sqrt {c x^{4} + a} c x^{2} - a \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2}}{\sqrt {c x^{4} + a}}\right )}{4 \, c}\right ] \]

[In]

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

[Out]

[1/8*(2*sqrt(c*x^4 + a)*c*x^2 + a*sqrt(c)*log(-2*c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a))/c, 1/4*(sqrt(c*x^
4 + a)*c*x^2 - a*sqrt(-c)*arctan(sqrt(-c)*x^2/sqrt(c*x^4 + a)))/c]

Sympy [A] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int x \sqrt {a+c x^4} \, dx=\frac {\sqrt {a} x^{2} \sqrt {1 + \frac {c x^{4}}{a}}}{4} + \frac {a \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {c}} \]

[In]

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

[Out]

sqrt(a)*x**2*sqrt(1 + c*x**4/a)/4 + a*asinh(sqrt(c)*x**2/sqrt(a))/(4*sqrt(c))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (38) = 76\).

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.56 \[ \int x \sqrt {a+c x^4} \, dx=-\frac {a \log \left (-\frac {\sqrt {c} - \frac {\sqrt {c x^{4} + a}}{x^{2}}}{\sqrt {c} + \frac {\sqrt {c x^{4} + a}}{x^{2}}}\right )}{8 \, \sqrt {c}} - \frac {\sqrt {c x^{4} + a} a}{4 \, {\left (c - \frac {c x^{4} + a}{x^{4}}\right )} x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.82 \[ \int x \sqrt {a+c x^4} \, dx=\frac {1}{4} \, \sqrt {c x^{4} + a} x^{2} - \frac {a \log \left ({\left | -\sqrt {c} x^{2} + \sqrt {c x^{4} + a} \right |}\right )}{4 \, \sqrt {c}} \]

[In]

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

[Out]

1/4*sqrt(c*x^4 + a)*x^2 - 1/4*a*log(abs(-sqrt(c)*x^2 + sqrt(c*x^4 + a)))/sqrt(c)

Mupad [F(-1)]

Timed out. \[ \int x \sqrt {a+c x^4} \, dx=\int x\,\sqrt {c\,x^4+a} \,d x \]

[In]

int(x*(a + c*x^4)^(1/2),x)

[Out]

int(x*(a + c*x^4)^(1/2), x)