\(\int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx\) [2]

Optimal result

Integrand size = 22, antiderivative size = 91 \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{2} x \sqrt [4]{a^2+2 a b x^2+b^2 x^4}+\frac {\sqrt {a} \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b} \sqrt {1+\frac {b x^2}{a}}} \]

[Out]

1/2*x*(b^2*x^4+2*a*b*x^2+a^2)^(1/4)+1/2*(b^2*x^4+2*a*b*x^2+a^2)^(1/4)*arcsinh(x*b^(1/2)/a^(1/2))*a^(1/2)/b^(1/
2)/(1+b*x^2/a)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1103, 201, 221} \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\sqrt {a} \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b} \sqrt {\frac {b x^2}{a}+1}}+\frac {1}{2} x \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4),x]

[Out]

(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4))/2 + (Sqrt[a]*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4)*ArcSinh[(Sqrt[b]*x)/Sqrt[
a]])/(2*Sqrt[b]*Sqrt[1 + (b*x^2)/a])

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 221

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

Rule 1103

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^2 + c*x^4)^FracPart[p]
/(1 + 2*c*(x^2/b))^(2*FracPart[p])), Int[(1 + 2*c*(x^2/b))^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2
- 4*a*c, 0] &&  !IntegerQ[2*p]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.65 \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{2} \sqrt [4]{\left (a+b x^2\right )^2} \left (x-\frac {a \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {b} \sqrt {a+b x^2}}\right ) \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4),x]

[Out]

(((a + b*x^2)^2)^(1/4)*(x - (a*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(Sqrt[b]*Sqrt[a + b*x^2])))/2

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.64

method	result	size
risch	\(\frac {x {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {1}{4}}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {1}{4}}}{2 \sqrt {b}\, \sqrt {b \,x^{2}+a}}\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.62 \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\left [\frac {a \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {b} x - a\right ) + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} b x}{4 \, b}, -\frac {a \sqrt {-b} \arctan \left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {-b} x}{b x^{2} + a}\right ) - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} b x}{2 \, b}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\int \sqrt [4]{a^{2} + 2 a b x^{2} + b^{2} x^{4}}\, dx \]

[In]

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

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(1/4), x)

Maxima [F]

\[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/4} \,d x \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/4),x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/4), x)