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

Optimal. Leaf size=91 \[ \frac {1}{2} x \sqrt [4]{a^2+2 a b x^2+b^2 x^4}+\frac {\sqrt {a} \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b} \sqrt {\frac {b x^2}{a}+1}} \]

[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)

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Rubi [A]  time = 0.02, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1089, 195, 215} \[ \frac {1}{2} x \sqrt [4]{a^2+2 a b x^2+b^2 x^4}+\frac {\sqrt {a} \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b} \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[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 195

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 215

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 1089

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

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Mathematica [A]  time = 0.04, size = 59, normalized size = 0.65 \[ \frac {1}{2} \sqrt [4]{\left (a+b x^2\right )^2} \left (\frac {a \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{\sqrt {b} \sqrt {a+b x^2}}+x\right ) \]

Antiderivative was successfully verified.

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

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fricas [A]  time = 0.97, size = 147, normalized size = 1.62 \[ \left [\frac {a \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {b} x - a\right ) + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} b x}{4 \, b}, -\frac {a \sqrt {-b} \arctan \left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {-b} x}{b x^{2} + a}\right ) - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} b x}{2 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 58, normalized size = 0.64 \[ \frac {\left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {1}{4}} a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b \,x^{2}+a}\, \sqrt {b}}+\frac {\left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {1}{4}} x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt [4]{a^{2} + 2 a b x^{2} + b^{2} x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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