Integrand size = 22, antiderivative size = 91 \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{2} x \sqrt [4]{a^2+2 a b x^2+b^2 x^4}+\frac {\sqrt {a} \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b} \sqrt {1+\frac {b x^2}{a}}} \] Output:
1/2*x*(b^2*x^4+2*a*b*x^2+a^2)^(1/4)+1/2*a^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^(1 /4)*arcsinh(b^(1/2)*x/a^(1/2))/b^(1/2)/(1+b*x^2/a)^(1/2)
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.65 \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{2} \sqrt [4]{\left (a+b x^2\right )^2} \left (x-\frac {a \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {b} \sqrt {a+b x^2}}\right ) \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4),x]
Output:
(((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
Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1385, 211, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx\) |
\(\Big \downarrow \) 1385 |
\(\displaystyle \frac {\sqrt [4]{a^2+2 a b x^2+b^2 x^4} \int \sqrt {\frac {b x^2}{a}+1}dx}{\sqrt {\frac {b x^2}{a}+1}}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\sqrt [4]{a^2+2 a b x^2+b^2 x^4} \left (\frac {1}{2} \int \frac {1}{\sqrt {\frac {b x^2}{a}+1}}dx+\frac {1}{2} x \sqrt {\frac {b x^2}{a}+1}\right )}{\sqrt {\frac {b x^2}{a}+1}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\sqrt [4]{a^2+2 a b x^2+b^2 x^4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {\frac {b x^2}{a}+1}\right )}{\sqrt {\frac {b x^2}{a}+1}}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4),x]
Output:
((a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4)*((x*Sqrt[1 + (b*x^2)/a])/2 + (Sqrt[a]*A rcSinh[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[b])))/Sqrt[1 + (b*x^2)/a]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* FracPart[p])) Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[2*p] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)]
Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.64
method | result | size |
risch | \(\frac {x {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {1}{4}}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {1}{4}}}{2 \sqrt {b}\, \sqrt {b \,x^{2}+a}}\) | \(58\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/2*x*((b*x^2+a)^2)^(1/4)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))*((b* x^2+a)^2)^(1/4)/(b*x^2+a)^(1/2)
Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.62 \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\left [\frac {a \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {b} x - a\right ) + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} b x}{4 \, b}, -\frac {a \sqrt {-b} \arctan \left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {-b} x}{b x^{2} + a}\right ) - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} b x}{2 \, b}\right ] \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(1/4),x, algorithm="fricas")
Output:
[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)*arc tan((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]
\[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\int \sqrt [4]{a^{2} + 2 a b x^{2} + b^{2} x^{4}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(1/4),x)
Output:
Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(1/4), x)
\[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \,d x } \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(1/4),x, algorithm="maxima")
Output:
integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4), x)
\[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \,d x } \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(1/4),x, algorithm="giac")
Output:
integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4), x)
Timed out. \[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/4} \,d x \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/4),x)
Output:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/4), x)
\[ \int \sqrt [4]{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{4}} x}{2}+\frac {\left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{4}}}{b \,x^{2}+a}d x \right ) a}{2} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(1/4),x)
Output:
((a**2 + 2*a*b*x**2 + b**2*x**4)**(1/4)*x + int((a**2 + 2*a*b*x**2 + b**2* x**4)**(1/4)/(a + b*x**2),x)*a)/2