Integrand size = 22, antiderivative size = 128 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\frac {1}{4} x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}+\frac {3 a x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}{8 \left (a+b x^2\right )}+\frac {3 \sqrt {a} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/2}} \]
1/4*x*(b^2*x^4+2*a*b*x^2+a^2)^(3/4)+3/8*a*x*(b^2*x^4+2*a*b*x^2+a^2)^(3/4)/ (b*x^2+a)+3/8*(b^2*x^4+2*a*b*x^2+a^2)^(3/4)*arcsinh(x*b^(1/2)/a^(1/2))*a^( 1/2)/(1+b*x^2/a)^(3/2)/b^(1/2)
Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.69 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\frac {\left (\left (a+b x^2\right )^2\right )^{3/4} \left (\sqrt {b} x \sqrt {a+b x^2} \left (5 a+2 b x^2\right )-3 a^2 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{8 \sqrt {b} \left (a+b x^2\right )^{3/2}} \]
(((a + b*x^2)^2)^(3/4)*(Sqrt[b]*x*Sqrt[a + b*x^2]*(5*a + 2*b*x^2) - 3*a^2* Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]]))/(8*Sqrt[b]*(a + b*x^2)^(3/2))
Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1385, 211, 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.
\(\displaystyle \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 1385 |
\(\displaystyle \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \int \left (\frac {b x^2}{a}+1\right )^{3/2}dx}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \left (\frac {3}{4} \int \sqrt {\frac {b x^2}{a}+1}dx+\frac {1}{4} x \left (\frac {b x^2}{a}+1\right )^{3/2}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {\frac {b x^2}{a}+1}}dx+\frac {1}{2} x \sqrt {\frac {b x^2}{a}+1}\right )+\frac {1}{4} x \left (\frac {b x^2}{a}+1\right )^{3/2}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {\frac {b x^2}{a}+1}\right )+\frac {1}{4} x \left (\frac {b x^2}{a}+1\right )^{3/2}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
((a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4)*((x*(1 + (b*x^2)/a)^(3/2))/4 + (3*((x*S qrt[1 + (b*x^2)/a])/2 + (Sqrt[a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[b]) ))/4))/(1 + (b*x^2)/a)^(3/2)
3.1.1.3.1 Defintions of rubi rules used
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 = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.60
method | result | size |
risch | \(\frac {x \left (2 b \,x^{2}+5 a \right ) \left (b \,x^{2}+a \right )}{8 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {1}{4}}}+\frac {3 a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) \sqrt {b \,x^{2}+a}}{8 \sqrt {b}\, {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {1}{4}}}\) | \(77\) |
1/8*x*(2*b*x^2+5*a)*(b*x^2+a)/((b*x^2+a)^2)^(1/4)+3/8*a^2*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)
Time = 0.25 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.38 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {b} x - a\right ) + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} {\left (2 \, b^{2} x^{3} + 5 \, a b x\right )}}{16 \, b}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {-b} x}{b x^{2} + a}\right ) - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} {\left (2 \, b^{2} x^{3} + 5 \, a b x\right )}}{8 \, b}\right ] \]
[1/16*(3*a^2*sqrt(b)*log(-2*b*x^2 - 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4)*sq rt(b)*x - a) + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4)*(2*b^2*x^3 + 5*a*b*x))/ b, -1/8*(3*a^2*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)*(2*b^2*x^3 + 5*a*b*x))/ b]
\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\int \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {3}{4}}\, dx \]
\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {3}{4}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.46 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=-\frac {1}{8} \, {\left (2 \, b x^{2} + 5 \, a\right )} \sqrt {-b x^{2} - a} x - \frac {3 \, a^{2} \log \left ({\left | -\sqrt {-b} x + \sqrt {-b x^{2} - a} \right |}\right )}{8 \, \sqrt {-b}} \]
-1/8*(2*b*x^2 + 5*a)*sqrt(-b*x^2 - a)*x - 3/8*a^2*log(abs(-sqrt(-b)*x + sq rt(-b*x^2 - a)))/sqrt(-b)
Timed out. \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4} \,d x \]