Integrand size = 28, antiderivative size = 98 \[ \int \frac {\left (-2 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {\left (8 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{2 x}+\frac {7}{4} \sqrt [4]{a} b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )-\frac {7}{4} \sqrt [4]{a} b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right ) \]
1/2*(a*x^2+8*b)*(a*x^4+b*x^2)^(1/4)/x+7/4*a^(1/4)*b*arctan(a^(1/4)*x/(a*x^ 4+b*x^2)^(1/4))-7/4*a^(1/4)*b*arctanh(a^(1/4)*x/(a*x^4+b*x^2)^(1/4))
Time = 0.38 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-2 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {x \left (b+a x^2\right )^{3/4} \left (2 \sqrt [4]{b+a x^2} \left (8 b+a x^2\right )+7 \sqrt [4]{a} b \sqrt {x} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )-7 \sqrt [4]{a} b \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )}{4 \left (x^2 \left (b+a x^2\right )\right )^{3/4}} \]
(x*(b + a*x^2)^(3/4)*(2*(b + a*x^2)^(1/4)*(8*b + a*x^2) + 7*a^(1/4)*b*Sqrt [x]*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] - 7*a^(1/4)*b*Sqrt[x]*ArcT anh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]))/(4*(x^2*(b + a*x^2))^(3/4))
Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.49, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1944, 1402, 248, 266, 854, 827, 216, 219}
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.
| \(\displaystyle \int \frac {\left (a x^2-2 b\right ) \sqrt [4]{a x^4+b x^2}}{x^2} \, dx\) |
| \(\Big \downarrow \) 1944 |
| \(\displaystyle \frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3}-7 a \int \sqrt [4]{a x^4+b x^2}dx\) |
| \(\Big \downarrow \) 1402 |
| \(\displaystyle \frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3}-\frac {7 a \sqrt [4]{a x^4+b x^2} \int \sqrt {x} \sqrt [4]{a x^2+b}dx}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3}-\frac {7 a \sqrt [4]{a x^4+b x^2} \left (\frac {1}{4} b \int \frac {\sqrt {x}}{\left (a x^2+b\right )^{3/4}}dx+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3}-\frac {7 a \sqrt [4]{a x^4+b x^2} \left (\frac {1}{2} b \int \frac {x}{\left (a x^2+b\right )^{3/4}}d\sqrt {x}+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle \frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3}-\frac {7 a \sqrt [4]{a x^4+b x^2} \left (\frac {1}{2} b \int \frac {x}{1-a x^2}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3}-\frac {7 a \sqrt [4]{a x^4+b x^2} \left (\frac {1}{2} b \left (\frac {\int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} x+1}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}\right )+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3}-\frac {7 a \sqrt [4]{a x^4+b x^2} \left (\frac {1}{2} b \left (\frac {\int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}\right )+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3}-\frac {7 a \sqrt [4]{a x^4+b x^2} \left (\frac {1}{2} b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}\right )+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
(4*(b*x^2 + a*x^4)^(5/4))/x^3 - (7*a*(b*x^2 + a*x^4)^(1/4)*((x^(3/2)*(b + a*x^2)^(1/4))/2 + (b*(-1/2*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]/a^( 3/4) + ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]/(2*a^(3/4))))/2))/(Sqr t[x]*(b + a*x^2)^(1/4))
3.14.61.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Int[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(b*x^2 + c*x^4)^p /(x^(2*p)*(b + c*x^2)^p) Int[x^(2*p)*(b + c*x^2)^p, x], x] /; FreeQ[{b, c , p}, x] && !IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 2.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.11
| method | result | size |
| pseudoelliptic | \(\frac {-7 b x \left (\ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )\right ) a^{\frac {1}{4}}+4 \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} \left (a \,x^{2}+8 b \right )}{8 x}\) | \(109\) |
1/8*(-7*b*x*(ln((-a^(1/4)*x-(x^2*(a*x^2+b))^(1/4))/(a^(1/4)*x-(x^2*(a*x^2+ b))^(1/4)))+2*arctan(1/a^(1/4)/x*(x^2*(a*x^2+b))^(1/4)))*a^(1/4)+4*(x^2*(a *x^2+b))^(1/4)*(a*x^2+8*b))/x
Timed out. \[ \int \frac {\left (-2 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (-2 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{2} - 2 b\right )}{x^{2}}\, dx \]
\[ \int \frac {\left (-2 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int { \frac {{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} - 2 \, b\right )}}{x^{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (78) = 156\).
Time = 0.29 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.26 \[ \int \frac {\left (-2 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {8 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} a b x^{2} - 14 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - 14 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - 7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right ) + 7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right ) + 64 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} b^{2}}{16 \, b} \]
1/16*(8*(a + b/x^2)^(1/4)*a*b*x^2 - 14*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*s qrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/(-a)^(1/4)) - 14*sqrt(2) *(-a)^(1/4)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^(1 /4))/(-a)^(1/4)) - 7*sqrt(2)*(-a)^(1/4)*b^2*log(sqrt(2)*(-a)^(1/4)*(a + b/ x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) + 7*sqrt(2)*(-a)^(1/4)*b^2*log(-s qrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) + 64*(a + b/x^2)^(1/4)*b^2)/b
Time = 6.82 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91 \[ \int \frac {\left (-2 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {2\,a\,x\,{\left (a\,x^4+b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {a\,x^2}{b}\right )}{3\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/4}}+\frac {4\,b\,{\left (a\,x^4+b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ -\frac {a\,x^2}{b}\right )}{x\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/4}} \]