Integrand size = 19, antiderivative size = 230 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {5}{2 b^2 \sqrt {x}}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}+\frac {5 \sqrt [4]{c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{c} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{c} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4}}+\frac {5 \sqrt [4]{c} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4}} \]
5/8*c^(1/4)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(9/4)*2^(1/2)-5/8* c^(1/4)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(9/4)*2^(1/2)-5/16*c^( 1/4)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(9/4)*2^(1/2) +5/16*c^(1/4)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(9/4 )*2^(1/2)-5/2/b^2/x^(1/2)+1/2/b/(c*x^2+b)/x^(1/2)
Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.60 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {-\frac {4 \sqrt [4]{b} \left (4 b+5 c x^2\right )}{\sqrt {x} \left (b+c x^2\right )}+5 \sqrt {2} \sqrt [4]{c} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+5 \sqrt {2} \sqrt [4]{c} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{8 b^{9/4}} \]
((-4*b^(1/4)*(4*b + 5*c*x^2))/(Sqrt[x]*(b + c*x^2)) + 5*Sqrt[2]*c^(1/4)*Ar cTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + 5*Sqrt[2]* c^(1/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/ (8*b^(9/4))
Time = 0.43 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {9, 253, 264, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {1}{x^{3/2} \left (b+c x^2\right )^2}dx\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {5 \int \frac {1}{x^{3/2} \left (c x^2+b\right )}dx}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {5 \left (-\frac {c \int \frac {\sqrt {x}}{c x^2+b}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \int \frac {x}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \left (\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \left (\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {5 \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{4 b}+\frac {1}{2 b \sqrt {x} \left (b+c x^2\right )}\) |
1/(2*b*Sqrt[x]*(b + c*x^2)) + (5*(-2/(b*Sqrt[x]) - (2*c*((-(ArcTan[1 - (Sq rt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4))) + ArcTan[1 + (S qrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[c]) - (-1/2*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]/(Sqrt[2]* b^(1/4)*c^(1/4)) + Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c] *x]/(2*Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[c])))/b))/(4*b)
3.4.34.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[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 + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && 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[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.59
| method | result | size |
| derivativedivides | \(-\frac {2}{b^{2} \sqrt {x}}-\frac {2 c \left (\frac {x^{\frac {3}{2}}}{4 c \,x^{2}+4 b}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{2}}\) | \(136\) |
| default | \(-\frac {2}{b^{2} \sqrt {x}}-\frac {2 c \left (\frac {x^{\frac {3}{2}}}{4 c \,x^{2}+4 b}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{2}}\) | \(136\) |
| risch | \(-\frac {2}{b^{2} \sqrt {x}}-\frac {c \left (\frac {x^{\frac {3}{2}}}{2 c \,x^{2}+2 b}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{2}}\) | \(136\) |
-2/b^2/x^(1/2)-2*c/b^2*(1/4*x^(3/2)/(c*x^2+b)+5/32/c/(b/c)^(1/4)*2^(1/2)*( ln((x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*x^(1/2)*2^(1 /2)+(b/c)^(1/2)))+2*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2) /(b/c)^(1/4)*x^(1/2)-1)))
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.95 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {5 \, {\left (b^{2} c x^{3} + b^{3} x\right )} \left (-\frac {c}{b^{9}}\right )^{\frac {1}{4}} \log \left (125 \, b^{7} \left (-\frac {c}{b^{9}}\right )^{\frac {3}{4}} + 125 \, c \sqrt {x}\right ) + 5 \, {\left (-i \, b^{2} c x^{3} - i \, b^{3} x\right )} \left (-\frac {c}{b^{9}}\right )^{\frac {1}{4}} \log \left (125 i \, b^{7} \left (-\frac {c}{b^{9}}\right )^{\frac {3}{4}} + 125 \, c \sqrt {x}\right ) + 5 \, {\left (i \, b^{2} c x^{3} + i \, b^{3} x\right )} \left (-\frac {c}{b^{9}}\right )^{\frac {1}{4}} \log \left (-125 i \, b^{7} \left (-\frac {c}{b^{9}}\right )^{\frac {3}{4}} + 125 \, c \sqrt {x}\right ) - 5 \, {\left (b^{2} c x^{3} + b^{3} x\right )} \left (-\frac {c}{b^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, b^{7} \left (-\frac {c}{b^{9}}\right )^{\frac {3}{4}} + 125 \, c \sqrt {x}\right ) + 4 \, {\left (5 \, c x^{2} + 4 \, b\right )} \sqrt {x}}{8 \, {\left (b^{2} c x^{3} + b^{3} x\right )}} \]
-1/8*(5*(b^2*c*x^3 + b^3*x)*(-c/b^9)^(1/4)*log(125*b^7*(-c/b^9)^(3/4) + 12 5*c*sqrt(x)) + 5*(-I*b^2*c*x^3 - I*b^3*x)*(-c/b^9)^(1/4)*log(125*I*b^7*(-c /b^9)^(3/4) + 125*c*sqrt(x)) + 5*(I*b^2*c*x^3 + I*b^3*x)*(-c/b^9)^(1/4)*lo g(-125*I*b^7*(-c/b^9)^(3/4) + 125*c*sqrt(x)) - 5*(b^2*c*x^3 + b^3*x)*(-c/b ^9)^(1/4)*log(-125*b^7*(-c/b^9)^(3/4) + 125*c*sqrt(x)) + 4*(5*c*x^2 + 4*b) *sqrt(x))/(b^2*c*x^3 + b^3*x)
Timed out. \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.90 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {5 \, c x^{2} + 4 \, b}{2 \, {\left (b^{2} c x^{\frac {5}{2}} + b^{3} \sqrt {x}\right )}} - \frac {5 \, c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, b^{2}} \]
-1/2*(5*c*x^2 + 4*b)/(b^2*c*x^(5/2) + b^3*sqrt(x)) - 5/16*c*(2*sqrt(2)*arc tan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b) *sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(-1/2*sqrt(2) *(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqr t(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^ (1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/b^2
Time = 0.27 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.91 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {5 \, c x^{2} + 4 \, b}{2 \, {\left (c x^{\frac {5}{2}} + b \sqrt {x}\right )} b^{2}} - \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3} c^{2}} - \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3} c^{2}} + \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{3} c^{2}} - \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{3} c^{2}} \]
-1/2*(5*c*x^2 + 4*b)/((c*x^(5/2) + b*sqrt(x))*b^2) - 5/8*sqrt(2)*(b*c^3)^( 3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b^ 3*c^2) - 5/8*sqrt(2)*(b*c^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4 ) - 2*sqrt(x))/(b/c)^(1/4))/(b^3*c^2) + 5/16*sqrt(2)*(b*c^3)^(3/4)*log(sqr t(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c^2) - 5/16*sqrt(2)*(b*c^3) ^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c^2)
Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.33 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {5\,{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{4\,b^{9/4}}-\frac {5\,{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{4\,b^{9/4}}-\frac {\frac {2}{b}+\frac {5\,c\,x^2}{2\,b^2}}{b\,\sqrt {x}+c\,x^{5/2}} \]