Integrand size = 19, antiderivative size = 215 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=-\frac {2}{5 b x^{5/2}}+\frac {2 c}{b^2 \sqrt {x}}-\frac {c^{5/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{9/4}}+\frac {c^{5/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{9/4}}+\frac {c^{5/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{9/4}}-\frac {c^{5/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{9/4}} \]
-2/5/b/x^(5/2)-1/2*c^(5/4)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(9/ 4)*2^(1/2)+1/2*c^(5/4)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(9/4)*2 ^(1/2)+1/4*c^(5/4)*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)-1/4*c^(5/4)*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)+2*c/b^2/x^(1/2)
Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {-\frac {4 \sqrt [4]{b} \left (b-5 c x^2\right )}{x^{5/2}}-5 \sqrt {2} c^{5/4} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-5 \sqrt {2} c^{5/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{10 b^{9/4}} \]
((-4*b^(1/4)*(b - 5*c*x^2))/x^(5/2) - 5*Sqrt[2]*c^(5/4)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] - 5*Sqrt[2]*c^(5/4)*ArcTanh[ (Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(10*b^(9/4))
Time = 0.42 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {9, 264, 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.
| \(\displaystyle \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {1}{x^{7/2} \left (b+c x^2\right )}dx\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle -\frac {c \int \frac {1}{x^{3/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle -\frac {c \left (-\frac {c \int \frac {\sqrt {x}}{c x^2+b}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \int \frac {x}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\) |
-2/(5*b*x^(5/2)) - (c*(-2/(b*Sqrt[x]) - (2*c*((-(ArcTan[1 - (Sqrt[2]*c^(1/ 4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4))) + ArcTan[1 + (Sqrt[2]*c^(1 /4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[c]) - (-1/2*Log[S qrt[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))/b
3.4.24.3.1 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)/(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 = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.58
| method | result | size |
| risch | \(-\frac {2 \left (-5 c \,x^{2}+b \right )}{5 b^{2} x^{\frac {5}{2}}}+\frac {c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\) | \(124\) |
| derivativedivides | \(-\frac {2}{5 b \,x^{\frac {5}{2}}}+\frac {2 c}{b^{2} \sqrt {x}}+\frac {c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\) | \(125\) |
| default | \(-\frac {2}{5 b \,x^{\frac {5}{2}}}+\frac {2 c}{b^{2} \sqrt {x}}+\frac {c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\) | \(125\) |
-2/5*(-5*c*x^2+b)/b^2/x^(5/2)+1/4*c/b^2/(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 = 0.26 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {5 \, b^{2} x^{3} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (b^{7} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {3}{4}} + c^{4} \sqrt {x}\right ) - 5 i \, b^{2} x^{3} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (i \, b^{7} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {3}{4}} + c^{4} \sqrt {x}\right ) + 5 i \, b^{2} x^{3} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-i \, b^{7} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {3}{4}} + c^{4} \sqrt {x}\right ) - 5 \, b^{2} x^{3} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-b^{7} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {3}{4}} + c^{4} \sqrt {x}\right ) + 4 \, {\left (5 \, c x^{2} - b\right )} \sqrt {x}}{10 \, b^{2} x^{3}} \]
1/10*(5*b^2*x^3*(-c^5/b^9)^(1/4)*log(b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) - 5*I*b^2*x^3*(-c^5/b^9)^(1/4)*log(I*b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) + 5*I*b^2*x^3*(-c^5/b^9)^(1/4)*log(-I*b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) - 5*b^2*x^3*(-c^5/b^9)^(1/4)*log(-b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) + 4*(5 *c*x^2 - b)*sqrt(x))/(b^2*x^3)
Time = 25.62 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2}{9 c x^{\frac {9}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: c = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} + \frac {c \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{2} \sqrt [4]{- \frac {b}{c}}} - \frac {c \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{2} \sqrt [4]{- \frac {b}{c}}} + \frac {c \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{b^{2} \sqrt [4]{- \frac {b}{c}}} + \frac {2 c}{b^{2} \sqrt {x}} & \text {otherwise} \end {cases} \]
Piecewise((zoo/x**(9/2), Eq(b, 0) & Eq(c, 0)), (-2/(9*c*x**(9/2)), Eq(b, 0 )), (-2/(5*b*x**(5/2)), Eq(c, 0)), (-2/(5*b*x**(5/2)) + c*log(sqrt(x) - (- b/c)**(1/4))/(2*b**2*(-b/c)**(1/4)) - c*log(sqrt(x) + (-b/c)**(1/4))/(2*b* *2*(-b/c)**(1/4)) + c*atan(sqrt(x)/(-b/c)**(1/4))/(b**2*(-b/c)**(1/4)) + 2 *c/(b**2*sqrt(x)), True))
Time = 0.28 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {c^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{4 \, b^{2}} + \frac {2 \, {\left (5 \, c x^{2} - b\right )}}{5 \, b^{2} x^{\frac {5}{2}}} \]
1/4*c^2*(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)))/(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)))/(sqrt(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)*l og(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4 )))/b^2 + 2/5*(5*c*x^2 - b)/(b^2*x^(5/2))
Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3} c} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3} c} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{3} c} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{3} c} + \frac {2 \, {\left (5 \, c x^{2} - b\right )}}{5 \, b^{2} x^{\frac {5}{2}}} \]
1/2*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) + 1/2*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) - 1/4*sqrt(2)*(b*c^ 3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) + 1/4*sq rt(2)*(b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3 *c) + 2/5*(5*c*x^2 - b)/(b^2*x^(5/2))
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {{\left (-c\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{9/4}}-\frac {{\left (-c\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{9/4}}-\frac {\frac {2}{5\,b}-\frac {2\,c\,x^2}{b^2}}{x^{5/2}} \]