Integrand size = 19, antiderivative size = 243 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx=-\frac {11}{14 b^2 x^{7/2}}+\frac {11 c}{6 b^3 x^{3/2}}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}-\frac {11 c^{7/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}+\frac {11 c^{7/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}-\frac {11 c^{7/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}+\frac {11 c^{7/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}} \]
-11/14/b^2/x^(7/2)+11/6*c/b^3/x^(3/2)+1/2/b/x^(7/2)/(c*x^2+b)-11/8*c^(7/4) *arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(15/4)*2^(1/2)+11/8*c^(7/4)*a rctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(15/4)*2^(1/2)-11/16*c^(7/4)*ln (b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(15/4)*2^(1/2)+11/16 *c^(7/4)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(15/4)*2^ (1/2)
Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 b^{3/4} \left (-12 b^2+44 b c x^2+77 c^2 x^4\right )}{x^{7/2} \left (b+c x^2\right )}-231 \sqrt {2} c^{7/4} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+231 \sqrt {2} c^{7/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{168 b^{15/4}} \]
((4*b^(3/4)*(-12*b^2 + 44*b*c*x^2 + 77*c^2*x^4))/(x^(7/2)*(b + c*x^2)) - 2 31*Sqrt[2]*c^(7/4)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*S qrt[x])] + 231*Sqrt[2]*c^(7/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/( Sqrt[b] + Sqrt[c]*x)])/(168*b^(15/4))
Time = 0.45 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.15, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {9, 253, 264, 264, 266, 755, 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.
| \(\displaystyle \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {1}{x^{9/2} \left (b+c x^2\right )^2}dx\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {11 \int \frac {1}{x^{9/2} \left (c x^2+b\right )}dx}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {11 \left (-\frac {c \int \frac {1}{x^{5/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {c \int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \int \frac {1}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\) |
1/(2*b*x^(7/2)*(b + c*x^2)) + (11*(-2/(7*b*x^(7/2)) - (c*(-2/(3*b*x^(3/2)) - (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[b]) + (-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[b])))/b ))/b))/(4*b)
3.4.37.3.1 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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) 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 = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.60
| method | result | size |
| derivativedivides | \(-\frac {2}{7 b^{2} x^{\frac {7}{2}}}+\frac {4 c}{3 b^{3} x^{\frac {3}{2}}}+\frac {2 c^{2} \left (\frac {\sqrt {x}}{4 c \,x^{2}+4 b}+\frac {11 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{b^{3}}\) | \(147\) |
| default | \(-\frac {2}{7 b^{2} x^{\frac {7}{2}}}+\frac {4 c}{3 b^{3} x^{\frac {3}{2}}}+\frac {2 c^{2} \left (\frac {\sqrt {x}}{4 c \,x^{2}+4 b}+\frac {11 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{b^{3}}\) | \(147\) |
| risch | \(-\frac {2 \left (-14 c \,x^{2}+3 b \right )}{21 b^{3} x^{\frac {7}{2}}}+\frac {c^{2} \left (\frac {\sqrt {x}}{2 c \,x^{2}+2 b}+\frac {11 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}\right )}{b^{3}}\) | \(147\) |
-2/7/b^2/x^(7/2)+4/3*c/b^3/x^(3/2)+2*c^2/b^3*(1/4*x^(1/2)/(c*x^2+b)+11/32* (b/c)^(1/4)/b*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 = 263, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx=\frac {231 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )} \left (-\frac {c^{7}}{b^{15}}\right )^{\frac {1}{4}} \log \left (11 \, b^{4} \left (-\frac {c^{7}}{b^{15}}\right )^{\frac {1}{4}} + 11 \, c^{2} \sqrt {x}\right ) - 231 \, {\left (-i \, b^{3} c x^{6} - i \, b^{4} x^{4}\right )} \left (-\frac {c^{7}}{b^{15}}\right )^{\frac {1}{4}} \log \left (11 i \, b^{4} \left (-\frac {c^{7}}{b^{15}}\right )^{\frac {1}{4}} + 11 \, c^{2} \sqrt {x}\right ) - 231 \, {\left (i \, b^{3} c x^{6} + i \, b^{4} x^{4}\right )} \left (-\frac {c^{7}}{b^{15}}\right )^{\frac {1}{4}} \log \left (-11 i \, b^{4} \left (-\frac {c^{7}}{b^{15}}\right )^{\frac {1}{4}} + 11 \, c^{2} \sqrt {x}\right ) - 231 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )} \left (-\frac {c^{7}}{b^{15}}\right )^{\frac {1}{4}} \log \left (-11 \, b^{4} \left (-\frac {c^{7}}{b^{15}}\right )^{\frac {1}{4}} + 11 \, c^{2} \sqrt {x}\right ) + 4 \, {\left (77 \, c^{2} x^{4} + 44 \, b c x^{2} - 12 \, b^{2}\right )} \sqrt {x}}{168 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} \]
1/168*(231*(b^3*c*x^6 + b^4*x^4)*(-c^7/b^15)^(1/4)*log(11*b^4*(-c^7/b^15)^ (1/4) + 11*c^2*sqrt(x)) - 231*(-I*b^3*c*x^6 - I*b^4*x^4)*(-c^7/b^15)^(1/4) *log(11*I*b^4*(-c^7/b^15)^(1/4) + 11*c^2*sqrt(x)) - 231*(I*b^3*c*x^6 + I*b ^4*x^4)*(-c^7/b^15)^(1/4)*log(-11*I*b^4*(-c^7/b^15)^(1/4) + 11*c^2*sqrt(x) ) - 231*(b^3*c*x^6 + b^4*x^4)*(-c^7/b^15)^(1/4)*log(-11*b^4*(-c^7/b^15)^(1 /4) + 11*c^2*sqrt(x)) + 4*(77*c^2*x^4 + 44*b*c*x^2 - 12*b^2)*sqrt(x))/(b^3 *c*x^6 + b^4*x^4)
Timed out. \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx=\frac {77 \, c^{2} x^{4} + 44 \, b c x^{2} - 12 \, b^{2}}{42 \, {\left (b^{3} c x^{\frac {11}{2}} + b^{4} x^{\frac {7}{2}}\right )}} + \frac {11 \, {\left (\frac {2 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} c^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} c^{\frac {7}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} c^{\frac {7}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{16 \, b^{3}} \]
1/42*(77*c^2*x^4 + 44*b*c*x^2 - 12*b^2)/(b^3*c*x^(11/2) + b^4*x^(7/2)) + 1 1/16*(2*sqrt(2)*c^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(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt (2)*c^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(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*c^(7/4)*l og(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/b^(3/4) - sqrt(2 )*c^(7/4)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/b^(3 /4))/b^3
Time = 0.27 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx=\frac {11 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{4}} + \frac {11 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{4}} + \frac {11 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{4}} - \frac {11 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{4}} + \frac {c^{2} \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} b^{3}} + \frac {2 \, {\left (14 \, c x^{2} - 3 \, b\right )}}{21 \, b^{3} x^{\frac {7}{2}}} \]
11/8*sqrt(2)*(b*c^3)^(1/4)*c*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*s qrt(x))/(b/c)^(1/4))/b^4 + 11/8*sqrt(2)*(b*c^3)^(1/4)*c*arctan(-1/2*sqrt(2 )*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^4 + 11/16*sqrt(2)*(b*c^ 3)^(1/4)*c*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 - 11/16*sq rt(2)*(b*c^3)^(1/4)*c*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^ 4 + 1/2*c^2*sqrt(x)/((c*x^2 + b)*b^3) + 2/21*(14*c*x^2 - 3*b)/(b^3*x^(7/2) )
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.36 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {22\,c\,x^2}{21\,b^2}-\frac {2}{7\,b}+\frac {11\,c^2\,x^4}{6\,b^3}}{b\,x^{7/2}+c\,x^{11/2}}+\frac {11\,{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{4\,b^{15/4}}+\frac {11\,{\left (-c\right )}^{7/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{4\,b^{15/4}} \]