\(\int \frac {1}{\sqrt {x} (b x^2+c x^4)^3} \, dx\) [152]

Optimal result

Integrand size = 19, antiderivative size = 226 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=-\frac {285}{176 b^3 x^{11/2}}+\frac {285 c}{112 b^4 x^{7/2}}-\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}+\frac {285 c^{11/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{23/4}} \] Output:

-285/176/b^3/x^(11/2)+285/112*c/b^4/x^(7/2)-95/16*c^2/b^5/x^(3/2)+1/4/b/x^ 
(11/2)/(c*x^2+b)^2+19/16/b^2/x^(11/2)/(c*x^2+b)+285/64*c^(11/4)*arctan(1-2 
^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(23/4)-285/64*c^(11/4)*arctan(1+ 
2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(23/4)-285/64*c^(11/4)*arctanh( 
2^(1/2)*b^(1/4)*c^(1/4)*x^(1/2)/(b^(1/2)+c^(1/2)*x))*2^(1/2)/b^(23/4)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 b^{3/4} \left (224 b^4-608 b^3 c x^2+3040 b^2 c^2 x^4+11495 b c^3 x^6+7315 c^4 x^8\right )}{x^{11/2} \left (b+c x^2\right )^2}+21945 \sqrt {2} c^{11/4} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-21945 \sqrt {2} c^{11/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4928 b^{23/4}} \] Input:

Integrate[1/(Sqrt[x]*(b*x^2 + c*x^4)^3),x]
 

Output:

((-4*b^(3/4)*(224*b^4 - 608*b^3*c*x^2 + 3040*b^2*c^2*x^4 + 11495*b*c^3*x^6 
 + 7315*c^4*x^8))/(x^(11/2)*(b + c*x^2)^2) + 21945*Sqrt[2]*c^(11/4)*ArcTan 
[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] - 21945*Sqrt[2]* 
c^(11/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)]) 
/(4928*b^(23/4))
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.45, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.789, Rules used = {9, 253, 253, 264, 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.

\(\displaystyle \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {1}{x^{13/2} \left (b+c x^2\right )^3}dx\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {19 \int \frac {1}{x^{13/2} \left (c x^2+b\right )^2}dx}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {19 \left (\frac {15 \int \frac {1}{x^{13/2} \left (c x^2+b\right )}dx}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \int \frac {1}{x^{9/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \int \frac {1}{x^{5/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {c \int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \int \frac {1}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {19 \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}\)

Input:

Int[1/(Sqrt[x]*(b*x^2 + c*x^4)^3),x]
 

Output:

1/(4*b*x^(11/2)*(b + c*x^2)^2) + (19*(1/(2*b*x^(11/2)*(b + c*x^2)) + (15*( 
-2/(11*b*x^(11/2)) - (c*(-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)*S 
qrt[x] + Sqrt[c]*x]/(2*Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[b])))/b))/b))/b)) 
/(4*b)))/(8*b)
 

Defintions of rubi rules used

rule 9

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]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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])
 

rule 253

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]
 

rule 264

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]
 

rule 266

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]
 

rule 755

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]]))
 

rule 1082

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]
 

rule 1103

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]
 

rule 1476

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]
 

rule 1479

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]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(-\frac {2 c^{3} \left (\frac {\frac {31 c \,x^{\frac {5}{2}}}{32}+\frac {35 b \sqrt {x}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 b}\right )}{b^{5}}-\frac {2}{11 b^{3} x^{\frac {11}{2}}}-\frac {4 c^{2}}{b^{5} x^{\frac {3}{2}}}+\frac {6 c}{7 b^{4} x^{\frac {7}{2}}}\)	\(167\)
default	\(-\frac {2 c^{3} \left (\frac {\frac {31 c \,x^{\frac {5}{2}}}{32}+\frac {35 b \sqrt {x}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 b}\right )}{b^{5}}-\frac {2}{11 b^{3} x^{\frac {11}{2}}}-\frac {4 c^{2}}{b^{5} x^{\frac {3}{2}}}+\frac {6 c}{7 b^{4} x^{\frac {7}{2}}}\)	\(167\)
risch	\(-\frac {2 \left (154 c^{2} x^{4}-33 b c \,x^{2}+7 b^{2}\right )}{77 b^{5} x^{\frac {11}{2}}}-\frac {c^{3} \left (\frac {\frac {31 c \,x^{\frac {5}{2}}}{16}+\frac {35 b \sqrt {x}}{16}}{\left (c \,x^{2}+b \right )^{2}}+\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b}\right )}{b^{5}}\)	\(169\)

Input:

int(1/x^(1/2)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

-2/b^5*c^3*((31/32*c*x^(5/2)+35/32*b*x^(1/2))/(c*x^2+b)^2+285/256*(1/c*b)^ 
(1/4)/b*2^(1/2)*(ln((x+(1/c*b)^(1/4)*x^(1/2)*2^(1/2)+(1/c*b)^(1/2))/(x-(1/ 
c*b)^(1/4)*x^(1/2)*2^(1/2)+(1/c*b)^(1/2)))+2*arctan(2^(1/2)/(1/c*b)^(1/4)* 
x^(1/2)+1)+2*arctan(2^(1/2)/(1/c*b)^(1/4)*x^(1/2)-1)))-2/11/b^3/x^(11/2)-4 
*c^2/b^5/x^(3/2)+6/7*c/b^4/x^(7/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=-\frac {21945 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} \log \left (285 \, b^{6} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} + 285 \, c^{3} \sqrt {x}\right ) + 21945 \, {\left (i \, b^{5} c^{2} x^{10} + 2 i \, b^{6} c x^{8} + i \, b^{7} x^{6}\right )} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} \log \left (285 i \, b^{6} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} + 285 \, c^{3} \sqrt {x}\right ) + 21945 \, {\left (-i \, b^{5} c^{2} x^{10} - 2 i \, b^{6} c x^{8} - i \, b^{7} x^{6}\right )} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} \log \left (-285 i \, b^{6} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} + 285 \, c^{3} \sqrt {x}\right ) - 21945 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} \log \left (-285 \, b^{6} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} + 285 \, c^{3} \sqrt {x}\right ) + 4 \, {\left (7315 \, c^{4} x^{8} + 11495 \, b c^{3} x^{6} + 3040 \, b^{2} c^{2} x^{4} - 608 \, b^{3} c x^{2} + 224 \, b^{4}\right )} \sqrt {x}}{4928 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )}} \] Input:

integrate(1/x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

-1/4928*(21945*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-c^11/b^23)^(1/4)*l 
og(285*b^6*(-c^11/b^23)^(1/4) + 285*c^3*sqrt(x)) + 21945*(I*b^5*c^2*x^10 + 
 2*I*b^6*c*x^8 + I*b^7*x^6)*(-c^11/b^23)^(1/4)*log(285*I*b^6*(-c^11/b^23)^ 
(1/4) + 285*c^3*sqrt(x)) + 21945*(-I*b^5*c^2*x^10 - 2*I*b^6*c*x^8 - I*b^7* 
x^6)*(-c^11/b^23)^(1/4)*log(-285*I*b^6*(-c^11/b^23)^(1/4) + 285*c^3*sqrt(x 
)) - 21945*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-c^11/b^23)^(1/4)*log(- 
285*b^6*(-c^11/b^23)^(1/4) + 285*c^3*sqrt(x)) + 4*(7315*c^4*x^8 + 11495*b* 
c^3*x^6 + 3040*b^2*c^2*x^4 - 608*b^3*c*x^2 + 224*b^4)*sqrt(x))/(b^5*c^2*x^ 
10 + 2*b^6*c*x^8 + b^7*x^6)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/x**(1/2)/(c*x**4+b*x**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=-\frac {7315 \, c^{4} x^{8} + 11495 \, b c^{3} x^{6} + 3040 \, b^{2} c^{2} x^{4} - 608 \, b^{3} c x^{2} + 224 \, b^{4}}{1232 \, {\left (b^{5} c^{2} x^{\frac {19}{2}} + 2 \, b^{6} c x^{\frac {15}{2}} + b^{7} x^{\frac {11}{2}}\right )}} - \frac {285 \, {\left (\frac {2 \, \sqrt {2} c^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} c^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} c^{\frac {11}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} c^{\frac {11}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{128 \, b^{5}} \] Input:

integrate(1/x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

-1/1232*(7315*c^4*x^8 + 11495*b*c^3*x^6 + 3040*b^2*c^2*x^4 - 608*b^3*c*x^2 
 + 224*b^4)/(b^5*c^2*x^(19/2) + 2*b^6*c*x^(15/2) + b^7*x^(11/2)) - 285/128 
*(2*sqrt(2)*c^3*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sq 
rt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)* 
c^3*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^(11/4)*log( 
sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/b^(3/4) - sqrt(2)*c 
^(11/4)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/b^(3/4 
))/b^5
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=-\frac {285 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{6}} - \frac {285 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{6}} - \frac {285 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c^{2} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{6}} + \frac {285 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c^{2} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{6}} - \frac {31 \, c^{4} x^{\frac {5}{2}} + 35 \, b c^{3} \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{5}} - \frac {2 \, {\left (154 \, c^{2} x^{4} - 33 \, b c x^{2} + 7 \, b^{2}\right )}}{77 \, b^{5} x^{\frac {11}{2}}} \] Input:

integrate(1/x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

-285/64*sqrt(2)*(b*c^3)^(1/4)*c^2*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) 
+ 2*sqrt(x))/(b/c)^(1/4))/b^6 - 285/64*sqrt(2)*(b*c^3)^(1/4)*c^2*arctan(-1 
/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^6 - 285/128*sq 
rt(2)*(b*c^3)^(1/4)*c^2*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b 
^6 + 285/128*sqrt(2)*(b*c^3)^(1/4)*c^2*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + 
x + sqrt(b/c))/b^6 - 1/16*(31*c^4*x^(5/2) + 35*b*c^3*sqrt(x))/((c*x^2 + b) 
^2*b^5) - 2/77*(154*c^2*x^4 - 33*b*c*x^2 + 7*b^2)/(b^5*x^(11/2))
 

Mupad [B] (verification not implemented)

Time = 18.41 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=\frac {285\,{\left (-c\right )}^{11/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{23/4}}-\frac {\frac {2}{11\,b}-\frac {38\,c\,x^2}{77\,b^2}+\frac {190\,c^2\,x^4}{77\,b^3}+\frac {1045\,c^3\,x^6}{112\,b^4}+\frac {95\,c^4\,x^8}{16\,b^5}}{b^2\,x^{11/2}+c^2\,x^{19/2}+2\,b\,c\,x^{15/2}}+\frac {285\,{\left (-c\right )}^{11/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{23/4}} \] Input:

int(1/(x^(1/2)*(b*x^2 + c*x^4)^3),x)
 

Output:

(285*(-c)^(11/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4)))/(32*b^(23/4)) - (2/(1 
1*b) - (38*c*x^2)/(77*b^2) + (190*c^2*x^4)/(77*b^3) + (1045*c^3*x^6)/(112* 
b^4) + (95*c^4*x^8)/(16*b^5))/(b^2*x^(11/2) + c^2*x^(19/2) + 2*b*c*x^(15/2 
)) + (285*(-c)^(11/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4)))/(32*b^(23/4))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 543, normalized size of antiderivative = 2.40 \[ \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(1/x^(1/2)/(c*x^4+b*x^2)^3,x)
 

Output:

(43890*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 
 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**2*c**2*x**5 + 87780*sq 
rt(x)*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x 
)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b*c**3*x**7 + 43890*sqrt(x)*c**(3/ 
4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/( 
c**(1/4)*b**(1/4)*sqrt(2)))*c**4*x**9 - 43890*sqrt(x)*c**(3/4)*b**(1/4)*sq 
rt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1 
/4)*sqrt(2)))*b**2*c**2*x**5 - 87780*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*ata 
n((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt( 
2)))*b*c**3*x**7 - 43890*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)* 
b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*c**4*x* 
*9 + 21945*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1 
/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b**2*c**2*x**5 + 43890*sqrt(x)*c**(3/4) 
*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqr 
t(c)*x)*b*c**3*x**7 + 21945*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt( 
x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*c**4*x**9 - 21945*sqrt 
(x)*c**(3/4)*b**(1/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt 
(b) + sqrt(c)*x)*b**2*c**2*x**5 - 43890*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)* 
log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b*c**3*x**7 - 
 21945*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*...