\(\int \frac {x^{21/2}}{(b x^2+c x^4)^3} \, dx\) [141]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 186 \[ \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {x^{7/2}}{4 c \left (b+c x^2\right )^2}-\frac {7 x^{3/2}}{16 c^2 \left (b+c x^2\right )}-\frac {21 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {21 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} \sqrt [4]{b} c^{11/4}}-\frac {21 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} \sqrt [4]{b} c^{11/4}} \] Output:

-1/4*x^(7/2)/c/(c*x^2+b)^2-7/16*x^(3/2)/c^2/(c*x^2+b)-21/64*arctan(1-2^(1/ 
2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(1/4)/c^(11/4)+21/64*arctan(1+2^(1/2 
)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(1/4)/c^(11/4)-21/64*arctanh(2^(1/2)* 
b^(1/4)*c^(1/4)*x^(1/2)/(b^(1/2)+c^(1/2)*x))*2^(1/2)/b^(1/4)/c^(11/4)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.74 \[ \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 c^{3/4} x^{3/2} \left (7 b+11 c x^2\right )}{\left (b+c x^2\right )^2}-\frac {21 \sqrt {2} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt [4]{b}}-\frac {21 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt [4]{b}}}{64 c^{11/4}} \] Input:

Integrate[x^(21/2)/(b*x^2 + c*x^4)^3,x]
 

Output:

((-4*c^(3/4)*x^(3/2)*(7*b + 11*c*x^2))/(b + c*x^2)^2 - (21*Sqrt[2]*ArcTan[ 
(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/b^(1/4) - (21*Sq 
rt[2]*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/b^ 
(1/4))/(64*c^(11/4))
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.45, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {9, 252, 252, 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.

\(\displaystyle \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {x^{9/2}}{\left (b+c x^2\right )^3}dx\)

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[x^(21/2)/(b*x^2 + c*x^4)^3,x]
 

Output:

-1/4*x^(7/2)/(c*(b + c*x^2)^2) + (7*(-1/2*x^(3/2)/(c*(b + c*x^2)) + (3*((- 
(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[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)*Sq 
rt[x] + Sqrt[c]*x]/(2*Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[c])))/(2*c)))/(8*c 
)
 

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 252
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[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 826
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]]))
 

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 = 136, normalized size of antiderivative = 0.73

method result size
derivativedivides \(\frac {-\frac {11 x^{\frac {7}{2}}}{16 c}-\frac {7 b \,x^{\frac {3}{2}}}{16 c^{2}}}{\left (c \,x^{2}+b \right )^{2}}+\frac {21 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{3} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\) \(136\)
default \(\frac {-\frac {11 x^{\frac {7}{2}}}{16 c}-\frac {7 b \,x^{\frac {3}{2}}}{16 c^{2}}}{\left (c \,x^{2}+b \right )^{2}}+\frac {21 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{3} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\) \(136\)

Input:

int(x^(21/2)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2*(-11/32*x^(7/2)/c-7/32*b*x^(3/2)/c^2)/(c*x^2+b)^2+21/128/c^3/(1/c*b)^(1/ 
4)*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))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.48 \[ \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {21 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )} \left (-\frac {1}{b c^{11}}\right )^{\frac {1}{4}} \log \left (b c^{8} \left (-\frac {1}{b c^{11}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 21 \, {\left (i \, c^{4} x^{4} + 2 i \, b c^{3} x^{2} + i \, b^{2} c^{2}\right )} \left (-\frac {1}{b c^{11}}\right )^{\frac {1}{4}} \log \left (i \, b c^{8} \left (-\frac {1}{b c^{11}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 21 \, {\left (-i \, c^{4} x^{4} - 2 i \, b c^{3} x^{2} - i \, b^{2} c^{2}\right )} \left (-\frac {1}{b c^{11}}\right )^{\frac {1}{4}} \log \left (-i \, b c^{8} \left (-\frac {1}{b c^{11}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 21 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )} \left (-\frac {1}{b c^{11}}\right )^{\frac {1}{4}} \log \left (-b c^{8} \left (-\frac {1}{b c^{11}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 4 \, {\left (11 \, c x^{3} + 7 \, b x\right )} \sqrt {x}}{64 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )}} \] Input:

integrate(x^(21/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

1/64*(21*(c^4*x^4 + 2*b*c^3*x^2 + b^2*c^2)*(-1/(b*c^11))^(1/4)*log(b*c^8*( 
-1/(b*c^11))^(3/4) + sqrt(x)) - 21*(I*c^4*x^4 + 2*I*b*c^3*x^2 + I*b^2*c^2) 
*(-1/(b*c^11))^(1/4)*log(I*b*c^8*(-1/(b*c^11))^(3/4) + sqrt(x)) - 21*(-I*c 
^4*x^4 - 2*I*b*c^3*x^2 - I*b^2*c^2)*(-1/(b*c^11))^(1/4)*log(-I*b*c^8*(-1/( 
b*c^11))^(3/4) + sqrt(x)) - 21*(c^4*x^4 + 2*b*c^3*x^2 + b^2*c^2)*(-1/(b*c^ 
11))^(1/4)*log(-b*c^8*(-1/(b*c^11))^(3/4) + sqrt(x)) - 4*(11*c*x^3 + 7*b*x 
)*sqrt(x))/(c^4*x^4 + 2*b*c^3*x^2 + b^2*c^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(21/2)/(c*x**4+b*x**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.17 \[ \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {11 \, c x^{\frac {7}{2}} + 7 \, b x^{\frac {3}{2}}}{16 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )}} + \frac {21 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{128 \, c^{2}} \] Input:

integrate(x^(21/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

-1/16*(11*c*x^(7/2) + 7*b*x^(3/2))/(c^4*x^4 + 2*b*c^3*x^2 + b^2*c^2) + 21/ 
128*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqr 
t(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*a 
rctan(-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)*log(- 
sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/ 
c^2
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.12 \[ \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {11 \, c x^{\frac {7}{2}} + 7 \, b x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} c^{2}} + \frac {21 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b c^{5}} + \frac {21 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b c^{5}} - \frac {21 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b c^{5}} + \frac {21 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b c^{5}} \] Input:

integrate(x^(21/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

-1/16*(11*c*x^(7/2) + 7*b*x^(3/2))/((c*x^2 + b)^2*c^2) + 21/64*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*c^5) + 21/64*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*c^5) - 21/128*sqrt(2)*(b*c^3)^(3/4)*l 
og(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^5) + 21/128*sqrt(2)*( 
b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^5)
 

Mupad [B] (verification not implemented)

Time = 17.74 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.47 \[ \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {21\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,{\left (-b\right )}^{1/4}\,c^{11/4}}-\frac {\frac {11\,x^{7/2}}{16\,c}+\frac {7\,b\,x^{3/2}}{16\,c^2}}{b^2+2\,b\,c\,x^2+c^2\,x^4}-\frac {21\,\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,{\left (-b\right )}^{1/4}\,c^{11/4}} \] Input:

int(x^(21/2)/(b*x^2 + c*x^4)^3,x)
 

Output:

(21*atan((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(32*(-b)^(1/4)*c^(11/4)) - ((11*x^ 
(7/2))/(16*c) + (7*b*x^(3/2))/(16*c^2))/(b^2 + c^2*x^4 + 2*b*c*x^2) - (21* 
atanh((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(32*(-b)^(1/4)*c^(11/4))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.57 \[ \int \frac {x^{21/2}}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^(21/2)/(c*x^4+b*x^2)^3,x)
 

Output:

( - 42*c**(1/4)*b**(3/4)*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 - 84*c**(1/4)*b**(3/4)*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*x**2 - 42*c**(1/4)*b**(3/4)*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**2*x**4 + 42* 
c**(1/4)*b**(3/4)*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 + 84*c**(1/4)*b**(3/4)*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*x**2 + 42*c**(1/4)*b**(3/4)*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**2*x**4 + 21*c**(1/4 
)*b**(3/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sq 
rt(c)*x)*b**2 + 42*c**(1/4)*b**(3/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1 
/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b*c*x**2 + 21*c**(1/4)*b**(3/4)*sqrt(2) 
*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*c**2*x**4 
 - 21*c**(1/4)*b**(3/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sq 
rt(b) + sqrt(c)*x)*b**2 - 42*c**(1/4)*b**(3/4)*sqrt(2)*log(sqrt(x)*c**(1/4 
)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b*c*x**2 - 21*c**(1/4)*b**(3/4)* 
sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*c**2* 
x**4 - 56*sqrt(x)*b**2*c*x - 88*sqrt(x)*b*c**2*x**3)/(128*b*c**3*(b**2 + 2 
*b*c*x**2 + c**2*x**4))