\(\int \frac {x^{19/2}}{(b x^2+c x^4)^2} \, dx\) [128]

Optimal result

Integrand size = 19, antiderivative size = 190 \[ \int \frac {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {9 b \sqrt {x}}{2 c^3}+\frac {9 x^{5/2}}{10 c^2}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}-\frac {9 b^{5/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{13/4}} \] Output:

-9/2*b*x^(1/2)/c^3+9/10*x^(5/2)/c^2-1/2*x^(9/2)/c/(c*x^2+b)-9/8*b^(5/4)*ar 
ctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/c^(13/4)+9/8*b^(5/4)*arcta 
n(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/c^(13/4)+9/8*b^(5/4)*arctanh( 
2^(1/2)*b^(1/4)*c^(1/4)*x^(1/2)/(b^(1/2)+c^(1/2)*x))*2^(1/2)/c^(13/4)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.78 \[ \int \frac {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{c} \sqrt {x} \left (-45 b^2-36 b c x^2+4 c^2 x^4\right )}{b+c x^2}-45 \sqrt {2} b^{5/4} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+45 \sqrt {2} b^{5/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{40 c^{13/4}} \] Input:

Integrate[x^(19/2)/(b*x^2 + c*x^4)^2,x]
 

Output:

((4*c^(1/4)*Sqrt[x]*(-45*b^2 - 36*b*c*x^2 + 4*c^2*x^4))/(b + c*x^2) - 45*S 
qrt[2]*b^(5/4)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[ 
x])] + 45*Sqrt[2]*b^(5/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[ 
b] + Sqrt[c]*x)])/(40*c^(13/4))
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.46, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {9, 252, 262, 262, 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.

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

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {x^{11/2}}{\left (b+c x^2\right )^2}dx\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {9 \int \frac {x^{7/2}}{c x^2+b}dx}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \int \frac {x^{3/2}}{c x^2+b}dx}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {b \int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \int \frac {1}{c x^2+b}d\sqrt {x}}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {9 \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 c}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}\)

Input:

Int[x^(19/2)/(b*x^2 + c*x^4)^2,x]
 

Output:

-1/2*x^(9/2)/(c*(b + c*x^2)) + (9*((2*x^(5/2))/(5*c) - (b*((2*Sqrt[x])/c - 
 (2*b*((-(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])))/c)) 
/c))/(4*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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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 = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {2 \left (-c \,x^{2}+10 b \right ) \sqrt {x}}{5 c^{3}}+\frac {b^{2} \left (-\frac {\sqrt {x}}{2 \left (c \,x^{2}+b \right )}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}\right )}{c^{3}}\)	\(147\)
derivativedivides	\(-\frac {2 \left (-\frac {c \,x^{\frac {5}{2}}}{5}+2 b \sqrt {x}\right )}{c^{3}}+\frac {2 b^{2} \left (-\frac {\sqrt {x}}{4 \left (c \,x^{2}+b \right )}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{c^{3}}\)	\(148\)
default	\(-\frac {2 \left (-\frac {c \,x^{\frac {5}{2}}}{5}+2 b \sqrt {x}\right )}{c^{3}}+\frac {2 b^{2} \left (-\frac {\sqrt {x}}{4 \left (c \,x^{2}+b \right )}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{c^{3}}\)	\(148\)

Input:

int(x^(19/2)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

-2/5*(-c*x^2+10*b)*x^(1/2)/c^3+b^2/c^3*(-1/2*x^(1/2)/(c*x^2+b)+9/16*(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)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.26 \[ \int \frac {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {45 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} \log \left (9 \, c^{3} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} + 9 \, b \sqrt {x}\right ) - 45 \, {\left (-i \, c^{4} x^{2} - i \, b c^{3}\right )} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} \log \left (9 i \, c^{3} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} + 9 \, b \sqrt {x}\right ) - 45 \, {\left (i \, c^{4} x^{2} + i \, b c^{3}\right )} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} \log \left (-9 i \, c^{3} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} + 9 \, b \sqrt {x}\right ) - 45 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} \log \left (-9 \, c^{3} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} + 9 \, b \sqrt {x}\right ) + 4 \, {\left (4 \, c^{2} x^{4} - 36 \, b c x^{2} - 45 \, b^{2}\right )} \sqrt {x}}{40 \, {\left (c^{4} x^{2} + b c^{3}\right )}} \] Input:

integrate(x^(19/2)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
 

Output:

1/40*(45*(c^4*x^2 + b*c^3)*(-b^5/c^13)^(1/4)*log(9*c^3*(-b^5/c^13)^(1/4) + 
 9*b*sqrt(x)) - 45*(-I*c^4*x^2 - I*b*c^3)*(-b^5/c^13)^(1/4)*log(9*I*c^3*(- 
b^5/c^13)^(1/4) + 9*b*sqrt(x)) - 45*(I*c^4*x^2 + I*b*c^3)*(-b^5/c^13)^(1/4 
)*log(-9*I*c^3*(-b^5/c^13)^(1/4) + 9*b*sqrt(x)) - 45*(c^4*x^2 + b*c^3)*(-b 
^5/c^13)^(1/4)*log(-9*c^3*(-b^5/c^13)^(1/4) + 9*b*sqrt(x)) + 4*(4*c^2*x^4 
- 36*b*c*x^2 - 45*b^2)*sqrt(x))/(c^4*x^2 + b*c^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**(19/2)/(c*x**4+b*x**2)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.14 \[ \int \frac {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {b^{2} \sqrt {x}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {2 \, {\left (c x^{\frac {5}{2}} - 10 \, b \sqrt {x}\right )}}{5 \, c^{3}} + \frac {9 \, {\left (\frac {2 \, \sqrt {2} b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} b^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} b^{\frac {5}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} b^{\frac {5}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}}\right )}}{16 \, c^{3}} \] Input:

integrate(x^(19/2)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
 

Output:

-1/2*b^2*sqrt(x)/(c^4*x^2 + b*c^3) + 2/5*(c*x^(5/2) - 10*b*sqrt(x))/c^3 + 
9/16*(2*sqrt(2)*b^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sq 
rt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/sqrt(sqrt(b)*sqrt(c)) + 2*sqrt(2)*b^ 
(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sq 
rt(sqrt(b)*sqrt(c)))/sqrt(sqrt(b)*sqrt(c)) + sqrt(2)*b^(5/4)*log(sqrt(2)*b 
^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4) - sqrt(2)*b^(5/4)*lo 
g(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4))/c^3
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.14 \[ \int \frac {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {9 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} b \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{4}} + \frac {9 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} b \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{4}} + \frac {9 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} b \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{4}} - \frac {9 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} b \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{4}} - \frac {b^{2} \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} c^{3}} + \frac {2 \, {\left (c^{8} x^{\frac {5}{2}} - 10 \, b c^{7} \sqrt {x}\right )}}{5 \, c^{10}} \] Input:

integrate(x^(19/2)/(c*x^4+b*x^2)^2,x, algorithm="giac")
 

Output:

9/8*sqrt(2)*(b*c^3)^(1/4)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sq 
rt(x))/(b/c)^(1/4))/c^4 + 9/8*sqrt(2)*(b*c^3)^(1/4)*b*arctan(-1/2*sqrt(2)* 
(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^4 + 9/16*sqrt(2)*(b*c^3)^ 
(1/4)*b*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 - 9/16*sqrt(2 
)*(b*c^3)^(1/4)*b*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 - 
1/2*b^2*sqrt(x)/((c*x^2 + b)*c^3) + 2/5*(c^8*x^(5/2) - 10*b*c^7*sqrt(x))/c 
^10
 

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.48 \[ \int \frac {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {2\,x^{5/2}}{5\,c^2}-\frac {b^2\,\sqrt {x}}{2\,\left (c^4\,x^2+b\,c^3\right )}-\frac {4\,b\,\sqrt {x}}{c^3}-\frac {9\,{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{4\,c^{13/4}}+\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,9{}\mathrm {i}}{4\,c^{13/4}} \] Input:

int(x^(19/2)/(b*x^2 + c*x^4)^2,x)
 

Output:

(2*x^(5/2))/(5*c^2) - (b^2*x^(1/2))/(2*(b*c^3 + c^4*x^2)) - (4*b*x^(1/2))/ 
c^3 - (9*(-b)^(5/4)*atan((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(4*c^(13/4)) + ((- 
b)^(5/4)*atan((c^(1/4)*x^(1/2)*1i)/(-b)^(1/4))*9i)/(4*c^(13/4))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.72 \[ \int \frac {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {-90 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )-90 c^{\frac {7}{4}} b^{\frac {5}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}+90 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )+90 c^{\frac {7}{4}} b^{\frac {5}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}-45 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )-45 c^{\frac {7}{4}} b^{\frac {5}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) x^{2}+45 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )+45 c^{\frac {7}{4}} b^{\frac {5}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) x^{2}-360 \sqrt {x}\, b^{2} c -288 \sqrt {x}\, b \,c^{2} x^{2}+32 \sqrt {x}\, c^{3} x^{4}}{80 c^{4} \left (c \,x^{2}+b \right )} \] Input:

int(x^(19/2)/(c*x^4+b*x^2)^2,x)
 

Output:

( - 90*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 - 90*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*x**2 + 90*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 + 90*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*x**2 - 45*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 - 45*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*x**2 + 45*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 + 45*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*x**2 - 36 
0*sqrt(x)*b**2*c - 288*sqrt(x)*b*c**2*x**2 + 32*sqrt(x)*c**3*x**4)/(80*c** 
4*(b + c*x**2))