\(\int \frac {x^{17/2}}{(b x^2+c x^4)^3} \, dx\) [143]

Optimal result

Integrand size = 19, antiderivative size = 189 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}} \] Output:

-1/4*x^(3/2)/c/(c*x^2+b)^2+3/16*x^(3/2)/b/c/(c*x^2+b)-3/64*arctan(1-2^(1/2 
)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(5/4)/c^(7/4)+3/64*arctan(1+2^(1/2)*c 
^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(5/4)/c^(7/4)-3/64*arctanh(2^(1/2)*b^(1/ 
4)*c^(1/4)*x^(1/2)/(b^(1/2)+c^(1/2)*x))*2^(1/2)/b^(5/4)/c^(7/4)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.72 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 \sqrt [4]{b} c^{3/4} x^{3/2} \left (b-3 c x^2\right )}{\left (b+c x^2\right )^2}-3 \sqrt {2} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{64 b^{5/4} c^{7/4}} \] Input:

Integrate[x^(17/2)/(b*x^2 + c*x^4)^3,x]
 

Output:

((-4*b^(1/4)*c^(3/4)*x^(3/2)*(b - 3*c*x^2))/(b + c*x^2)^2 - 3*Sqrt[2]*ArcT 
an[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] - 3*Sqrt[2]*Ar 
cTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(64*b^(5/4 
)*c^(7/4))
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.43, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {9, 252, 253, 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.

\(\displaystyle \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 253

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[x^(17/2)/(b*x^2 + c*x^4)^3,x]
 

Output:

-1/4*x^(3/2)/(c*(b + c*x^2)^2) + (3*(x^(3/2)/(2*b*(b + c*x^2)) + ((-(ArcTa 
n[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4))) + ArcT 
an[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sq 
rt[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)*Sqrt[x] 
+ Sqrt[c]*x]/(2*Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[c]))/(2*b)))/(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 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 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 = 138, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\frac {3 x^{\frac {7}{2}}}{16 b}-\frac {x^{\frac {3}{2}}}{16 c}}{\left (c \,x^{2}+b \right )^{2}}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{2} b \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(138\)
default	\(\frac {\frac {3 x^{\frac {7}{2}}}{16 b}-\frac {x^{\frac {3}{2}}}{16 c}}{\left (c \,x^{2}+b \right )^{2}}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{2} b \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(138\)

Input:

int(x^(17/2)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2*(3/32/b*x^(7/2)-1/32*x^(3/2)/c)/(c*x^2+b)^2+3/128/c^2/b/(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 = 289, normalized size of antiderivative = 1.53 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (b^{4} c^{5} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (i \, b c^{3} x^{4} + 2 i \, b^{2} c^{2} x^{2} + i \, b^{3} c\right )} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (i \, b^{4} c^{5} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (-i \, b c^{3} x^{4} - 2 i \, b^{2} c^{2} x^{2} - i \, b^{3} c\right )} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (-i \, b^{4} c^{5} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (-b^{4} c^{5} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) + 4 \, {\left (3 \, c x^{3} - b x\right )} \sqrt {x}}{64 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )}} \] Input:

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

1/64*(3*(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c)*(-1/(b^5*c^7))^(1/4)*log(b^4*c 
^5*(-1/(b^5*c^7))^(3/4) + sqrt(x)) - 3*(I*b*c^3*x^4 + 2*I*b^2*c^2*x^2 + I* 
b^3*c)*(-1/(b^5*c^7))^(1/4)*log(I*b^4*c^5*(-1/(b^5*c^7))^(3/4) + sqrt(x)) 
- 3*(-I*b*c^3*x^4 - 2*I*b^2*c^2*x^2 - I*b^3*c)*(-1/(b^5*c^7))^(1/4)*log(-I 
*b^4*c^5*(-1/(b^5*c^7))^(3/4) + sqrt(x)) - 3*(b*c^3*x^4 + 2*b^2*c^2*x^2 + 
b^3*c)*(-1/(b^5*c^7))^(1/4)*log(-b^4*c^5*(-1/(b^5*c^7))^(3/4) + sqrt(x)) + 
 4*(3*c*x^3 - b*x)*sqrt(x))/(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(17/2)/(c*x**4+b*x**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.17 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, c x^{\frac {7}{2}} - b x^{\frac {3}{2}}}{16 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{128 \, b c} \] Input:

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

1/16*(3*c*x^(7/2) - b*x^(3/2))/(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c) + 3/128 
*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x 
))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arct 
an(-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(-sqr 
t(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/(b* 
c)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.12 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, c x^{\frac {7}{2}} - b x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} b c} + \frac {3 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{4}} + \frac {3 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{4}} - \frac {3 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{4}} + \frac {3 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{4}} \] Input:

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 17.52 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.45 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {3\,x^{7/2}}{16\,b}-\frac {x^{3/2}}{16\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4}-\frac {3\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,{\left (-b\right )}^{5/4}\,c^{7/4}}+\frac {3\,\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,{\left (-b\right )}^{5/4}\,c^{7/4}} \] Input:

int(x^(17/2)/(b*x^2 + c*x^4)^3,x)
 

Output:

((3*x^(7/2))/(16*b) - x^(3/2)/(16*c))/(b^2 + c^2*x^4 + 2*b*c*x^2) - (3*ata 
n((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(32*(-b)^(5/4)*c^(7/4)) + (3*atanh((c^(1/ 
4)*x^(1/2))/(-b)^(1/4)))/(32*(-b)^(5/4)*c^(7/4))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.53 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^(17/2)/(c*x^4+b*x^2)^3,x)
 

Output:

( - 6*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 - 12*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 - 6*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*s 
qrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*c**2*x**4 + 6*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 + 12*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 + 6*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 + 3*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**2 + 6*c**(1/4)*b**(3/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sq 
rt(2) + sqrt(b) + sqrt(c)*x)*b*c*x**2 + 3*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 - 3*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**2 - 6*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 - 3*c**(1/4)*b**(3/4)*sqrt(2)*lo 
g(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*c**2*x**4 - 8*s 
qrt(x)*b**2*c*x + 24*sqrt(x)*b*c**2*x**3)/(128*b**2*c**2*(b**2 + 2*b*c*x** 
2 + c**2*x**4))