\(\int \frac {x^{15/2}}{(b x^2+c x^4)^2} \, dx\) [130]

Optimal result

Integrand size = 19, antiderivative size = 177 \[ \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {5 \sqrt {x}}{2 c^2}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}+\frac {5 \sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{9/4}} \] Output:

5/2*x^(1/2)/c^2-1/2*x^(5/2)/c/(c*x^2+b)+5/8*b^(1/4)*arctan(1-2^(1/2)*c^(1/ 
4)*x^(1/2)/b^(1/4))*2^(1/2)/c^(9/4)-5/8*b^(1/4)*arctan(1+2^(1/2)*c^(1/4)*x 
^(1/2)/b^(1/4))*2^(1/2)/c^(9/4)-5/8*b^(1/4)*arctanh(2^(1/2)*b^(1/4)*c^(1/4 
)*x^(1/2)/(b^(1/2)+c^(1/2)*x))*2^(1/2)/c^(9/4)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.78 \[ \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{c} \sqrt {x} \left (5 b+4 c x^2\right )}{b+c x^2}+5 \sqrt {2} \sqrt [4]{b} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-5 \sqrt {2} \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{8 c^{9/4}} \] Input:

Integrate[x^(15/2)/(b*x^2 + c*x^4)^2,x]
 

Output:

((4*c^(1/4)*Sqrt[x]*(5*b + 4*c*x^2))/(b + c*x^2) + 5*Sqrt[2]*b^(1/4)*ArcTa 
n[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] - 5*Sqrt[2]*b^( 
1/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(8* 
c^(9/4))
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.46, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {9, 252, 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.

\(\displaystyle \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[x^(15/2)/(b*x^2 + c*x^4)^2,x]
 

Output:

-1/2*x^(5/2)/(c*(b + c*x^2)) + (5*((2*Sqrt[x])/c - (2*b*((-(ArcTan[1 - (Sq 
rt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4))) + ArcTan[1 + (S 
qrt[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))/(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 = 0.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {2 \sqrt {x}}{c^{2}}-\frac {2 b \left (-\frac {\sqrt {x}}{4 \left (c \,x^{2}+b \right )}+\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{c^{2}}\)	\(136\)
default	\(\frac {2 \sqrt {x}}{c^{2}}-\frac {2 b \left (-\frac {\sqrt {x}}{4 \left (c \,x^{2}+b \right )}+\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{c^{2}}\)	\(136\)
risch	\(\frac {2 \sqrt {x}}{c^{2}}-\frac {b \left (-\frac {\sqrt {x}}{2 \left (c \,x^{2}+b \right )}+\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}\right )}{c^{2}}\)	\(136\)

Input:

int(x^(15/2)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

2*x^(1/2)/c^2-2*b/c^2*(-1/4*x^(1/2)/(c*x^2+b)+5/32*(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*ar 
ctan(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 = 0.17 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.18 \[ \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {5 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} \log \left (5 \, c^{2} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {x}\right ) + 5 \, {\left (i \, c^{3} x^{2} + i \, b c^{2}\right )} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} \log \left (5 i \, c^{2} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {x}\right ) + 5 \, {\left (-i \, c^{3} x^{2} - i \, b c^{2}\right )} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} \log \left (-5 i \, c^{2} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {x}\right ) - 5 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} \log \left (-5 \, c^{2} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {x}\right ) - 4 \, {\left (4 \, c x^{2} + 5 \, b\right )} \sqrt {x}}{8 \, {\left (c^{3} x^{2} + b c^{2}\right )}} \] Input:

integrate(x^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
 

Output:

-1/8*(5*(c^3*x^2 + b*c^2)*(-b/c^9)^(1/4)*log(5*c^2*(-b/c^9)^(1/4) + 5*sqrt 
(x)) + 5*(I*c^3*x^2 + I*b*c^2)*(-b/c^9)^(1/4)*log(5*I*c^2*(-b/c^9)^(1/4) + 
 5*sqrt(x)) + 5*(-I*c^3*x^2 - I*b*c^2)*(-b/c^9)^(1/4)*log(-5*I*c^2*(-b/c^9 
)^(1/4) + 5*sqrt(x)) - 5*(c^3*x^2 + b*c^2)*(-b/c^9)^(1/4)*log(-5*c^2*(-b/c 
^9)^(1/4) + 5*sqrt(x)) - 4*(4*c*x^2 + 5*b)*sqrt(x))/(c^3*x^2 + b*c^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**(15/2)/(c*x**4+b*x**2)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.16 \[ \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {b \sqrt {x}}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} - \frac {5 \, {\left (\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}}\right )}}{16 \, c^{2}} + \frac {2 \, \sqrt {x}}{c^{2}} \] Input:

integrate(x^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
 

Output:

1/2*b*sqrt(x)/(c^3*x^2 + b*c^2) - 5/16*(2*sqrt(2)*sqrt(b)*arctan(1/2*sqrt( 
2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/sq 
rt(sqrt(b)*sqrt(c)) + 2*sqrt(2)*sqrt(b)*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(2)*b^(1/4)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqr 
t(b))/c^(1/4) - sqrt(2)*b^(1/4)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqr 
t(c)*x + sqrt(b))/c^(1/4))/c^2 + 2*sqrt(x)/c^2
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.11 \[ \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{3}} - \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{3}} - \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{3}} + \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{3}} + \frac {b \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} c^{2}} + \frac {2 \, \sqrt {x}}{c^{2}} \] Input:

integrate(x^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="giac")
 

Output:

-5/8*sqrt(2)*(b*c^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqr 
t(x))/(b/c)^(1/4))/c^3 - 5/8*sqrt(2)*(b*c^3)^(1/4)*arctan(-1/2*sqrt(2)*(sq 
rt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^3 - 5/16*sqrt(2)*(b*c^3)^(1/ 
4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^3 + 5/16*sqrt(2)*(b* 
c^3)^(1/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^3 + 1/2*b*s 
qrt(x)/((c*x^2 + b)*c^2) + 2*sqrt(x)/c^2
 

Mupad [B] (verification not implemented)

Time = 18.58 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.45 \[ \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {2\,\sqrt {x}}{c^2}-\frac {5\,{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{4\,c^{9/4}}+\frac {b\,\sqrt {x}}{2\,\left (c^3\,x^2+b\,c^2\right )}+\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,5{}\mathrm {i}}{4\,c^{9/4}} \] Input:

int(x^(15/2)/(b*x^2 + c*x^4)^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.77 \[ \int \frac {x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx=\frac {10 c^{\frac {3}{4}} b^{\frac {5}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )+10 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}-10 c^{\frac {3}{4}} b^{\frac {5}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )-10 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}+5 c^{\frac {3}{4}} b^{\frac {5}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )+5 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) x^{2}-5 c^{\frac {3}{4}} b^{\frac {5}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )-5 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) x^{2}+40 \sqrt {x}\, b c +32 \sqrt {x}\, c^{2} x^{2}}{16 c^{3} \left (c \,x^{2}+b \right )} \] Input:

int(x^(15/2)/(c*x^4+b*x^2)^2,x)
 

Output:

(10*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 + 10*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)))*c*x**2 - 10*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 - 10*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*x**2 + 5*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 + 5*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*x**2 
- 5*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 - 5*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*x**2 + 40*sqrt(x)*b*c + 32*sqrt(x)*c 
**2*x**2)/(16*c**3*(b + c*x**2))