\(\int \frac {x^{7/2} (A+B x^2)}{b x^2+c x^4} \, dx\) [120]

Optimal result

Integrand size = 26, antiderivative size = 191 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{5/2}}{5 c}-\frac {\sqrt [4]{b} (b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} c^{9/4}} \] Output:

-2*(-A*c+B*b)*x^(1/2)/c^2+2/5*B*x^(5/2)/c-1/2*b^(1/4)*(-A*c+B*b)*arctan(1- 
2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/c^(9/4)+1/2*b^(1/4)*(-A*c+B*b)*ar 
ctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/c^(9/4)+1/2*b^(1/4)*(-A*c+ 
B*b)*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 = 0.18 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.79 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {4 \sqrt [4]{c} \sqrt {x} \left (-5 b B+5 A c+B c x^2\right )-5 \sqrt {2} \sqrt [4]{b} (b B-A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+5 \sqrt {2} \sqrt [4]{b} (b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{10 c^{9/4}} \] Input:

Integrate[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]
 

Output:

(4*c^(1/4)*Sqrt[x]*(-5*b*B + 5*A*c + B*c*x^2) - 5*Sqrt[2]*b^(1/4)*(b*B - A 
*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + 5*Sq 
rt[2]*b^(1/4)*(b*B - A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[ 
b] + Sqrt[c]*x)])/(10*c^(9/4))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.34, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {9, 363, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 363

\(\displaystyle \frac {2 B x^{5/2}}{5 c}-\frac {(b B-A c) \int \frac {x^{3/2}}{c x^2+b}dx}{c}\)

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]
 

Output:

(2*B*x^(5/2))/(5*c) - ((b*B - A*c)*((2*Sqrt[x])/c - (2*b*((-(ArcTan[1 - (S 
qrt[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
 

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 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 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

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 = 0.44 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {2 \left (B c \,x^{2}+5 A c -5 B b \right ) \sqrt {x}}{5 c^{2}}-\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2}}\)	\(138\)
derivativedivides	\(\frac {\frac {2 B c \,x^{\frac {5}{2}}}{5}+2 A c \sqrt {x}-2 B b \sqrt {x}}{c^{2}}-\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2}}\)	\(141\)
default	\(\frac {\frac {2 B c \,x^{\frac {5}{2}}}{5}+2 A c \sqrt {x}-2 B b \sqrt {x}}{c^{2}}-\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2}}\)	\(141\)

Input:

int(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)
 

Output:

2/5*(B*c*x^2+5*A*c-5*B*b)*x^(1/2)/c^2-1/4*(A*c-B*b)/c^2*(b/c)^(1/4)*2^(1/2 
)*(ln((x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2 
^(1/2)+(b/c)^(1/2)))+2*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+2*arctan(2^(1 
/2)/(b/c)^(1/4)*x^(1/2)-1))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 597, normalized size of antiderivative = 3.13 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {5 \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) + 5 i \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (i \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 5 i \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (-i \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 5 \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (-c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 4 \, {\left (B c x^{2} - 5 \, B b + 5 \, A c\right )} \sqrt {x}}{10 \, c^{2}} \] Input:

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")
 

Output:

-1/10*(5*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2* 
c^3 + A^4*b*c^4)/c^9)^(1/4)*log(c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2 
*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x)) 
+ 5*I*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 
 + A^4*b*c^4)/c^9)^(1/4)*log(I*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2* 
b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x)) - 
 5*I*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 
+ A^4*b*c^4)/c^9)^(1/4)*log(-I*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2* 
b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x)) - 
 5*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + 
A^4*b*c^4)/c^9)^(1/4)*log(-c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3* 
c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x)) - 4*( 
B*c*x^2 - 5*B*b + 5*A*c)*sqrt(x))/c^2
 

Sympy [A] (verification not implemented)

Time = 46.44 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.44 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {9}{2}}}{9}}{b} & \text {for}\: c = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{c} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x}}{c} + \frac {A \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 c} - \frac {A \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 c} - \frac {A \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{c} - \frac {2 B b \sqrt {x}}{c^{2}} - \frac {B b \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 c^{2}} + \frac {B b \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 c^{2}} + \frac {B b \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{c^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 c} & \text {otherwise} \end {cases} \] Input:

integrate(x**(7/2)*(B*x**2+A)/(c*x**4+b*x**2),x)
 

Output:

Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(5/2)/5), Eq(b, 0) & Eq(c, 0)), ((2*A 
*x**(5/2)/5 + 2*B*x**(9/2)/9)/b, Eq(c, 0)), ((2*A*sqrt(x) + 2*B*x**(5/2)/5 
)/c, Eq(b, 0)), (2*A*sqrt(x)/c + A*(-b/c)**(1/4)*log(sqrt(x) - (-b/c)**(1/ 
4))/(2*c) - A*(-b/c)**(1/4)*log(sqrt(x) + (-b/c)**(1/4))/(2*c) - A*(-b/c)* 
*(1/4)*atan(sqrt(x)/(-b/c)**(1/4))/c - 2*B*b*sqrt(x)/c**2 - B*b*(-b/c)**(1 
/4)*log(sqrt(x) - (-b/c)**(1/4))/(2*c**2) + B*b*(-b/c)**(1/4)*log(sqrt(x) 
+ (-b/c)**(1/4))/(2*c**2) + B*b*(-b/c)**(1/4)*atan(sqrt(x)/(-b/c)**(1/4))/ 
c**2 + 2*B*x**(5/2)/(5*c), True))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.23 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {{\left (\frac {2 \, \sqrt {2} {\left (B b - A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (B b - A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (B b - A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (B b - A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )} b}{4 \, c^{2}} + \frac {2 \, {\left (B c x^{\frac {5}{2}} - 5 \, {\left (B b - A c\right )} \sqrt {x}\right )}}{5 \, c^{2}} \] Input:

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")
 

Output:

1/4*(2*sqrt(2)*(B*b - A*c)*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))) + 
 2*sqrt(2)*(B*b - A*c)*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(b)*sqrt(sqrt(b)*sqrt(c))) + sq 
rt(2)*(B*b - A*c)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b 
))/(b^(3/4)*c^(1/4)) - sqrt(2)*(B*b - A*c)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sq 
rt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))*b/c^2 + 2/5*(B*c*x^(5/2) - 
 5*(B*b - A*c)*sqrt(x))/c^2
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.38 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{3}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{3}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{3}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{3}} + \frac {2 \, {\left (B c^{4} x^{\frac {5}{2}} - 5 \, B b c^{3} \sqrt {x} + 5 \, A c^{4} \sqrt {x}\right )}}{5 \, c^{5}} \] Input:

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")
 

Output:

1/2*sqrt(2)*((b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*arctan(1/2*sqrt(2)*(sq 
rt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^3 + 1/2*sqrt(2)*((b*c^3)^(1/ 
4)*B*b - (b*c^3)^(1/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*s 
qrt(x))/(b/c)^(1/4))/c^3 + 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)* 
A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^3 - 1/4*sqrt(2)*(( 
b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x 
 + sqrt(b/c))/c^3 + 2/5*(B*c^4*x^(5/2) - 5*B*b*c^3*sqrt(x) + 5*A*c^4*sqrt( 
x))/c^5
 

Mupad [B] (verification not implemented)

Time = 9.16 (sec) , antiderivative size = 789, normalized size of antiderivative = 4.13 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx =\text {Too large to display} \] Input:

int((x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4),x)
 

Output:

x^(1/2)*((2*A)/c - (2*B*b)/c^2) + (2*B*x^(5/2))/(5*c) - ((-b)^(1/4)*atan(( 
((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c) 
)/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4)))*1i 
)/(2*c^(9/4)) + ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^ 
2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b)) 
/(2*c^(9/4)))*1i)/(2*c^(9/4)))/(((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2* 
b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3 
*c)*(A*c - B*b))/(2*c^(9/4))))/(2*c^(9/4)) - ((-b)^(1/4)*(A*c - B*b)*((16* 
x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A*b^2*c 
^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4))))/(2*c^(9/4))))*(A*c - B*b)*1i)/ 
c^(9/4) - ((-b)^(1/4)*atan((((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 
+ A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)* 
(A*c - B*b)*1i)/(2*c^(9/4))))/(2*c^(9/4)) + ((-b)^(1/4)*(A*c - B*b)*((16*x 
^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A*b^2*c^ 
2 - 32*B*b^3*c)*(A*c - B*b)*1i)/(2*c^(9/4))))/(2*c^(9/4)))/(((-b)^(1/4)*(A 
*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1 
/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b)*1i)/(2*c^(9/4)))*1i)/(2*c^(9/4 
)) - ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b 
^3*c))/c + ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b)*1i)/(2*c^(9 
/4)))*1i)/(2*c^(9/4))))*(A*c - B*b))/c^(9/4)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.61 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {10 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a -10 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )-10 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a +10 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )+5 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a -5 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )-5 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a +5 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )+40 \sqrt {x}\, a \,c^{2}-40 \sqrt {x}\, b^{2} c +8 \sqrt {x}\, b \,c^{2} x^{2}}{20 c^{3}} \] Input:

int(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^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)))*a*c - 10*c**(3/4)*b**(1/4)*sqrt(2)*a 
tan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqr 
t(2)))*b**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)))*a*c + 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**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)*a*c - 5*c**(3/4)*b**(1/4)*sqr 
t(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b**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)*a*c + 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**2 + 40*sqrt(x)*a*c**2 - 40*sqrt(x 
)*b**2*c + 8*sqrt(x)*b*c**2*x**2)/(20*c**3)