\(\int \frac {(a+b x^2)^2}{x^{7/2} (c+d x^2)} \, dx\) [744]

Optimal result

Integrand size = 24, antiderivative size = 202 \[ \int \frac {\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )} \, dx=-\frac {2 a^2}{5 c x^{5/2}}-\frac {2 a (2 b c-a d)}{c^2 \sqrt {x}}-\frac {(b c-a d)^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{9/4} d^{3/4}}+\frac {(b c-a d)^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{9/4} d^{3/4}}-\frac {(b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{\sqrt {2} c^{9/4} d^{3/4}} \] Output:

-2/5*a^2/c/x^(5/2)-2*a*(-a*d+2*b*c)/c^2/x^(1/2)-1/2*(-a*d+b*c)^2*arctan(1- 
2^(1/2)*d^(1/4)*x^(1/2)/c^(1/4))*2^(1/2)/c^(9/4)/d^(3/4)+1/2*(-a*d+b*c)^2* 
arctan(1+2^(1/2)*d^(1/4)*x^(1/2)/c^(1/4))*2^(1/2)/c^(9/4)/d^(3/4)-1/2*(-a* 
d+b*c)^2*arctanh(2^(1/2)*c^(1/4)*d^(1/4)*x^(1/2)/(c^(1/2)+d^(1/2)*x))*2^(1 
/2)/c^(9/4)/d^(3/4)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )} \, dx=\frac {-\frac {4 a \sqrt [4]{c} \left (10 b c x^2+a \left (c-5 d x^2\right )\right )}{x^{5/2}}-\frac {5 \sqrt {2} (b c-a d)^2 \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )}{d^{3/4}}-\frac {5 \sqrt {2} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{d^{3/4}}}{10 c^{9/4}} \] Input:

Integrate[(a + b*x^2)^2/(x^(7/2)*(c + d*x^2)),x]
 

Output:

((-4*a*c^(1/4)*(10*b*c*x^2 + a*(c - 5*d*x^2)))/x^(5/2) - (5*Sqrt[2]*(b*c - 
 a*d)^2*ArcTan[(Sqrt[c] - Sqrt[d]*x)/(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])])/d 
^(3/4) - (5*Sqrt[2]*(b*c - a*d)^2*ArcTanh[(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] 
)/(Sqrt[c] + Sqrt[d]*x)])/d^(3/4))/(10*c^(9/4))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.33, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {365, 27, 359, 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.

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

\(\Big \downarrow \) 365

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 359

\(\displaystyle \frac {\frac {(b c-a d)^2 \int \frac {\sqrt {x}}{d x^2+c}dx}{c}-\frac {2 a (2 b c-a d)}{c \sqrt {x}}}{c}-\frac {2 a^2}{5 c x^{5/2}}\)

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(a + b*x^2)^2/(x^(7/2)*(c + d*x^2)),x]
 

Output:

(-2*a^2)/(5*c*x^(5/2)) + ((-2*a*(2*b*c - a*d))/(c*Sqrt[x]) + (2*(b*c - a*d 
)^2*((-(ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)]/(Sqrt[2]*c^(1/4)*d^( 
1/4))) + ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)]/(Sqrt[2]*c^(1/4)*d^ 
(1/4)))/(2*Sqrt[d]) - (-1/2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] 
+ Sqrt[d]*x]/(Sqrt[2]*c^(1/4)*d^(1/4)) + Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^( 
1/4)*Sqrt[x] + Sqrt[d]*x]/(2*Sqrt[2]*c^(1/4)*d^(1/4)))/(2*Sqrt[d])))/c)/c
 

Defintions of rubi rules used

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 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 359

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

rule 365

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

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 = 0.48 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {2 \left (-5 a d \,x^{2}+10 x^{2} b c +a c \right ) a}{5 c^{2} x^{\frac {5}{2}}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2} d \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(158\)
derivativedivides	\(-\frac {2 a^{2}}{5 c \,x^{\frac {5}{2}}}+\frac {2 a \left (a d -2 b c \right )}{c^{2} \sqrt {x}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2} d \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(159\)
default	\(-\frac {2 a^{2}}{5 c \,x^{\frac {5}{2}}}+\frac {2 a \left (a d -2 b c \right )}{c^{2} \sqrt {x}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2} d \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(159\)

Input:

int((b*x^2+a)^2/x^(7/2)/(d*x^2+c),x,method=_RETURNVERBOSE)
 

Output:

-2/5*(-5*a*d*x^2+10*b*c*x^2+a*c)*a/c^2/x^(5/2)+1/4*(a^2*d^2-2*a*b*c*d+b^2* 
c^2)/c^2/d/(c/d)^(1/4)*2^(1/2)*(ln((x-(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1 
/2))/(x+(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2)))+2*arctan(2^(1/2)/(c/d)^( 
1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 1352, normalized size of antiderivative = 6.69 \[ \int \frac {\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate((b*x^2+a)^2/x^(7/2)/(d*x^2+c),x, algorithm="fricas")
 

Output:

1/10*(5*c^2*x^3*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b 
^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 
- 8*a^7*b*c*d^7 + a^8*d^8)/(c^9*d^3))^(1/4)*log(c^7*d^2*(-(b^8*c^8 - 8*a*b 
^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 
56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^9*d^ 
3))^(3/4) + (b^6*c^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c^3 
*d^3 + 15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*sqrt(x)) - 5*I*c^2*x^ 
3*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 7 
0*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^ 
7 + a^8*d^8)/(c^9*d^3))^(1/4)*log(I*c^7*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 2 
8*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c 
^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^9*d^3))^(3/4) + 
(b^6*c^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c^3*d^3 + 15*a^ 
4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*sqrt(x)) + 5*I*c^2*x^3*(-(b^8*c^8 
 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^ 
4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8) 
/(c^9*d^3))^(1/4)*log(-I*c^7*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c 
^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28 
*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^9*d^3))^(3/4) + (b^6*c^6 - 
6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c^3*d^3 + 15*a^4*b^2*c^...
 

Sympy [A] (verification not implemented)

Time = 55.22 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.84 \[ \int \frac {\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )} \, dx=a^{2} \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: c = 0 \wedge d = 0 \\- \frac {2}{9 d x^{\frac {9}{2}}} & \text {for}\: c = 0 \\- \frac {2}{5 c x^{\frac {5}{2}}} & \text {for}\: d = 0 \\- \frac {2}{5 c x^{\frac {5}{2}}} + \frac {d \log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 c^{2} \sqrt [4]{- \frac {c}{d}}} - \frac {d \log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 c^{2} \sqrt [4]{- \frac {c}{d}}} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{c^{2} \sqrt [4]{- \frac {c}{d}}} + \frac {2 d}{c^{2} \sqrt {x}} & \text {otherwise} \end {cases}\right ) + 2 a b \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: c = 0 \wedge d = 0 \\- \frac {2}{5 d x^{\frac {5}{2}}} & \text {for}\: c = 0 \\- \frac {2}{c \sqrt {x}} & \text {for}\: d = 0 \\- \frac {\log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 c \sqrt [4]{- \frac {c}{d}}} + \frac {\log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 c \sqrt [4]{- \frac {c}{d}}} - \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{c \sqrt [4]{- \frac {c}{d}}} - \frac {2}{c \sqrt {x}} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: c = 0 \wedge d = 0 \\\frac {2 x^{\frac {3}{2}}}{3 c} & \text {for}\: d = 0 \\- \frac {2}{d \sqrt {x}} & \text {for}\: c = 0 \\\frac {\log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 d \sqrt [4]{- \frac {c}{d}}} - \frac {\log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 d \sqrt [4]{- \frac {c}{d}}} + \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{d \sqrt [4]{- \frac {c}{d}}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((b*x**2+a)**2/x**(7/2)/(d*x**2+c),x)
 

Output:

a**2*Piecewise((zoo/x**(9/2), Eq(c, 0) & Eq(d, 0)), (-2/(9*d*x**(9/2)), Eq 
(c, 0)), (-2/(5*c*x**(5/2)), Eq(d, 0)), (-2/(5*c*x**(5/2)) + d*log(sqrt(x) 
 - (-c/d)**(1/4))/(2*c**2*(-c/d)**(1/4)) - d*log(sqrt(x) + (-c/d)**(1/4))/ 
(2*c**2*(-c/d)**(1/4)) + d*atan(sqrt(x)/(-c/d)**(1/4))/(c**2*(-c/d)**(1/4) 
) + 2*d/(c**2*sqrt(x)), True)) + 2*a*b*Piecewise((zoo/x**(5/2), Eq(c, 0) & 
 Eq(d, 0)), (-2/(5*d*x**(5/2)), Eq(c, 0)), (-2/(c*sqrt(x)), Eq(d, 0)), (-l 
og(sqrt(x) - (-c/d)**(1/4))/(2*c*(-c/d)**(1/4)) + log(sqrt(x) + (-c/d)**(1 
/4))/(2*c*(-c/d)**(1/4)) - atan(sqrt(x)/(-c/d)**(1/4))/(c*(-c/d)**(1/4)) - 
 2/(c*sqrt(x)), True)) + b**2*Piecewise((zoo/sqrt(x), Eq(c, 0) & Eq(d, 0)) 
, (2*x**(3/2)/(3*c), Eq(d, 0)), (-2/(d*sqrt(x)), Eq(c, 0)), (log(sqrt(x) - 
 (-c/d)**(1/4))/(2*d*(-c/d)**(1/4)) - log(sqrt(x) + (-c/d)**(1/4))/(2*d*(- 
c/d)**(1/4)) + atan(sqrt(x)/(-c/d)**(1/4))/(d*(-c/d)**(1/4)), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )} \, dx=\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{4 \, c^{2}} - \frac {2 \, {\left (a^{2} c + 5 \, {\left (2 \, a b c - a^{2} d\right )} x^{2}\right )}}{5 \, c^{2} x^{\frac {5}{2}}} \] Input:

integrate((b*x^2+a)^2/x^(7/2)/(d*x^2+c),x, algorithm="maxima")
 

Output:

1/4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) 
*c^(1/4)*d^(1/4) + 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c) 
*sqrt(d))*sqrt(d)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4 
) - 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))*sqrt( 
d)) - sqrt(2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/( 
c^(1/4)*d^(3/4)) + sqrt(2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)* 
x + sqrt(c))/(c^(1/4)*d^(3/4)))/c^2 - 2/5*(a^2*c + 5*(2*a*b*c - a^2*d)*x^2 
)/(c^2*x^(5/2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (151) = 302\).

Time = 0.14 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.75 \[ \int \frac {\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )} \, dx=-\frac {2 \, {\left (10 \, a b c x^{2} - 5 \, a^{2} d x^{2} + a^{2} c\right )}}{5 \, c^{2} x^{\frac {5}{2}}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c^{3} d^{3}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c^{3} d^{3}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c^{3} d^{3}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c^{3} d^{3}} \] Input:

integrate((b*x^2+a)^2/x^(7/2)/(d*x^2+c),x, algorithm="giac")
 

Output:

-2/5*(10*a*b*c*x^2 - 5*a^2*d*x^2 + a^2*c)/(c^2*x^(5/2)) + 1/2*sqrt(2)*((c* 
d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*arct 
an(1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1/4))/(c^3*d^3) + 
1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/ 
4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1 
/4))/(c^3*d^3) - 1/4*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b* 
c*d + (c*d^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/ 
d))/(c^3*d^3) + 1/4*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c 
*d + (c*d^3)^(3/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/ 
d))/(c^3*d^3)
 

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.06 \[ \int \frac {\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (16\,a^4\,c^7\,d^6-64\,a^3\,b\,c^8\,d^5+96\,a^2\,b^2\,c^9\,d^4-64\,a\,b^3\,c^{10}\,d^3+16\,b^4\,c^{11}\,d^2\right )}{{\left (-c\right )}^{9/4}\,d^{3/4}\,\left (16\,a^6\,c^5\,d^7-96\,a^5\,b\,c^6\,d^6+240\,a^4\,b^2\,c^7\,d^5-320\,a^3\,b^3\,c^8\,d^4+240\,a^2\,b^4\,c^9\,d^3-96\,a\,b^5\,c^{10}\,d^2+16\,b^6\,c^{11}\,d\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (-c\right )}^{9/4}\,d^{3/4}}-\frac {\frac {2\,a^2}{5\,c}-\frac {2\,a\,x^2\,\left (a\,d-2\,b\,c\right )}{c^2}}{x^{5/2}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (16\,a^4\,c^7\,d^6-64\,a^3\,b\,c^8\,d^5+96\,a^2\,b^2\,c^9\,d^4-64\,a\,b^3\,c^{10}\,d^3+16\,b^4\,c^{11}\,d^2\right )}{{\left (-c\right )}^{9/4}\,d^{3/4}\,\left (16\,a^6\,c^5\,d^7-96\,a^5\,b\,c^6\,d^6+240\,a^4\,b^2\,c^7\,d^5-320\,a^3\,b^3\,c^8\,d^4+240\,a^2\,b^4\,c^9\,d^3-96\,a\,b^5\,c^{10}\,d^2+16\,b^6\,c^{11}\,d\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (-c\right )}^{9/4}\,d^{3/4}} \] Input:

int((a + b*x^2)^2/(x^(7/2)*(c + d*x^2)),x)
 

Output:

(atan((x^(1/2)*(a*d - b*c)^2*(16*a^4*c^7*d^6 + 16*b^4*c^11*d^2 - 64*a*b^3* 
c^10*d^3 - 64*a^3*b*c^8*d^5 + 96*a^2*b^2*c^9*d^4))/((-c)^(9/4)*d^(3/4)*(16 
*b^6*c^11*d + 16*a^6*c^5*d^7 - 96*a*b^5*c^10*d^2 - 96*a^5*b*c^6*d^6 + 240* 
a^2*b^4*c^9*d^3 - 320*a^3*b^3*c^8*d^4 + 240*a^4*b^2*c^7*d^5)))*(a*d - b*c) 
^2)/((-c)^(9/4)*d^(3/4)) - ((2*a^2)/(5*c) - (2*a*x^2*(a*d - 2*b*c))/c^2)/x 
^(5/2) - (atanh((x^(1/2)*(a*d - b*c)^2*(16*a^4*c^7*d^6 + 16*b^4*c^11*d^2 - 
 64*a*b^3*c^10*d^3 - 64*a^3*b*c^8*d^5 + 96*a^2*b^2*c^9*d^4))/((-c)^(9/4)*d 
^(3/4)*(16*b^6*c^11*d + 16*a^6*c^5*d^7 - 96*a*b^5*c^10*d^2 - 96*a^5*b*c^6* 
d^6 + 240*a^2*b^4*c^9*d^3 - 320*a^3*b^3*c^8*d^4 + 240*a^4*b^2*c^7*d^5)))*( 
a*d - b*c)^2)/((-c)^(9/4)*d^(3/4))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 545, normalized size of antiderivative = 2.70 \[ \int \frac {\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )} \, dx =\text {Too large to display} \] Input:

int((b*x^2+a)^2/x^(7/2)/(d*x^2+c),x)
 

Output:

( - 10*sqrt(x)*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 
 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*d**2*x**2 + 20*sqrt( 
x)*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(x)*s 
qrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*a*b*c*d*x**2 - 10*sqrt(x)*d**(1/4)*c* 
*(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(d))/(d**(1 
/4)*c**(1/4)*sqrt(2)))*b**2*c**2*x**2 + 10*sqrt(x)*d**(1/4)*c**(3/4)*sqrt( 
2)*atan((d**(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4) 
*sqrt(2)))*a**2*d**2*x**2 - 20*sqrt(x)*d**(1/4)*c**(3/4)*sqrt(2)*atan((d** 
(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*a 
*b*c*d*x**2 + 10*sqrt(x)*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4) 
*sqrt(2) + 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*b**2*c**2*x**2 
+ 5*sqrt(x)*d**(1/4)*c**(3/4)*sqrt(2)*log( - sqrt(x)*d**(1/4)*c**(1/4)*sqr 
t(2) + sqrt(c) + sqrt(d)*x)*a**2*d**2*x**2 - 10*sqrt(x)*d**(1/4)*c**(3/4)* 
sqrt(2)*log( - sqrt(x)*d**(1/4)*c**(1/4)*sqrt(2) + sqrt(c) + sqrt(d)*x)*a* 
b*c*d*x**2 + 5*sqrt(x)*d**(1/4)*c**(3/4)*sqrt(2)*log( - sqrt(x)*d**(1/4)*c 
**(1/4)*sqrt(2) + sqrt(c) + sqrt(d)*x)*b**2*c**2*x**2 - 5*sqrt(x)*d**(1/4) 
*c**(3/4)*sqrt(2)*log(sqrt(x)*d**(1/4)*c**(1/4)*sqrt(2) + sqrt(c) + sqrt(d 
)*x)*a**2*d**2*x**2 + 10*sqrt(x)*d**(1/4)*c**(3/4)*sqrt(2)*log(sqrt(x)*d** 
(1/4)*c**(1/4)*sqrt(2) + sqrt(c) + sqrt(d)*x)*a*b*c*d*x**2 - 5*sqrt(x)*d** 
(1/4)*c**(3/4)*sqrt(2)*log(sqrt(x)*d**(1/4)*c**(1/4)*sqrt(2) + sqrt(c) ...