\(\int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx\) [449]

Optimal result

Integrand size = 28, antiderivative size = 204 \[ \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}-\frac {3 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}}+\frac {3 d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}}-\frac {3 d^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}}{\sqrt {d} \left (\sqrt {a}+\sqrt {b} x\right )}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}} \] Output:

-1/2*d*(d*x)^(3/2)/b/(b*x^2+a)-3/8*d^(5/2)*arctan(1-2^(1/2)*b^(1/4)*(d*x)^ 
(1/2)/a^(1/4)/d^(1/2))*2^(1/2)/a^(1/4)/b^(7/4)+3/8*d^(5/2)*arctan(1+2^(1/2 
)*b^(1/4)*(d*x)^(1/2)/a^(1/4)/d^(1/2))*2^(1/2)/a^(1/4)/b^(7/4)-3/8*d^(5/2) 
*arctanh(2^(1/2)*a^(1/4)*b^(1/4)*(d*x)^(1/2)/d^(1/2)/(a^(1/2)+b^(1/2)*x))* 
2^(1/2)/a^(1/4)/b^(7/4)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.75 \[ \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {(d x)^{5/2} \left (4 \sqrt [4]{a} b^{3/4} x^{3/2}+3 \sqrt {2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )+3 \sqrt {2} \left (a+b x^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )\right )}{8 \sqrt [4]{a} b^{7/4} x^{5/2} \left (a+b x^2\right )} \] Input:

Integrate[(d*x)^(5/2)/(a^2 + 2*a*b*x^2 + b^2*x^4),x]
 

Output:

-1/8*((d*x)^(5/2)*(4*a^(1/4)*b^(3/4)*x^(3/2) + 3*Sqrt[2]*(a + b*x^2)*ArcTa 
n[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] + 3*Sqrt[2]*(a 
+ b*x^2)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)]) 
)/(a^(1/4)*b^(7/4)*x^(5/2)*(a + b*x^2))
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.48, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {1380, 27, 252, 266, 27, 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.

\(\displaystyle \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx\)

\(\Big \downarrow \) 1380

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(d*x)^(5/2)/(a^2 + 2*a*b*x^2 + b^2*x^4),x]
 

Output:

-1/2*(d*(d*x)^(3/2))/(b*(a + b*x^2)) + (3*d^3*((-(ArcTan[1 - (Sqrt[2]*b^(1 
/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])]/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d])) + Arc 
Tan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])]/(Sqrt[2]*a^(1/4)*b^ 
(1/4)*Sqrt[d]))/(2*Sqrt[b]) - (-1/2*Log[Sqrt[a]*d + Sqrt[b]*d*x - Sqrt[2]* 
a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x]]/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]) + Log 
[Sqrt[a]*d + Sqrt[b]*d*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x]]/(2*S 
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]))/(2*Sqrt[b])))/(2*b)
 

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 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 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 1380

Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^p   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

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 = 0.43 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(2 d^{3} \left (-\frac {\left (d x \right )^{\frac {3}{2}}}{4 b \left (b \,d^{2} x^{2}+a \,d^{2}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b^{2} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )\)	\(171\)
default	\(2 d^{3} \left (-\frac {\left (d x \right )^{\frac {3}{2}}}{4 b \left (b \,d^{2} x^{2}+a \,d^{2}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b^{2} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )\)	\(171\)
pseudoelliptic	\(\frac {\left (-8 x \sqrt {d x}\, b \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}+3 d \sqrt {2}\, \left (b \,x^{2}+a \right ) \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )\right )\right ) d^{2}}{16 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (b \,x^{2}+a \right ) b^{2}}\)	\(203\)

Input:

int((d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x,method=_RETURNVERBOSE)
 

Output:

2*d^3*(-1/4/b*(d*x)^(3/2)/(b*d^2*x^2+a*d^2)+3/32/b^2/(a*d^2/b)^(1/4)*2^(1/ 
2)*(ln((d*x-(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2))/(d*x+(a*d 
^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2)))+2*arctan(2^(1/2)/(a*d^2/ 
b)^(1/4)*(d*x)^(1/2)+1)+2*arctan(2^(1/2)/(a*d^2/b)^(1/4)*(d*x)^(1/2)-1)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.27 \[ \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {4 \, \sqrt {d x} d^{2} x - 3 \, {\left (b^{2} x^{2} + a b\right )} \left (-\frac {d^{10}}{a b^{7}}\right )^{\frac {1}{4}} \log \left (27 \, \sqrt {d x} d^{7} + 27 \, \left (-\frac {d^{10}}{a b^{7}}\right )^{\frac {3}{4}} a b^{5}\right ) + 3 \, {\left (i \, b^{2} x^{2} + i \, a b\right )} \left (-\frac {d^{10}}{a b^{7}}\right )^{\frac {1}{4}} \log \left (27 \, \sqrt {d x} d^{7} + 27 i \, \left (-\frac {d^{10}}{a b^{7}}\right )^{\frac {3}{4}} a b^{5}\right ) + 3 \, {\left (-i \, b^{2} x^{2} - i \, a b\right )} \left (-\frac {d^{10}}{a b^{7}}\right )^{\frac {1}{4}} \log \left (27 \, \sqrt {d x} d^{7} - 27 i \, \left (-\frac {d^{10}}{a b^{7}}\right )^{\frac {3}{4}} a b^{5}\right ) + 3 \, {\left (b^{2} x^{2} + a b\right )} \left (-\frac {d^{10}}{a b^{7}}\right )^{\frac {1}{4}} \log \left (27 \, \sqrt {d x} d^{7} - 27 \, \left (-\frac {d^{10}}{a b^{7}}\right )^{\frac {3}{4}} a b^{5}\right )}{8 \, {\left (b^{2} x^{2} + a b\right )}} \] Input:

integrate((d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")
 

Output:

-1/8*(4*sqrt(d*x)*d^2*x - 3*(b^2*x^2 + a*b)*(-d^10/(a*b^7))^(1/4)*log(27*s 
qrt(d*x)*d^7 + 27*(-d^10/(a*b^7))^(3/4)*a*b^5) + 3*(I*b^2*x^2 + I*a*b)*(-d 
^10/(a*b^7))^(1/4)*log(27*sqrt(d*x)*d^7 + 27*I*(-d^10/(a*b^7))^(3/4)*a*b^5 
) + 3*(-I*b^2*x^2 - I*a*b)*(-d^10/(a*b^7))^(1/4)*log(27*sqrt(d*x)*d^7 - 27 
*I*(-d^10/(a*b^7))^(3/4)*a*b^5) + 3*(b^2*x^2 + a*b)*(-d^10/(a*b^7))^(1/4)* 
log(27*sqrt(d*x)*d^7 - 27*(-d^10/(a*b^7))^(3/4)*a*b^5))/(b^2*x^2 + a*b)
 

Sympy [F]

\[ \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx=\int \frac {\left (d x\right )^{\frac {5}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \] Input:

integrate((d*x)**(5/2)/(b**2*x**4+2*a*b*x**2+a**2),x)
 

Output:

Integral((d*x)**(5/2)/(a + b*x**2)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.25 \[ \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {\frac {8 \, \left (d x\right )^{\frac {3}{2}} d^{4}}{b^{2} d^{2} x^{2} + a b d^{2}} - \frac {3 \, d^{4} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{b}}{16 \, d} \] Input:

integrate((d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")
 

Output:

-1/16*(8*(d*x)^(3/2)*d^4/(b^2*d^2*x^2 + a*b*d^2) - 3*d^4*(2*sqrt(2)*arctan 
(1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) + 2*sqrt(d*x)*sqrt(b))/sqrt(sq 
rt(a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sqrt(b)) + 2*sqrt(2)*arctan(-1/ 
2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) - 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt( 
a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sqrt(b)) - sqrt(2)*log(sqrt(b)*d*x 
 + sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^( 
3/4)) + sqrt(2)*log(sqrt(b)*d*x - sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) 
+ sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)))/b)/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (136) = 272\).

Time = 0.12 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.36 \[ \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {1}{16} \, {\left (\frac {8 \, \sqrt {d x} d^{2} x}{{\left (b d^{2} x^{2} + a d^{2}\right )} b} - \frac {6 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{4} d} - \frac {6 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{4} d} + \frac {3 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{4} d} - \frac {3 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{4} d}\right )} d^{2} \] Input:

integrate((d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")
 

Output:

-1/16*(8*sqrt(d*x)*d^2*x/((b*d^2*x^2 + a*d^2)*b) - 6*sqrt(2)*(a*b^3*d^2)^( 
3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4) + 2*sqrt(d*x))/(a*d^2/b)^ 
(1/4))/(a*b^4*d) - 6*sqrt(2)*(a*b^3*d^2)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2 
)*(a*d^2/b)^(1/4) - 2*sqrt(d*x))/(a*d^2/b)^(1/4))/(a*b^4*d) + 3*sqrt(2)*(a 
*b^3*d^2)^(3/4)*log(d*x + sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b 
))/(a*b^4*d) - 3*sqrt(2)*(a*b^3*d^2)^(3/4)*log(d*x - sqrt(2)*(a*d^2/b)^(1/ 
4)*sqrt(d*x) + sqrt(a*d^2/b))/(a*b^4*d))*d^2
 

Mupad [B] (verification not implemented)

Time = 18.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.45 \[ \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {3\,d^{5/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{4\,{\left (-a\right )}^{1/4}\,b^{7/4}}-\frac {3\,d^{5/2}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{4\,{\left (-a\right )}^{1/4}\,b^{7/4}}-\frac {d^3\,{\left (d\,x\right )}^{3/2}}{2\,b\,\left (b\,d^2\,x^2+a\,d^2\right )} \] Input:

int((d*x)^(5/2)/(a^2 + b^2*x^4 + 2*a*b*x^2),x)
 

Output:

(3*d^(5/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(4*(-a)^(1/4) 
*b^(7/4)) - (3*d^(5/2)*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/ 
(4*(-a)^(1/4)*b^(7/4)) - (d^3*(d*x)^(3/2))/(2*b*(a*d^2 + b*d^2*x^2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.53 \[ \int \frac {(d x)^{5/2}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\sqrt {d}\, d^{2} \left (-6 b^{\frac {1}{4}} a^{\frac {7}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-6 b^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}+6 b^{\frac {1}{4}} a^{\frac {7}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )+6 b^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}+3 b^{\frac {1}{4}} a^{\frac {7}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right )+3 b^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right ) x^{2}-3 b^{\frac {1}{4}} a^{\frac {7}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right )-3 b^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right ) x^{2}-8 \sqrt {x}\, a b x \right )}{16 a \,b^{2} \left (b \,x^{2}+a \right )} \] Input:

int((d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x)
 

Output:

(sqrt(d)*d**2*( - 6*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt 
(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a - 6*b**(1/4)*a**(3 
/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4) 
*a**(1/4)*sqrt(2)))*b*x**2 + 6*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a* 
*(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a + 6*b** 
(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b) 
)/(b**(1/4)*a**(1/4)*sqrt(2)))*b*x**2 + 3*b**(1/4)*a**(3/4)*sqrt(2)*log( - 
 sqrt(x)*b**(1/4)*a**(1/4)*sqrt(2) + sqrt(a) + sqrt(b)*x)*a + 3*b**(1/4)*a 
**(3/4)*sqrt(2)*log( - sqrt(x)*b**(1/4)*a**(1/4)*sqrt(2) + sqrt(a) + sqrt( 
b)*x)*b*x**2 - 3*b**(1/4)*a**(3/4)*sqrt(2)*log(sqrt(x)*b**(1/4)*a**(1/4)*s 
qrt(2) + sqrt(a) + sqrt(b)*x)*a - 3*b**(1/4)*a**(3/4)*sqrt(2)*log(sqrt(x)* 
b**(1/4)*a**(1/4)*sqrt(2) + sqrt(a) + sqrt(b)*x)*b*x**2 - 8*sqrt(x)*a*b*x) 
)/(16*a*b**2*(a + b*x**2))