\(\int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [637]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 351 \[ \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {2 \left (a+b x^2\right )}{5 a d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 b \left (a+b x^2\right )}{a^2 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{5/4} \left (a+b x^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{5/4} \left (a+b x^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{5/4} \left (a+b x^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}}{\sqrt {d} \left (\sqrt {a}+\sqrt {b} x\right )}\right )}{\sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:

1/5*(-2*b*x^2-2*a)/a/d/(d*x)^(5/2)/((b*x^2+a)^2)^(1/2)+2*b*(b*x^2+a)/a^2/d 
^3/(d*x)^(1/2)/((b*x^2+a)^2)^(1/2)-1/2*b^(5/4)*(b*x^2+a)*arctan(1-2^(1/2)* 
b^(1/4)*(d*x)^(1/2)/a^(1/4)/d^(1/2))*2^(1/2)/a^(9/4)/d^(7/2)/((b*x^2+a)^2) 
^(1/2)+1/2*b^(5/4)*(b*x^2+a)*arctan(1+2^(1/2)*b^(1/4)*(d*x)^(1/2)/a^(1/4)/ 
d^(1/2))*2^(1/2)/a^(9/4)/d^(7/2)/((b*x^2+a)^2)^(1/2)-1/2*b^(5/4)*(b*x^2+a) 
*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^(9/4)/d^(7/2)/((b*x^2+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.46 \[ \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {x \left (a+b x^2\right ) \left (4 \sqrt [4]{a} \left (a-5 b x^2\right )+5 \sqrt {2} b^{5/4} x^{5/2} \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )+5 \sqrt {2} b^{5/4} x^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )\right )}{10 a^{9/4} (d x)^{7/2} \sqrt {\left (a+b x^2\right )^2}} \] Input:

Integrate[1/((d*x)^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]
 

Output:

-1/10*(x*(a + b*x^2)*(4*a^(1/4)*(a - 5*b*x^2) + 5*Sqrt[2]*b^(5/4)*x^(5/2)* 
ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] + 5*Sqrt[2 
]*b^(5/4)*x^(5/2)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqr 
t[b]*x)]))/(a^(9/4)*(d*x)^(7/2)*Sqrt[(a + b*x^2)^2])
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 348, normalized size of antiderivative = 0.99, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {1384, 27, 264, 264, 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.

\(\displaystyle \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\)

\(\Big \downarrow \) 1384

\(\displaystyle \frac {b \left (a+b x^2\right ) \int \frac {1}{b (d x)^{7/2} \left (b x^2+a\right )}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{(d x)^{7/2} \left (b x^2+a\right )}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {b \int \frac {1}{(d x)^{3/2} \left (b x^2+a\right )}dx}{a d^2}-\frac {2}{5 a d (d x)^{5/2}}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[1/((d*x)^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]
 

Output:

((a + b*x^2)*(-2/(5*a*d*(d*x)^(5/2)) - (b*(-2/(a*d*Sqrt[d*x]) - (2*b*((-(A 
rcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])]/(Sqrt[2]*a^(1/4)* 
b^(1/4)*Sqrt[d])) + 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]))/(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*Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]))/(2*Sqrt[b])))/(a*d 
)))/(a*d^2)))/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
 

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 264
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 826
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1384
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac 
Part[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 - 1/2] && NeQ[u, x^(n 
- 1)] && NeQ[u, x^(2*n - 1)] &&  !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
 

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 = 0.18 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.59

method result size
risch \(-\frac {2 \left (-5 b \,x^{2}+a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{5 a^{2} \sqrt {d x}\, x^{2} d^{3} \left (b \,x^{2}+a \right )}+\frac {b \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 a^{2} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} d^{3} \left (b \,x^{2}+a \right )}\) \(208\)
default \(-\frac {\left (b \,x^{2}+a \right ) \left (-5 b \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} \ln \left (-\frac {\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}-d x -\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )-10 b \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )-10 b \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )-40 b \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} d^{2} x^{2}+8 a \,d^{2} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}\right )}{20 d^{3} \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (d x \right )^{\frac {5}{2}}}\) \(251\)

Input:

int(1/(d*x)^(7/2)/((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {5 \, a^{2} d^{4} x^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}} \log \left (a^{7} d^{11} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {3}{4}} + \sqrt {d x} b^{4}\right ) - 5 i \, a^{2} d^{4} x^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}} \log \left (i \, a^{7} d^{11} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {3}{4}} + \sqrt {d x} b^{4}\right ) + 5 i \, a^{2} d^{4} x^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}} \log \left (-i \, a^{7} d^{11} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {3}{4}} + \sqrt {d x} b^{4}\right ) - 5 \, a^{2} d^{4} x^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}} \log \left (-a^{7} d^{11} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {3}{4}} + \sqrt {d x} b^{4}\right ) + 4 \, {\left (5 \, b x^{2} - a\right )} \sqrt {d x}}{10 \, a^{2} d^{4} x^{3}} \] Input:

integrate(1/(d*x)^(7/2)/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

1/10*(5*a^2*d^4*x^3*(-b^5/(a^9*d^14))^(1/4)*log(a^7*d^11*(-b^5/(a^9*d^14)) 
^(3/4) + sqrt(d*x)*b^4) - 5*I*a^2*d^4*x^3*(-b^5/(a^9*d^14))^(1/4)*log(I*a^ 
7*d^11*(-b^5/(a^9*d^14))^(3/4) + sqrt(d*x)*b^4) + 5*I*a^2*d^4*x^3*(-b^5/(a 
^9*d^14))^(1/4)*log(-I*a^7*d^11*(-b^5/(a^9*d^14))^(3/4) + sqrt(d*x)*b^4) - 
 5*a^2*d^4*x^3*(-b^5/(a^9*d^14))^(1/4)*log(-a^7*d^11*(-b^5/(a^9*d^14))^(3/ 
4) + sqrt(d*x)*b^4) + 4*(5*b*x^2 - a)*sqrt(d*x))/(a^2*d^4*x^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\text {Timed out} \] Input:

integrate(1/(d*x)**(7/2)/((b*x**2+a)**2)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\frac {5 \, b^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{a^{2} d^{2}} + \frac {8 \, {\left (5 \, b d^{2} x^{2} - a d^{2}\right )}}{\left (d x\right )^{\frac {5}{2}} a^{2} d^{2}}}{20 \, d} \] Input:

integrate(1/(d*x)^(7/2)/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

1/20*(5*b^2*(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)*sq 
rt(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)))/(a^2*d^ 
2) + 8*(5*b*d^2*x^2 - a*d^2)/((d*x)^(5/2)*a^2*d^2))/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {1}{20} \, {\left (\frac {10 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} b d^{5}} + \frac {10 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} b d^{5}} - \frac {5 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{3} b d^{5}} + \frac {5 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{3} b d^{5}} + \frac {8 \, {\left (5 \, b d^{2} x^{2} - a d^{2}\right )}}{\sqrt {d x} a^{2} d^{5} x^{2}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:

integrate(1/(d*x)^(7/2)/((b*x^2+a)^2)^(1/2),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

1/20*(10*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^3*b*d^5) + 10*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^3*b*d^5) - 5*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^3*b*d^5) + 5*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^3*b* 
d^5) + 8*(5*b*d^2*x^2 - a*d^2)/(sqrt(d*x)*a^2*d^5*x^2))*sgn(b*x^2 + a)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{{\left (d\,x\right )}^{7/2}\,\sqrt {{\left (b\,x^2+a\right )}^2}} \,d x \] Input:

int(1/((d*x)^(7/2)*((a + b*x^2)^2)^(1/2)),x)
 

Output:

int(1/((d*x)^(7/2)*((a + b*x^2)^2)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\sqrt {d}\, \left (-10 \sqrt {x}\, b^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}+10 \sqrt {x}\, b^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}+5 \sqrt {x}\, b^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right ) x^{2}-5 \sqrt {x}\, b^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right ) x^{2}-8 a^{2}+40 a b \,x^{2}\right )}{20 \sqrt {x}\, a^{3} d^{4} x^{2}} \] Input:

int(1/(d*x)^(7/2)/((b*x^2+a)^2)^(1/2),x)
 

Output:

(sqrt(d)*( - 10*sqrt(x)*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 + 10*sqrt 
(x)*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 + 5*sqrt(x)*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 - 5*sqrt(x)*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*a**2 + 40*a*b*x**2))/(20*sqrt(x) 
*a**3*d**4*x**2)