\(\int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [632]

Optimal result

Integrand size = 30, antiderivative size = 302 \[ \int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {2 d \sqrt {d x} \left (a+b x^2\right )}{b \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\sqrt [4]{a} d^{3/2} \left (a+b x^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\sqrt [4]{a} d^{3/2} \left (a+b x^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\sqrt [4]{a} d^{3/2} \left (a+b x^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}}{\sqrt {d} \left (\sqrt {a}+\sqrt {b} x\right )}\right )}{\sqrt {2} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.50 \[ \int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {(d x)^{3/2} \left (a+b x^2\right ) \left (4 \sqrt [4]{b} \sqrt {x}+\sqrt {2} \sqrt [4]{a} \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )-\sqrt {2} \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )\right )}{2 b^{5/4} x^{3/2} \sqrt {\left (a+b x^2\right )^2}} \] Input:

Integrate[(d*x)^(3/2)/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]
 

Output:

((d*x)^(3/2)*(a + b*x^2)*(4*b^(1/4)*Sqrt[x] + Sqrt[2]*a^(1/4)*ArcTan[(Sqrt 
[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] - Sqrt[2]*a^(1/4)*ArcT 
anh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)]))/(2*b^(5/4)* 
x^(3/2)*Sqrt[(a + b*x^2)^2])
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {1384, 27, 262, 266, 755, 27, 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.

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

\(\Big \downarrow \) 1384

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(d*x)^(3/2)/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]
 

Output:

((a + b*x^2)*((2*d*Sqrt[d*x])/b - (2*a*d*((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])) + ArcTa 
n[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[a]) + (d*(-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]) + L 
og[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[a])))/b))/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 262

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

rule 266

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

rule 755

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

rule 1082

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

rule 1103

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

rule 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 = 0.17 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.65

method	result	size
risch	\(\frac {2 x \,d^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \sqrt {d x}\, \left (b \,x^{2}+a \right )}-\frac {d \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 b \left (b \,x^{2}+a \right )}\)	\(195\)
default	\(-\frac {\left (b \,x^{2}+a \right ) d \left (\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+2 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )-8 \sqrt {d x}\right )}{4 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b}\)	\(214\)

Input:

int((d*x)^(3/2)/((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/b*x/(d*x)^(1/2)*d^2*((b*x^2+a)^2)^(1/2)/(b*x^2+a)-1/4/b*d*(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))*((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 = 160, normalized size of antiderivative = 0.53 \[ \int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\left (-\frac {a d^{6}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\sqrt {d x} d + \left (-\frac {a d^{6}}{b^{5}}\right )^{\frac {1}{4}} b\right ) + i \, \left (-\frac {a d^{6}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\sqrt {d x} d + i \, \left (-\frac {a d^{6}}{b^{5}}\right )^{\frac {1}{4}} b\right ) - i \, \left (-\frac {a d^{6}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\sqrt {d x} d - i \, \left (-\frac {a d^{6}}{b^{5}}\right )^{\frac {1}{4}} b\right ) - \left (-\frac {a d^{6}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\sqrt {d x} d - \left (-\frac {a d^{6}}{b^{5}}\right )^{\frac {1}{4}} b\right ) - 4 \, \sqrt {d x} d}{2 \, b} \] Input:

integrate((d*x)^(3/2)/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

-1/2*((-a*d^6/b^5)^(1/4)*b*log(sqrt(d*x)*d + (-a*d^6/b^5)^(1/4)*b) + I*(-a 
*d^6/b^5)^(1/4)*b*log(sqrt(d*x)*d + I*(-a*d^6/b^5)^(1/4)*b) - I*(-a*d^6/b^ 
5)^(1/4)*b*log(sqrt(d*x)*d - I*(-a*d^6/b^5)^(1/4)*b) - (-a*d^6/b^5)^(1/4)* 
b*log(sqrt(d*x)*d - (-a*d^6/b^5)^(1/4)*b) - 4*sqrt(d*x)*d)/b
 

Sympy [F]

\[ \int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {\left (d x\right )^{\frac {3}{2}}}{\sqrt {\left (a + b x^{2}\right )^{2}}}\, dx \] Input:

integrate((d*x)**(3/2)/((b*x**2+a)**2)**(1/2),x)
 

Output:

Integral((d*x)**(3/2)/sqrt((a + b*x**2)**2), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.83 \[ \int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\frac {8 \, \sqrt {d x} d^{2}}{b} - \frac {{\left (\frac {\sqrt {2} d^{4} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} d^{4} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}} b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a}} + \frac {2 \, \sqrt {2} d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a}}\right )} a}{b}}{4 \, d} \] Input:

integrate((d*x)^(3/2)/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

1/4*(8*sqrt(d*x)*d^2/b - (sqrt(2)*d^4*log(sqrt(b)*d*x + sqrt(2)*(a*d^2)^(1 
/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(3/4)*b^(1/4)) - sqrt(2)*d^4*l 
og(sqrt(b)*d*x - sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a* 
d^2)^(3/4)*b^(1/4)) + 2*sqrt(2)*d^3*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(a)) + 2*sqrt(2)*d^3*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(a)))*a/b)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.84 \[ \int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {{\left (\frac {2 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} d^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{b^{2}} + \frac {2 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} d^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{b^{2}} + \frac {\sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} d^{2} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{b^{2}} - \frac {\sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} d^{2} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{b^{2}} - \frac {8 \, \sqrt {d x} d^{2}}{b}\right )} \mathrm {sgn}\left (b x^{2} + a\right )}{4 \, d} \] Input:

integrate((d*x)^(3/2)/((b*x^2+a)^2)^(1/2),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {{\left (d\,x\right )}^{3/2}}{\sqrt {{\left (b\,x^2+a\right )}^2}} \,d x \] Input:

int((d*x)^(3/2)/((a + b*x^2)^2)^(1/2),x)
 

Output:

int((d*x)^(3/2)/((a + b*x^2)^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.49 \[ \int \frac {(d x)^{3/2}}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\sqrt {d}\, d \left (2 b^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-2 b^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )+b^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right )-b^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right )+8 \sqrt {x}\, b \right )}{4 b^{2}} \] Input:

int((d*x)^(3/2)/((b*x^2+a)^2)^(1/2),x)
 

Output:

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