\(\int \frac {x^{5/2}}{(a+b x^2)^2} \, dx\) [128]

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

Optimal result

Integrand size = 15, antiderivative size = 126 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {x^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 \left (\arctan \left (\frac {\sqrt {2} \sqrt [4]{\frac {a}{b}} \sqrt {x}}{\sqrt {\frac {a}{b}}-x}\right )-\log \left (\frac {\sqrt {\frac {a}{b}}+\sqrt {2} \sqrt [4]{\frac {a}{b}} \sqrt {x}+x}{\sqrt {a+b x^2}}\right )\right )}{4 \sqrt {2} \sqrt [4]{\frac {a}{b}} b^2} \]

[Out]

-1/2*x^(3/2)/b/(b*x^2+a)+3/8/b^2/(a/b)^(1/4)*2^(1/2)*(-ln((x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(b*x^2+a
)^(1/2))+arctan((a/b)^(1/4)*2^(1/2)*x^(1/2)/((a/b)^(1/2)-x)))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.73, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {294, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}}+\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}}+\frac {3 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{7/4}}-\frac {3 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{7/4}}-\frac {x^{3/2}}{2 b \left (a+b x^2\right )} \]

[In]

Int[x^(5/2)/(a + b*x^2)^2,x]

[Out]

-1/2*x^(3/2)/(b*(a + b*x^2)) - (3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(7/4)) +
 (3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(7/4)) + (3*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(7/4)) - (3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[
x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(7/4))

Rule 210

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 294

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

Rule 303

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
 && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 \int \frac {\sqrt {x}}{a+b x^2} \, dx}{4 b} \\ & = -\frac {x^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 \text {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 b} \\ & = -\frac {x^{3/2}}{2 b \left (a+b x^2\right )}-\frac {3 \text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 b^{3/2}}+\frac {3 \text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 b^{3/2}} \\ & = -\frac {x^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{8 b^2}+\frac {3 \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{8 b^2}+\frac {3 \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} \sqrt [4]{a} b^{7/4}}+\frac {3 \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} \sqrt [4]{a} b^{7/4}} \\ & = -\frac {x^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{7/4}}-\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{7/4}}+\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}}-\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}} \\ & = -\frac {x^{3/2}}{2 b \left (a+b x^2\right )}-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}}+\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{7/4}}+\frac {3 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{7/4}}-\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{7/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {-\frac {4 b^{3/4} x^{3/2}}{a+b x^2}-\frac {3 \sqrt {2} \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt [4]{a}}-\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt [4]{a}}}{8 b^{7/4}} \]

[In]

Integrate[x^(5/2)/(a + b*x^2)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98

method result size
derivativedivides \(-\frac {x^{\frac {3}{2}}}{2 b \left (x^{2} b +a \right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\) \(124\)
default \(-\frac {x^{\frac {3}{2}}}{2 b \left (x^{2} b +a \right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\) \(124\)

[In]

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

[Out]

-1/2*x^(3/2)/b/(b*x^2+a)+3/16/b^2/(a/b)^(1/4)*2^(1/2)*(ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x+(a/b)
^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2)))+2*arctan(1/(a/b)^(1/4)*2^(1/2)*x^(1/2)+1)+2*arctan(1/(a/b)^(1/4)*2^(1/2)*
x^(1/2)-1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.59 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {3 \, {\left (b^{2} x^{2} + a b\right )} \left (-\frac {1}{a b^{7}}\right )^{\frac {1}{4}} \log \left (a b^{5} \left (-\frac {1}{a b^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (i \, b^{2} x^{2} + i \, a b\right )} \left (-\frac {1}{a b^{7}}\right )^{\frac {1}{4}} \log \left (i \, a b^{5} \left (-\frac {1}{a b^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (-i \, b^{2} x^{2} - i \, a b\right )} \left (-\frac {1}{a b^{7}}\right )^{\frac {1}{4}} \log \left (-i \, a b^{5} \left (-\frac {1}{a b^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (b^{2} x^{2} + a b\right )} \left (-\frac {1}{a b^{7}}\right )^{\frac {1}{4}} \log \left (-a b^{5} \left (-\frac {1}{a b^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 4 \, x^{\frac {3}{2}}}{8 \, {\left (b^{2} x^{2} + a b\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (97) = 194\).

Time = 51.81 (sec) , antiderivative size = 393, normalized size of antiderivative = 3.12 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx=\begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a^{2}} & \text {for}\: b = 0 \\- \frac {2}{b^{2} \sqrt {x}} & \text {for}\: a = 0 \\\frac {3 a \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{8 a b^{2} \sqrt [4]{- \frac {a}{b}} + 8 b^{3} x^{2} \sqrt [4]{- \frac {a}{b}}} - \frac {3 a \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{8 a b^{2} \sqrt [4]{- \frac {a}{b}} + 8 b^{3} x^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {6 a \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{8 a b^{2} \sqrt [4]{- \frac {a}{b}} + 8 b^{3} x^{2} \sqrt [4]{- \frac {a}{b}}} - \frac {4 b x^{\frac {3}{2}} \sqrt [4]{- \frac {a}{b}}}{8 a b^{2} \sqrt [4]{- \frac {a}{b}} + 8 b^{3} x^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {3 b x^{2} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{8 a b^{2} \sqrt [4]{- \frac {a}{b}} + 8 b^{3} x^{2} \sqrt [4]{- \frac {a}{b}}} - \frac {3 b x^{2} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{8 a b^{2} \sqrt [4]{- \frac {a}{b}} + 8 b^{3} x^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {6 b x^{2} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{8 a b^{2} \sqrt [4]{- \frac {a}{b}} + 8 b^{3} x^{2} \sqrt [4]{- \frac {a}{b}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo/sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a**2), Eq(b, 0)), (-2/(b**2*sqrt(x)), Eq(a, 0)),
(3*a*log(sqrt(x) - (-a/b)**(1/4))/(8*a*b**2*(-a/b)**(1/4) + 8*b**3*x**2*(-a/b)**(1/4)) - 3*a*log(sqrt(x) + (-a
/b)**(1/4))/(8*a*b**2*(-a/b)**(1/4) + 8*b**3*x**2*(-a/b)**(1/4)) + 6*a*atan(sqrt(x)/(-a/b)**(1/4))/(8*a*b**2*(
-a/b)**(1/4) + 8*b**3*x**2*(-a/b)**(1/4)) - 4*b*x**(3/2)*(-a/b)**(1/4)/(8*a*b**2*(-a/b)**(1/4) + 8*b**3*x**2*(
-a/b)**(1/4)) + 3*b*x**2*log(sqrt(x) - (-a/b)**(1/4))/(8*a*b**2*(-a/b)**(1/4) + 8*b**3*x**2*(-a/b)**(1/4)) - 3
*b*x**2*log(sqrt(x) + (-a/b)**(1/4))/(8*a*b**2*(-a/b)**(1/4) + 8*b**3*x**2*(-a/b)**(1/4)) + 6*b*x**2*atan(sqrt
(x)/(-a/b)**(1/4))/(8*a*b**2*(-a/b)**(1/4) + 8*b**3*x**2*(-a/b)**(1/4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.55 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {x^{\frac {3}{2}}}{2 \, {\left (b^{2} x^{2} + a b\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{16 \, b} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (99) = 198\).

Time = 0.26 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.58 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {x^{\frac {3}{2}}}{2 \, {\left (b x^{2} + a\right )} b} + \frac {3 \, \sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{4}} + \frac {3 \, \sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{4}} - \frac {3 \, \sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{4}} + \frac {3 \, \sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.51 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {3\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{4\,{\left (-a\right )}^{1/4}\,b^{7/4}}-\frac {x^{3/2}}{2\,b\,\left (b\,x^2+a\right )}-\frac {3\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{4\,{\left (-a\right )}^{1/4}\,b^{7/4}} \]

[In]

int(x^(5/2)/(a + b*x^2)^2,x)

[Out]

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