3.3.99 \(\int \frac {\sqrt {x}}{(a+b x^2)^2} \, dx\) [299]

Optimal. Leaf size=218 \[ \frac {x^{3/2}}{2 a \left (a+b x^2\right )}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}-\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}} \]

[Out]

1/2*x^(3/2)/a/(b*x^2+a)-1/8*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(5/4)/b^(3/4)*2^(1/2)+1/8*arctan(1+b^(
1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(5/4)/b^(3/4)*2^(1/2)+1/16*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2
))/a^(5/4)/b^(3/4)*2^(1/2)-1/16*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(5/4)/b^(3/4)*2^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {296, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}-\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}+\frac {x^{3/2}}{2 a \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a + b*x^2)^2,x]

[Out]

x^(3/2)/(2*a*(a + b*x^2)) - ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]/(4*Sqrt[2]*a^(5/4)*b^(3/4)) + ArcTan
[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]/(4*Sqrt[2]*a^(5/4)*b^(3/4)) + Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sq
rt[x] + Sqrt[b]*x]/(8*Sqrt[2]*a^(5/4)*b^(3/4)) - Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x]/(8
*Sqrt[2]*a^(5/4)*b^(3/4))

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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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*} \int \frac {\sqrt {x}}{\left (a+b x^2\right )^2} \, dx &=\frac {x^{3/2}}{2 a \left (a+b x^2\right )}+\frac {\int \frac {\sqrt {x}}{a+b x^2} \, dx}{4 a}\\ &=\frac {x^{3/2}}{2 a \left (a+b x^2\right )}+\frac {\text {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 a}\\ &=\frac {x^{3/2}}{2 a \left (a+b x^2\right )}-\frac {\text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a \sqrt {b}}+\frac {\text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a \sqrt {b}}\\ &=\frac {x^{3/2}}{2 a \left (a+b x^2\right )}+\frac {\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 a b}+\frac {\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 a b}+\frac {\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} a^{5/4} b^{3/4}}+\frac {\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} a^{5/4} b^{3/4}}\\ &=\frac {x^{3/2}}{2 a \left (a+b x^2\right )}+\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}-\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\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} a^{5/4} b^{3/4}}-\frac {\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} a^{5/4} b^{3/4}}\\ &=\frac {x^{3/2}}{2 a \left (a+b x^2\right )}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}-\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 128, normalized size = 0.59 \begin {gather*} \frac {\frac {4 \sqrt [4]{a} x^{3/2}}{a+b x^2}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{b^{3/4}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{b^{3/4}}}{8 a^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a + b*x^2)^2,x]

[Out]

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

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Maple [A]
time = 0.05, size = 127, normalized size = 0.58

method	result	size
derivativedivides	\(\frac {x^{\frac {3}{2}}}{2 a \left (b \,x^{2}+a \right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 a b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(127\)
default	\(\frac {x^{\frac {3}{2}}}{2 a \left (b \,x^{2}+a \right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 a b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.54, size = 194, normalized size = 0.89 \begin {gather*} \frac {x^{\frac {3}{2}}}{2 \, {\left (a b x^{2} + a^{2}\right )}} + \frac {\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}}}}{16 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(3/2)/(a*b*x^2 + a^2) + 1/16*(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)))/a

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Fricas [A]
time = 0.87, size = 182, normalized size = 0.83 \begin {gather*} -\frac {4 \, {\left (a b x^{2} + a^{2}\right )} \left (-\frac {1}{a^{5} b^{3}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {-a^{3} b \sqrt {-\frac {1}{a^{5} b^{3}}} + x} a b \left (-\frac {1}{a^{5} b^{3}}\right )^{\frac {1}{4}} - a b \sqrt {x} \left (-\frac {1}{a^{5} b^{3}}\right )^{\frac {1}{4}}\right ) - {\left (a b x^{2} + a^{2}\right )} \left (-\frac {1}{a^{5} b^{3}}\right )^{\frac {1}{4}} \log \left (a^{4} b^{2} \left (-\frac {1}{a^{5} b^{3}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) + {\left (a b x^{2} + a^{2}\right )} \left (-\frac {1}{a^{5} b^{3}}\right )^{\frac {1}{4}} \log \left (-a^{4} b^{2} \left (-\frac {1}{a^{5} b^{3}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 4 \, x^{\frac {3}{2}}}{8 \, {\left (a b x^{2} + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (197) = 394\).
time = 16.59, size = 400, normalized size = 1.83 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{5 b^{2} x^{\frac {5}{2}}} & \text {for}\: a = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a^{2}} & \text {for}\: b = 0 \\\frac {a \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{8 a^{2} b \sqrt [4]{- \frac {a}{b}} + 8 a b^{2} x^{2} \sqrt [4]{- \frac {a}{b}}} - \frac {a \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{8 a^{2} b \sqrt [4]{- \frac {a}{b}} + 8 a b^{2} x^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{8 a^{2} b \sqrt [4]{- \frac {a}{b}} + 8 a b^{2} x^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {4 b x^{\frac {3}{2}} \sqrt [4]{- \frac {a}{b}}}{8 a^{2} b \sqrt [4]{- \frac {a}{b}} + 8 a b^{2} x^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {b x^{2} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{8 a^{2} b \sqrt [4]{- \frac {a}{b}} + 8 a b^{2} x^{2} \sqrt [4]{- \frac {a}{b}}} - \frac {b x^{2} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{8 a^{2} b \sqrt [4]{- \frac {a}{b}} + 8 a b^{2} x^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {2 b x^{2} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{8 a^{2} b \sqrt [4]{- \frac {a}{b}} + 8 a b^{2} x^{2} \sqrt [4]{- \frac {a}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(5/2), Eq(a, 0) & Eq(b, 0)), (-2/(5*b**2*x**(5/2)), Eq(a, 0)), (2*x**(3/2)/(3*a**2), Eq(b, 0
)), (a*log(sqrt(x) - (-a/b)**(1/4))/(8*a**2*b*(-a/b)**(1/4) + 8*a*b**2*x**2*(-a/b)**(1/4)) - a*log(sqrt(x) + (
-a/b)**(1/4))/(8*a**2*b*(-a/b)**(1/4) + 8*a*b**2*x**2*(-a/b)**(1/4)) + 2*a*atan(sqrt(x)/(-a/b)**(1/4))/(8*a**2
*b*(-a/b)**(1/4) + 8*a*b**2*x**2*(-a/b)**(1/4)) + 4*b*x**(3/2)*(-a/b)**(1/4)/(8*a**2*b*(-a/b)**(1/4) + 8*a*b**
2*x**2*(-a/b)**(1/4)) + b*x**2*log(sqrt(x) - (-a/b)**(1/4))/(8*a**2*b*(-a/b)**(1/4) + 8*a*b**2*x**2*(-a/b)**(1
/4)) - b*x**2*log(sqrt(x) + (-a/b)**(1/4))/(8*a**2*b*(-a/b)**(1/4) + 8*a*b**2*x**2*(-a/b)**(1/4)) + 2*b*x**2*a
tan(sqrt(x)/(-a/b)**(1/4))/(8*a**2*b*(-a/b)**(1/4) + 8*a*b**2*x**2*(-a/b)**(1/4)), True))

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Giac [A]
time = 0.87, size = 199, normalized size = 0.91 \begin {gather*} \frac {x^{\frac {3}{2}}}{2 \, {\left (b x^{2} + a\right )} a} + \frac {\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^{2} b^{3}} + \frac {\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^{2} b^{3}} - \frac {\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^{2} b^{3}} + \frac {\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^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.09, size = 64, normalized size = 0.29 \begin {gather*} \frac {x^{3/2}}{2\,a\,\left (b\,x^2+a\right )}-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{4\,{\left (-a\right )}^{5/4}\,b^{3/4}}+\frac {\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{4\,{\left (-a\right )}^{5/4}\,b^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(a + b*x^2)^2,x)

[Out]

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

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