3.298 \(\int \frac {x^{3/2}}{(a+b x^2)^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.15, 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 = {288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{5/4}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{5/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} b^{5/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{3/4} b^{5/4}}-\frac {\sqrt {x}}{2 b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a + b*x^2)^2,x]

[Out]

-Sqrt[x]/(2*b*(a + b*x^2)) - ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]/(4*Sqrt[2]*a^(3/4)*b^(5/4)) + ArcTa
n[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]/(4*Sqrt[2]*a^(3/4)*b^(5/4)) - Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*S
qrt[x] + Sqrt[b]*x]/(8*Sqrt[2]*a^(3/4)*b^(5/4)) + Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x]/(
8*Sqrt[2]*a^(3/4)*b^(5/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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 288

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 329

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 617

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 628

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 1162

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 1165

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 {x^{3/2}}{\left (a+b x^2\right )^2} \, dx &=-\frac {\sqrt {x}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{4 b}\\ &=-\frac {\sqrt {x}}{2 b \left (a+b x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 b}\\ &=-\frac {\sqrt {x}}{2 b \left (a+b x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 \sqrt {a} b}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 \sqrt {a} b}\\ &=-\frac {\sqrt {x}}{2 b \left (a+b x^2\right )}+\frac {\operatorname {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 \sqrt {a} b^{3/2}}+\frac {\operatorname {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 \sqrt {a} b^{3/2}}-\frac {\operatorname {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^{3/4} b^{5/4}}-\frac {\operatorname {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^{3/4} b^{5/4}}\\ &=-\frac {\sqrt {x}}{2 b \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^{3/4} b^{5/4}}+\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{5/4}}+\frac {\operatorname {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^{3/4} b^{5/4}}-\frac {\operatorname {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^{3/4} b^{5/4}}\\ &=-\frac {\sqrt {x}}{2 b \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^{3/4} b^{5/4}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} b^{5/4}}-\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{5/4}}+\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 198, normalized size = 0.91 \[ \frac {-\frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4}}+\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4}}-\frac {2 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac {2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{a^{3/4}}-\frac {8 \sqrt [4]{b} \sqrt {x}}{a+b x^2}}{16 b^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a + b*x^2)^2,x]

[Out]

((-8*b^(1/4)*Sqrt[x])/(a + b*x^2) - (2*Sqrt[2]*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) + (2*Sqr
t[2]*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) - (Sqrt[2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*S
qrt[x] + Sqrt[b]*x])/a^(3/4) + (Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4))/(
16*b^(5/4))

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fricas [A]  time = 0.95, size = 187, normalized size = 0.86 \[ \frac {4 \, {\left (b^{2} x^{2} + a b\right )} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {a^{2} b^{2} \sqrt {-\frac {1}{a^{3} b^{5}}} + x} a^{2} b^{4} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {3}{4}} - a^{2} b^{4} \sqrt {x} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {3}{4}}\right ) + {\left (b^{2} x^{2} + a b\right )} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} \log \left (a b \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - {\left (b^{2} x^{2} + a b\right )} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} \log \left (-a b \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - 4 \, \sqrt {x}}{8 \, {\left (b^{2} x^{2} + a b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.65, size = 199, normalized size = 0.91 \[ \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{2}} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{2}} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{2}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{2}} - \frac {\sqrt {x}}{2 \, {\left (b x^{2} + a\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 158, normalized size = 0.72 \[ -\frac {\sqrt {x}}{2 \left (b \,x^{2}+a \right ) b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 a b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 a b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.88, size = 195, normalized size = 0.89 \[ \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}}{16 \, b} - \frac {\sqrt {x}}{2 \, {\left (b^{2} x^{2} + a b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.65, size = 64, normalized size = 0.29 \[ -\frac {\sqrt {x}}{2\,b\,\left (b\,x^2+a\right )}-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{4\,{\left (-a\right )}^{3/4}\,b^{5/4}}-\frac {\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{4\,{\left (-a\right )}^{3/4}\,b^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a + b*x^2)^2,x)

[Out]

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

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sympy [A]  time = 78.16, size = 440, normalized size = 2.02 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {3}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{3 b^{2} x^{\frac {3}{2}}} & \text {for}\: a = 0 \\\frac {2 x^{\frac {5}{2}}}{5 a^{2}} & \text {for}\: b = 0 \\- \frac {\sqrt [4]{-1} a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b + 8 a b^{2} x^{2}} + \frac {\sqrt [4]{-1} a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b + 8 a b^{2} x^{2}} - \frac {2 \sqrt [4]{-1} a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{8 a^{2} b + 8 a b^{2} x^{2}} - \frac {\sqrt [4]{-1} \sqrt [4]{a} b x^{2} \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b + 8 a b^{2} x^{2}} + \frac {\sqrt [4]{-1} \sqrt [4]{a} b x^{2} \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b + 8 a b^{2} x^{2}} - \frac {2 \sqrt [4]{-1} \sqrt [4]{a} b x^{2} \sqrt [4]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{8 a^{2} b + 8 a b^{2} x^{2}} - \frac {4 a \sqrt {x}}{8 a^{2} b + 8 a b^{2} x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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