\(\int \frac {A+B x^2}{x^{3/2} (a+b x^2)} \, dx\) [372]

Optimal result

Integrand size = 22, antiderivative size = 235 \[ \int \frac {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx=-\frac {2 A}{a \sqrt {x}}+\frac {(A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {(A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {(A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{5/4} b^{3/4}}+\frac {(A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{5/4} b^{3/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {464, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx=\frac {(A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {(A b-a B) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {(A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{5/4} b^{3/4}}+\frac {(A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{5/4} b^{3/4}}-\frac {2 A}{a \sqrt {x}} \]

[In]

Int[(A + B*x^2)/(x^(3/2)*(a + b*x^2)),x]

[Out]

(-2*A)/(a*Sqrt[x]) + ((A*b - a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(5/4)*b^(3/4)) - (
(A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(5/4)*b^(3/4)) - ((A*b - a*B)*Log[Sqrt[a
] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(5/4)*b^(3/4)) + ((A*b - a*B)*Log[Sqrt[a] + Sqr
t[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*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 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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.57 \[ \int \frac {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx=-\frac {2 A}{a \sqrt {x}}-\frac {(-A b+a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {(-A b+a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt {2} a^{5/4} b^{3/4}} \]

[In]

Integrate[(A + B*x^2)/(x^(3/2)*(a + b*x^2)),x]

[Out]

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

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.54

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

[In]

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

[Out]

-1/4*(A*b-B*a)/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))-2*A/a/x
^(1/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 705, normalized size of antiderivative = 3.00 \[ \int \frac {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx=-\frac {a x \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{3}}\right )^{\frac {1}{4}} \log \left (a^{4} b^{2} \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{3}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} - 3 \, A B^{2} a^{2} b + 3 \, A^{2} B a b^{2} - A^{3} b^{3}\right )} \sqrt {x}\right ) - i \, a x \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{3}}\right )^{\frac {1}{4}} \log \left (i \, a^{4} b^{2} \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{3}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} - 3 \, A B^{2} a^{2} b + 3 \, A^{2} B a b^{2} - A^{3} b^{3}\right )} \sqrt {x}\right ) + i \, a x \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{3}}\right )^{\frac {1}{4}} \log \left (-i \, a^{4} b^{2} \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{3}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} - 3 \, A B^{2} a^{2} b + 3 \, A^{2} B a b^{2} - A^{3} b^{3}\right )} \sqrt {x}\right ) - a x \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{3}}\right )^{\frac {1}{4}} \log \left (-a^{4} b^{2} \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{3}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} - 3 \, A B^{2} a^{2} b + 3 \, A^{2} B a b^{2} - A^{3} b^{3}\right )} \sqrt {x}\right ) + 4 \, A \sqrt {x}}{2 \, a x} \]

[In]

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

[Out]

-1/2*(a*x*(-(B^4*a^4 - 4*A*B^3*a^3*b + 6*A^2*B^2*a^2*b^2 - 4*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^3))^(1/4)*log(a^4*b
^2*(-(B^4*a^4 - 4*A*B^3*a^3*b + 6*A^2*B^2*a^2*b^2 - 4*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^3))^(3/4) - (B^3*a^3 - 3*A
*B^2*a^2*b + 3*A^2*B*a*b^2 - A^3*b^3)*sqrt(x)) - I*a*x*(-(B^4*a^4 - 4*A*B^3*a^3*b + 6*A^2*B^2*a^2*b^2 - 4*A^3*
B*a*b^3 + A^4*b^4)/(a^5*b^3))^(1/4)*log(I*a^4*b^2*(-(B^4*a^4 - 4*A*B^3*a^3*b + 6*A^2*B^2*a^2*b^2 - 4*A^3*B*a*b
^3 + A^4*b^4)/(a^5*b^3))^(3/4) - (B^3*a^3 - 3*A*B^2*a^2*b + 3*A^2*B*a*b^2 - A^3*b^3)*sqrt(x)) + I*a*x*(-(B^4*a
^4 - 4*A*B^3*a^3*b + 6*A^2*B^2*a^2*b^2 - 4*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^3))^(1/4)*log(-I*a^4*b^2*(-(B^4*a^4 -
4*A*B^3*a^3*b + 6*A^2*B^2*a^2*b^2 - 4*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^3))^(3/4) - (B^3*a^3 - 3*A*B^2*a^2*b + 3*
A^2*B*a*b^2 - A^3*b^3)*sqrt(x)) - a*x*(-(B^4*a^4 - 4*A*B^3*a^3*b + 6*A^2*B^2*a^2*b^2 - 4*A^3*B*a*b^3 + A^4*b^4
)/(a^5*b^3))^(1/4)*log(-a^4*b^2*(-(B^4*a^4 - 4*A*B^3*a^3*b + 6*A^2*B^2*a^2*b^2 - 4*A^3*B*a*b^3 + A^4*b^4)/(a^5
*b^3))^(3/4) - (B^3*a^3 - 3*A*B^2*a^2*b + 3*A^2*B*a*b^2 - A^3*b^3)*sqrt(x)) + 4*A*sqrt(x))/(a*x)

Sympy [A] (verification not implemented)

Time = 12.76 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx=A \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: a = 0 \\- \frac {2}{a \sqrt {x}} & \text {for}\: b = 0 \\- \frac {\log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 a \sqrt [4]{- \frac {a}{b}}} + \frac {\log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 a \sqrt [4]{- \frac {a}{b}}} - \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{a \sqrt [4]{- \frac {a}{b}}} - \frac {2}{a \sqrt {x}} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a} & \text {for}\: b = 0 \\- \frac {2}{b \sqrt {x}} & \text {for}\: a = 0 \\\frac {\log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 b \sqrt [4]{- \frac {a}{b}}} - \frac {\log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 b \sqrt [4]{- \frac {a}{b}}} + \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b \sqrt [4]{- \frac {a}{b}}} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx=\frac {{\left (B a - A b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{4 \, a} - \frac {2 \, A}{a \sqrt {x}} \]

[In]

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

[Out]

1/4*(B*a - A*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)) + 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) + sq
rt(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 - 2*A/(a*sqrt(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx=-\frac {2 \, A}{a \sqrt {x}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{2} b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{2} b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.30 \[ \int \frac {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx=\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b-B\,a\right )}{{\left (-a\right )}^{5/4}\,b^{3/4}}-\frac {2\,A}{a\,\sqrt {x}}-\frac {\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b-B\,a\right )}{{\left (-a\right )}^{5/4}\,b^{3/4}} \]

[In]

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

[Out]

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