\(\int \frac {A+B x^2}{\sqrt {x} (a+b x^2)} \, dx\) [371]

Optimal result

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

[Out]

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

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 = {470, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {A+B x^2}{\sqrt {x} \left (a+b x^2\right )} \, dx=-\frac {(A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{3/4} b^{5/4}}+\frac {(A b-a B) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{3/4} b^{5/4}}-\frac {(A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{3/4} b^{5/4}}+\frac {(A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{3/4} b^{5/4}}+\frac {2 B \sqrt {x}}{b} \]

[In]

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

[Out]

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

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 217

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 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 470

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.57 \[ \int \frac {A+B x^2}{\sqrt {x} \left (a+b x^2\right )} \, dx=\frac {2 B \sqrt {x}}{b}+\frac {(-A b+a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt {2} a^{3/4} b^{5/4}}-\frac {(-A b+a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt {2} a^{3/4} b^{5/4}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.54

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 570, normalized size of antiderivative = 2.43 \[ \int \frac {A+B x^2}{\sqrt {x} \left (a+b x^2\right )} \, dx=\frac {b \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{5}}\right )^{\frac {1}{4}} \log \left (a b \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{5}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) + i \, b \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{5}}\right )^{\frac {1}{4}} \log \left (i \, a b \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{5}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) - i \, b \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{5}}\right )^{\frac {1}{4}} \log \left (-i \, a b \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{5}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) - b \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{5}}\right )^{\frac {1}{4}} \log \left (-a b \left (-\frac {B^{4} a^{4} - 4 \, A B^{3} a^{3} b + 6 \, A^{2} B^{2} a^{2} b^{2} - 4 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{5}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) + 4 \, B \sqrt {x}}{2 \, b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.71 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x^2}{\sqrt {x} \left (a+b x^2\right )} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} + 2 B \sqrt {x}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} + 2 B \sqrt {x}}{b} & \text {for}\: a = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{a} & \text {for}\: b = 0 \\- \frac {A \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 a} + \frac {A \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 a} + \frac {A \sqrt [4]{- \frac {a}{b}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{a} + \frac {2 B \sqrt {x}}{b} + \frac {B \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 b} - \frac {B \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 b} - \frac {B \sqrt [4]{- \frac {a}{b}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x^2}{\sqrt {x} \left (a+b x^2\right )} \, dx=\frac {2 \, B \sqrt {x}}{b} - \frac {\frac {2 \, \sqrt {2} {\left (B a - A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (B a - A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (B a - A b\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (B a - A b\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}}{4 \, b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 739, normalized size of antiderivative = 3.14 \[ \int \frac {A+B x^2}{\sqrt {x} \left (a+b x^2\right )} \, dx=\frac {2\,B\,\sqrt {x}}{b}-\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,b-B\,a\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^3-32\,A\,B\,a\,b^2+16\,B^2\,a^2\,b\right )-\frac {\left (32\,B\,a^2\,b^2-32\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}\right )\,1{}\mathrm {i}}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}+\frac {\left (A\,b-B\,a\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^3-32\,A\,B\,a\,b^2+16\,B^2\,a^2\,b\right )+\frac {\left (32\,B\,a^2\,b^2-32\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}\right )\,1{}\mathrm {i}}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}}{\frac {\left (A\,b-B\,a\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^3-32\,A\,B\,a\,b^2+16\,B^2\,a^2\,b\right )-\frac {\left (32\,B\,a^2\,b^2-32\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}\right )}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}-\frac {\left (A\,b-B\,a\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^3-32\,A\,B\,a\,b^2+16\,B^2\,a^2\,b\right )+\frac {\left (32\,B\,a^2\,b^2-32\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}\right )}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}}\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{{\left (-a\right )}^{3/4}\,b^{5/4}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,b-B\,a\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^3-32\,A\,B\,a\,b^2+16\,B^2\,a^2\,b\right )-\frac {\left (32\,B\,a^2\,b^2-32\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}\right )}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}+\frac {\left (A\,b-B\,a\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^3-32\,A\,B\,a\,b^2+16\,B^2\,a^2\,b\right )+\frac {\left (32\,B\,a^2\,b^2-32\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}\right )}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}}{\frac {\left (A\,b-B\,a\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^3-32\,A\,B\,a\,b^2+16\,B^2\,a^2\,b\right )-\frac {\left (32\,B\,a^2\,b^2-32\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}\right )\,1{}\mathrm {i}}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}-\frac {\left (A\,b-B\,a\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^3-32\,A\,B\,a\,b^2+16\,B^2\,a^2\,b\right )+\frac {\left (32\,B\,a^2\,b^2-32\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}\right )\,1{}\mathrm {i}}{2\,{\left (-a\right )}^{3/4}\,b^{5/4}}}\right )\,\left (A\,b-B\,a\right )}{{\left (-a\right )}^{3/4}\,b^{5/4}} \]

[In]

int((A + B*x^2)/(x^(1/2)*(a + b*x^2)),x)

[Out]

(2*B*x^(1/2))/b - (atan((((A*b - B*a)*(x^(1/2)*(16*A^2*b^3 + 16*B^2*a^2*b - 32*A*B*a*b^2) - ((32*B*a^2*b^2 - 3
2*A*a*b^3)*(A*b - B*a))/(2*(-a)^(3/4)*b^(5/4)))*1i)/(2*(-a)^(3/4)*b^(5/4)) + ((A*b - B*a)*(x^(1/2)*(16*A^2*b^3
+ 16*B^2*a^2*b - 32*A*B*a*b^2) + ((32*B*a^2*b^2 - 32*A*a*b^3)*(A*b - B*a))/(2*(-a)^(3/4)*b^(5/4)))*1i)/(2*(-a
)^(3/4)*b^(5/4)))/(((A*b - B*a)*(x^(1/2)*(16*A^2*b^3 + 16*B^2*a^2*b - 32*A*B*a*b^2) - ((32*B*a^2*b^2 - 32*A*a*
b^3)*(A*b - B*a))/(2*(-a)^(3/4)*b^(5/4))))/(2*(-a)^(3/4)*b^(5/4)) - ((A*b - B*a)*(x^(1/2)*(16*A^2*b^3 + 16*B^2
*a^2*b - 32*A*B*a*b^2) + ((32*B*a^2*b^2 - 32*A*a*b^3)*(A*b - B*a))/(2*(-a)^(3/4)*b^(5/4))))/(2*(-a)^(3/4)*b^(5
/4))))*(A*b - B*a)*1i)/((-a)^(3/4)*b^(5/4)) - (atan((((A*b - B*a)*(x^(1/2)*(16*A^2*b^3 + 16*B^2*a^2*b - 32*A*B
*a*b^2) - ((32*B*a^2*b^2 - 32*A*a*b^3)*(A*b - B*a)*1i)/(2*(-a)^(3/4)*b^(5/4))))/(2*(-a)^(3/4)*b^(5/4)) + ((A*b
- B*a)*(x^(1/2)*(16*A^2*b^3 + 16*B^2*a^2*b - 32*A*B*a*b^2) + ((32*B*a^2*b^2 - 32*A*a*b^3)*(A*b - B*a)*1i)/(2*
(-a)^(3/4)*b^(5/4))))/(2*(-a)^(3/4)*b^(5/4)))/(((A*b - B*a)*(x^(1/2)*(16*A^2*b^3 + 16*B^2*a^2*b - 32*A*B*a*b^2
) - ((32*B*a^2*b^2 - 32*A*a*b^3)*(A*b - B*a)*1i)/(2*(-a)^(3/4)*b^(5/4)))*1i)/(2*(-a)^(3/4)*b^(5/4)) - ((A*b -
B*a)*(x^(1/2)*(16*A^2*b^3 + 16*B^2*a^2*b - 32*A*B*a*b^2) + ((32*B*a^2*b^2 - 32*A*a*b^3)*(A*b - B*a)*1i)/(2*(-a
)^(3/4)*b^(5/4)))*1i)/(2*(-a)^(3/4)*b^(5/4))))*(A*b - B*a))/((-a)^(3/4)*b^(5/4))