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

Optimal. Leaf size=235 \[ -\frac {(A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{5/4} b^{3/4}}+\frac {(A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{5/4} b^{3/4}}+\frac {(A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {(A b-a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {2 A}{a \sqrt {x}} \]

[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)

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Rubi [A]  time = 0.18, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {453, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {(A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{5/4} b^{3/4}}+\frac {(A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{5/4} b^{3/4}}+\frac {(A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {(A b-a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{5/4} b^{3/4}}-\frac {2 A}{a \sqrt {x}} \]

Antiderivative was successfully verified.

[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 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 297

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 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 453

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 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 {A+B x^2}{x^{3/2} \left (a+b x^2\right )} \, dx &=-\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 ) \operatorname {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) \operatorname {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) \operatorname {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) \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 )}{2 a b}-\frac {(A b-a B) \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 )}{2 a b}-\frac {(A b-a B) \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 )}{2 \sqrt {2} a^{5/4} b^{3/4}}-\frac {(A b-a B) \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 )}{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) \operatorname {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) \operatorname {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*}

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Mathematica [A]  time = 0.08, size = 74, normalized size = 0.31 \[ \frac {\frac {(a B-A b) \left (\tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )+\tanh ^{-1}\left (\frac {a \sqrt [4]{b} \sqrt {x}}{(-a)^{5/4}}\right )\right )}{\sqrt [4]{-a} b^{3/4}}-\frac {2 A}{\sqrt {x}}}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.52, size = 843, normalized size = 3.59 \[ \frac {4 \, 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}} \arctan \left (\frac {\sqrt {{\left (B^{6} a^{6} - 6 \, A B^{5} a^{5} b + 15 \, A^{2} B^{4} a^{4} b^{2} - 20 \, A^{3} B^{3} a^{3} b^{3} + 15 \, A^{4} B^{2} a^{2} b^{4} - 6 \, A^{5} B a b^{5} + A^{6} b^{6}\right )} x - {\left (B^{4} a^{7} b - 4 \, A B^{3} a^{6} b^{2} + 6 \, A^{2} B^{2} a^{5} b^{3} - 4 \, A^{3} B a^{4} b^{4} + A^{4} a^{3} b^{5}\right )} \sqrt {-\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}}}} 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^{5} b^{3}}\right )^{\frac {1}{4}} + {\left (B^{3} a^{4} b - 3 \, A B^{2} a^{3} b^{2} + 3 \, A^{2} B a^{2} b^{3} - A^{3} a b^{4}\right )} \sqrt {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}}}{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}}\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 ) + 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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*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)*arctan((
sqrt((B^6*a^6 - 6*A*B^5*a^5*b + 15*A^2*B^4*a^4*b^2 - 20*A^3*B^3*a^3*b^3 + 15*A^4*B^2*a^2*b^4 - 6*A^5*B*a*b^5 +
A^6*b^6)*x - (B^4*a^7*b - 4*A*B^3*a^6*b^2 + 6*A^2*B^2*a^5*b^3 - 4*A^3*B*a^4*b^4 + A^4*a^3*b^5)*sqrt(-(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)))*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^5*b^3))^(1/4) + (B^3*a^4*b - 3*A*B^2*a^3*b^2 + 3*A^2*B*a^2*b^3
- A^3*a*b^4)*sqrt(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
))/(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*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)) + 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)

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giac [A]  time = 0.37, size = 251, normalized size = 1.07 \[ -\frac {2 \, A}{a \sqrt {x}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{2} b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 277, normalized size = 1.18 \[ -\frac {\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}-\frac {\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}-\frac {\sqrt {2}\, A \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}+\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, B \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}-\frac {2 A}{a \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.42, size = 194, normalized size = 0.83 \[ \frac {{\left (B a - A b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{4 \, a} - \frac {2 \, A}{a \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.16, size = 71, normalized size = 0.30 \[ \frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b-B\,a\right )}{{\left (-a\right )}^{5/4}\,b^{3/4}}-\frac {2\,A}{a\,\sqrt {x}}-\frac {\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b-B\,a\right )}{{\left (-a\right )}^{5/4}\,b^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 19.84, size = 206, normalized size = 0.88 \[ A \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: a = 0 \\- \frac {2}{a \sqrt {x}} & \text {for}\: b = 0 \\- \frac {2}{a \sqrt {x}} + \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}}} - \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}}} - \frac {\left (-1\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}}} & \text {otherwise} \end {cases}\right ) + 2 B \operatorname {RootSum} {\left (256 t^{4} a b^{3} + 1, \left (t \mapsto t \log {\left (64 t^{3} a b^{2} + \sqrt {x} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)), (-
2/(a*sqrt(x)) + (-1)**(3/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(5/4)*(1/b)**(1/4)) - (-1)
**(3/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(5/4)*(1/b)**(1/4)) - (-1)**(3/4)*atan((-1)**(3
/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/(a**(5/4)*(1/b)**(1/4)), True)) + 2*B*RootSum(256*_t**4*a*b**3 + 1, Lambd
a(_t, _t*log(64*_t**3*a*b**2 + sqrt(x))))

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