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

Optimal. Leaf size=261 \[ \frac {(3 a B+A b) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{7/4}}-\frac {(3 a B+A b) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{7/4}}-\frac {(3 a B+A b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}+\frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}+\frac {x^{3/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.18, antiderivative size = 261, 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 = {457, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {(3 a B+A b) \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^{7/4}}-\frac {(3 a B+A b) \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^{7/4}}-\frac {(3 a B+A b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}+\frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}+\frac {x^{3/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^(3/2))/(2*a*b*(a + b*x^2)) - ((A*b + 3*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*S
qrt[2]*a^(5/4)*b^(7/4)) + ((A*b + 3*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(5/4)*b^(
7/4)) + ((A*b + 3*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(5/4)*b^(7/4))
- ((A*b + 3*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(5/4)*b^(7/4))

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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(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 {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac {(A b-a B) x^{3/2}}{2 a b \left (a+b x^2\right )}+\frac {\left (\frac {A b}{2}+\frac {3 a B}{2}\right ) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{2 a b}\\ &=\frac {(A b-a B) x^{3/2}}{2 a b \left (a+b x^2\right )}+\frac {\left (\frac {A b}{2}+\frac {3 a B}{2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a b}\\ &=\frac {(A b-a B) x^{3/2}}{2 a b \left (a+b x^2\right )}-\frac {(A b+3 a B) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a b^{3/2}}+\frac {(A b+3 a B) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a b^{3/2}}\\ &=\frac {(A b-a B) x^{3/2}}{2 a b \left (a+b x^2\right )}+\frac {(A b+3 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 )}{8 a b^2}+\frac {(A b+3 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 )}{8 a b^2}+\frac {(A b+3 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 )}{8 \sqrt {2} a^{5/4} b^{7/4}}+\frac {(A b+3 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 )}{8 \sqrt {2} a^{5/4} b^{7/4}}\\ &=\frac {(A b-a B) x^{3/2}}{2 a b \left (a+b x^2\right )}+\frac {(A b+3 a B) \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^{7/4}}-\frac {(A b+3 a B) \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^{7/4}}+\frac {(A b+3 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 )}{4 \sqrt {2} a^{5/4} b^{7/4}}-\frac {(A b+3 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 )}{4 \sqrt {2} a^{5/4} b^{7/4}}\\ &=\frac {(A b-a B) x^{3/2}}{2 a b \left (a+b x^2\right )}-\frac {(A b+3 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}+\frac {(A b+3 a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}+\frac {(A b+3 a B) \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^{7/4}}-\frac {(A b+3 a B) \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^{7/4}}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 95, normalized size = 0.36 \begin {gather*} \frac {2 x^{3/2} (A b-a B) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {b x^2}{a}\right )}{3 a^2 b}+\frac {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^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(ArcTan[(b^(1/4)*Sqrt[x])/(-a)^(1/4)] + ArcTanh[(a*b^(1/4)*Sqrt[x])/(-a)^(5/4)]))/((-a)^(1/4)*b^(7/4)) + (2
*(A*b - a*B)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((b*x^2)/a)])/(3*a^2*b)

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IntegrateAlgebraic [A]  time = 0.57, size = 160, normalized size = 0.61 \begin {gather*} -\frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}-\frac {(3 a B+A b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}-\frac {x^{3/2} (a B-A b)}{2 a b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((-(A*b) + a*B)*x^(3/2))/(a*b*(a + b*x^2)) - ((A*b + 3*a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)
*b^(1/4)*Sqrt[x])])/(4*Sqrt[2]*a^(5/4)*b^(7/4)) - ((A*b + 3*a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sq
rt[a] + Sqrt[b]*x)])/(4*Sqrt[2]*a^(5/4)*b^(7/4))

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fricas [B]  time = 1.21, size = 912, normalized size = 3.49 \begin {gather*} -\frac {4 \, {\left (B a - A b\right )} x^{\frac {3}{2}} + 4 \, {\left (a b^{2} x^{2} + a^{2} b\right )} \left (-\frac {81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{7}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (729 \, B^{6} a^{6} + 1458 \, A B^{5} a^{5} b + 1215 \, A^{2} B^{4} a^{4} b^{2} + 540 \, A^{3} B^{3} a^{3} b^{3} + 135 \, A^{4} B^{2} a^{2} b^{4} + 18 \, A^{5} B a b^{5} + A^{6} b^{6}\right )} x - {\left (81 \, B^{4} a^{7} b^{3} + 108 \, A B^{3} a^{6} b^{4} + 54 \, A^{2} B^{2} a^{5} b^{5} + 12 \, A^{3} B a^{4} b^{6} + A^{4} a^{3} b^{7}\right )} \sqrt {-\frac {81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{7}}}} a b^{2} \left (-\frac {81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{7}}\right )^{\frac {1}{4}} - {\left (27 \, B^{3} a^{4} b^{2} + 27 \, A B^{2} a^{3} b^{3} + 9 \, A^{2} B a^{2} b^{4} + A^{3} a b^{5}\right )} \sqrt {x} \left (-\frac {81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{7}}\right )^{\frac {1}{4}}}{81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}\right ) - {\left (a b^{2} x^{2} + a^{2} b\right )} \left (-\frac {81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{7}}\right )^{\frac {1}{4}} \log \left (a^{4} b^{5} \left (-\frac {81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{7}}\right )^{\frac {3}{4}} + {\left (27 \, B^{3} a^{3} + 27 \, A B^{2} a^{2} b + 9 \, A^{2} B a b^{2} + A^{3} b^{3}\right )} \sqrt {x}\right ) + {\left (a b^{2} x^{2} + a^{2} b\right )} \left (-\frac {81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{7}}\right )^{\frac {1}{4}} \log \left (-a^{4} b^{5} \left (-\frac {81 \, B^{4} a^{4} + 108 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 12 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{7}}\right )^{\frac {3}{4}} + {\left (27 \, B^{3} a^{3} + 27 \, A B^{2} a^{2} b + 9 \, A^{2} B a b^{2} + A^{3} b^{3}\right )} \sqrt {x}\right )}{8 \, {\left (a b^{2} x^{2} + a^{2} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*(B*a - A*b)*x^(3/2) + 4*(a*b^2*x^2 + a^2*b)*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*
A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4)*arctan((sqrt((729*B^6*a^6 + 1458*A*B^5*a^5*b + 1215*A^2*B^4*a^4*b^2 +
540*A^3*B^3*a^3*b^3 + 135*A^4*B^2*a^2*b^4 + 18*A^5*B*a*b^5 + A^6*b^6)*x - (81*B^4*a^7*b^3 + 108*A*B^3*a^6*b^4
+ 54*A^2*B^2*a^5*b^5 + 12*A^3*B*a^4*b^6 + A^4*a^3*b^7)*sqrt(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^
2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7)))*a*b^2*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*
B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4) - (27*B^3*a^4*b^2 + 27*A*B^2*a^3*b^3 + 9*A^2*B*a^2*b^4 + A^3*a*b^5)*sqrt(x
)*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4))/(81*B^4*a
^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a*b^3 + A^4*b^4)) - (a*b^2*x^2 + a^2*b)*(-(81*B^4*a^4 + 1
08*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4)*log(a^4*b^5*(-(81*B^4*a^4 + 1
08*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(3/4) + (27*B^3*a^3 + 27*A*B^2*a^2*
b + 9*A^2*B*a*b^2 + A^3*b^3)*sqrt(x)) + (a*b^2*x^2 + a^2*b)*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b
^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4)*log(-a^4*b^5*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*
b^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(3/4) + (27*B^3*a^3 + 27*A*B^2*a^2*b + 9*A^2*B*a*b^2 + A^3*b^3)*sqr
t(x)))/(a*b^2*x^2 + a^2*b)

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giac [A]  time = 0.37, size = 273, normalized size = 1.05 \begin {gather*} -\frac {B a x^{\frac {3}{2}} - A b x^{\frac {3}{2}}}{2 \, {\left (b x^{2} + a\right )} a b} + \frac {\sqrt {2} {\left (3 \, \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 )}{8 \, a^{2} b^{4}} + \frac {\sqrt {2} {\left (3 \, \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 )}{8 \, a^{2} b^{4}} - \frac {\sqrt {2} {\left (3 \, \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 )}{16 \, a^{2} b^{4}} + \frac {\sqrt {2} {\left (3 \, \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 )}{16 \, a^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*a*x^(3/2) - A*b*x^(3/2))/((b*x^2 + a)*a*b) + 1/8*sqrt(2)*(3*(a*b^3)^(3/4)*B*a + (a*b^3)^(3/4)*A*b)*arc
tan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^2*b^4) + 1/8*sqrt(2)*(3*(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^4) - 1/16*sqrt(2)
*(3*(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^4) + 1/16*s
qrt(2)*(3*(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^4)

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maple [A]  time = 0.02, size = 305, normalized size = 1.17 \begin {gather*} \frac {\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} a b}+\frac {\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} a b}+\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 )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} a b}+\frac {3 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{2}}+\frac {3 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{2}}+\frac {3 \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 )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{2}}+\frac {\left (A b -B a \right ) x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(A*b-B*a)*x^(3/2)/a/b/(b*x^2+a)+1/8/a/b/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+1/8/a/
b/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+1/16/a/b/(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)))+3/8/b^2/(a/b)^(1/4)*2^(1/2)*B*arcta
n(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+3/8/b^2/(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+3/16/b^2/
(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
)))

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maxima [A]  time = 2.28, size = 217, normalized size = 0.83 \begin {gather*} -\frac {{\left (B a - A b\right )} x^{\frac {3}{2}}}{2 \, {\left (a b^{2} x^{2} + a^{2} b\right )}} + \frac {{\left (3 \, 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 )}}{16 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.31, size = 91, normalized size = 0.35 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b+3\,B\,a\right )}{4\,{\left (-a\right )}^{5/4}\,b^{7/4}}-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b+3\,B\,a\right )}{4\,{\left (-a\right )}^{5/4}\,b^{7/4}}+\frac {x^{3/2}\,\left (A\,b-B\,a\right )}{2\,a\,b\,\left (b\,x^2+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((b^(1/4)*x^(1/2))/(-a)^(1/4))*(A*b + 3*B*a))/(4*(-a)^(5/4)*b^(7/4)) - (atan((b^(1/4)*x^(1/2))/(-a)^(1/4
))*(A*b + 3*B*a))/(4*(-a)^(5/4)*b^(7/4)) + (x^(3/2)*(A*b - B*a))/(2*a*b*(a + b*x^2))

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sympy [A]  time = 32.13, size = 162, normalized size = 0.62 \begin {gather*} \frac {2 A x^{\frac {3}{2}}}{4 a^{2} + 4 a b x^{2}} + 2 A \operatorname {RootSum} {\left (65536 t^{4} a^{5} b^{3} + 1, \left (t \mapsto t \log {\left (4096 t^{3} a^{4} b^{2} + \sqrt {x} \right )} \right )\right )} - \frac {2 B a x^{\frac {3}{2}}}{4 a^{2} b + 4 a b^{2} x^{2}} - \frac {2 B a \operatorname {RootSum} {\left (65536 t^{4} a^{5} b^{3} + 1, \left (t \mapsto t \log {\left (4096 t^{3} a^{4} b^{2} + \sqrt {x} \right )} \right )\right )}}{b} + \frac {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 )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*x**(3/2)/(4*a**2 + 4*a*b*x**2) + 2*A*RootSum(65536*_t**4*a**5*b**3 + 1, Lambda(_t, _t*log(4096*_t**3*a**4*
b**2 + sqrt(x)))) - 2*B*a*x**(3/2)/(4*a**2*b + 4*a*b**2*x**2) - 2*B*a*RootSum(65536*_t**4*a**5*b**3 + 1, Lambd
a(_t, _t*log(4096*_t**3*a**4*b**2 + sqrt(x))))/b + 2*B*RootSum(256*_t**4*a*b**3 + 1, Lambda(_t, _t*log(64*_t**
3*a*b**2 + sqrt(x))))/b

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