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

Optimal. Leaf size=287 \[ \frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\sqrt {x} (A+B x)}{2 a \left (a+c x^2\right )} \]

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Rubi [A]  time = 0.25, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {823, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\sqrt {x} (A+B x)}{2 a \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(A + B*x))/(2*a*(a + c*x^2)) - ((Sqrt[a]*B + 3*A*Sqrt[c])*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4
)])/(4*Sqrt[2]*a^(7/4)*c^(3/4)) + ((Sqrt[a]*B + 3*A*Sqrt[c])*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(4
*Sqrt[2]*a^(7/4)*c^(3/4)) + ((Sqrt[a]*B - 3*A*Sqrt[c])*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]
*x])/(8*Sqrt[2]*a^(7/4)*c^(3/4)) - ((Sqrt[a]*B - 3*A*Sqrt[c])*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] +
Sqrt[c]*x])/(8*Sqrt[2]*a^(7/4)*c^(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 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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 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]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rubi steps

\begin {align*} \int \frac {A+B x}{\sqrt {x} \left (a+c x^2\right )^2} \, dx &=\frac {\sqrt {x} (A+B x)}{2 a \left (a+c x^2\right )}-\frac {\int \frac {-\frac {3}{2} a A c-\frac {1}{2} a B c x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{2 a^2 c}\\ &=\frac {\sqrt {x} (A+B x)}{2 a \left (a+c x^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {3}{2} a A c-\frac {1}{2} a B c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a^2 c}\\ &=\frac {\sqrt {x} (A+B x)}{2 a \left (a+c x^2\right )}-\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 a^{3/2} c}+\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}+c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 a^{3/2} c}\\ &=\frac {\sqrt {x} (A+B x)}{2 a \left (a+c x^2\right )}+\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{8 a^{3/2} c}+\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{8 a^{3/2} c}+\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}\\ &=\frac {\sqrt {x} (A+B x)}{2 a \left (a+c x^2\right )}+\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}\\ &=\frac {\sqrt {x} (A+B x)}{2 a \left (a+c x^2\right )}-\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 304, normalized size = 1.06 \begin {gather*} \frac {\frac {8 a A \sqrt {x}}{a+c x^2}-\frac {3 \sqrt {2} \sqrt [4]{a} A \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{\sqrt [4]{c}}+\frac {3 \sqrt {2} \sqrt [4]{a} A \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{\sqrt [4]{c}}-\frac {6 \sqrt {2} \sqrt [4]{a} A \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{c}}+\frac {6 \sqrt {2} \sqrt [4]{a} A \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{c}}-\frac {4 (-a)^{3/4} B \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac {4 (-a)^{3/4} B \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac {8 a B x^{3/2}}{a+c x^2}}{16 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*A*Sqrt[x])/(a + c*x^2) + (8*a*B*x^(3/2))/(a + c*x^2) - (6*Sqrt[2]*a^(1/4)*A*ArcTan[1 - (Sqrt[2]*c^(1/4)*
Sqrt[x])/a^(1/4)])/c^(1/4) + (6*Sqrt[2]*a^(1/4)*A*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(1/4) - (4*
(-a)^(3/4)*B*ArcTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(3/4) + (4*(-a)^(3/4)*B*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^(1
/4)])/c^(3/4) - (3*Sqrt[2]*a^(1/4)*A*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4) + (3*
Sqrt[2]*a^(1/4)*A*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4))/(16*a^2)

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IntegrateAlgebraic [A]  time = 0.68, size = 175, normalized size = 0.61 \begin {gather*} -\frac {\left (\sqrt {a} B+3 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} B-3 A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{4 \sqrt {2} a^{7/4} c^{3/4}}+\frac {A \sqrt {x}+B x^{3/2}}{2 a \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 877, normalized size = 3.06 \begin {gather*} \frac {{\left (a c x^{2} + a^{2}\right )} \sqrt {-\frac {a^{3} c \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} + 6 \, A B}{a^{3} c}} \log \left (-{\left (B^{4} a^{2} - 81 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (B a^{6} c^{2} \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} - 3 \, A B^{2} a^{3} c + 27 \, A^{3} a^{2} c^{2}\right )} \sqrt {-\frac {a^{3} c \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} + 6 \, A B}{a^{3} c}}\right ) - {\left (a c x^{2} + a^{2}\right )} \sqrt {-\frac {a^{3} c \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} + 6 \, A B}{a^{3} c}} \log \left (-{\left (B^{4} a^{2} - 81 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (B a^{6} c^{2} \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} - 3 \, A B^{2} a^{3} c + 27 \, A^{3} a^{2} c^{2}\right )} \sqrt {-\frac {a^{3} c \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} + 6 \, A B}{a^{3} c}}\right ) - {\left (a c x^{2} + a^{2}\right )} \sqrt {\frac {a^{3} c \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} - 6 \, A B}{a^{3} c}} \log \left (-{\left (B^{4} a^{2} - 81 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (B a^{6} c^{2} \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} + 3 \, A B^{2} a^{3} c - 27 \, A^{3} a^{2} c^{2}\right )} \sqrt {\frac {a^{3} c \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} - 6 \, A B}{a^{3} c}}\right ) + {\left (a c x^{2} + a^{2}\right )} \sqrt {\frac {a^{3} c \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} - 6 \, A B}{a^{3} c}} \log \left (-{\left (B^{4} a^{2} - 81 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (B a^{6} c^{2} \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} + 3 \, A B^{2} a^{3} c - 27 \, A^{3} a^{2} c^{2}\right )} \sqrt {\frac {a^{3} c \sqrt {-\frac {B^{4} a^{2} - 18 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a^{7} c^{3}}} - 6 \, A B}{a^{3} c}}\right ) + 4 \, {\left (B x + A\right )} \sqrt {x}}{8 \, {\left (a c x^{2} + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*c*x^2 + a^2)*sqrt(-(a^3*c*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3)) + 6*A*B)/(a^3*c))*l
og(-(B^4*a^2 - 81*A^4*c^2)*sqrt(x) + (B*a^6*c^2*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3)) - 3*A
*B^2*a^3*c + 27*A^3*a^2*c^2)*sqrt(-(a^3*c*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3)) + 6*A*B)/(a
^3*c))) - (a*c*x^2 + a^2)*sqrt(-(a^3*c*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3)) + 6*A*B)/(a^3*
c))*log(-(B^4*a^2 - 81*A^4*c^2)*sqrt(x) - (B*a^6*c^2*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3))
- 3*A*B^2*a^3*c + 27*A^3*a^2*c^2)*sqrt(-(a^3*c*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3)) + 6*A*
B)/(a^3*c))) - (a*c*x^2 + a^2)*sqrt((a^3*c*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3)) - 6*A*B)/(
a^3*c))*log(-(B^4*a^2 - 81*A^4*c^2)*sqrt(x) + (B*a^6*c^2*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^
3)) + 3*A*B^2*a^3*c - 27*A^3*a^2*c^2)*sqrt((a^3*c*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3)) - 6
*A*B)/(a^3*c))) + (a*c*x^2 + a^2)*sqrt((a^3*c*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3)) - 6*A*B
)/(a^3*c))*log(-(B^4*a^2 - 81*A^4*c^2)*sqrt(x) - (B*a^6*c^2*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7
*c^3)) + 3*A*B^2*a^3*c - 27*A^3*a^2*c^2)*sqrt((a^3*c*sqrt(-(B^4*a^2 - 18*A^2*B^2*a*c + 81*A^4*c^2)/(a^7*c^3))
- 6*A*B)/(a^3*c))) + 4*(B*x + A)*sqrt(x))/(a*c*x^2 + a^2)

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giac [A]  time = 0.22, size = 273, normalized size = 0.95 \begin {gather*} \frac {B x^{\frac {3}{2}} + A \sqrt {x}}{2 \, {\left (c x^{2} + a\right )} a} + \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} + \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{8 \, a^{2} c^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} + \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{8 \, a^{2} c^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} - \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{16 \, a^{2} c^{3}} - \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} - \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{16 \, a^{2} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 313, normalized size = 1.09 \begin {gather*} \frac {B \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+a \right ) a}+\frac {A \sqrt {x}}{2 \left (c \,x^{2}+a \right ) a}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{8 a^{2}}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{8 a^{2}}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \ln \left (\frac {x +\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}{x -\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}\right )}{16 a^{2}}+\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{c}\right )^{\frac {1}{4}} a c}+\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{c}\right )^{\frac {1}{4}} a c}+\frac {\sqrt {2}\, B \ln \left (\frac {x -\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}{x +\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}\right )}{16 \left (\frac {a}{c}\right )^{\frac {1}{4}} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*A*x^(1/2)/a/(c*x^2+a)+3/16*A/a^2*(a/c)^(1/4)*2^(1/2)*ln((x+(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2))/(x-(a/
c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2)))+3/8*A/a^2*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+3/8
*A/a^2*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+1/2*B*x^(3/2)/a/(c*x^2+a)+1/16*B/a/c/(a/c)^(1
/4)*2^(1/2)*ln((x-(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2))/(x+(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2)))+1/8*B/
a/c/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+1/8*B/a/c/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/
c)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 1.14, size = 256, normalized size = 0.89 \begin {gather*} \frac {B x^{\frac {3}{2}} + A \sqrt {x}}{2 \, {\left (a c x^{2} + a^{2}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (B \sqrt {a} + 3 \, A \sqrt {c}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} {\left (B \sqrt {a} + 3 \, A \sqrt {c}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} {\left (B \sqrt {a} - 3 \, A \sqrt {c}\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (B \sqrt {a} - 3 \, A \sqrt {c}\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}}}{16 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*x^(3/2) + A*sqrt(x))/(a*c*x^2 + a^2) + 1/16*(2*sqrt(2)*(B*sqrt(a) + 3*A*sqrt(c))*arctan(1/2*sqrt(2)*(sq
rt(2)*a^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))*sqrt(c)) + 2*
sqrt(2)*(B*sqrt(a) + 3*A*sqrt(c))*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(
a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))*sqrt(c)) - sqrt(2)*(B*sqrt(a) - 3*A*sqrt(c))*log(sqrt(2)*a^(1/4)*c
^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(a))/(a^(3/4)*c^(3/4)) + sqrt(2)*(B*sqrt(a) - 3*A*sqrt(c))*log(-sqrt(2)*a^(1/
4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(a))/(a^(3/4)*c^(3/4)))/a

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mupad [B]  time = 1.28, size = 649, normalized size = 2.26 \begin {gather*} \frac {\frac {A\,\sqrt {x}}{2\,a}+\frac {B\,x^{3/2}}{2\,a}}{c\,x^2+a}-2\,\mathrm {atanh}\left (\frac {2\,B^2\,c^2\,\sqrt {x}\,\sqrt {\frac {B^2\,\sqrt {-a^7\,c^3}}{64\,a^6\,c^3}-\frac {9\,A^2\,\sqrt {-a^7\,c^3}}{64\,a^7\,c^2}-\frac {3\,A\,B}{32\,a^3\,c}}}{\frac {B^3\,c}{4\,a}+\frac {3\,A\,B^2\,\sqrt {-a^7\,c^3}}{4\,a^5}-\frac {27\,A^3\,c\,\sqrt {-a^7\,c^3}}{4\,a^6}-\frac {9\,A^2\,B\,c^2}{4\,a^2}}-\frac {18\,A^2\,c^3\,\sqrt {x}\,\sqrt {\frac {B^2\,\sqrt {-a^7\,c^3}}{64\,a^6\,c^3}-\frac {9\,A^2\,\sqrt {-a^7\,c^3}}{64\,a^7\,c^2}-\frac {3\,A\,B}{32\,a^3\,c}}}{\frac {B^3\,c}{4}+\frac {3\,A\,B^2\,\sqrt {-a^7\,c^3}}{4\,a^4}-\frac {27\,A^3\,c\,\sqrt {-a^7\,c^3}}{4\,a^5}-\frac {9\,A^2\,B\,c^2}{4\,a}}\right )\,\sqrt {-\frac {9\,A^2\,c\,\sqrt {-a^7\,c^3}-B^2\,a\,\sqrt {-a^7\,c^3}+6\,A\,B\,a^4\,c^2}{64\,a^7\,c^3}}-2\,\mathrm {atanh}\left (\frac {2\,B^2\,c^2\,\sqrt {x}\,\sqrt {\frac {9\,A^2\,\sqrt {-a^7\,c^3}}{64\,a^7\,c^2}-\frac {3\,A\,B}{32\,a^3\,c}-\frac {B^2\,\sqrt {-a^7\,c^3}}{64\,a^6\,c^3}}}{\frac {B^3\,c}{4\,a}-\frac {3\,A\,B^2\,\sqrt {-a^7\,c^3}}{4\,a^5}+\frac {27\,A^3\,c\,\sqrt {-a^7\,c^3}}{4\,a^6}-\frac {9\,A^2\,B\,c^2}{4\,a^2}}-\frac {18\,A^2\,c^3\,\sqrt {x}\,\sqrt {\frac {9\,A^2\,\sqrt {-a^7\,c^3}}{64\,a^7\,c^2}-\frac {3\,A\,B}{32\,a^3\,c}-\frac {B^2\,\sqrt {-a^7\,c^3}}{64\,a^6\,c^3}}}{\frac {B^3\,c}{4}-\frac {3\,A\,B^2\,\sqrt {-a^7\,c^3}}{4\,a^4}+\frac {27\,A^3\,c\,\sqrt {-a^7\,c^3}}{4\,a^5}-\frac {9\,A^2\,B\,c^2}{4\,a}}\right )\,\sqrt {-\frac {B^2\,a\,\sqrt {-a^7\,c^3}-9\,A^2\,c\,\sqrt {-a^7\,c^3}+6\,A\,B\,a^4\,c^2}{64\,a^7\,c^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*x^(1/2))/(2*a) + (B*x^(3/2))/(2*a))/(a + c*x^2) - 2*atanh((2*B^2*c^2*x^(1/2)*((B^2*(-a^7*c^3)^(1/2))/(64*a
^6*c^3) - (9*A^2*(-a^7*c^3)^(1/2))/(64*a^7*c^2) - (3*A*B)/(32*a^3*c))^(1/2))/((B^3*c)/(4*a) + (3*A*B^2*(-a^7*c
^3)^(1/2))/(4*a^5) - (27*A^3*c*(-a^7*c^3)^(1/2))/(4*a^6) - (9*A^2*B*c^2)/(4*a^2)) - (18*A^2*c^3*x^(1/2)*((B^2*
(-a^7*c^3)^(1/2))/(64*a^6*c^3) - (9*A^2*(-a^7*c^3)^(1/2))/(64*a^7*c^2) - (3*A*B)/(32*a^3*c))^(1/2))/((B^3*c)/4
+ (3*A*B^2*(-a^7*c^3)^(1/2))/(4*a^4) - (27*A^3*c*(-a^7*c^3)^(1/2))/(4*a^5) - (9*A^2*B*c^2)/(4*a)))*(-(9*A^2*c
*(-a^7*c^3)^(1/2) - B^2*a*(-a^7*c^3)^(1/2) + 6*A*B*a^4*c^2)/(64*a^7*c^3))^(1/2) - 2*atanh((2*B^2*c^2*x^(1/2)*(
(9*A^2*(-a^7*c^3)^(1/2))/(64*a^7*c^2) - (3*A*B)/(32*a^3*c) - (B^2*(-a^7*c^3)^(1/2))/(64*a^6*c^3))^(1/2))/((B^3
*c)/(4*a) - (3*A*B^2*(-a^7*c^3)^(1/2))/(4*a^5) + (27*A^3*c*(-a^7*c^3)^(1/2))/(4*a^6) - (9*A^2*B*c^2)/(4*a^2))
- (18*A^2*c^3*x^(1/2)*((9*A^2*(-a^7*c^3)^(1/2))/(64*a^7*c^2) - (3*A*B)/(32*a^3*c) - (B^2*(-a^7*c^3)^(1/2))/(64
*a^6*c^3))^(1/2))/((B^3*c)/4 - (3*A*B^2*(-a^7*c^3)^(1/2))/(4*a^4) + (27*A^3*c*(-a^7*c^3)^(1/2))/(4*a^5) - (9*A
^2*B*c^2)/(4*a)))*(-(B^2*a*(-a^7*c^3)^(1/2) - 9*A^2*c*(-a^7*c^3)^(1/2) + 6*A*B*a^4*c^2)/(64*a^7*c^3))^(1/2)

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sympy [A]  time = 77.78, size = 1294, normalized size = 4.51

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(a, 0) & Eq(c, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(
5/2)))/c**2, Eq(a, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/a**2, Eq(c, 0)), (4*(-1)**(1/4)*A*a**(5/4)*c*sqrt(x)*(
1/c)**(1/4)/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) - 3*I*A*a
**(3/2)*c*sqrt(1/c)*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4)
+ 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) + 3*I*A*a**(3/2)*c*sqrt(1/c)*log((-1)**(1/4)*a**(1/4)*(1/c)**
(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) - 6*
I*A*a**(3/2)*c*sqrt(1/c)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(
1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) - 3*I*A*sqrt(a)*c**2*x**2*sqrt(1/c)*log(-(-1)**(1/4)*a**
(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)
**(1/4)) + 3*I*A*sqrt(a)*c**2*x**2*sqrt(1/c)*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a
**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) - 6*I*A*sqrt(a)*c**2*x**2*sqrt(1/c)*a
tan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9
/4)*c**2*x**2*(1/c)**(1/4)) + 4*(-1)**(1/4)*B*a**(5/4)*c*x**(3/2)*(1/c)**(1/4)/(8*(-1)**(1/4)*a**(13/4)*c*(1/c
)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) + B*a**2*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sq
rt(x))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) - B*a**2*log((
-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c
**2*x**2*(1/c)**(1/4)) - 2*B*a**2*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(8*(-1)**(1/4)*a**(13/4)*c
*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) + B*a*c*x**2*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(
1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) - B*a
*c*x**2*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/
4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)) - 2*B*a*c*x**2*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(8*(-1)**
(1/4)*a**(13/4)*c*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(9/4)*c**2*x**2*(1/c)**(1/4)), True))

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