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

Optimal. Leaf size=254 \[ \frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}-\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{3/4} c^{3/4}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{3/4} c^{3/4}} \]

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Rubi [A]  time = 0.19, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}-\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{3/4} c^{3/4}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{3/4} c^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.06, size = 262, normalized size = 1.03 \begin {gather*} \frac {-\sqrt {2} \sqrt [4]{a} A \sqrt {c} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )+\sqrt {2} \sqrt [4]{a} A \sqrt {c} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )-2 \sqrt {2} \sqrt [4]{a} A \sqrt {c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )+2 \sqrt {2} \sqrt [4]{a} A \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )-4 (-a)^{3/4} B \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )+4 (-a)^{3/4} B \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{4 a c^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*a^(1/4)*A*Sqrt[c]*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)] + 2*Sqrt[2]*a^(1/4)*A*Sqrt[c]*ArcT
an[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)] - 4*(-a)^(3/4)*B*ArcTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)] + 4*(-a)^(3/4)
*B*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^(1/4)] - Sqrt[2]*a^(1/4)*A*Sqrt[c]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqr
t[x] + Sqrt[c]*x] + Sqrt[2]*a^(1/4)*A*Sqrt[c]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(4*a
*c^(3/4))

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IntegrateAlgebraic [A]  time = 0.23, size = 139, normalized size = 0.55 \begin {gather*} -\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} a^{3/4} c^{3/4}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{\sqrt {2} a^{3/4} c^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 775, normalized size = 3.05 \begin {gather*} \frac {1}{2} \, \sqrt {-\frac {a c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} + 2 \, A B}{a c}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} + {\left (B a^{3} c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} - A B^{2} a^{2} c + A^{3} a c^{2}\right )} \sqrt {-\frac {a c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} + 2 \, A B}{a c}}\right ) - \frac {1}{2} \, \sqrt {-\frac {a c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} + 2 \, A B}{a c}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} - {\left (B a^{3} c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} - A B^{2} a^{2} c + A^{3} a c^{2}\right )} \sqrt {-\frac {a c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} + 2 \, A B}{a c}}\right ) - \frac {1}{2} \, \sqrt {\frac {a c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} - 2 \, A B}{a c}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} + {\left (B a^{3} c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} + A B^{2} a^{2} c - A^{3} a c^{2}\right )} \sqrt {\frac {a c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} - 2 \, A B}{a c}}\right ) + \frac {1}{2} \, \sqrt {\frac {a c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} - 2 \, A B}{a c}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} - {\left (B a^{3} c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} + A B^{2} a^{2} c - A^{3} a c^{2}\right )} \sqrt {\frac {a c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{3}}} - 2 \, A B}{a c}}\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),x, algorithm="fricas")

[Out]

1/2*sqrt(-(a*c*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) + 2*A*B)/(a*c))*log(-(B^4*a^2 - A^4*c^2)*s
qrt(x) + (B*a^3*c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) - A*B^2*a^2*c + A^3*a*c^2)*sqrt(-(a*c
*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) + 2*A*B)/(a*c))) - 1/2*sqrt(-(a*c*sqrt(-(B^4*a^2 - 2*A^2
*B^2*a*c + A^4*c^2)/(a^3*c^3)) + 2*A*B)/(a*c))*log(-(B^4*a^2 - A^4*c^2)*sqrt(x) - (B*a^3*c^2*sqrt(-(B^4*a^2 -
2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) - A*B^2*a^2*c + A^3*a*c^2)*sqrt(-(a*c*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4
*c^2)/(a^3*c^3)) + 2*A*B)/(a*c))) - 1/2*sqrt((a*c*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) - 2*A*B
)/(a*c))*log(-(B^4*a^2 - A^4*c^2)*sqrt(x) + (B*a^3*c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) +
A*B^2*a^2*c - A^3*a*c^2)*sqrt((a*c*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) - 2*A*B)/(a*c))) + 1/2
*sqrt((a*c*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) - 2*A*B)/(a*c))*log(-(B^4*a^2 - A^4*c^2)*sqrt(
x) - (B*a^3*c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) + A*B^2*a^2*c - A^3*a*c^2)*sqrt((a*c*sqrt
(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^3*c^3)) - 2*A*B)/(a*c)))

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giac [A]  time = 0.18, size = 244, normalized size = 0.96 \begin {gather*} \frac {\sqrt {2} {\left (\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 )}{2 \, a c^{3}} + \frac {\sqrt {2} {\left (\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 )}{2 \, a c^{3}} + \frac {\sqrt {2} {\left (\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 )}{4 \, a c^{3}} - \frac {\sqrt {2} {\left (\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 )}{4 \, a 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),x, algorithm="giac")

[Out]

1/2*sqrt(2)*((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*c^3) + 1/2*sqrt(2)*((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*c^3) + 1/4*sqrt(2)*((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*c^3) - 1/4*sqrt(2)*((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*c^3)

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maple [A]  time = 0.07, size = 268, normalized size = 1.06 \begin {gather*} \frac {\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 )}{2 a}+\frac {\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 )}{2 a}+\frac {\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 )}{4 a}+\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{c}\right )^{\frac {1}{4}} c}+\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{c}\right )^{\frac {1}{4}} 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 )}{4 \left (\frac {a}{c}\right )^{\frac {1}{4}} 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),x)

[Out]

1/4*A*(a/c)^(1/4)/a*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)))+1/2*A*(a/c)^(1/4)/a*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+1/2*A*(a/c)^(1/4)/a*2^(1/2)*arctan
(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+1/4*B/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)))+1/2*B/c/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+1/
2*B/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 = 1.15, size = 224, normalized size = 0.88 \begin {gather*} \frac {\sqrt {2} {\left (B \sqrt {a} + 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 )}{2 \, \sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {\sqrt {2} {\left (B \sqrt {a} + 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 )}{2 \, \sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} {\left (B \sqrt {a} - A \sqrt {c}\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{4 \, a^{\frac {3}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (B \sqrt {a} - A \sqrt {c}\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{4 \, a^{\frac {3}{4}} c^{\frac {3}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*(B*sqrt(a) + 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)) + 1/2*sqrt(2)*(B*sqrt(a) + 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))
- 1/4*sqrt(2)*(B*sqrt(a) - 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)) + 1/4*sqrt(2)*(B*sqrt(a) - 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))

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mupad [B]  time = 1.27, size = 607, normalized size = 2.39 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {32\,A^2\,c^3\,\sqrt {x}\,\sqrt {\frac {B^2\,\sqrt {-a^3\,c^3}}{4\,a^2\,c^3}-\frac {A^2\,\sqrt {-a^3\,c^3}}{4\,a^3\,c^2}-\frac {A\,B}{2\,a\,c}}}{16\,A^2\,B\,c^2-16\,B^3\,a\,c-\frac {16\,A\,B^2\,\sqrt {-a^3\,c^3}}{a}+\frac {16\,A^3\,c\,\sqrt {-a^3\,c^3}}{a^2}}-\frac {32\,B^2\,a\,c^2\,\sqrt {x}\,\sqrt {\frac {B^2\,\sqrt {-a^3\,c^3}}{4\,a^2\,c^3}-\frac {A^2\,\sqrt {-a^3\,c^3}}{4\,a^3\,c^2}-\frac {A\,B}{2\,a\,c}}}{16\,A^2\,B\,c^2-16\,B^3\,a\,c-\frac {16\,A\,B^2\,\sqrt {-a^3\,c^3}}{a}+\frac {16\,A^3\,c\,\sqrt {-a^3\,c^3}}{a^2}}\right )\,\sqrt {-\frac {A^2\,c\,\sqrt {-a^3\,c^3}-B^2\,a\,\sqrt {-a^3\,c^3}+2\,A\,B\,a^2\,c^2}{4\,a^3\,c^3}}-2\,\mathrm {atanh}\left (\frac {32\,A^2\,c^3\,\sqrt {x}\,\sqrt {\frac {A^2\,\sqrt {-a^3\,c^3}}{4\,a^3\,c^2}-\frac {A\,B}{2\,a\,c}-\frac {B^2\,\sqrt {-a^3\,c^3}}{4\,a^2\,c^3}}}{16\,A^2\,B\,c^2-16\,B^3\,a\,c+\frac {16\,A\,B^2\,\sqrt {-a^3\,c^3}}{a}-\frac {16\,A^3\,c\,\sqrt {-a^3\,c^3}}{a^2}}-\frac {32\,B^2\,a\,c^2\,\sqrt {x}\,\sqrt {\frac {A^2\,\sqrt {-a^3\,c^3}}{4\,a^3\,c^2}-\frac {A\,B}{2\,a\,c}-\frac {B^2\,\sqrt {-a^3\,c^3}}{4\,a^2\,c^3}}}{16\,A^2\,B\,c^2-16\,B^3\,a\,c+\frac {16\,A\,B^2\,\sqrt {-a^3\,c^3}}{a}-\frac {16\,A^3\,c\,\sqrt {-a^3\,c^3}}{a^2}}\right )\,\sqrt {-\frac {B^2\,a\,\sqrt {-a^3\,c^3}-A^2\,c\,\sqrt {-a^3\,c^3}+2\,A\,B\,a^2\,c^2}{4\,a^3\,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)),x)

[Out]

- 2*atanh((32*A^2*c^3*x^(1/2)*((B^2*(-a^3*c^3)^(1/2))/(4*a^2*c^3) - (A^2*(-a^3*c^3)^(1/2))/(4*a^3*c^2) - (A*B)
/(2*a*c))^(1/2))/(16*A^2*B*c^2 - 16*B^3*a*c - (16*A*B^2*(-a^3*c^3)^(1/2))/a + (16*A^3*c*(-a^3*c^3)^(1/2))/a^2)
- (32*B^2*a*c^2*x^(1/2)*((B^2*(-a^3*c^3)^(1/2))/(4*a^2*c^3) - (A^2*(-a^3*c^3)^(1/2))/(4*a^3*c^2) - (A*B)/(2*a
*c))^(1/2))/(16*A^2*B*c^2 - 16*B^3*a*c - (16*A*B^2*(-a^3*c^3)^(1/2))/a + (16*A^3*c*(-a^3*c^3)^(1/2))/a^2))*(-(
A^2*c*(-a^3*c^3)^(1/2) - B^2*a*(-a^3*c^3)^(1/2) + 2*A*B*a^2*c^2)/(4*a^3*c^3))^(1/2) - 2*atanh((32*A^2*c^3*x^(1
/2)*((A^2*(-a^3*c^3)^(1/2))/(4*a^3*c^2) - (A*B)/(2*a*c) - (B^2*(-a^3*c^3)^(1/2))/(4*a^2*c^3))^(1/2))/(16*A^2*B
*c^2 - 16*B^3*a*c + (16*A*B^2*(-a^3*c^3)^(1/2))/a - (16*A^3*c*(-a^3*c^3)^(1/2))/a^2) - (32*B^2*a*c^2*x^(1/2)*(
(A^2*(-a^3*c^3)^(1/2))/(4*a^3*c^2) - (A*B)/(2*a*c) - (B^2*(-a^3*c^3)^(1/2))/(4*a^2*c^3))^(1/2))/(16*A^2*B*c^2
- 16*B^3*a*c + (16*A*B^2*(-a^3*c^3)^(1/2))/a - (16*A^3*c*(-a^3*c^3)^(1/2))/a^2))*(-(B^2*a*(-a^3*c^3)^(1/2) - A
^2*c*(-a^3*c^3)^(1/2) + 2*A*B*a^2*c^2)/(4*a^3*c^3))^(1/2)

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sympy [A]  time = 7.03, size = 348, normalized size = 1.37 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}\right ) & \text {for}\: a = 0 \wedge c = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{c} & \text {for}\: a = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}}{a} & \text {for}\: c = 0 \\- \frac {\sqrt [4]{-1} A \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {3}{4}}} + \frac {\sqrt [4]{-1} A \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {3}{4}}} - \frac {\sqrt [4]{-1} A \sqrt [4]{\frac {1}{c}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{c}}} \right )}}{a^{\frac {3}{4}}} - \frac {\left (-1\right )^{\frac {3}{4}} B \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 \sqrt [4]{a} c \sqrt [4]{\frac {1}{c}}} + \frac {\left (-1\right )^{\frac {3}{4}} B \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 \sqrt [4]{a} c \sqrt [4]{\frac {1}{c}}} + \frac {\left (-1\right )^{\frac {3}{4}} B \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{c}}} \right )}}{\sqrt [4]{a} c \sqrt [4]{\frac {1}{c}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(3*x**(3/2)) - 2*B/sqrt(x)), Eq(a, 0) & Eq(c, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/c,
Eq(a, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/a, Eq(c, 0)), (-(-1)**(1/4)*A*(1/c)**(1/4)*log(-(-1)**(1/4)*a**(1/4
)*(1/c)**(1/4) + sqrt(x))/(2*a**(3/4)) + (-1)**(1/4)*A*(1/c)**(1/4)*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sq
rt(x))/(2*a**(3/4)) - (-1)**(1/4)*A*(1/c)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/a**(3/4) -
(-1)**(3/4)*B*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*a**(1/4)*c*(1/c)**(1/4)) + (-1)**(3/4)*B*lo
g((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*a**(1/4)*c*(1/c)**(1/4)) + (-1)**(3/4)*B*atan((-1)**(3/4)*sq
rt(x)/(a**(1/4)*(1/c)**(1/4)))/(a**(1/4)*c*(1/c)**(1/4)), True))

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