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

Optimal. Leaf size=265 \[ \frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{a} c^{5/4}}-\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{a} c^{5/4}}+\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} c^{5/4}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} c^{5/4}}+\frac {2 B \sqrt {x}}{c} \]

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Rubi [A]  time = 0.23, antiderivative size = 265, 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 = {825, 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} \sqrt [4]{a} c^{5/4}}-\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} \sqrt [4]{a} c^{5/4}}+\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} \sqrt [4]{a} c^{5/4}}-\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} \sqrt [4]{a} c^{5/4}}+\frac {2 B \sqrt {x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 825

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g*(d + e*x)^m)/
(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x])/(a + c*x^2), x], x] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 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 {\sqrt {x} (A+B x)}{a+c x^2} \, dx &=\frac {2 B \sqrt {x}}{c}+\frac {\int \frac {-a B+A c x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{c}\\ &=\frac {2 B \sqrt {x}}{c}+\frac {2 \operatorname {Subst}\left (\int \frac {-a B+A c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{c}\\ &=\frac {2 B \sqrt {x}}{c}+\frac {\left (A-\frac {\sqrt {a} B}{\sqrt {c}}\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 (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{c}\\ &=\frac {2 B \sqrt {x}}{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} \sqrt [4]{a} c^{5/4}}+\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} \sqrt [4]{a} c^{5/4}}+\frac {\left (A-\frac {\sqrt {a} B}{\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 )}{2 c}+\frac {\left (A-\frac {\sqrt {a} B}{\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 )}{2 c}\\ &=\frac {2 B \sqrt {x}}{c}+\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} \sqrt [4]{a} c^{5/4}}-\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} \sqrt [4]{a} c^{5/4}}+\frac {\left (A-\frac {\sqrt {a} B}{\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} \sqrt [4]{a} c^{3/4}}-\frac {\left (A-\frac {\sqrt {a} B}{\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} \sqrt [4]{a} c^{3/4}}\\ &=\frac {2 B \sqrt {x}}{c}-\frac {\left (A-\frac {\sqrt {a} B}{\sqrt {c}}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} c^{3/4}}+\frac {\left (A-\frac {\sqrt {a} B}{\sqrt {c}}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} 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} \sqrt [4]{a} c^{5/4}}-\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} \sqrt [4]{a} c^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 266, normalized size = 1.00 \begin {gather*} \frac {\sqrt {2} a^{5/4} B \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )-\sqrt {2} a^{5/4} B \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )+2 \sqrt {2} a^{5/4} B \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )-2 \sqrt {2} a^{5/4} B \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )-4 (-a)^{3/4} A \sqrt {c} \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )+4 (-a)^{3/4} A \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )+8 a B \sqrt [4]{c} \sqrt {x}}{4 a c^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.28, size = 149, normalized size = 0.56 \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} \sqrt [4]{a} c^{5/4}}-\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} \sqrt [4]{a} c^{5/4}}+\frac {2 B \sqrt {x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*Sqrt[x])/c + ((Sqrt[a]*B - A*Sqrt[c])*ArcTan[(Sqrt[a] - Sqrt[c]*x)/(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])])/(S
qrt[2]*a^(1/4)*c^(5/4)) - ((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^(1/4)*c^(5/4))

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fricas [B]  time = 0.43, size = 764, normalized size = 2.88 \begin {gather*} \frac {c \sqrt {\frac {c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} + 2 \, A B}{c^{2}}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} + {\left (A a c^{4} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} + B^{3} a^{2} c - A^{2} B a c^{2}\right )} \sqrt {\frac {c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} + 2 \, A B}{c^{2}}}\right ) - c \sqrt {\frac {c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} + 2 \, A B}{c^{2}}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} - {\left (A a c^{4} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} + B^{3} a^{2} c - A^{2} B a c^{2}\right )} \sqrt {\frac {c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} + 2 \, A B}{c^{2}}}\right ) - c \sqrt {-\frac {c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} - 2 \, A B}{c^{2}}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} + {\left (A a c^{4} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} - B^{3} a^{2} c + A^{2} B a c^{2}\right )} \sqrt {-\frac {c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} - 2 \, A B}{c^{2}}}\right ) + c \sqrt {-\frac {c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} - 2 \, A B}{c^{2}}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} - {\left (A a c^{4} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} - B^{3} a^{2} c + A^{2} B a c^{2}\right )} \sqrt {-\frac {c^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a c^{5}}} - 2 \, A B}{c^{2}}}\right ) + 4 \, B \sqrt {x}}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*sqrt((c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a*c^5)) + 2*A*B)/c^2)*log(-(B^4*a^2 - A^4*c^2)*sqr
t(x) + (A*a*c^4*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a*c^5)) + B^3*a^2*c - A^2*B*a*c^2)*sqrt((c^2*sqrt(-
(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a*c^5)) + 2*A*B)/c^2)) - c*sqrt((c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4
*c^2)/(a*c^5)) + 2*A*B)/c^2)*log(-(B^4*a^2 - A^4*c^2)*sqrt(x) - (A*a*c^4*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*
c^2)/(a*c^5)) + B^3*a^2*c - A^2*B*a*c^2)*sqrt((c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a*c^5)) + 2*A*B)
/c^2)) - c*sqrt(-(c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a*c^5)) - 2*A*B)/c^2)*log(-(B^4*a^2 - A^4*c^2
)*sqrt(x) + (A*a*c^4*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a*c^5)) - B^3*a^2*c + A^2*B*a*c^2)*sqrt(-(c^2*
sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a*c^5)) - 2*A*B)/c^2)) + c*sqrt(-(c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*
c + A^4*c^2)/(a*c^5)) - 2*A*B)/c^2)*log(-(B^4*a^2 - A^4*c^2)*sqrt(x) - (A*a*c^4*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c
+ A^4*c^2)/(a*c^5)) - B^3*a^2*c + A^2*B*a*c^2)*sqrt(-(c^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a*c^5))
- 2*A*B)/c^2)) + 4*B*sqrt(x))/c

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giac [A]  time = 0.19, size = 249, normalized size = 0.94 \begin {gather*} \frac {2 \, B \sqrt {x}}{c} - \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} B a c - \left (a c^{3}\right )^{\frac {3}{4}} A\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}} B a c - \left (a c^{3}\right )^{\frac {3}{4}} A\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}} B a c + \left (a c^{3}\right )^{\frac {3}{4}} A\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}} B a c + \left (a c^{3}\right )^{\frac {3}{4}} A\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(x^(1/2)*(B*x+A)/(c*x^2+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 277, normalized size = 1.05 \begin {gather*} \frac {\sqrt {2}\, A \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}\, A \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}\, 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 \left (\frac {a}{c}\right )^{\frac {1}{4}} c}-\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{2 c}-\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{2 c}-\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \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 c}+\frac {2 B \sqrt {x}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B/c*x^(1/2)-1/4/c*B*(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/c*B*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)-1/2/c*B*(a/c)^(1/4)*
2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+1/4/c*A/(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/c*A/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)
*x^(1/2)+1)+1/2/c*A/(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 = 1.45, size = 246, normalized size = 0.93 \begin {gather*} \frac {2 \, B \sqrt {x}}{c} - \frac {\frac {2 \, \sqrt {2} {\left (B a \sqrt {c} - A \sqrt {a} 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 a \sqrt {c} - A \sqrt {a} 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 a \sqrt {c} + A \sqrt {a} 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 a \sqrt {c} + A \sqrt {a} 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}}}}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.26, size = 566, normalized size = 2.14 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {32\,A^2\,a\,c^2\,\sqrt {x}\,\sqrt {\frac {A\,B}{2\,c^2}-\frac {B^2\,\sqrt {-a\,c^5}}{4\,c^5}+\frac {A^2\,\sqrt {-a\,c^5}}{4\,a\,c^4}}}{16\,A\,B^2\,a^2-16\,A^3\,a\,c-\frac {16\,B^3\,a^2\,\sqrt {-a\,c^5}}{c^3}+\frac {16\,A^2\,B\,a\,\sqrt {-a\,c^5}}{c^2}}-\frac {32\,B^2\,a^2\,c\,\sqrt {x}\,\sqrt {\frac {A\,B}{2\,c^2}-\frac {B^2\,\sqrt {-a\,c^5}}{4\,c^5}+\frac {A^2\,\sqrt {-a\,c^5}}{4\,a\,c^4}}}{16\,A\,B^2\,a^2-16\,A^3\,a\,c-\frac {16\,B^3\,a^2\,\sqrt {-a\,c^5}}{c^3}+\frac {16\,A^2\,B\,a\,\sqrt {-a\,c^5}}{c^2}}\right )\,\sqrt {\frac {A^2\,c\,\sqrt {-a\,c^5}-B^2\,a\,\sqrt {-a\,c^5}+2\,A\,B\,a\,c^3}{4\,a\,c^5}}+2\,\mathrm {atanh}\left (\frac {32\,A^2\,a\,c^2\,\sqrt {x}\,\sqrt {\frac {A\,B}{2\,c^2}+\frac {B^2\,\sqrt {-a\,c^5}}{4\,c^5}-\frac {A^2\,\sqrt {-a\,c^5}}{4\,a\,c^4}}}{16\,A\,B^2\,a^2-16\,A^3\,a\,c+\frac {16\,B^3\,a^2\,\sqrt {-a\,c^5}}{c^3}-\frac {16\,A^2\,B\,a\,\sqrt {-a\,c^5}}{c^2}}-\frac {32\,B^2\,a^2\,c\,\sqrt {x}\,\sqrt {\frac {A\,B}{2\,c^2}+\frac {B^2\,\sqrt {-a\,c^5}}{4\,c^5}-\frac {A^2\,\sqrt {-a\,c^5}}{4\,a\,c^4}}}{16\,A\,B^2\,a^2-16\,A^3\,a\,c+\frac {16\,B^3\,a^2\,\sqrt {-a\,c^5}}{c^3}-\frac {16\,A^2\,B\,a\,\sqrt {-a\,c^5}}{c^2}}\right )\,\sqrt {\frac {B^2\,a\,\sqrt {-a\,c^5}-A^2\,c\,\sqrt {-a\,c^5}+2\,A\,B\,a\,c^3}{4\,a\,c^5}}+\frac {2\,B\,\sqrt {x}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*atanh((32*A^2*a*c^2*x^(1/2)*((A*B)/(2*c^2) - (B^2*(-a*c^5)^(1/2))/(4*c^5) + (A^2*(-a*c^5)^(1/2))/(4*a*c^4))^
(1/2))/(16*A*B^2*a^2 - 16*A^3*a*c - (16*B^3*a^2*(-a*c^5)^(1/2))/c^3 + (16*A^2*B*a*(-a*c^5)^(1/2))/c^2) - (32*B
^2*a^2*c*x^(1/2)*((A*B)/(2*c^2) - (B^2*(-a*c^5)^(1/2))/(4*c^5) + (A^2*(-a*c^5)^(1/2))/(4*a*c^4))^(1/2))/(16*A*
B^2*a^2 - 16*A^3*a*c - (16*B^3*a^2*(-a*c^5)^(1/2))/c^3 + (16*A^2*B*a*(-a*c^5)^(1/2))/c^2))*((A^2*c*(-a*c^5)^(1
/2) - B^2*a*(-a*c^5)^(1/2) + 2*A*B*a*c^3)/(4*a*c^5))^(1/2) + 2*atanh((32*A^2*a*c^2*x^(1/2)*((A*B)/(2*c^2) + (B
^2*(-a*c^5)^(1/2))/(4*c^5) - (A^2*(-a*c^5)^(1/2))/(4*a*c^4))^(1/2))/(16*A*B^2*a^2 - 16*A^3*a*c + (16*B^3*a^2*(
-a*c^5)^(1/2))/c^3 - (16*A^2*B*a*(-a*c^5)^(1/2))/c^2) - (32*B^2*a^2*c*x^(1/2)*((A*B)/(2*c^2) + (B^2*(-a*c^5)^(
1/2))/(4*c^5) - (A^2*(-a*c^5)^(1/2))/(4*a*c^4))^(1/2))/(16*A*B^2*a^2 - 16*A^3*a*c + (16*B^3*a^2*(-a*c^5)^(1/2)
)/c^3 - (16*A^2*B*a*(-a*c^5)^(1/2))/c^2))*((B^2*a*(-a*c^5)^(1/2) - A^2*c*(-a*c^5)^(1/2) + 2*A*B*a*c^3)/(4*a*c^
5))^(1/2) + (2*B*x^(1/2))/c

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sympy [A]  time = 6.48, size = 359, normalized size = 1.35 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}\right ) & \text {for}\: a = 0 \wedge c = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{c} & \text {for}\: a = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{a} & \text {for}\: c = 0 \\- \frac {\left (-1\right )^{\frac {3}{4}} A \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}} A \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}} A \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}}} + \frac {\sqrt [4]{-1} B \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c} - \frac {\sqrt [4]{-1} B \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c} + \frac {\sqrt [4]{-1} B \sqrt [4]{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 )}}{c} + \frac {2 B \sqrt {x}}{c} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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