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

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

[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))

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Rubi [A]  time = 0.234568, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {825, 827, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\sqrt{a} B+A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{a} c^{5/4}}-\frac{\left (\sqrt{a} B+A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{a} c^{5/4}}+\frac{\left (\sqrt{a} B-A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} \sqrt [4]{a} c^{5/4}}-\frac{\left (\sqrt{a} B-A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} \sqrt [4]{a} c^{5/4}}+\frac{2 B \sqrt{x}}{c} \]

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 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 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)]

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 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 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 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 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]

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*}

Mathematica [A]  time = 0.0664204, size = 266, normalized size = 1. \[ \frac{\sqrt{2} a^{5/4} B \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )-\sqrt{2} a^{5/4} B \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )+2 \sqrt{2} a^{5/4} B \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )-2 \sqrt{2} a^{5/4} B \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )-4 (-a)^{3/4} A \sqrt{c} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )+4 (-a)^{3/4} A \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )+8 a B \sqrt [4]{c} \sqrt{x}}{4 a c^{5/4}} \]

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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Maple [A]  time = 0.006, size = 277, normalized size = 1.1 \begin{align*} 2\,{\frac{B\sqrt{x}}{c}}-{\frac{B\sqrt{2}}{2\,c}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }-{\frac{B\sqrt{2}}{2\,c}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }-{\frac{B\sqrt{2}}{4\,c}\sqrt [4]{{\frac{a}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ) }+{\frac{A\sqrt{2}}{4\,c}\ln \left ({ \left ( x-\sqrt [4]{{\frac{a}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ( x+\sqrt [4]{{\frac{a}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+{\frac{A\sqrt{2}}{2\,c}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+{\frac{A\sqrt{2}}{2\,c}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}} \end{align*}

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*x^(1/2)/c-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)*ar
ctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)-1/4/c*B*(a/c)^(1/4)*2^(1/2)*ln((x+(a/c)^(1/4)*x^(1/2)*2^(1/2)+(a/c)^(1/2))
/(x-(a/c)^(1/4)*x^(1/2)*2^(1/2)+(a/c)^(1/2)))+1/4/c*A/(a/c)^(1/4)*2^(1/2)*ln((x-(a/c)^(1/4)*x^(1/2)*2^(1/2)+(a
/c)^(1/2))/(x+(a/c)^(1/4)*x^(1/2)*2^(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.53447, size = 1530, normalized size = 5.77 \begin{align*} \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{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.28881, size = 350, normalized size = 1.32 \begin{align*} \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 )} \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^{3} - \left (a c^{3}\right )^{\frac{3}{4}} A c^{2}\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^{5}} - \frac{\sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} B a c^{3} + \left (a c^{3}\right )^{\frac{3}{4}} A c^{2}\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{a}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{c}}\right )}{4 \, a c^{5}} \end{align*}

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