3.1030 \(\int \frac{(A+B x) \sqrt{a+b x+c x^2}}{\sqrt{x}} \, dx\)

Optimal. Leaf size=373 \[ -\frac{\sqrt [4]{a} \left (2 \sqrt{a} \sqrt{c}+b\right ) \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (-3 \sqrt{a} B \sqrt{c}-5 A c+2 b B\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 c^{7/4} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{x} \sqrt{a+b x+c x^2} \left (-6 a B c-5 A b c+2 b^2 B\right )}{15 c^{3/2} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 \sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (-6 a B c-5 A b c+2 b^2 B\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 c^{7/4} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{x} \sqrt{a+b x+c x^2} (5 A c+b B+3 B c x)}{15 c} \]

[Out]

(-2*(2*b^2*B - 5*A*b*c - 6*a*B*c)*Sqrt[x]*Sqrt[a + b*x + c*x^2])/(15*c^(3/2)*(Sqrt[a] + Sqrt[c]*x)) + (2*Sqrt[
x]*(b*B + 5*A*c + 3*B*c*x)*Sqrt[a + b*x + c*x^2])/(15*c) + (2*a^(1/4)*(2*b^2*B - 5*A*b*c - 6*a*B*c)*(Sqrt[a] +
Sqrt[c]*x)*Sqrt[(a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2
- b/(Sqrt[a]*Sqrt[c]))/4])/(15*c^(7/4)*Sqrt[a + b*x + c*x^2]) - (a^(1/4)*(b + 2*Sqrt[a]*Sqrt[c])*(2*b*B - 3*Sq
rt[a]*B*Sqrt[c] - 5*A*c)*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*Arc
Tan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(15*c^(7/4)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.345108, antiderivative size = 373, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {814, 839, 1197, 1103, 1195} \[ -\frac{2 \sqrt{x} \sqrt{a+b x+c x^2} \left (-6 a B c-5 A b c+2 b^2 B\right )}{15 c^{3/2} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 \sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (-6 a B c-5 A b c+2 b^2 B\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 c^{7/4} \sqrt{a+b x+c x^2}}-\frac{\sqrt [4]{a} \left (2 \sqrt{a} \sqrt{c}+b\right ) \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (-3 \sqrt{a} B \sqrt{c}-5 A c+2 b B\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 c^{7/4} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{x} \sqrt{a+b x+c x^2} (5 A c+b B+3 B c x)}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*b^2*B - 5*A*b*c - 6*a*B*c)*Sqrt[x]*Sqrt[a + b*x + c*x^2])/(15*c^(3/2)*(Sqrt[a] + Sqrt[c]*x)) + (2*Sqrt[
x]*(b*B + 5*A*c + 3*B*c*x)*Sqrt[a + b*x + c*x^2])/(15*c) + (2*a^(1/4)*(2*b^2*B - 5*A*b*c - 6*a*B*c)*(Sqrt[a] +
Sqrt[c]*x)*Sqrt[(a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2
- b/(Sqrt[a]*Sqrt[c]))/4])/(15*c^(7/4)*Sqrt[a + b*x + c*x^2]) - (a^(1/4)*(b + 2*Sqrt[a]*Sqrt[c])*(2*b*B - 3*Sq
rt[a]*B*Sqrt[c] - 5*A*c)*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*Arc
Tan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(15*c^(7/4)*Sqrt[a + b*x + c*x^2])

Rule 814

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

Rule 839

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f +
g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{a+b x+c x^2}}{\sqrt{x}} \, dx &=\frac{2 \sqrt{x} (b B+5 A c+3 B c x) \sqrt{a+b x+c x^2}}{15 c}-\frac{2 \int \frac{\frac{1}{2} a (b B-10 A c)+\frac{1}{2} \left (2 b^2 B-5 A b c-6 a B c\right ) x}{\sqrt{x} \sqrt{a+b x+c x^2}} \, dx}{15 c}\\ &=\frac{2 \sqrt{x} (b B+5 A c+3 B c x) \sqrt{a+b x+c x^2}}{15 c}-\frac{4 \operatorname{Subst}\left (\int \frac{\frac{1}{2} a (b B-10 A c)+\frac{1}{2} \left (2 b^2 B-5 A b c-6 a B c\right ) x^2}{\sqrt{a+b x^2+c x^4}} \, dx,x,\sqrt{x}\right )}{15 c}\\ &=\frac{2 \sqrt{x} (b B+5 A c+3 B c x) \sqrt{a+b x+c x^2}}{15 c}-\frac{\left (2 \sqrt{a} \left (b+2 \sqrt{a} \sqrt{c}\right ) \left (2 b B-3 \sqrt{a} B \sqrt{c}-5 A c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx,x,\sqrt{x}\right )}{15 c^{3/2}}+\frac{\left (2 \sqrt{a} \left (2 b^2 B-5 A b c-6 a B c\right )\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx,x,\sqrt{x}\right )}{15 c^{3/2}}\\ &=-\frac{2 \left (2 b^2 B-5 A b c-6 a B c\right ) \sqrt{x} \sqrt{a+b x+c x^2}}{15 c^{3/2} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 \sqrt{x} (b B+5 A c+3 B c x) \sqrt{a+b x+c x^2}}{15 c}+\frac{2 \sqrt [4]{a} \left (2 b^2 B-5 A b c-6 a B c\right ) \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 c^{7/4} \sqrt{a+b x+c x^2}}-\frac{\sqrt [4]{a} \left (b+2 \sqrt{a} \sqrt{c}\right ) \left (2 b B-3 \sqrt{a} B \sqrt{c}-5 A c\right ) \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 c^{7/4} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 4.01687, size = 550, normalized size = 1.47 \[ \frac{\frac{4 \sqrt{x} (a+x (b+c x)) (5 A c+b B+3 B c x)}{c}+\frac{x \left (\frac{i \sqrt{\frac{4 a}{x \left (\sqrt{b^2-4 a c}+b\right )}+2} \sqrt{\frac{-x \sqrt{b^2-4 a c}+2 a+b x}{b x-x \sqrt{b^2-4 a c}}} \left (-b^2 \left (2 B \sqrt{b^2-4 a c}+5 A c\right )+b \left (5 A c \sqrt{b^2-4 a c}-8 a B c\right )+2 a c \left (3 B \sqrt{b^2-4 a c}+10 A c\right )+2 b^3 B\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}{\sqrt{x}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}-\frac{4 (a+x (b+c x)) \left (-6 a B c-5 A b c+2 b^2 B\right )}{x^{3/2}}+\frac{i \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{4 a}{x \left (\sqrt{b^2-4 a c}+b\right )}+2} \sqrt{\frac{-x \sqrt{b^2-4 a c}+2 a+b x}{b x-x \sqrt{b^2-4 a c}}} \left (-6 a B c-5 A b c+2 b^2 B\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a}{b+\sqrt{b^2-4 a c}}}}{\sqrt{x}}\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}\right )}{c^2}}{30 \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*Sqrt[x]*(b*B + 5*A*c + 3*B*c*x)*(a + x*(b + c*x)))/c + (x*((-4*(2*b^2*B - 5*A*b*c - 6*a*B*c)*(a + x*(b + c
*x)))/x^(3/2) + (I*(2*b^2*B - 5*A*b*c - 6*a*B*c)*(-b + Sqrt[b^2 - 4*a*c])*Sqrt[2 + (4*a)/((b + Sqrt[b^2 - 4*a*
c])*x)]*Sqrt[(2*a + b*x - Sqrt[b^2 - 4*a*c]*x)/(b*x - Sqrt[b^2 - 4*a*c]*x)]*EllipticE[I*ArcSinh[(Sqrt[2]*Sqrt[
a/(b + Sqrt[b^2 - 4*a*c])])/Sqrt[x]], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/Sqrt[a/(b + Sqrt[b^2 -
4*a*c])] + (I*(2*b^3*B - b^2*(5*A*c + 2*B*Sqrt[b^2 - 4*a*c]) + 2*a*c*(10*A*c + 3*B*Sqrt[b^2 - 4*a*c]) + b*(-8
*a*B*c + 5*A*c*Sqrt[b^2 - 4*a*c]))*Sqrt[2 + (4*a)/((b + Sqrt[b^2 - 4*a*c])*x)]*Sqrt[(2*a + b*x - Sqrt[b^2 - 4*
a*c]*x)/(b*x - Sqrt[b^2 - 4*a*c]*x)]*EllipticF[I*ArcSinh[(Sqrt[2]*Sqrt[a/(b + Sqrt[b^2 - 4*a*c])])/Sqrt[x]], (
b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/Sqrt[a/(b + Sqrt[b^2 - 4*a*c])]))/c^2)/(30*Sqrt[a + x*(b + c*
x)])

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Maple [B]  time = 0.036, size = 2012, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(10*A*((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+
b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticF(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^
(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*(-4*a*c+b^2)^(1/2)*a*c^2+20*A*((2
*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1
/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticE(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),
1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*a*b*c^2-5*A*((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-
4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2))
)^(1/2)*EllipticE(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/
2))/(-4*a*c+b^2)^(1/2))^(1/2))*b^3*c-5*A*((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+
(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticE(((2*c*x+(-4*a*c+
b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*(-4
*a*c+b^2)^(1/2)*b^2*c-12*B*((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(
1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticF(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/
(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*a^2*c^2+3*B*((2*c
*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2
)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticF(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/
2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*a*b^2*c-B*((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*
c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1
/2)*EllipticF(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/
(-4*a*c+b^2)^(1/2))^(1/2))*(-4*a*c+b^2)^(1/2)*a*b*c+24*B*((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))
^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticE(
((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(
1/2))^(1/2))*a^2*c^2-14*B*((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1
/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticE(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(
b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*a*b^2*c-6*B*((2*c*
x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)
*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticE(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2
*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*(-4*a*c+b^2)^(1/2)*a*b*c+2*B*((2*c*x+(-4*a*c+b^2)^
(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a
*c+b^2)^(1/2)))^(1/2)*EllipticE(((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-
4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*b^4+2*B*((2*c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/
2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(-c*x/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*EllipticE(((2*
c*x+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)
)^(1/2))*(-4*a*c+b^2)^(1/2)*b^3+6*B*x^4*c^4+10*A*x^3*c^4+8*B*x^3*b*c^3+10*A*x^2*b*c^3+6*B*x^2*a*c^3+2*B*x^2*b^
2*c^2+10*A*x*a*c^3+2*B*x*a*b*c^2)/(c*x^2+b*x+a)^(1/2)/x^(1/2)/c^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x + a}{\left (B x + A\right )}}{\sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)/sqrt(x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + b x + a}{\left (B x + A\right )}}{\sqrt{x}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*(B*x + A)/sqrt(x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \sqrt{a + b x + c x^{2}}}{\sqrt{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*sqrt(a + b*x + c*x**2)/sqrt(x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x + a}{\left (B x + A\right )}}{\sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)/sqrt(x), x)