3.1032 \(\int \frac{(A+B x) \sqrt{a+b x+c x^2}}{x^{5/2}} \, dx\)

Optimal. Leaf size=353 \[ \frac{\left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\sqrt{c} (6 a B+A b)+\sqrt{a} (2 A c+3 b B)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{3/4} \sqrt [4]{c} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x\right ) (6 a B+A b) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{3/4} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{c} \sqrt{x} (6 a B+A b) \sqrt{a+b x+c x^2}}{3 a \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 \sqrt{a+b x+c x^2} (x (3 a B+A b)+a A)}{3 a x^{3/2}} \]

[Out]

(-2*(a*A + (A*b + 3*a*B)*x)*Sqrt[a + b*x + c*x^2])/(3*a*x^(3/2)) + (2*(A*b + 6*a*B)*Sqrt[c]*Sqrt[x]*Sqrt[a + b
*x + c*x^2])/(3*a*(Sqrt[a] + Sqrt[c]*x)) - (2*(A*b + 6*a*B)*c^(1/4)*(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])/(3*a^
(3/4)*Sqrt[a + b*x + c*x^2]) + (((A*b + 6*a*B)*Sqrt[c] + Sqrt[a]*(3*b*B + 2*A*c))*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(
a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[
c]))/4])/(3*a^(3/4)*c^(1/4)*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.333279, antiderivative size = 353, 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 = {810, 839, 1197, 1103, 1195} \[ \frac{\left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\sqrt{c} (6 a B+A b)+\sqrt{a} (2 A c+3 b B)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{3/4} \sqrt [4]{c} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x\right ) (6 a B+A b) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{3/4} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{c} \sqrt{x} (6 a B+A b) \sqrt{a+b x+c x^2}}{3 a \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 \sqrt{a+b x+c x^2} (x (3 a B+A b)+a A)}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^(5/2),x]

[Out]

(-2*(a*A + (A*b + 3*a*B)*x)*Sqrt[a + b*x + c*x^2])/(3*a*x^(3/2)) + (2*(A*b + 6*a*B)*Sqrt[c]*Sqrt[x]*Sqrt[a + b
*x + c*x^2])/(3*a*(Sqrt[a] + Sqrt[c]*x)) - (2*(A*b + 6*a*B)*c^(1/4)*(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])/(3*a^
(3/4)*Sqrt[a + b*x + c*x^2]) + (((A*b + 6*a*B)*Sqrt[c] + Sqrt[a]*(3*b*B + 2*A*c))*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(
a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[
c]))/4])/(3*a^(3/4)*c^(1/4)*Sqrt[a + b*x + c*x^2])

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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

Mathematica [C]  time = 2.84854, size = 499, normalized size = 1.41 \[ \frac{-4 (a+x (b+c x)) (a (A+3 B x)+A b x)+\frac{x \left (i x^{3/2} \sqrt{\frac{2 a}{x \left (\sqrt{b^2-4 a c}+b\right )}+1} \sqrt{\frac{-2 x \sqrt{b^2-4 a c}+4 a+2 b x}{b x-x \sqrt{b^2-4 a c}}} \left (A \left (b \sqrt{b^2-4 a c}+4 a c-b^2\right )+6 a B \sqrt{b^2-4 a c}\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}{\sqrt{x}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+i x^{3/2} \left (b-\sqrt{b^2-4 a c}\right ) (6 a B+A b) \sqrt{\frac{2 a}{x \left (\sqrt{b^2-4 a c}+b\right )}+1} \sqrt{\frac{-2 x \sqrt{b^2-4 a c}+4 a+2 b x}{b x-x \sqrt{b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a}{b+\sqrt{b^2-4 a c}}}}{\sqrt{x}}\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )+4 \sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}} (6 a B+A b) (a+x (b+c x))\right )}{\sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}}{6 a x^{3/2} \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^(5/2),x]

[Out]

(-4*(A*b*x + a*(A + 3*B*x))*(a + x*(b + c*x)) + (x*(4*(A*b + 6*a*B)*Sqrt[a/(b + Sqrt[b^2 - 4*a*c])]*(a + x*(b
+ c*x)) + I*(A*b + 6*a*B)*(b - Sqrt[b^2 - 4*a*c])*Sqrt[1 + (2*a)/((b + Sqrt[b^2 - 4*a*c])*x)]*x^(3/2)*Sqrt[(4*
a + 2*b*x - 2*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])] + I*(6*a*B*Sqrt[b^2 - 4*a*c] + A*(-
b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c]))*Sqrt[1 + (2*a)/((b + Sqrt[b^2 - 4*a*c])*x)]*x^(3/2)*Sqrt[(4*a + 2*b*x - 2*
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])])/(6*a*x^(3/2)*S
qrt[a + x*(b + c*x)])

________________________________________________________________________________________

Maple [B]  time = 0.036, size = 1687, normalized size = 4.8 \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^(5/2),x)

[Out]

1/3*(2*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)*x*a*c+4*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))*x*a*b*c-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))*x*b^3-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)*x*b^2-12*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))*x*a^2*c+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))*x*a*b^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)*E
llipticF(((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)*x*a*b+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))*x*a^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))*x*a*b^2-6*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)*x*a*b-2*A*x^3*b*c^2-6*a*B*c^2*x^3-2*a
*A*c^2*x^2-2*A*x^2*b^2*c-6*B*x^2*a*b*c-4*A*a*b*c*x-6*a^2*B*c*x-2*A*a^2*c)/(c*x^2+b*x+a)^(1/2)/x^(3/2)/a/c

________________________________________________________________________________________

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 )}}{x^{\frac{5}{2}}}\,{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^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(5/2), x)

________________________________________________________________________________________

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 )}}{x^{\frac{5}{2}}}, 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^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(5/2), x)

________________________________________________________________________________________

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}}}{x^{\frac{5}{2}}}\, 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**(5/2),x)

[Out]

Integral((A + B*x)*sqrt(a + b*x + c*x**2)/x**(5/2), x)

________________________________________________________________________________________

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 )}}{x^{\frac{5}{2}}}\,{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^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(5/2), x)