3.1031 \(\int \frac{(A+B x) \sqrt{a+b x+c x^2}}{x^{3/2}} \, dx\)

Optimal. Leaf size=341 \[ \frac{\left (2 \sqrt{a} \sqrt{c}+b\right ) \left (\sqrt{a}+\sqrt{c} x\right ) \left (\sqrt{a} B+3 A \sqrt{c}\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 \sqrt [4]{a} c^{3/4} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} (6 A c+b B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 c^{3/4} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{x} \sqrt{a+b x+c x^2} (6 A c+b B)}{3 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 (3 A-B x) \sqrt{a+b x+c x^2}}{3 \sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.315616, antiderivative size = 341, 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 = {812, 839, 1197, 1103, 1195} \[ \frac{\left (2 \sqrt{a} \sqrt{c}+b\right ) \left (\sqrt{a}+\sqrt{c} x\right ) \left (\sqrt{a} B+3 A \sqrt{c}\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 \sqrt [4]{a} c^{3/4} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} (6 A c+b B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 c^{3/4} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{x} \sqrt{a+b x+c x^2} (6 A c+b B)}{3 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 (3 A-B x) \sqrt{a+b x+c x^2}}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^(3/2),x]

[Out]

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

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*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] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 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}}{x^{3/2}} \, dx &=-\frac{2 (3 A-B x) \sqrt{a+b x+c x^2}}{3 \sqrt{x}}-\frac{2}{3} \int \frac{\frac{1}{2} (-3 A b-2 a B)-\frac{1}{2} (b B+6 A c) x}{\sqrt{x} \sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 (3 A-B x) \sqrt{a+b x+c x^2}}{3 \sqrt{x}}-\frac{4}{3} \operatorname{Subst}\left (\int \frac{\frac{1}{2} (-3 A b-2 a B)+\frac{1}{2} (-b B-6 A c) x^2}{\sqrt{a+b x^2+c x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 (3 A-B x) \sqrt{a+b x+c x^2}}{3 \sqrt{x}}+\frac{\left (2 \left (b+2 \sqrt{a} \sqrt{c}\right ) \left (\sqrt{a} B+3 A \sqrt{c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx,x,\sqrt{x}\right )}{3 \sqrt{c}}-\frac{\left (2 \sqrt{a} (b B+6 A 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{c}}\\ &=-\frac{2 (3 A-B x) \sqrt{a+b x+c x^2}}{3 \sqrt{x}}+\frac{2 (b B+6 A c) \sqrt{x} \sqrt{a+b x+c x^2}}{3 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 \sqrt [4]{a} (b B+6 A 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 c^{3/4} \sqrt{a+b x+c x^2}}+\frac{\left (b+2 \sqrt{a} \sqrt{c}\right ) \left (\sqrt{a} B+3 A \sqrt{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} c^{3/4} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 2.43451, size = 491, normalized size = 1.44 \[ \frac{\frac{i x \sqrt{\frac{4 a}{x \left (\sqrt{b^2-4 a c}+b\right )}+2} \sqrt{\frac{-x \sqrt{b^2-4 a c}+2 a+b x}{b x-x \sqrt{b^2-4 a c}}} \left (6 A c \sqrt{b^2-4 a c}+b B \sqrt{b^2-4 a c}+4 a B c+b^2 (-B)\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}{\sqrt{x}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )}{c \sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}-\frac{i x \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{4 a}{x \left (\sqrt{b^2-4 a c}+b\right )}+2} \sqrt{\frac{-x \sqrt{b^2-4 a c}+2 a+b x}{b x-x \sqrt{b^2-4 a c}}} (6 A c+b B) E\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a}{b+\sqrt{b^2-4 a c}}}}{\sqrt{x}}\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{c \sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}+\frac{4 (B x-3 A) (a+x (b+c x))}{\sqrt{x}}+\frac{4 (a+x (b+c x)) (6 A c+b B)}{c \sqrt{x}}}{6 \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^(3/2),x]

[Out]

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

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Maple [B]  time = 0.037, size = 1652, 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^(3/2),x)

[Out]

-1/3*(12*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))*a*c^2-3*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)*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))*b^2*c-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)*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)*b*c-24*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*c^2+6
*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^2*c+6*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*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)*Ellip
ticF(((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-4*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*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)*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+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^2-2*B*c^3*x^3+6*A*x^2*c^3-2*B*x^2*b*c^2+6*A*b*c^2*x-
2*B*a*c^2*x+6*a*A*c^2)/(c*x^2+b*x+a)^(1/2)/x^(1/2)/c^2

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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 )}}{x^{\frac{3}{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^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(3/2), 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 )}}{x^{\frac{3}{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^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(3/2), 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}}}{x^{\frac{3}{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**(3/2),x)

[Out]

Integral((A + B*x)*sqrt(a + b*x + c*x**2)/x**(3/2), 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 )}}{x^{\frac{3}{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^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(3/2), x)