3.1033 \(\int \frac{(A+B x) \sqrt{a+b x+c x^2}}{x^{7/2}} \, dx\)
Optimal. Leaf size=421 \[ -\frac{\sqrt [4]{c} \left (2 \sqrt{a} \sqrt{c}+b\right ) \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (-3 \sqrt{a} A \sqrt{c}-5 a B+2 A b\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 a^{7/4} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{a+b x+c x^2} \left (-6 a A c-5 a b B+2 A b^2\right )}{15 a^2 \sqrt{x}}+\frac{2 \sqrt{c} \sqrt{x} \sqrt{a+b x+c x^2} \left (5 a b B-2 A \left (b^2-3 a c\right )\right )}{15 a^2 \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (5 a b B-2 A \left (b^2-3 a c\right )\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 a^{7/4} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{a+b x+c x^2} (x (5 a B+A b)+3 a A)}{15 a x^{5/2}} \]
[Out]
(2*(2*A*b^2 - 5*a*b*B - 6*a*A*c)*Sqrt[a + b*x + c*x^2])/(15*a^2*Sqrt[x]) - (2*(3*a*A + (A*b + 5*a*B)*x)*Sqrt[a
+ b*x + c*x^2])/(15*a*x^(5/2)) + (2*Sqrt[c]*(5*a*b*B - 2*A*(b^2 - 3*a*c))*Sqrt[x]*Sqrt[a + b*x + c*x^2])/(15*
a^2*(Sqrt[a] + Sqrt[c]*x)) - (2*c^(1/4)*(5*a*b*B - 2*A*(b^2 - 3*a*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
*a^(7/4)*Sqrt[a + b*x + c*x^2]) - ((b + 2*Sqrt[a]*Sqrt[c])*(2*A*b - 5*a*B - 3*Sqrt[a]*A*Sqrt[c])*c^(1/4)*(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])/(15*a^(7/4)*Sqrt[a + b*x + c*x^2])
________________________________________________________________________________________
Rubi [A] time = 0.438679, antiderivative size = 421, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used =
{810, 834, 839, 1197, 1103, 1195} \[ \frac{2 \sqrt{a+b x+c x^2} \left (-6 a A c-5 a b B+2 A b^2\right )}{15 a^2 \sqrt{x}}+\frac{2 \sqrt{c} \sqrt{x} \sqrt{a+b x+c x^2} \left (5 a b B-2 A \left (b^2-3 a c\right )\right )}{15 a^2 \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (5 a b B-2 A \left (b^2-3 a c\right )\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 a^{7/4} \sqrt{a+b x+c x^2}}-\frac{\sqrt [4]{c} \left (2 \sqrt{a} \sqrt{c}+b\right ) \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+b x+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (-3 \sqrt{a} A \sqrt{c}-5 a B+2 A b\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 a^{7/4} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{a+b x+c x^2} (x (5 a B+A b)+3 a A)}{15 a x^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^(7/2),x]
[Out]
(2*(2*A*b^2 - 5*a*b*B - 6*a*A*c)*Sqrt[a + b*x + c*x^2])/(15*a^2*Sqrt[x]) - (2*(3*a*A + (A*b + 5*a*B)*x)*Sqrt[a
+ b*x + c*x^2])/(15*a*x^(5/2)) + (2*Sqrt[c]*(5*a*b*B - 2*A*(b^2 - 3*a*c))*Sqrt[x]*Sqrt[a + b*x + c*x^2])/(15*
a^2*(Sqrt[a] + Sqrt[c]*x)) - (2*c^(1/4)*(5*a*b*B - 2*A*(b^2 - 3*a*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
*a^(7/4)*Sqrt[a + b*x + c*x^2]) - ((b + 2*Sqrt[a]*Sqrt[c])*(2*A*b - 5*a*B - 3*Sqrt[a]*A*Sqrt[c])*c^(1/4)*(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])/(15*a^(7/4)*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 834
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (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^{7/2}} \, dx &=-\frac{2 (3 a A+(A b+5 a B) x) \sqrt{a+b x+c x^2}}{15 a x^{5/2}}-\frac{2 \int \frac{\frac{1}{2} \left (-5 a b B+2 A \left (b^2-3 a c\right )\right )+\frac{1}{2} (A b-10 a B) c x}{x^{3/2} \sqrt{a+b x+c x^2}} \, dx}{15 a}\\ &=\frac{2 \left (2 A b^2-5 a b B-6 a A c\right ) \sqrt{a+b x+c x^2}}{15 a^2 \sqrt{x}}-\frac{2 (3 a A+(A b+5 a B) x) \sqrt{a+b x+c x^2}}{15 a x^{5/2}}+\frac{4 \int \frac{-\frac{1}{4} a (A b-10 a B) c+\frac{1}{4} c \left (5 a b B-2 A \left (b^2-3 a c\right )\right ) x}{\sqrt{x} \sqrt{a+b x+c x^2}} \, dx}{15 a^2}\\ &=\frac{2 \left (2 A b^2-5 a b B-6 a A c\right ) \sqrt{a+b x+c x^2}}{15 a^2 \sqrt{x}}-\frac{2 (3 a A+(A b+5 a B) x) \sqrt{a+b x+c x^2}}{15 a x^{5/2}}+\frac{8 \operatorname{Subst}\left (\int \frac{-\frac{1}{4} a (A b-10 a B) c+\frac{1}{4} c \left (5 a b B-2 A \left (b^2-3 a c\right )\right ) x^2}{\sqrt{a+b x^2+c x^4}} \, dx,x,\sqrt{x}\right )}{15 a^2}\\ &=\frac{2 \left (2 A b^2-5 a b B-6 a A c\right ) \sqrt{a+b x+c x^2}}{15 a^2 \sqrt{x}}-\frac{2 (3 a A+(A b+5 a B) x) \sqrt{a+b x+c x^2}}{15 a x^{5/2}}-\frac{\left (2 \left (b+2 \sqrt{a} \sqrt{c}\right ) \left (2 A b-5 a B-3 \sqrt{a} A \sqrt{c}\right ) \sqrt{c}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx,x,\sqrt{x}\right )}{15 a^{3/2}}-\frac{\left (2 \sqrt{c} \left (5 a b B-2 A \left (b^2-3 a c\right )\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 a^{3/2}}\\ &=\frac{2 \left (2 A b^2-5 a b B-6 a A c\right ) \sqrt{a+b x+c x^2}}{15 a^2 \sqrt{x}}-\frac{2 (3 a A+(A b+5 a B) x) \sqrt{a+b x+c x^2}}{15 a x^{5/2}}-\frac{2 \sqrt{c} \left (2 A b^2-5 a b B-6 a A c\right ) \sqrt{x} \sqrt{a+b x+c x^2}}{15 a^2 \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 \sqrt [4]{c} \left (2 A b^2-5 a b B-6 a 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}} 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 a^{7/4} \sqrt{a+b x+c x^2}}-\frac{\left (b+2 \sqrt{a} \sqrt{c}\right ) \left (2 A b-5 a B-3 \sqrt{a} A \sqrt{c}\right ) \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}} 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 a^{7/4} \sqrt{a+b x+c x^2}}\\ \end{align*}
Mathematica [C] time = 4.08669, size = 576, normalized size = 1.37 \[ \frac{-4 (a+x (b+c x)) \left (a^2 (3 A+5 B x)+a x (A (b+6 c x)+5 b B x)-2 A b^2 x^2\right )+\frac{x^2 \left (-i x^{3/2} \sqrt{\frac{2 a}{x \left (\sqrt{b^2-4 a c}+b\right )}+1} \sqrt{\frac{-2 x \sqrt{b^2-4 a c}+4 a+2 b x}{b x-x \sqrt{b^2-4 a c}}} \left (2 A \left (b^2 \sqrt{b^2-4 a c}-3 a c \sqrt{b^2-4 a c}+4 a b c-b^3\right )+5 a B \left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right )\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}{\sqrt{x}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+i x^{3/2} \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{2 a}{x \left (\sqrt{b^2-4 a c}+b\right )}+1} \sqrt{\frac{-2 x \sqrt{b^2-4 a c}+4 a+2 b x}{b x-x \sqrt{b^2-4 a c}}} \left (2 A \left (b^2-3 a c\right )-5 a b B\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a}{b+\sqrt{b^2-4 a c}}}}{\sqrt{x}}\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )+4 \sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}} (a+x (b+c x)) \left (6 a A c+5 a b B-2 A b^2\right )\right )}{\sqrt{\frac{a}{\sqrt{b^2-4 a c}+b}}}}{30 a^2 x^{5/2} \sqrt{a+x (b+c x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^(7/2),x]
[Out]
(-4*(a + x*(b + c*x))*(-2*A*b^2*x^2 + a^2*(3*A + 5*B*x) + a*x*(5*b*B*x + A*(b + 6*c*x))) + (x^2*(4*(-2*A*b^2 +
5*a*b*B + 6*a*A*c)*Sqrt[a/(b + Sqrt[b^2 - 4*a*c])]*(a + x*(b + c*x)) + I*(-b + Sqrt[b^2 - 4*a*c])*(-5*a*b*B +
2*A*(b^2 - 3*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*(5*a*B*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) + 2*A*(-b^3 + 4*a*
b*c + b^2*Sqrt[b^2 - 4*a*c] - 3*a*c*Sqrt[b^2 - 4*a*c]))*Sqrt[1 + (2*a)/((b + Sqrt[b^2 - 4*a*c])*x)]*x^(3/2)*Sq
rt[(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])])/(30*a^2*x^(5/2)*Sqrt[a + x*(b + c*x)])
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Maple [B] time = 0.039, size = 2109, normalized size = 5. \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^(7/2),x)
[Out]
1/15*(-6*A*a^3*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)*x^2*
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)*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^2*a*b*c+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)*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^2*b^4-10*B*x^4*a*b*c^2-10*a^3*B*c*x-18*a^2*A*
c^2*x^2-10*a^2*B*c^2*x^3+4*A*x^4*b^2*c^2+4*A*x^3*b^3*c-12*a*A*c^3*x^4-14*A*x^3*a*b*c^2-10*B*x^3*a*b^2*c+2*A*x^
2*a*b^2*c-20*B*x^2*a^2*b*c-8*A*x*a^2*b*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))
*x^2*a*b^2*c-14*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^2*a*b^2*c+10*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)*x^2*a^2*c-5*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^2*a*b^2+20*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^2*a^2*b*c-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))*x^2*a^2*c^2+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)*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^2*b^3+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))*x^2*a^2*c^2-5*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^2*a*b^3)/(c*x^2+b*x+a)^(1/2)/x^(5/2)/a^2/c
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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{7}{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^(7/2),x, algorithm="maxima")
[Out]
integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(7/2), 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{7}{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^(7/2),x, algorithm="fricas")
[Out]
integral(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(7/2), 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{7}{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**(7/2),x)
[Out]
Integral((A + B*x)*sqrt(a + b*x + c*x**2)/x**(7/2), 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{7}{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^(7/2),x, algorithm="giac")
[Out]
integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)/x^(7/2), x)