3.1646 \(\int \frac{(b+2 c x) \sqrt{d+e x}}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=216 \[ \frac{2 \sqrt{2} e \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 e \sqrt{b^2-4 a c}}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{c \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{d+e x}}{\sqrt{a+b x+c x^2}} \]

[Out]

(-2*Sqrt[d + e*x])/Sqrt[a + b*x + c*x^2] + (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqr
t[b^2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c]
+ 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(c*Sqrt[d
+ e*x]*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.109953, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {768, 718, 419} \[ \frac{2 \sqrt{2} e \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{c \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{d+e x}}{\sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x])/Sqrt[a + b*x + c*x^2] + (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqr
t[b^2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c]
+ 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(c*Sqrt[d
+ e*x]*Sqrt[a + b*x + c*x^2])

Rule 768

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

Rule 718

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

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rubi steps

\begin{align*} \int \frac{(b+2 c x) \sqrt{d+e x}}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 \sqrt{d+e x}}{\sqrt{a+b x+c x^2}}+e \int \frac{1}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 \sqrt{d+e x}}{\sqrt{a+b x+c x^2}}+\frac{\left (2 \sqrt{2} \sqrt{b^2-4 a c} e \sqrt{\frac{c (d+e x)}{2 c d-b e-\sqrt{b^2-4 a c} e}} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+\frac{2 \sqrt{b^2-4 a c} e x^2}{2 c d-b e-\sqrt{b^2-4 a c} e}}} \, dx,x,\frac{\sqrt{\frac{b+\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )}{c \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ &=-\frac{2 \sqrt{d+e x}}{\sqrt{a+b x+c x^2}}+\frac{2 \sqrt{2} \sqrt{b^2-4 a c} e \sqrt{\frac{c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{c \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 2.48523, size = 318, normalized size = 1.47 \[ \frac{-2 \sqrt{d+e x}+\frac{i (d+e x) \sqrt{2-\frac{4 \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (\sqrt{e^2 \left (b^2-4 a c\right )}-b e+2 c d\right )}} \sqrt{\frac{2 \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (\sqrt{e^2 \left (b^2-4 a c\right )}+b e-2 c d\right )}+1} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a e^2-b d e+c d^2}{\sqrt{e^2 \left (b^2-4 a c\right )}+b e-2 c d}}}{\sqrt{d+e x}}\right ),-\frac{\sqrt{e^2 \left (b^2-4 a c\right )}+b e-2 c d}{\sqrt{e^2 \left (b^2-4 a c\right )}-b e+2 c d}\right )}{\sqrt{\frac{e (a e-b d)+c d^2}{\sqrt{e^2 \left (b^2-4 a c\right )}+b e-2 c d}}}}{\sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x] + (I*(d + e*x)*Sqrt[2 - (4*(c*d^2 + e*(-(b*d) + a*e)))/((2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^
2])*(d + e*x))]*Sqrt[1 + (2*(c*d^2 + e*(-(b*d) + a*e)))/((-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*
EllipticF[I*ArcSinh[(Sqrt[2]*Sqrt[(c*d^2 - b*d*e + a*e^2)/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])])/Sqrt[d +
e*x]], -((-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2]))])/Sqrt[(c*d^2 + e*(
-(b*d) + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])])/Sqrt[a + x*(b + c*x)]

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Maple [B]  time = 0.044, size = 705, normalized size = 3.3 \begin{align*}{\frac{1}{ \left ( ce{x}^{3}+be{x}^{2}+cd{x}^{2}+aex+bdx+ad \right ) c} \left ( -\sqrt{2}{\it EllipticF} \left ( \sqrt{2}\sqrt{-{ \left ( ex+d \right ) c \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}},\sqrt{-{ \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd \right ) ^{-1}}} \right ) be\sqrt{-{ \left ( ex+d \right ) c \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}}\sqrt{{e \left ( -2\,cx+\sqrt{-4\,ac+{b}^{2}}-b \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd \right ) ^{-1}}}\sqrt{{e \left ( 2\,cx+\sqrt{-4\,ac+{b}^{2}}+b \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}}+2\,\sqrt{2}{\it EllipticF} \left ( \sqrt{2}\sqrt{-{\frac{ \left ( ex+d \right ) c}{e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd}}},\sqrt{-{\frac{e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd}{e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd}}} \right ) cd\sqrt{-{\frac{ \left ( ex+d \right ) c}{e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd}}}\sqrt{{\frac{ \left ( -2\,cx+\sqrt{-4\,ac+{b}^{2}}-b \right ) e}{e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd}}}\sqrt{{\frac{ \left ( 2\,cx+\sqrt{-4\,ac+{b}^{2}}+b \right ) e}{e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd}}}-\sqrt{2}{\it EllipticF} \left ( \sqrt{2}\sqrt{-{ \left ( ex+d \right ) c \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}},\sqrt{-{ \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd \right ) ^{-1}}} \right ) e\sqrt{-4\,ac+{b}^{2}}\sqrt{-{ \left ( ex+d \right ) c \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}}\sqrt{{e \left ( -2\,cx+\sqrt{-4\,ac+{b}^{2}}-b \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd \right ) ^{-1}}}\sqrt{{e \left ( 2\,cx+\sqrt{-4\,ac+{b}^{2}}+b \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}}-2\,cex-2\,cd \right ) \sqrt{ex+d}\sqrt{c{x}^{2}+bx+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(e*x+d)^(1/2)/(c*x^2+b*x+a)^(3/2),x)

[Out]

(-2^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)*c/(e*(-4*a*c+b^2)^(1/2)+b*e-2*c*d))^(1/2),(-(e*(-4*a*c+b^2)^(1/2)+b*e-2*
c*d)/(e*(-4*a*c+b^2)^(1/2)-b*e+2*c*d))^(1/2))*b*e*(-(e*x+d)*c/(e*(-4*a*c+b^2)^(1/2)+b*e-2*c*d))^(1/2)*((-2*c*x
+(-4*a*c+b^2)^(1/2)-b)*e/(e*(-4*a*c+b^2)^(1/2)-b*e+2*c*d))^(1/2)*((2*c*x+(-4*a*c+b^2)^(1/2)+b)*e/(e*(-4*a*c+b^
2)^(1/2)+b*e-2*c*d))^(1/2)+2*2^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)*c/(e*(-4*a*c+b^2)^(1/2)+b*e-2*c*d))^(1/2),(-(
e*(-4*a*c+b^2)^(1/2)+b*e-2*c*d)/(e*(-4*a*c+b^2)^(1/2)-b*e+2*c*d))^(1/2))*c*d*(-(e*x+d)*c/(e*(-4*a*c+b^2)^(1/2)
+b*e-2*c*d))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)*e/(e*(-4*a*c+b^2)^(1/2)-b*e+2*c*d))^(1/2)*((2*c*x+(-4*a*c+b^
2)^(1/2)+b)*e/(e*(-4*a*c+b^2)^(1/2)+b*e-2*c*d))^(1/2)-2^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)*c/(e*(-4*a*c+b^2)^(1
/2)+b*e-2*c*d))^(1/2),(-(e*(-4*a*c+b^2)^(1/2)+b*e-2*c*d)/(e*(-4*a*c+b^2)^(1/2)-b*e+2*c*d))^(1/2))*e*(-4*a*c+b^
2)^(1/2)*(-(e*x+d)*c/(e*(-4*a*c+b^2)^(1/2)+b*e-2*c*d))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)*e/(e*(-4*a*c+b^2)^
(1/2)-b*e+2*c*d))^(1/2)*((2*c*x+(-4*a*c+b^2)^(1/2)+b)*e/(e*(-4*a*c+b^2)^(1/2)+b*e-2*c*d))^(1/2)-2*c*e*x-2*c*d)
*(e*x+d)^(1/2)*(c*x^2+b*x+a)^(1/2)/(c*e*x^3+b*e*x^2+c*d*x^2+a*e*x+b*d*x+a*d)/c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c x + b\right )} \sqrt{e x + d}}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(1/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((2*c*x + b)*sqrt(e*x + d)/(c*x^2 + b*x + a)^(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 (2 \, c x + b\right )} \sqrt{e x + d}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(1/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(e*x + d)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 +
a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b + 2 c x\right ) \sqrt{d + e x}}{\left (a + b x + c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)**(1/2)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((b + 2*c*x)*sqrt(d + e*x)/(a + b*x + c*x**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c x + b\right )} \sqrt{e x + d}}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(1/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate((2*c*x + b)*sqrt(e*x + d)/(c*x^2 + b*x + a)^(3/2), x)