3.1553 \(\int \frac{(b+2 c x) \sqrt{a+b x+c x^2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=280 \[ -\frac{\sqrt{a+b x+c x^2} \left (e x \left (-4 c e (3 b d-2 a e)+b^2 e^2+12 c^2 d^2\right )-2 c d e (3 b d-2 a e)-b e^2 (b d-2 a e)+8 c^2 d^3\right )}{4 e^2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac{(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{8 e^3 \left (a e^2-b d e+c d^2\right )^{3/2}}+\frac{2 c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e^3} \]

[Out]

-((8*c^2*d^3 - b*e^2*(b*d - 2*a*e) - 2*c*d*e*(3*b*d - 2*a*e) + e*(12*c^2*d^2 + b^2*e^2 - 4*c*e*(3*b*d - 2*a*e)
)*x)*Sqrt[a + b*x + c*x^2])/(4*e^2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) + (2*c^(3/2)*ArcTanh[(b + 2*c*x)/(2*Sq
rt[c]*Sqrt[a + b*x + c*x^2])])/e^3 - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 - 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b*d
- 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(8*e^3*(c*d^2 - b*d*e + a*
e^2)^(3/2))

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Rubi [A]  time = 0.321095, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {810, 843, 621, 206, 724} \[ -\frac{\sqrt{a+b x+c x^2} \left (e x \left (-4 c e (3 b d-2 a e)+b^2 e^2+12 c^2 d^2\right )-2 c d e (3 b d-2 a e)-b e^2 (b d-2 a e)+8 c^2 d^3\right )}{4 e^2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac{(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{8 e^3 \left (a e^2-b d e+c d^2\right )^{3/2}}+\frac{2 c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*c^2*d^3 - b*e^2*(b*d - 2*a*e) - 2*c*d*e*(3*b*d - 2*a*e) + e*(12*c^2*d^2 + b^2*e^2 - 4*c*e*(3*b*d - 2*a*e)
)*x)*Sqrt[a + b*x + c*x^2])/(4*e^2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) + (2*c^(3/2)*ArcTanh[(b + 2*c*x)/(2*Sq
rt[c]*Sqrt[a + b*x + c*x^2])])/e^3 - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 - 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b*d
- 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(8*e^3*(c*d^2 - b*d*e + a*
e^2)^(3/2))

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 843

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

Rule 621

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

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

Rubi steps

\begin{align*} \int \frac{(b+2 c x) \sqrt{a+b x+c x^2}}{(d+e x)^3} \, dx &=-\frac{\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{\int \frac{\frac{1}{2} \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )-8 c^2 \left (c d^2-b d e+a e^2\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{4 e^2 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac{\left (2 c^2\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{e^3}-\frac{\left (8 c^2 d \left (c d^2-b d e+a e^2\right )+\frac{1}{2} e \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{4 e^3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac{\left (4 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{e^3}+\frac{\left (8 c^2 d \left (c d^2-b d e+a e^2\right )+\frac{1}{2} e \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac{-b d+2 a e-(2 c d-b e) x}{\sqrt{a+b x+c x^2}}\right )}{2 e^3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac{2 c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e^3}-\frac{(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \tanh ^{-1}\left (\frac{b d-2 a e+(2 c d-b e) x}{2 \sqrt{c d^2-b d e+a e^2} \sqrt{a+b x+c x^2}}\right )}{8 e^3 \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.26781, size = 389, normalized size = 1.39 \[ \frac{\frac{2 (a+x (b+c x))^{3/2} \left (4 c e (b d-2 a e)+b^2 e^2-4 c^2 d^2\right )}{d+e x}+\frac{-2 c e \sqrt{a+x (b+c x)} \left (-2 c^2 e (2 a e (2 e x-3 d)+b d (7 d-2 e x))+b c e^2 (-10 a e+5 b d+b e x)+b^3 e^3+4 c^3 d^2 (2 d-e x)\right )+c (2 c d-b e) \sqrt{e (a e-b d)+c d^2} \left (4 c e (3 a e-2 b d)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )+16 c^{5/2} \left (e (a e-b d)+c d^2\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{c e^3}+\frac{4 (a+x (b+c x))^{3/2} (2 c d-b e) \left (e (a e-b d)+c d^2\right )}{(d+e x)^2}}{8 \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))*(a + x*(b + c*x))^(3/2))/(d + e*x)^2 + (2*(-4*c^2*d^2 + b^2*e^2 +
4*c*e*(b*d - 2*a*e))*(a + x*(b + c*x))^(3/2))/(d + e*x) + (-2*c*e*Sqrt[a + x*(b + c*x)]*(b^3*e^3 + 4*c^3*d^2*
(2*d - e*x) + b*c*e^2*(5*b*d - 10*a*e + b*e*x) - 2*c^2*e*(b*d*(7*d - 2*e*x) + 2*a*e*(-3*d + 2*e*x))) + 16*c^(5
/2)*(c*d^2 + e*(-(b*d) + a*e))^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + c*(2*c*d - b*e)*Sqrt
[c*d^2 + e*(-(b*d) + a*e)]*(8*c^2*d^2 - b^2*e^2 + 4*c*e*(-2*b*d + 3*a*e))*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x +
b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(c*e^3))/(8*(c*d^2 + e*(-(b*d) + a*e))^2)

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Maple [B]  time = 0.014, size = 5046, normalized size = 18. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError