3.1574 \(\int \frac{(b+2 c x) (d+e x)^2}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=148 \[ \frac{\sqrt{a+b x+c x^2} \left (-2 c e (4 a e+3 b d)+3 b^2 e^2+2 c e x (2 c d-b e)+8 c^2 d^2\right )}{6 c^2}+\frac{e \left (b^2-4 a c\right ) (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{4 c^{5/2}}+\frac{2}{3} (d+e x)^2 \sqrt{a+b x+c x^2} \]

[Out]

(2*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/3 + ((8*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(3*b*d + 4*a*e) + 2*c*e*(2*c*d - b*e
)*x)*Sqrt[a + b*x + c*x^2])/(6*c^2) + ((b^2 - 4*a*c)*e*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b
*x + c*x^2])])/(4*c^(5/2))

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Rubi [A]  time = 0.199393, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {832, 779, 621, 206} \[ \frac{\sqrt{a+b x+c x^2} \left (-2 c e (4 a e+3 b d)+3 b^2 e^2+2 c e x (2 c d-b e)+8 c^2 d^2\right )}{6 c^2}+\frac{e \left (b^2-4 a c\right ) (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{4 c^{5/2}}+\frac{2}{3} (d+e x)^2 \sqrt{a+b x+c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/3 + ((8*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(3*b*d + 4*a*e) + 2*c*e*(2*c*d - b*e
)*x)*Sqrt[a + b*x + c*x^2])/(6*c^2) + ((b^2 - 4*a*c)*e*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b
*x + c*x^2])])/(4*c^(5/2))

Rule 832

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))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*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] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 779

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

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])

Rubi steps

\begin{align*} \int \frac{(b+2 c x) (d+e x)^2}{\sqrt{a+b x+c x^2}} \, dx &=\frac{2}{3} (d+e x)^2 \sqrt{a+b x+c x^2}+\frac{\int \frac{(d+e x) (2 c (b d-2 a e)+2 c (2 c d-b e) x)}{\sqrt{a+b x+c x^2}} \, dx}{3 c}\\ &=\frac{2}{3} (d+e x)^2 \sqrt{a+b x+c x^2}+\frac{\left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{6 c^2}+\frac{\left (\left (b^2-4 a c\right ) e (2 c d-b e)\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{4 c^2}\\ &=\frac{2}{3} (d+e x)^2 \sqrt{a+b x+c x^2}+\frac{\left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{6 c^2}+\frac{\left (\left (b^2-4 a c\right ) e (2 c d-b e)\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 )}{2 c^2}\\ &=\frac{2}{3} (d+e x)^2 \sqrt{a+b x+c x^2}+\frac{\left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{6 c^2}+\frac{\left (b^2-4 a c\right ) e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{4 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.19355, size = 185, normalized size = 1.25 \[ \frac{-8 a^2 c e^2+a \left (3 b^2 e^2-2 b c e (3 d+5 e x)+4 c^2 \left (3 d^2+3 d e x-e^2 x^2\right )\right )+x (b+c x) \left (3 b^2 e^2-2 b c e (3 d+e x)+4 c^2 \left (3 d^2+3 d e x+e^2 x^2\right )\right )}{6 c^2 \sqrt{a+x (b+c x)}}-\frac{e \left (b^2-4 a c\right ) (b e-2 c d) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{4 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*c*e^2 + a*(3*b^2*e^2 - 2*b*c*e*(3*d + 5*e*x) + 4*c^2*(3*d^2 + 3*d*e*x - e^2*x^2)) + x*(b + c*x)*(3*b^2
*e^2 - 2*b*c*e*(3*d + e*x) + 4*c^2*(3*d^2 + 3*d*e*x + e^2*x^2)))/(6*c^2*Sqrt[a + x*(b + c*x)]) - ((b^2 - 4*a*c
)*e*(-2*c*d + b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(4*c^(5/2))

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Maple [B]  time = 0.008, size = 280, normalized size = 1.9 \begin{align*}{\frac{2\,{e}^{2}{x}^{2}}{3}\sqrt{c{x}^{2}+bx+a}}-{\frac{b{e}^{2}x}{3\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{{b}^{2}{e}^{2}}{2\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{{b}^{3}{e}^{2}}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}+{ab{e}^{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{4\,a{e}^{2}}{3\,c}\sqrt{c{x}^{2}+bx+a}}+2\,x\sqrt{c{x}^{2}+bx+a}de-{\frac{bde}{c}\sqrt{c{x}^{2}+bx+a}}+{\frac{{b}^{2}de}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}-2\,{\frac{ade}{\sqrt{c}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) }+2\,\sqrt{c{x}^{2}+bx+a}{d}^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*e^2*x^2*(c*x^2+b*x+a)^(1/2)-1/3/c*e^2*b*x*(c*x^2+b*x+a)^(1/2)+1/2/c^2*e^2*b^2*(c*x^2+b*x+a)^(1/2)-1/4/c^(5
/2)*e^2*b^3*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/c^(3/2)*e^2*b*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)
^(1/2))-4/3/c*e^2*a*(c*x^2+b*x+a)^(1/2)+2*x*(c*x^2+b*x+a)^(1/2)*d*e-b/c*(c*x^2+b*x+a)^(1/2)*d*e+1/2*b^2/c^(3/2
)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e-2*a/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e+
2*(c*x^2+b*x+a)^(1/2)*d^2

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.16736, size = 757, normalized size = 5.11 \begin{align*} \left [\frac{3 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e -{\left (b^{3} - 4 \, a b c\right )} e^{2}\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) + 4 \,{\left (4 \, c^{3} e^{2} x^{2} + 12 \, c^{3} d^{2} - 6 \, b c^{2} d e +{\left (3 \, b^{2} c - 8 \, a c^{2}\right )} e^{2} + 2 \,{\left (6 \, c^{3} d e - b c^{2} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{24 \, c^{3}}, -\frac{3 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e -{\left (b^{3} - 4 \, a b c\right )} e^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \,{\left (4 \, c^{3} e^{2} x^{2} + 12 \, c^{3} d^{2} - 6 \, b c^{2} d e +{\left (3 \, b^{2} c - 8 \, a c^{2}\right )} e^{2} + 2 \,{\left (6 \, c^{3} d e - b c^{2} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{12 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(3*(2*(b^2*c - 4*a*c^2)*d*e - (b^3 - 4*a*b*c)*e^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2
+ b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(4*c^3*e^2*x^2 + 12*c^3*d^2 - 6*b*c^2*d*e + (3*b^2*c - 8*a*c^2)*e
^2 + 2*(6*c^3*d*e - b*c^2*e^2)*x)*sqrt(c*x^2 + b*x + a))/c^3, -1/12*(3*(2*(b^2*c - 4*a*c^2)*d*e - (b^3 - 4*a*b
*c)*e^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(4*c^3*e^
2*x^2 + 12*c^3*d^2 - 6*b*c^2*d*e + (3*b^2*c - 8*a*c^2)*e^2 + 2*(6*c^3*d*e - b*c^2*e^2)*x)*sqrt(c*x^2 + b*x + a
))/c^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.98264, size = 197, normalized size = 1.33 \begin{align*} \frac{1}{6} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (2 \, x e^{2} + \frac{6 \, c^{2} d e - b c e^{2}}{c^{2}}\right )} x + \frac{12 \, c^{2} d^{2} - 6 \, b c d e + 3 \, b^{2} e^{2} - 8 \, a c e^{2}}{c^{2}}\right )} - \frac{{\left (2 \, b^{2} c d e - 8 \, a c^{2} d e - b^{3} e^{2} + 4 \, a b c e^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{4 \, c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(c*x^2 + b*x + a)*(2*(2*x*e^2 + (6*c^2*d*e - b*c*e^2)/c^2)*x + (12*c^2*d^2 - 6*b*c*d*e + 3*b^2*e^2 - 8
*a*c*e^2)/c^2) - 1/4*(2*b^2*c*d*e - 8*a*c^2*d*e - b^3*e^2 + 4*a*b*c*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))*sqrt(c) - b))/c^(5/2)