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

Optimal. Leaf size=162 \[ \frac{(b+2 c x) \sqrt{b x+c x^2} \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{64 c^3}-\frac{b^2 \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}+\frac{5 e \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{24 c^2}+\frac{e \left (b x+c x^2\right )^{3/2} (d+e x)}{4 c} \]

[Out]

((16*c^2*d^2 - 16*b*c*d*e + 5*b^2*e^2)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) + (5*e*(2*c*d - b*e)*(b*x + c*x
^2)^(3/2))/(24*c^2) + (e*(d + e*x)*(b*x + c*x^2)^(3/2))/(4*c) - (b^2*(16*c^2*d^2 - 16*b*c*d*e + 5*b^2*e^2)*Arc
Tanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.12639, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {742, 640, 612, 620, 206} \[ \frac{(b+2 c x) \sqrt{b x+c x^2} \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{64 c^3}-\frac{b^2 \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}+\frac{5 e \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{24 c^2}+\frac{e \left (b x+c x^2\right )^{3/2} (d+e x)}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*Sqrt[b*x + c*x^2],x]

[Out]

((16*c^2*d^2 - 16*b*c*d*e + 5*b^2*e^2)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) + (5*e*(2*c*d - b*e)*(b*x + c*x
^2)^(3/2))/(24*c^2) + (e*(d + e*x)*(b*x + c*x^2)^(3/2))/(4*c) - (b^2*(16*c^2*d^2 - 16*b*c*d*e + 5*b^2*e^2)*Arc
Tanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(7/2))

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, 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]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 640

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

Rule 612

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

Rule 620

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

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 (d+e x)^2 \sqrt{b x+c x^2} \, dx &=\frac{e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}+\frac{\int \left (\frac{1}{2} d (8 c d-3 b e)+\frac{5}{2} e (2 c d-b e) x\right ) \sqrt{b x+c x^2} \, dx}{4 c}\\ &=\frac{5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}+\frac{\left (c d (8 c d-3 b e)-\frac{5}{2} b e (2 c d-b e)\right ) \int \sqrt{b x+c x^2} \, dx}{8 c^2}\\ &=\frac{\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}+\frac{5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}-\frac{\left (b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right )\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{128 c^3}\\ &=\frac{\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}+\frac{5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}-\frac{\left (b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{64 c^3}\\ &=\frac{\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}+\frac{5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}-\frac{b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.298318, size = 164, normalized size = 1.01 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-2 b^2 c e (24 d+5 e x)+15 b^3 e^2+8 b c^2 \left (6 d^2+4 d e x+e^2 x^2\right )+16 c^3 x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )-\frac{3 b^{3/2} \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{192 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*e^2 - 2*b^2*c*e*(24*d + 5*e*x) + 8*b*c^2*(6*d^2 + 4*d*e*x + e^2*x^2) + 16*
c^3*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2)) - (3*b^(3/2)*(16*c^2*d^2 - 16*b*c*d*e + 5*b^2*e^2)*ArcSinh[(Sqrt[c]*Sqrt[
x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(7/2))

________________________________________________________________________________________

Maple [B]  time = 0.052, size = 287, normalized size = 1.8 \begin{align*}{\frac{{e}^{2}x}{4\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{e}^{2}b}{24\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}{e}^{2}x}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,{b}^{3}{e}^{2}}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,{e}^{2}{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{2\,de}{3\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{bdex}{2\,c}\sqrt{c{x}^{2}+bx}}-{\frac{{b}^{2}de}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{de{b}^{3}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{{d}^{2}x}{2}\sqrt{c{x}^{2}+bx}}+{\frac{{d}^{2}b}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{{b}^{2}{d}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^2*x*(c*x^2+b*x)^(3/2)/c-5/24*e^2*b/c^2*(c*x^2+b*x)^(3/2)+5/32*e^2*b^2/c^2*x*(c*x^2+b*x)^(1/2)+5/64*e^2*b
^3/c^3*(c*x^2+b*x)^(1/2)-5/128*e^2*b^4/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+2/3*d*e*(c*x^2+b*x)^(
3/2)/c-1/2*d*e*b/c*x*(c*x^2+b*x)^(1/2)-1/4*d*e*b^2/c^2*(c*x^2+b*x)^(1/2)+1/8*d*e*b^3/c^(5/2)*ln((1/2*b+c*x)/c^
(1/2)+(c*x^2+b*x)^(1/2))+1/2*d^2*x*(c*x^2+b*x)^(1/2)+1/4*d^2/c*(c*x^2+b*x)^(1/2)*b-1/8*d^2*b^2/c^(3/2)*ln((1/2
*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.66614, size = 757, normalized size = 4.67 \begin{align*} \left [\frac{3 \,{\left (16 \, b^{2} c^{2} d^{2} - 16 \, b^{3} c d e + 5 \, b^{4} e^{2}\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (48 \, c^{4} e^{2} x^{3} + 48 \, b c^{3} d^{2} - 48 \, b^{2} c^{2} d e + 15 \, b^{3} c e^{2} + 8 \,{\left (16 \, c^{4} d e + b c^{3} e^{2}\right )} x^{2} + 2 \,{\left (48 \, c^{4} d^{2} + 16 \, b c^{3} d e - 5 \, b^{2} c^{2} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{384 \, c^{4}}, \frac{3 \,{\left (16 \, b^{2} c^{2} d^{2} - 16 \, b^{3} c d e + 5 \, b^{4} e^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (48 \, c^{4} e^{2} x^{3} + 48 \, b c^{3} d^{2} - 48 \, b^{2} c^{2} d e + 15 \, b^{3} c e^{2} + 8 \,{\left (16 \, c^{4} d e + b c^{3} e^{2}\right )} x^{2} + 2 \,{\left (48 \, c^{4} d^{2} + 16 \, b c^{3} d e - 5 \, b^{2} c^{2} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{192 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(3*(16*b^2*c^2*d^2 - 16*b^3*c*d*e + 5*b^4*e^2)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2
*(48*c^4*e^2*x^3 + 48*b*c^3*d^2 - 48*b^2*c^2*d*e + 15*b^3*c*e^2 + 8*(16*c^4*d*e + b*c^3*e^2)*x^2 + 2*(48*c^4*d
^2 + 16*b*c^3*d*e - 5*b^2*c^2*e^2)*x)*sqrt(c*x^2 + b*x))/c^4, 1/192*(3*(16*b^2*c^2*d^2 - 16*b^3*c*d*e + 5*b^4*
e^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (48*c^4*e^2*x^3 + 48*b*c^3*d^2 - 48*b^2*c^2*d*e + 15*
b^3*c*e^2 + 8*(16*c^4*d*e + b*c^3*e^2)*x^2 + 2*(48*c^4*d^2 + 16*b*c^3*d*e - 5*b^2*c^2*e^2)*x)*sqrt(c*x^2 + b*x
))/c^4]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.3201, size = 232, normalized size = 1.43 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (6 \, x e^{2} + \frac{16 \, c^{3} d e + b c^{2} e^{2}}{c^{3}}\right )} x + \frac{48 \, c^{3} d^{2} + 16 \, b c^{2} d e - 5 \, b^{2} c e^{2}}{c^{3}}\right )} x + \frac{3 \,{\left (16 \, b c^{2} d^{2} - 16 \, b^{2} c d e + 5 \, b^{3} e^{2}\right )}}{c^{3}}\right )} + \frac{{\left (16 \, b^{2} c^{2} d^{2} - 16 \, b^{3} c d e + 5 \, b^{4} e^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{128 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*x*e^2 + (16*c^3*d*e + b*c^2*e^2)/c^3)*x + (48*c^3*d^2 + 16*b*c^2*d*e - 5*b^2*
c*e^2)/c^3)*x + 3*(16*b*c^2*d^2 - 16*b^2*c*d*e + 5*b^3*e^2)/c^3) + 1/128*(16*b^2*c^2*d^2 - 16*b^3*c*d*e + 5*b^
4*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)