3.1163 \(\int (A+B x) (d+e x)^2 \sqrt{b x+c x^2} \, dx\)

Optimal. Leaf size=267 \[ \frac{\left (b x+c x^2\right )^{3/2} \left (6 c e x (10 A c e-7 b B e+4 B c d)+10 A c e (16 c d-5 b e)+B \left (35 b^2 e^2-100 b c d e+32 c^2 d^2\right )\right )}{240 c^3}+\frac{(b+2 c x) \sqrt{b x+c x^2} \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{128 c^4}-\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{128 c^{9/2}}+\frac{B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c} \]

[Out]

((32*A*c^3*d^2 - 7*b^3*B*e^2 + 10*b^2*c*e*(2*B*d + A*e) - 16*b*c^2*d*(B*d + 2*A*e))*(b + 2*c*x)*Sqrt[b*x + c*x
^2])/(128*c^4) + (B*(d + e*x)^2*(b*x + c*x^2)^(3/2))/(5*c) + ((10*A*c*e*(16*c*d - 5*b*e) + B*(32*c^2*d^2 - 100
*b*c*d*e + 35*b^2*e^2) + 6*c*e*(4*B*c*d - 7*b*B*e + 10*A*c*e)*x)*(b*x + c*x^2)^(3/2))/(240*c^3) - (b^2*(32*A*c
^3*d^2 - 7*b^3*B*e^2 + 10*b^2*c*e*(2*B*d + A*e) - 16*b*c^2*d*(B*d + 2*A*e))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x
^2]])/(128*c^(9/2))

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Rubi [A]  time = 0.274054, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {832, 779, 612, 620, 206} \[ \frac{\left (b x+c x^2\right )^{3/2} \left (6 c e x (10 A c e-7 b B e+4 B c d)+10 A c e (16 c d-5 b e)+B \left (35 b^2 e^2-100 b c d e+32 c^2 d^2\right )\right )}{240 c^3}+\frac{(b+2 c x) \sqrt{b x+c x^2} \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{128 c^4}-\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{128 c^{9/2}}+\frac{B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((32*A*c^3*d^2 - 7*b^3*B*e^2 + 10*b^2*c*e*(2*B*d + A*e) - 16*b*c^2*d*(B*d + 2*A*e))*(b + 2*c*x)*Sqrt[b*x + c*x
^2])/(128*c^4) + (B*(d + e*x)^2*(b*x + c*x^2)^(3/2))/(5*c) + ((10*A*c*e*(16*c*d - 5*b*e) + B*(32*c^2*d^2 - 100
*b*c*d*e + 35*b^2*e^2) + 6*c*e*(4*B*c*d - 7*b*B*e + 10*A*c*e)*x)*(b*x + c*x^2)^(3/2))/(240*c^3) - (b^2*(32*A*c
^3*d^2 - 7*b^3*B*e^2 + 10*b^2*c*e*(2*B*d + A*e) - 16*b*c^2*d*(B*d + 2*A*e))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x
^2]])/(128*c^(9/2))

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 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 (A+B x) (d+e x)^2 \sqrt{b x+c x^2} \, dx &=\frac{B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\int (d+e x) \left (-\frac{1}{2} (3 b B-10 A c) d+\frac{1}{2} (4 B c d-7 b B e+10 A c e) x\right ) \sqrt{b x+c x^2} \, dx}{5 c}\\ &=\frac{B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}+\frac{\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) \int \sqrt{b x+c x^2} \, dx}{32 c^3}\\ &=\frac{\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac{\left (b^2 \left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right )\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{256 c^4}\\ &=\frac{\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac{\left (b^2 \left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{128 c^4}\\ &=\frac{\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac{b^2 \left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.613463, size = 293, normalized size = 1.1 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-4 b^2 c^2 \left (5 A e (24 d+5 e x)+2 B \left (30 d^2+25 d e x+7 e^2 x^2\right )\right )+10 b^3 c e (15 A e+30 B d+7 B e x)+16 b c^3 \left (5 A \left (6 d^2+4 d e x+e^2 x^2\right )+B x \left (10 d^2+10 d e x+3 e^2 x^2\right )\right )+32 c^4 x \left (5 A \left (6 d^2+8 d e x+3 e^2 x^2\right )+2 B x \left (10 d^2+15 d e x+6 e^2 x^2\right )\right )-105 b^4 B e^2\right )+\frac{15 b^{3/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right ) \left (-10 b^2 c e (A e+2 B d)+16 b c^2 d (2 A e+B d)-32 A c^3 d^2+7 b^3 B e^2\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{1920 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4*B*e^2 + 10*b^3*c*e*(30*B*d + 15*A*e + 7*B*e*x) + 16*b*c^3*(5*A*(6*d^2 +
4*d*e*x + e^2*x^2) + B*x*(10*d^2 + 10*d*e*x + 3*e^2*x^2)) + 32*c^4*x*(5*A*(6*d^2 + 8*d*e*x + 3*e^2*x^2) + 2*B*
x*(10*d^2 + 15*d*e*x + 6*e^2*x^2)) - 4*b^2*c^2*(5*A*e*(24*d + 5*e*x) + 2*B*(30*d^2 + 25*d*e*x + 7*e^2*x^2))) +
(15*b^(3/2)*(-32*A*c^3*d^2 + 7*b^3*B*e^2 - 10*b^2*c*e*(2*B*d + A*e) + 16*b*c^2*d*(B*d + 2*A*e))*ArcSinh[(Sqrt
[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(9/2))

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Maple [B]  time = 0.01, size = 671, normalized size = 2.5 \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)*(e*x+d)^2*(c*x^2+b*x)^(1/2),x)

[Out]

1/5*B*e^2*x^2*(c*x^2+b*x)^(3/2)/c+7/48*B*e^2*b^2/c^3*(c*x^2+b*x)^(3/2)-7/128*B*e^2*b^4/c^4*(c*x^2+b*x)^(1/2)+7
/256*B*e^2*b^5/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+2/3*(c*x^2+b*x)^(3/2)/c*A*d*e-1/8*b^2/c^2*(c*
x^2+b*x)^(1/2)*B*d^2+1/16*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d^2+1/4*x*(c*x^2+b*x)^(3/2)/
c*A*e^2-5/24*b/c^2*(c*x^2+b*x)^(3/2)*A*e^2+5/64*b^3/c^3*(c*x^2+b*x)^(1/2)*A*e^2-5/128*b^4/c^(7/2)*ln((1/2*b+c*
x)/c^(1/2)+(c*x^2+b*x)^(1/2))*A*e^2+1/4*A*d^2/c*(c*x^2+b*x)^(1/2)*b-1/8*A*d^2*b^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/
2)+(c*x^2+b*x)^(1/2))+5/16*b^2/c^2*(c*x^2+b*x)^(1/2)*x*B*d*e-1/2*b/c*(c*x^2+b*x)^(1/2)*x*A*d*e+1/2*A*d^2*(c*x^
2+b*x)^(1/2)*x+1/3*(c*x^2+b*x)^(3/2)/c*B*d^2+1/8*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*A*d*e-5
/64*b^4/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d*e-1/4*b/c*(c*x^2+b*x)^(1/2)*x*B*d^2-1/4*b^2/c^2*
(c*x^2+b*x)^(1/2)*A*d*e-5/12*b/c^2*(c*x^2+b*x)^(3/2)*B*d*e+5/32*b^2/c^2*(c*x^2+b*x)^(1/2)*x*A*e^2+5/32*b^3/c^3
*(c*x^2+b*x)^(1/2)*B*d*e+1/2*x*(c*x^2+b*x)^(3/2)/c*B*d*e-7/40*B*e^2*b/c^2*x*(c*x^2+b*x)^(3/2)-7/64*B*e^2*b^3/c
^3*(c*x^2+b*x)^(1/2)*x

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6254, size = 1508, normalized size = 5.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3840*(15*(16*(B*b^3*c^2 - 2*A*b^2*c^3)*d^2 - 4*(5*B*b^4*c - 8*A*b^3*c^2)*d*e + (7*B*b^5 - 10*A*b^4*c)*e^2)
*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*B*c^5*e^2*x^4 + 48*(20*B*c^5*d*e + (B*b*c^4 + 1
0*A*c^5)*e^2)*x^3 - 240*(B*b^2*c^3 - 2*A*b*c^4)*d^2 + 60*(5*B*b^3*c^2 - 8*A*b^2*c^3)*d*e - 15*(7*B*b^4*c - 10*
A*b^3*c^2)*e^2 + 8*(80*B*c^5*d^2 + 20*(B*b*c^4 + 8*A*c^5)*d*e - (7*B*b^2*c^3 - 10*A*b*c^4)*e^2)*x^2 + 10*(16*(
B*b*c^4 + 6*A*c^5)*d^2 - 4*(5*B*b^2*c^3 - 8*A*b*c^4)*d*e + (7*B*b^3*c^2 - 10*A*b^2*c^3)*e^2)*x)*sqrt(c*x^2 + b
*x))/c^5, -1/1920*(15*(16*(B*b^3*c^2 - 2*A*b^2*c^3)*d^2 - 4*(5*B*b^4*c - 8*A*b^3*c^2)*d*e + (7*B*b^5 - 10*A*b^
4*c)*e^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (384*B*c^5*e^2*x^4 + 48*(20*B*c^5*d*e + (B*b*c^4
+ 10*A*c^5)*e^2)*x^3 - 240*(B*b^2*c^3 - 2*A*b*c^4)*d^2 + 60*(5*B*b^3*c^2 - 8*A*b^2*c^3)*d*e - 15*(7*B*b^4*c -
10*A*b^3*c^2)*e^2 + 8*(80*B*c^5*d^2 + 20*(B*b*c^4 + 8*A*c^5)*d*e - (7*B*b^2*c^3 - 10*A*b*c^4)*e^2)*x^2 + 10*(
16*(B*b*c^4 + 6*A*c^5)*d^2 - 4*(5*B*b^2*c^3 - 8*A*b*c^4)*d*e + (7*B*b^3*c^2 - 10*A*b^2*c^3)*e^2)*x)*sqrt(c*x^2
+ b*x))/c^5]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.37207, size = 471, normalized size = 1.76 \begin{align*} \frac{1}{1920} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (6 \,{\left (8 \, B x e^{2} + \frac{20 \, B c^{4} d e + B b c^{3} e^{2} + 10 \, A c^{4} e^{2}}{c^{4}}\right )} x + \frac{80 \, B c^{4} d^{2} + 20 \, B b c^{3} d e + 160 \, A c^{4} d e - 7 \, B b^{2} c^{2} e^{2} + 10 \, A b c^{3} e^{2}}{c^{4}}\right )} x + \frac{5 \,{\left (16 \, B b c^{3} d^{2} + 96 \, A c^{4} d^{2} - 20 \, B b^{2} c^{2} d e + 32 \, A b c^{3} d e + 7 \, B b^{3} c e^{2} - 10 \, A b^{2} c^{2} e^{2}\right )}}{c^{4}}\right )} x - \frac{15 \,{\left (16 \, B b^{2} c^{2} d^{2} - 32 \, A b c^{3} d^{2} - 20 \, B b^{3} c d e + 32 \, A b^{2} c^{2} d e + 7 \, B b^{4} e^{2} - 10 \, A b^{3} c e^{2}\right )}}{c^{4}}\right )} - \frac{{\left (16 \, B b^{3} c^{2} d^{2} - 32 \, A b^{2} c^{3} d^{2} - 20 \, B b^{4} c d e + 32 \, A b^{3} c^{2} d e + 7 \, B b^{5} e^{2} - 10 \, A b^{4} c e^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*x*e^2 + (20*B*c^4*d*e + B*b*c^3*e^2 + 10*A*c^4*e^2)/c^4)*x + (80*B*c^4*
d^2 + 20*B*b*c^3*d*e + 160*A*c^4*d*e - 7*B*b^2*c^2*e^2 + 10*A*b*c^3*e^2)/c^4)*x + 5*(16*B*b*c^3*d^2 + 96*A*c^4
*d^2 - 20*B*b^2*c^2*d*e + 32*A*b*c^3*d*e + 7*B*b^3*c*e^2 - 10*A*b^2*c^2*e^2)/c^4)*x - 15*(16*B*b^2*c^2*d^2 - 3
2*A*b*c^3*d^2 - 20*B*b^3*c*d*e + 32*A*b^2*c^2*d*e + 7*B*b^4*e^2 - 10*A*b^3*c*e^2)/c^4) - 1/256*(16*B*b^3*c^2*d
^2 - 32*A*b^2*c^3*d^2 - 20*B*b^4*c*d*e + 32*A*b^3*c^2*d*e + 7*B*b^5*e^2 - 10*A*b^4*c*e^2)*log(abs(-2*(sqrt(c)*
x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)